William's Quantumplations

Problem #\(1\): What is the Landau-Ginzburg free energy functional \(F[m]\) for the Ising model?

Solution #\(1\): It is defined implicitly through:

\[e^{-\beta F[m]}=\sum_{\{\sigma_i\}\to_{\text{c.g.}}m(\textbf x)}e^{-\beta E_{\{\sigma_i\}}}\]

where \(E_{\sigma_i}=-E_{\text{ext}}\sum_{i=1}^N\sigma_i-E_{\text{int}}\sum_{\langle i,j\rangle}\sigma_i\sigma_j\) is the energy of a given spin microstate \(\{\sigma_i\}\) and “\(\text{c.g.}\)” is short for coarse graining the underlying \(d\)-dimensional lattice \(\Lambda\to\textbf R^d\).

Problem #\(2\): Describe how a saddle-point approximation can be used to evaluate the canonical partition function \(Z\) of a Landau-Ginzburg theory.

Solution #\(2\): In the canonical ensemble, the probability density functional \(p[\phi]\) of finding the system in a given configuration \(\phi=\phi(\textbf x)\) is given by the Boltzmann distribution:

\[p[\phi]=\frac{e^{-\beta F[\phi]}}{Z}\]

where, to ensure normalization \(\int\mathcal D\phi p[\phi]=1\) over the space of all local order parameter configurations \(\phi\), the canonical partition function \(Z\) is given by the path integral:

\[Z=\int\mathcal D\phi e^{-\beta F[\phi]}\]

In general, functional integrals (so-called because the integrand \(e^{-\beta F[\phi]}\) is a functional) are difficult to evaluate (partly because they are hard to even rigorously define!). However, whether one is doing integrals over \(\textbf R,\textbf C\) or function spaces, as long as one’s integrand looks like \(e^{-\text{something}}\), it’s always worth trying a saddle-point approximation, which in this case looks like:

\[Z\approx e^{-\beta F[\phi_*]}\]

where \(\phi_*\) is the order parameter configuration minimizing the free energy \(F=F[\phi]\). In other words, \(\phi_*\) is a stationary point of \(F[\phi]\) so that the functional derivative \(\frac{\delta F}{\delta\phi^*}=0\) vanishes.

Landau mean field theory is the special case of this saddle-point approximation in which all fluctuations \(\phi_*=\phi_*(\textbf x)\) are completely ignored, yielding a homogeneous mean field order parameter.

Problem #\(3\): Explain why, as with many other functionals in physics (e.g. the action \(S[\textbf x]\)), the Landau-Ginzburg free energy functional \(F[\phi]\) must take the form:

\[F[\phi]=\int d^d\textbf x \mathcal F\left(\phi(\textbf x),\frac{\partial\phi}{\partial\textbf x},…\right)\]

Combining this with the saddle-point approximation in Solution #\(2\), what can one conclude?

Solution #\(3\): The presence of the integral \(\int d^d\textbf x\) simply reflects the extensive nature of the free energy \(F\), while the dependence of the integrand \(\mathcal F\) on \(\phi\) and its derivatives only reflects locality.

In the special case that the free energy density \(\mathcal F\) depends only on the field \(\phi(\textbf x)\) and its gradient \(\frac{\partial\phi}{\partial\textbf x}\) (but no higher derivatives), and it obeys suitable boundary conditions, one then has the usual Euler-Lagrange equations:

\[\frac{\partial}{\partial\textbf x}\cdot\frac{\partial \mathcal F}{\partial (\partial\phi/\partial\textbf x)}=\frac{\partial\mathcal F}{\partial\textbf x}\]

or equivalently, because of the lack of explicit \(\textbf x\)-dependence in \(\mathcal F\), one has the equivalent Beltrami identity:

\[\frac{\partial \mathcal F}{\partial (\partial\phi/\partial\textbf x)}\cdot\frac{\partial\phi}{\partial\textbf x}-\mathcal F=-\mathcal F_0\]

for some constant \(\mathcal F_0\in\textbf R\).

Problem #\(3\): Suppose that free energy density \(\mathcal F(\phi)\) of a particular system (e.g. the Ising ferromagnet) is taken (on locality, analyticity and suitable symmetry grounds) to be of the form:

\[\mathcal F\left(\phi,\frac{\partial\phi}{\partial\textbf x}\right)=\frac{\alpha_2}{2}\phi^2+\frac{\alpha_4}{4}\phi^4+\frac{\gamma}{2}\biggr|\frac{\partial\phi}{\partial\textbf x}\biggr|^2\]

where the phenomenological coupling constants \(\alpha_2,\alpha_4,\gamma\) each may carry some \(T\)-dependence, though as far as the study of second-order phase transitions at critical points is concerned, only the \(T\)-dependence \(\alpha_2(T)\sim T-T_c\) on the quadratic \(\phi^2\) term matters, and all that needs to be assumed about the other coupling constants is their sign for all temperatures \(T\), in this case \(\alpha_4,\gamma>0\). Show that in the subcritical regime \(T<T_c\), the \(\textbf Z_2\) symmetry \(F[-\phi]=F[\phi]\) of the theory is spontaneously broken via a bifurcation into \(2\) degenerate ground states (also called vacua in analogy with QFT) representing mean-field homogeneous/ordered phases/configurations \(\phi_*(\textbf x)=\phi_0\). Show that there is also a more interesting domain wall soliton:

\[\phi_*^{\text{DW}}(x)=\phi_0\tanh\left(\sqrt{-\frac{\alpha_2}{2\gamma}}x\right)\]

that emerges upon imposing Dirichlet boundary conditions \(\lim_{x\to\pm\infty}\phi_*(\textbf x)=\pm\phi_0\) implementing a smooth transition interpolating between the two ground state phases \(\pm\phi_0\).

Solution #\(3\): The Euler-Lagrange equations for this particular free energy density \(\mathcal F\) yield a Poisson/Helmholtz-like (but nonlinear!) PDE for the on-shell field \(\phi_*(\textbf x)\):

\[\gamma\biggr|\frac{\partial}{\partial\textbf x}\biggr|^2\phi_*=\alpha_2\phi_*+\alpha_4\phi_*^3\]

Looking for a homogeneous ansatz \(\phi_*(\textbf x)=\phi_0\) yields the \(2\) degenerate ground states \(\phi_0=\sqrt{-\alpha_2/\alpha_4}\) with free energy density \(\mathcal F_0:=\mathcal F(\pm\phi_0)=-\alpha_2^2/4\alpha_4\) and corresponding Landau-Ginzburg free energy \(F_0:=F[\pm\phi_0]=L^d\mathcal F_0\) (putting the system in a box \([-L/2,L/2]^d\) to regularize the obvious IR divergence that would otherwise arise).

By contrast, reverting now to the Beltrami identity, assuming that \(\phi_*^{\text{DW}}(\textbf x)=\phi_*^{\text{DW}}(x)\) varies only along the \(x\)-direction, the PDE reduces to the nonlinear separable first-order ODE:

\[\frac{\gamma}{2}\left(\frac{d\phi_*^{\text{DW}}}{dx}\right)^2-\frac{\alpha_2}{2}(\phi_*^{\text{DW}})^2-\frac{\alpha_4}{4}(\phi_*^{\text{DW}})^4=-\mathcal F_0\]

In particular, placing the domain wall at the origin \(x=0\) so that \(\phi(0)=0\), one obtains the soliton described (for a domain wall at some other location \(x_0\in\textbf R\), just translate \(x\mapsto x-x_0\) in the \(\tanh\) function). In particular; the width of the domain wall is \(\Delta x=\sqrt{-2\gamma/\alpha_2}\) which is pretty intuitive (c.f. the formula \(\omega_0=\sqrt{k/m}\) for a mass \(m\) on a spring \(k\)). Unlike the homogeneous ground states \(\pm\phi_0\), this domain wall soliton is a non-MF stationary point of the Landau-Ginzburg free energy functional \(F\).

Problem #\(4\): By definition, the ground state(s) of any system are global minima of its energy. In particular, it is clear that the domain wall soliton \(\phi_*^{\text{DW}}(x)\) is not a ground state, having free energy \(F[\phi_*^{\text{DW}}]>F_0\); precisely how much free energy \(\Delta F_{\text{DW}}:=F[\phi_*^{\text{DW}}]-F_0\) does it cost to create such a domain wall from a homogeneous ground state phase?

Solution #\(4\): Differential equations can (and should!) often be thought of as a dance/tension between conflicting characters. Even without doing any of the math in Solution #\(3\), it should be clear that the domain wall transition cannot happen instantaneously or the free energy cost \(\int d^d\textbf x\frac{\gamma}{2}\left(\frac{d\phi}{dx}\right)^2\) from the “kinetic” term would be too great, but neither can it proceed too slowly otherwise the free energy cost \(\int d^d\textbf x\left(\frac{\alpha_2}{2}\phi^2+\frac{\alpha_4}{4}\phi^4\right)\) from the “potential” term would be too great; the domain wall \(\phi_*^{\text{DW}}(x)\) must therefore strike a balance between these two free energy costs while satisfying the boundary conditions \(\lim_{x\to\pm\infty}\phi_*^{\text{DW}}(x)=\pm\phi_0\) (cf. the virial theorem in classical mechanics). In other words, \(0<\Delta x<\infty\).

(easy to forget, but remember one is working in the subcritical \(T<T_c\) regime where \(\alpha_2<0,\alpha_4>0\) so the potential term \(\frac{\alpha_2}{2}\phi^2+\frac{\alpha_4}{4}\phi^4\) is not positive semi-definite, in particular its minimum does not lie at \(\phi_0=0\) but rather has degenerate minima at \(\phi_0=\pm\sqrt{-\alpha_2/\alpha_4}\)! This means that anywhere \(\textbf x\in\textbf R^d\) that \(\phi_*^{\text{DW}}(\textbf x)\) strays too far away from the bottom of the potential wells at \(\pm\phi_0\) costs free energy, so for this reason the domain wall cannot take too long to climb over the hump between the two minima).

Indeed, the Beltrami identity quantifies this free energy balance and one can exploit it to quickly calculate the free energy of the domain wall soliton:

\[F[\phi_*^{\text{DW}}]=\int d^d\textbf x\left(\frac{\alpha_2}{2}(\phi_*^{\text{DW}})^2+\frac{\alpha_4}{4}(\phi_*^{\text{DW}})^4+\frac{\gamma}{2}\left(\frac{d\phi_*^{\text{DW}}}{dx}\right)^2\right)=\int d^d\textbf x\left(\gamma\left(\frac{d\phi_*^{\text{DW}}}{dx}\right)^2+f_0\right)\]

the latter term is recognized as just the free energy \(F_0=\int d^d\textbf x f_0=L^df_0\) of the ground state(s) so the excess free energy cost \(\Delta F_{\text{DW}}\) of creating the domain wall is (for \(L\gg\Delta x\)):

\[\Delta F_{\text{DW}}=\gamma\int d^d\textbf x\left(\frac{d\phi_*^{\text{DW}}}{dx}\right)^2=\gamma L^{d-1}\left(\frac{\phi_0}{\Delta x}\right)^2\int_{-L/2}^{L/2}dx\space\text{sech}^4\left(\frac{x}{\Delta x}\right)\]

\[\approx\frac{\gamma L^{d-1}\phi_0^2}{\Delta x}\int_{-\infty}^{\infty}d\varphi\space\text{sech}^4\varphi=\frac{4}{3\sqrt{2}}\frac{\sqrt{-\gamma\alpha_2^3}}{\alpha_4}L^{d-1}\]

but the key point is that \(\Delta F_{\text{DW}}\sim L^{d-1}\) scales with the (hyper)area of the domain wall. Well actually, another important scaling to note is that \(\Delta F_{\text{DW}}\sim(-\alpha_2)^{3/2}\) which suggests that near criticality where \(\alpha_2\to 0\), the free energy cost \(\Delta F_{\text{DW}}\to 0\) of creating a domain wall also vanishes, while its width \(\Delta x\sim(-\alpha_2)^{-1/2}\) diverges. This suggests that domain walls become important near critical points.

Problem #\(5\): Working with the same Landau-Ginzburg system as above, estimate in \(d=1\) dimension the probability \(p_N\) that thermal fluctuations will spontaneously break the \(\textbf Z_2\) symmetry of the free energy \(F\) via the creation of \(N\ll L/\Delta x\) domain walls anywhere, and comment on the implication of this for the lower critical dimension \(d_{\ell}\) of this system.

Solution #\(5\): Because the ODE arising from the Beltrami identity was nonlinear, the superposition of a bunch of domain walls at different locations is not strictly speaking a stationary point of the free energy \(F\), but nonetheless one can sweep this under the rug and assume it is still an approximate solution provided the \(N\) domain walls are well-separated. In particular, this means their total free energy \(\approx N\Delta F_{\text{DW}}+F_0\) is also approximately additive. If one imagines discretizing the \(d=1\) line \([-L/2,L/2]\) into \(L/\Delta x\) bins each of width \(\Delta x\), then there are \(L/\Delta x\choose{N}\) choices for where to put the domain walls in (ignoring the fact that in some cases, they may not be so well-separated), so the probability of having any configuration of \(N\) domain walls is approximately:

\[p_N\approx {{L/\Delta x}\choose{N}}\frac{e^{-N\beta\Delta F_{\text{DW}}+F_0}}{Z}\approx\frac{(L/\Delta x)^N}{N!}\frac{e^{-N\beta\Delta F_{\text{DW}}+F_0}}{Z}\]

where the sparse approximation has been used \(N\ll L/\Delta x\). Alternatively, normalizing with respect to the \(N=0\) “vacuum” probability \(p_0=e^{-\beta F_0}/Z\):

\[\frac{p_N}{p_0}=\frac{(L/\Delta x)^N}{N!}e^{-N\beta\Delta F_{\text{DW}}}\]

Importantly, in \(d=1\) domain walls are free since \(\Delta F_{\text{DW}}\sim L^{1-1}=L^0\). So all the \(L\)-dependence in the expression above for \(p_N/p_0\) is contained in the entropic factor \((L/\Delta x)^N/N!\) and shows that as one takes the infinite system limit \(L\to\infty\), the probability \(p_N/p_0\) receives no exponential suppression from the \(e^{-N\beta\Delta F_{\text{DW}}}\) and instead grows unbounded. The probability that thermal fluctuations produce an even number \(N\in 2\textbf N\) of domain walls is:

\[\sum_{N=2,4,6,…}p_N=\cosh\left(\frac{Le^{-\beta\Delta F_{\text{DW}}}}{\Delta x}\right)p_0\]

and similarly for odd \(N\in 2\textbf N+1\):

\[\sum_{N=1,3,5,…}p_N=\sinh\left(\frac{Le^{-\beta\Delta F_{\text{DW}}}}{\Delta x}\right)p_0\]

and since \(\lim_{\varphi\to\infty}\tanh\varphi=1\), these two probabilities both approach the same \(e^{Le^{-\beta\Delta F_{\text{DW}}}/\Delta x}p_0/2\) as \(L\to\infty\).

More generally, any Landau-Ginzburg theory with a discrete symmetry (like the \(\textbf Z_2\) symmetry of this particular \(F\)) will possess a bunch of disconnected, degenerate ground states/vacua (like the \(\phi(\textbf x)=\pm\phi_0\) in this case) that spontaneously break that discrete symmetry, and all have \(d_{\ell}=1\) because there is very high probability that thermal fluctuations will proliferate domain walls that toggle between the degenerate ground states, preventing ordered phases from forming. Hence for instance the Ising model has no phase transitions in \(d=1\).

Problem #\(6\):

Solution #\(6\):

Problem #\(7\):

Solution #\(7\):

Note that if one does not work in natural units, then the heat capacity is instead \(C=k_B\beta^2\frac{\partial^2\ln Z}{\partial\beta^2}\) so all specific heat capacities should come with an additional factor of \(k_B\).

Problem #\(8\):

Solution #\(8\):

Problem #\(9\):

Solution #\(9\):

(NEED TO COME BACK TO THIS QUESTION!)

Problem #\(10\): Describe the quadratic approximation to the Landau-Ginzburg free energy density \(\mathcal F\) governing a \(\textbf Z_2\)-symmetric system/theory (e.g. Ising ferromagnet) described by a single, real scalar order parameter \(\phi\) in the absence of any external coupling \(E_{\text{ext}}=0\).

Solution #\(10\): On LAS (locality, analyticity, symmetry) grounds, the exact free energy density \(\mathcal F\) that knows about the all detailed microscopic physics must take the phenomenological form:

\[\mathcal F=\frac{\alpha_2}{2}\phi^2+\frac{\alpha_4}{4}\phi^4+\frac{\gamma}{2}\biggr|\frac{\partial\phi}{\partial\textbf x}\biggr|^2+…\]

where in principle there are also coupling constants for \(\phi^6,\phi^5\biggr|\frac{\partial\phi}{\partial\textbf x}\biggr|^2\biggr|\frac{\partial}{\partial\textbf x}\biggr|^2\phi\) etc. though not for couplings like \(\phi^3\) or \(1/\phi^2\). In addition, one also has to import from mean-field theory the assumption that \(\alpha_2\sim T-T_c\) (i.e. \(\alpha_2\to 0\) as \(T\to T_c\) in a linear/critical exponent \(1\) manner) and that \(\gamma,\alpha_4>0\) for all \(T\) in a neighbourhood of \(T_c\).

The quadratic approximation to \(\mathcal F\) looks slightly different depending on whether one is working in the supercritical \(T>T_c\) regime (where \(\alpha_2(T)>0\)) or the subcritical \(T<T_c\) regime (where \(\alpha_2(T)<0\)).

In the supercritical \(T>T_c\) regime, the quadratic approximation does what it sounds like, in fact drop not only all quartic, sextic, octic, decic, etc. couplings like \(\phi^4,\biggr|\frac{\partial\phi}{\partial\textbf x}\biggr|^{22}\) but even quadratic couplings containing Laplacians, third derivatives, biharmonics, and all higher-derivative couplings. So at \(T>T_c\), this amounts to keeping only \(2\) couplings, a “kinetic” coupling \(\frac{\gamma}{2}\biggr|\frac{\partial\phi}{\partial\textbf x}\biggr|^2\) and a “Hookean quadratic potential” coupling \(\frac{\alpha_2}{2}\phi^2\):

\[\mathcal F\approx\frac{\alpha_2}{2}\phi^2+\frac{\gamma}{2}\biggr|\frac{\partial\phi}{\partial\textbf x}\biggr|^2\]

Notice in this case one can also replace \(\phi\mapsto\delta\phi\) everywhere:

\[=\frac{\alpha_2}{2}\delta\phi^2+\frac{\gamma}{2}\biggr|\frac{\partial\delta\phi}{\partial\textbf x}\biggr|^2\]

where the fluctuation \(\delta\phi(\textbf x):=\phi(\textbf x)-\langle\phi(\textbf x)\rangle\). This is simply because, at \(T>T_c\), \(\langle\phi(\textbf x)\rangle=0\) vanishes homogeneously (from mean-field theory).

By contrast, in the subcritical \(T<T_c\) regime, the quadratic approximation consists of \(2\) separate approximation steps:

Step #\(1\): Keep not only the zeroth and first-order quadratic couplings that were retained in the supercritical \(T>T_c\) case, but also keep the quartic \(\phi^4\) coupling:

\[\mathcal F\approx\frac{\alpha_2}{2}\phi^2+\frac{\gamma}{2}\biggr|\frac{\partial\phi}{\partial\textbf x}\biggr|^2\]

The reason for this is that for \(T<T_c\), \(\alpha_2(T)<0\) so the free energy density \(\mathcal F(\phi)=-\frac{|\alpha_2|}{2}\phi^2+…\) would be unbounded from below, yielding an unstable theory. By including the quartic coupling, one instead has \(2\) degenerate \(\textbf Z_2\) symmetry breaking homogeneous ordered phases \(\phi(\textbf x)=\pm\phi_0\) with \(\phi_0=\sqrt{-\alpha_2/\alpha_4}\).

However, beyond the presence of \(\alpha_4\) in \(\phi_0\), one would otherwise like to remove all other vestiges of it in \(\mathcal F\) in order to get a more quadratic-looking free energy density like in the \(T>T_c\) case. Thus:

Step #\(2\): Insert the “Reynolds decomposition” \(\phi=\phi_0+\delta\phi\) into \(\mathcal F\) and notice that the term linear in \(\delta\phi\) vanishes because \(\phi_0\) is on-shell:

\[\mathcal F\approx\mathcal F[\phi_0]-\alpha_2\delta\phi^2+\frac{\gamma}{2}\biggr|\frac{\partial\delta\phi}{\partial\textbf x}\biggr|^2+O(\delta\phi^3)\]

where the cubic and quartic couplings \(O(\delta\phi^3)=2\alpha_4\phi_0\delta\phi^3+\frac{\alpha_4}{2}\delta\phi^4\) at the level of the fluctuations \(\delta\phi\) from the mean field \(\phi_0\) are assumed negligible in this second step of the quadratic approximation for \(T<T_c\). Finally, note that the constant term \(\mathcal F[\phi_0]\) would drop out when differentiating \(\ln Z\) to compute Boltzmannian cumulants, so can safely be omitted.

In this way, the name “quadratic approximation” is justified because both the \(T>T_c\) and \(T<T_c\) cases can be unified by writing the free energy as:

\[\mathcal F\approx\frac{1}{2}\left(\mu^2\delta\phi^2+\gamma\biggr|\frac{\partial\delta\phi}{\partial\textbf x}\biggr|^2\right)\]

where the “mass coupling” \(\mu^2\geq 0\) is defined piecewise to capture both the subcritical and supercritical regimes:

\[\mu^2(T) =\begin{cases}
\alpha_{2}(T), & T \geq T_c\\
-2\,\alpha_{2}(T), & T \leq T_c
\end{cases}\]

cf. the Klein-Gordon Lagrangian density when written in natural units:

\[\mathcal L=\frac{1}{2}\partial^{\mu}\partial_{\mu}\phi-\frac{1}{2}m^2\phi^2\]

Problem #\(11\): Working within the quadratic approximation to the free energy density \(\mathcal F\) outlined in Solution #\(11\), compute the partition function \(Z\).

Solution #\(11\): This is pretty much the only case where the path integral underlying \(Z\) can be computed analytically thanks to the fact that Gaussian integrals are straightforward to do. Here is a sketch of the computation:

Step #\(1\): The free energy \(F=\int d^d\textbf x\mathcal F(\delta\phi)\) is certainly extensive so one must add up \(\int\) the chunks of free energy \(d^d\textbf x\mathcal F(\delta\phi)\) due to fluctuations \(\delta\phi\) of the order parameter \(\phi\) at each point \(\textbf x\in\textbf R^d\) in space (or \(\textbf x\in V\subset\textbf R^d\) in some suitably large volume). But conceptually, some of these fluctuations might be more short-range, rapid oscillations across \(\textbf x\), while others may be more long-range, slow envelopes. Nonetheless, it should be intuitively clear that, rather than taking the local approach of stepping through each \(\textbf x\in\textbf R^d\) and adding up the energies of all fluctuations at that point \(\textbf x\), one can take a more global approach of adding up the energies of all fluctuations across the entire space \(\textbf R^d\) that have a given wavelength \(\lambda\) (or equivalently wavenumber \(k=2\pi/\lambda\)), and stepping through all possible wavelengths, from the long (“IR”) wavelengths all the way down to the short (“UV”) wavelengths. This intuitive picture can of course be formalized by explicitly writing \(\delta\phi(\textbf x)\) as a superposition of plane wave excitations:

\[\delta\phi(\textbf x)=\int\frac{d^d\textbf k}{(2\pi)^d}\delta\phi_{\textbf k}e^{i\textbf k\cdot\textbf x}\]

whereupon, one obtains what is essentially just Plancherel’s theorem:

\[F=\frac{1}{2}\int\frac{d^d\textbf k}{(2\pi)^d}|\delta\phi_{\textbf k}|^2(\mu^2+\gamma|\textbf k|^2)\]

which uses the fact that for \(\delta\phi(\textbf x)\in\textbf R\), the Fourier transform satisfies \(\delta\phi_{-\textbf k}=\delta\phi^{\dagger}_{\textbf k}\). By viewing the system as having some large but finite volume \(V\), one can replace on dimensional grounds:

\[\int\frac{d^d\textbf k}{(2\pi)^d}\Leftrightarrow\frac{1}{V}\sum_{\textbf k}\]

where \(\sum_{\textbf k}\) means over all \(\textbf k=\frac{2\pi}{L}\textbf n\) for \(\textbf n\in\textbf Z^d\) since these are the only wavevectors compatible with periodic boundary conditions on \(\delta\phi(\textbf x)\) in a box \([-L/2,L/2]^d\) of volume \(V=L^d\). Put another way, it reduces Plancherel’s theorem for the Fourier transform to Parseval’s theorem for Fourier series (since \(\delta\phi(\textbf x)\) is now \(L\)-periodic in all \(d\) dimensions).

Finally, the existence of the plane wave basis also allows one to “rigorously” define the measure \(\mathcal D\phi=\mathcal D\delta\phi\) in the path integral for \(Z=\int\mathcal D\delta\phi e^{-\beta F[\delta\phi]}\). Intuitively, the path integral \(\int\mathcal D\delta\phi\) wants to sum over all possible fluctuations \(\delta\phi\) of the field about the mean field. But the Fourier transform allows one to explicitly parameterize this abstract space! Simply integrate over all possible choices of the “Fourier knobs/coefficients” \(\phi_{\textbf k}\) which can span any fluctuation \(\delta\phi\):

\[\int\mathcal D\delta\phi\sim\int\prod_{\textbf k}d\delta\phi_{\textbf k}\sim\int\prod_{\textbf k}d\Re \delta\phi_{\textbf k}d\Im \delta\phi_{\textbf k}\]

where the product \(\prod_{\textbf k}\) is over the same countably infinite lattice of \(\textbf k\)-wavevectors as the earlier sum \(\sum_{\textbf k}\) (actually, strictly speaking, one should only take the product over half of the entire \(\textbf k\)-space, for instance imposing \(k_x>0\). This is because if \(\delta\phi_{\textbf k}\in\textbf C\) is already known for some \(\textbf k\), then \(\delta\phi_{-\textbf k}\) is also automatically known by virtue of the reality criterion \(\delta\phi_{-\textbf k}=\delta\phi^{\dagger}_{\textbf k}\), cf. \(\sin(kx)=\frac{1}{2i}e^{ikx}+?e^{-ikx}\) where, knowing that \(\sin(kx)\in\textbf R\), one can immediately conclude that \(?=(\frac{1}{2i})^{\dagger}=-\frac{1}{2i}\). So \(\delta\phi_{\textbf k}\) and \(\delta\phi_{-\textbf k}\) are not independent knobs (imagine a “complex conjugation gear” between them), but the path integral \(\int\mathcal D\delta\phi\) only wants to integrate over independent degrees of freedom since double-counting the same fluctuation is obviously not desired. That being said, this only leads to a factor of \(2\) discrepancy which doesn’t affect any physical quantities, and the measure \(\mathcal D\delta\phi\) is only defined up to some “normalization” anyways).

At this point one writes \(|\delta\phi_{\textbf k}|^2=\Re^2\delta\phi_{\textbf k}+\Im^2\delta\phi_{\textbf k}\) and decouples all the Gaussian integrals to obtain the final result for \(Z\):

\[Z\sim\prod_{\textbf k}\sqrt{\frac{\pi V}{\beta(\mu^2+\gamma|\textbf k|^2)}}\]

Problem #\(12\): Using this \(Z\), and making the specific assumption that \(\alpha_2(T)=k_B(T-T_c)\) and that \(\gamma\) is independent of \(T\), compute the specific heat capacity \(c\) of this Landau-Ginzburg system in the supercritical \(T>T_c\) regime.

Solution #\(12\): Simply use the formula:

\[c=\frac{k_B}{V}\beta^2\frac{\partial^2\ln Z}{\partial\beta^2}\]

with cumulant generating function (ignoring the temperature-independent parts):

\[\ln Z\sim-\frac{1}{2}\sum_{\textbf k}\ln\beta(\mu^2+\gamma|\textbf k|^2)\]

So, noting that \(\partial\mu^2/\partial\beta=-1/\beta^2\):

\[c=-\frac{k_B}{2}\beta^2\frac{\partial^2}{\partial\beta^2}\int\frac{d^d\textbf k}{(2\pi)^d}\ln\beta(\mu^2+\gamma|\textbf k|^2)\]

\[=-\frac{k_B}{2}\int\frac{d^d\textbf k}{(2\pi)^d}\beta^2\frac{\partial^2}{\partial\beta^2}\left(\ln\beta+\ln(\mu^2+\gamma|\textbf k|^2)\right)\]

\[=\frac{k_B}{2}\int\frac{d^d\textbf k}{(2\pi)^d}\left(1-\frac{2}{\beta(\mu^2+\gamma |\textbf k|^2)}+\frac{1}{\beta^2(\mu^2+\gamma |\textbf k|^2)^2}\right)\]

where the \(\frac{k_B}{2}\times 1\) part reflects equipartition of quadratic degrees of freedom and is simply due to the \(\beta\) temperature dependence in \(e^{-\beta F}\) (fluctuation-dissipation theorem?). By contrast, the other terms are additional contributions to the heat capacity arising from the \(T\)-dependence in \(F\) instead, specifically in \(\mu^2=k_B(T-T_c)\).

Problem #\(13\): Clearly, the specific heat capacity \(c\) as it’s currently written suffers from a \(|\textbf k|\to\infty\) UV divergence, since it is that part of the integrand that causes the integral to diverge to \(\infty\). What should one make of this?

Solution #\(13\): To regularize this UV divergence, one has to impose a UV cutoff \(k^*\) with the property that the Fourier transform \(\delta\phi_{\textbf k}\) is only supported for \(|\textbf k|\leq k^*\) in the \(k^*\)-ball. This UV cutoff should be chosen so that \(k^*\sim 1/\Delta x\), with \(\Delta x\) being the lattice spacing of some underlying microscopic structure which has been coarse-grained away. By implementing a UV cutoff, the specific heat capacity \(c\) is made finite:

\[c=\frac{k_B}{2(2\pi)^d}|S^{d-1}|\left(\frac{(k^*)^d}{d}-\frac{2}{\beta}\int_0^{k^*}dk\frac{k^{d-1}}{\mu^2+\gamma k^2}+\frac{1}{\beta^2}\int_0^{k^*}dk\frac{k^{d-1}}{(\mu^2+\gamma k^2)^2}\right)\]

where the (hyper)surface area of the unit \(d-1\)-sphere is \(|S^{d-1}|=2\pi^{d/2}/\Gamma(d/2)\).

(aside: to prove this, notice that the \(d\)-dimensional isotropic Gaussian integral \(\int_{\textbf x\in\textbf R^d} d^d\textbf x e^{-|\textbf x|^2}\) evaluates to \(\pi^{d/2}\) when separated in Cartesian coordinates, or equivalently \(|S^{d-1}|\int_0^{\infty}d|\textbf x||\textbf x|^{d-1}e^{-|\textbf x|^2}\) in spherical coordinates. The latter integral can then be massaged into the form of a gamma function \(\Gamma(d/2)/2\) via the substitution \(z:=|\textbf x|^2\)).

For some small dimensions \(d=1,2,3,4,5\), the \(2\) integrals can be evaluated analytically, with results compiled in the table below:

or in hindsight many of these can be written more compactly via the (yet unmotivated) correlation length \(\xi:=\sqrt{\gamma}/\mu\).

Problem #\(14\): Using the results of Problem #\(14\), comment on how \(c\) behaves in various dimensions \(d\) as one approaches the critical point \(T\to T_c^+\) from above (since the results above were all computed within the supercritical \(T>T_c\) regime, though the analysis is the same in the subcritical \(T<T_c\) regime).

Solution #\(14\): As \(T\to T_c^+\), the quadratic coupling constant \(\mu^2\to 0\) vanishes. For \(d\geq 5\), both of the integrals above converge to a finite value determined by the choice of UV cutoff wavenumber \(k^*\), so \(c\) is also finite, and more precisely \(c\sim(k^*)^{d-2}\) from the first integral. For \(d=4,3\), only the first integral diverges while the second one converges, whereas for \(d=2,1\), both integrals diverge. For \(d\leq 3\), this divergence goes like \(c\sim\mu^{d-4}\) whereas for \(d=4\) it is a logarithmic divergence \(c\sim\ln\mu\) (both of these being determined in this case by the second integral).

In the case \(d\leq 3\), since \(\mu^2\sim T-T_c\) as \(T\to T_c^+\), this implies that \(c\sim (T-T_c)^{(d-4)/2}\) but in general the critical exponent \(\alpha\) is defined by the property that \(c\sim |T-T_c|^{-\alpha}\). Although this analysis was only for \(T>T_c\), one can check that for \(T<T_c\) one would have mirror behavior, so this analysis shows that, at least for \(d\leq 3\), the critical exponent is \(\alpha=2-d/2\). Thus, the contribution of fluctuations causes the critical exponent to differ from the mean-field prediction \(\alpha=0\).

Problem #\(15\): At each \(\textbf x\in\textbf R^d\), one can associate a continuous random variable \(\phi(\textbf x)\) which draws an order parameter configuration \(\phi\) from a thermal Boltzmann distribution \(p[\phi]=e^{-\beta F[\phi]}/Z\) and returns the evaluation of \(\phi\) at \(\textbf x\). Given two arbitrary positions \(\textbf x,\textbf x’\in\textbf R^d\), each giving rise to its own random variable \(\phi(\textbf x),\phi(\textbf x’)\), define the connected \(2\)-point correlation propagator \(\langle\delta\phi(\textbf x)\delta\phi(\textbf x’)\rangle\) between \(\textbf x,\textbf x’\).

Solution #\(15\): The connected \(2\)-point correlation propagator is simply the cross-covariance of the random variables \(\phi(\textbf x),\phi(\textbf x’)\). It is thus related to the cross-correlation \(\langle\phi(\textbf x)\phi(\textbf x’)\rangle\) between \(\textbf x,\textbf x’\) and their individual expected values \(\langle\phi(\textbf x)\rangle,\langle\phi(\textbf x’)\rangle\) by the usual parallel axis theorem (sort of…):

\[\langle\delta\phi(\textbf x)\delta\phi(\textbf x’)\rangle=\langle\phi(\textbf x)\phi(\textbf x’)\rangle-\langle\phi(\textbf x)\rangle\langle\phi(\textbf x’)\rangle\]

where, just to flesh it out explicitly, these thermal Boltzmann canonical ensemble averages look like configuration space functional integrals:

\[\langle\phi(\textbf x)\rangle=\int\mathcal D\phi\phi(\textbf x)e^{-\beta F[\phi]}\]

(remember that evaluation at \(\textbf x\) is a functional \(\phi(\textbf x)=\text{eval}_{\textbf x}[\phi]\)).

Problem #\(16\): Define the functional derivative of a functional. Evaluate the following functional derivatives:

\[\frac{\delta}{\delta f(\textbf x)}\int d^d\textbf x’\cos f(\textbf x’)\]

\[\frac{\delta}{\delta f(\textbf x)}\int d^d\textbf x’ d^d\textbf x^{\prime\prime}\frac{f(\textbf x’)f(\textbf x^{\prime\prime})}{|\textbf x’-\textbf {x}^{\prime\prime}|}\]

\[\frac{\delta}{\delta f(\textbf x)}\exp{\int d^d\textbf x’\left(\frac{1}{2}f(\textbf x’)^2+\frac{1}{2}\biggr|\frac{\partial f}{\partial\textbf x’}\biggr|^2\right)}\]

(comment: the \(1\)st and \(3\)rd functionals are local whereas the \(2\)nd functional one is nonlocal).

Solution #\(16\): Functions \(f(\textbf x)\) are like \(\infty\)-dimensional vectors whose components are indexed not by a discrete label \(i\) but rather by a continuous label \(\textbf x\in\textbf R^d\); one may as well write \(f_{\textbf x}\) or \(\langle\textbf x|f\rangle\) rather than \(f(\textbf x)\) to stress this point. The functional derivative is then conceptually no different from a partial derivative with respect to one of these infinitely many components \(f(\textbf x)\) of \(f\) and just ends up returning a vanilla function of \(\textbf x\).

Just as for a function \(F(f_1,f_2,…)\) of some real variables \(f_1,f_2,…\in\textbf R\) one has the total differential:

\[dF=\sum_i\frac{\partial F}{\partial f_i}df_i\]

So simply replacing the sum by an integral \(\sum_i\mapsto\int d^d\textbf x\), the functional derivative of a functional \(F[f]\) with respect to the variable \(f(\textbf x)\) is defined by requiring:

\[\delta F=\int d^d\textbf x\frac{\delta F}{\delta f(\textbf x)}\delta f(\textbf x)\]

where typically one takes \(\delta\phi(\textbf x)=0\) for \(\textbf x\) on the boundary of the domain of integration in \(\textbf R^d\).

Notice that functional derivatives are much easier to compute than functional integrals, reflecting the general trend from single-variable calculus that differentiation is easier than integration. In particular, for functionals which are either directly (as in the \(1\)st and \(2\)nd examples) or indirectly (as in the \(3\)rd example) related to some integral of the function, functional differentiation just boils down to partial differentiation of the integrand. Keeping in mind that \(\delta f(\textbf x)=0\) for \(\textbf x\in\partial\), the key identity in this regard, generalizing the Euler-Lagrange equations, is:

\[\frac{\delta}{\delta f(\textbf x)}\int d^d\textbf x\mathcal F\left(f,\frac{\partial f}{\partial\textbf x},\frac{\partial^2 f}{\partial\textbf x^2},…\right)=\frac{\partial\mathcal F}{\partial f}-\frac{\partial}{\partial\textbf x}\cdot\frac{\partial\mathcal F}{\partial(\partial f/\partial \textbf x)}+\frac{\partial^2}{\partial\textbf x^2}\cdot\frac{\partial^2\mathcal F}{\partial(\partial^2 f/\partial \textbf x^2)}-+…\]

Problem #\(17\): Another functional differentiation problem for fun. For any function \(f(\textbf x)\), the “orthonormality condition”:

\[\frac{\delta f(\textbf x’)}{\delta f(\textbf x)}=\delta^d(\textbf x-\textbf x’)\]

Solution #\(17\): The idea, mentioned before, is to view \(f(\textbf x’)=\text{eval}_{\textbf x’}[f]\) as an evaluation functional so that it really can be interpreted as a functional derivative. Then, express the evaluation functional in the integral form that one is most comfortable with by convolving with a delta:

\[f(\textbf x’)=\int d^d\textbf x f(\textbf x)\delta^d(\textbf x-\textbf x’)\]

So the result follows from the considerations in Solution #\(16\).

Problem #\(18\): In general, what are the dimensions of the functional derivative \(\frac{\delta F}{\delta f}\)? Look back at both Problems #\(16,17\) and check this.

Solution #\(18\): Although one is used to intuitively reading off dimensions from regular derivatives like \([\partial f/\partial x]=[f]/[x]\), for functional derivatives the “densitized” nature of the continuum \(d^d\textbf x\) means that actually:

\[\biggr[\frac{\delta F}{\delta f(\textbf x)}\biggr]=\frac{[F]}{[f][\textbf x]^d}\]

which in particular is not equal to the naive \([F]/[f]\) (unless \(d=0\) which is boring). This is consistent with the functional derivatives in both Solutions #\(16,17\).

Problem #\(19\): Show that by “applying an external magnetic field” \(E_{\text{ext}}(\textbf x)\) to the system, one has for the original exact Landau-Ginzburg free energy bifunctional \(F[\phi,E_{\text{ext}}]\) (i.e. not the quadratic approximation \(F=F[\delta\phi]\)) the functional derivatives:

\[\langle\phi(\textbf x)\rangle=\frac{\delta\ln Z}{\delta\beta E_{\text{ext}}(\textbf x)}\]

\[\langle\delta\phi(\textbf x)\delta\phi(\textbf x’)\rangle=\frac{\delta^2\ln Z}{\delta\beta E_{\text{ext}}(\textbf x’)\delta\beta E_{\text{ext}}(\textbf x)}\]

Solution #\(19\): The Zeeman coupling is linear, so mathematically it is a chemist’s Legendre transform of the free energy density \(\mathcal F\) from \(\phi\to E_{\text{ext}}\):

\[\mathcal F=-E_{\text{ext}}\phi+\frac{\mu^2}{2}\phi^2+\frac{\gamma}{2}\biggr|\frac{\partial\phi}{\partial\textbf x}\biggr|^2+…\]\]

Proving the \(1\)st identity concerning \(\langle\phi(\textbf x)\rangle\) is easy, and it can be directly substituted into the \(2\)nd identity to give:

where at the end one can also turn off \(E_{\text{ext}}(\textbf x)=0\) to get the ensemble average and \(2\)-point correlator in the original \(\textbf Z_2\) theory.

Problem #\(20\): Using the results of Solution #\(17\), show that, reverting back to the quadratic approximation of \(\mathcal F\):

\[\langle\phi(\textbf x)\rangle=(E_{\text{ext}}*G)(\textbf x)=\int d^d\textbf x’E_{\text{ext}}(\textbf x’)G(\textbf x-\textbf x’)\]

\[\langle\delta\phi(\textbf x)\delta\phi(\textbf x’)\rangle=\beta^{-1}G(\textbf x-\textbf x’)\]

where \(G(\textbf x)\) is almost a stationary point of the quadratic free energy \(F\) through being the fundamental Green’s function of the Helmholtz operator \(\mu^2-\gamma\biggr|\frac{\partial}{\partial\textbf x}\biggr|^2\) on \(\textbf R^d\):

\[G(\textbf x)=\int\frac{d^d\textbf k}{(2\pi)^d}\frac{e^{-i\textbf k\cdot\textbf x}}{k^2+1/\xi^2}\]

Solution #\(20\): Essentially the same as Solution #\(11\) except that need to first complete the square in the free energy (which uses the reality of the magnetic field \(E_{\text{ext}}(\textbf x)\in\textbf R\Leftrightarrow E_{\text{ext}}^{-\textbf k}=\left(E_{\text{ext}}^{\textbf k}\right)^{\dagger}\))

\[F=\int\frac{d^d\textbf k}{(2\pi)^d}\left(\frac{\mu^2+\gamma |\textbf k|^2}{2}\biggr|\phi_{\textbf k}-\frac{E_{\text{ext}}^{\textbf k}}{\mu^2+\gamma |\textbf k|^2}\biggr|^2-\frac{|E_{\text{ext}}^{\textbf k}|^2}{2(\mu^2+\gamma |\textbf k|^2)}\right)\]

and then do the Gaussian path integral to get the original partition function (when \(E_{\text{ext}}=0\)) with an additional Plancherelian contribution from \(E_{\text{ext}}\neq 0\):

\[\ln Z=\frac{1}{2}\sum_{\textbf k}\ln\frac{\pi V}{\beta(\mu^2+\gamma |\textbf k|^2)}+\frac{\beta}{2}\int\frac{d^d\textbf k}{(2\pi)^d}\frac{|E_{\text{ext}}^{\textbf k}|^2}{\mu^2+\gamma |\textbf k|^2}\]

Or, substituting \(E_{\text{ext}}^{\textbf k}=\int d^d\textbf x E_{\text{ext}}(\textbf x)e^{-i\textbf k\cdot\textbf x}\) to revert from \(\textbf k\mapsto\textbf x\) (in anticipation that one would like to take functional derivatives with respect to \(E_{\text{ext}}(\textbf x)\)):

\[\ln Z=\frac{1}{2}\sum_{\textbf k}\ln\frac{\pi V}{\beta(\mu^2+\gamma |\textbf k|^2)}+\frac{\beta}{2}\int d^d\textbf x d^d\textbf x’ E_{\text{ext}}(\textbf x)E_{\text{ext}}(\textbf x’)G(\textbf x-\textbf x’)\]

From here, the functional derivatives are straightforward to take (and only the \(2\)nd term above matters):

(aside: it seems that \(E_{\text{ext}}(\textbf x)\) can be interpreted as some kind of fluctuation distribution which, upon convolving with the \(2\)-point correlator, solves the inhomogeneous Helmholtz equation:

\[\left(\mu^2-\gamma\biggr|\frac{\partial}{\partial\textbf x}\biggr|^2\right)\langle\phi(\textbf x)\rangle=E_{\text{ext}}(\textbf x)\]

is there an intuitive interpretation of this?)

Problem #\(21\): Check explicitly that:

\[\left(\mu^2-\gamma\biggr|\frac{\partial}{\partial\textbf x}\biggr|^2\right)G(\textbf x)=\delta^d(\textbf x)\]

Solution #\(21\):

Problem #\(22\): Show that the isotropic \(2\)-point correlator has the Ornstein-Zernicke asymptotics:

\[\beta\langle\delta\phi(\textbf x)\delta\phi(\textbf x’)\rangle\sim\begin{cases}2^{-3d/2}\pi^{(d-1)/2}e^{-d/8}d^{(d-4)/2}\gamma^{-1}\frac{1}{r^{d-2}}, & r\ll\xi\\ 2^{-(d+3)/2}\pi^{(1-d)/2}\gamma^{-1}\frac{e^{-r/\xi}}{\xi^{(d-3)/2}r^{(d-1)/2}}, & r\gg\xi
\end{cases}\]

where \(r:=|\textbf x-\textbf x’|\) the identity \(\frac{d-3}{2}+\frac{d-1}{2}=d-2\) ensures dimensional consistency and is a good way to remember it (and indeed the \(d-2\) follows on dimensional analysis grounds).

Solution #\(22\): The derivation involves a clever saddle-point approximation:

Problem #\(23\): What are the critical exponents \(\nu,\eta\) in the mean-field, quadratic approximation to the full Landau-Ginzburg theory?

Solution #\(23\): The critical point is defined to be where the quadratic coupling \(\mu^2=0\) vanishes. So get any kind of critical exponent, this always just means: write down a formula for the quantity of interest in terms of \(\mu\sim |T-T_c|^{1/2}\). For the correlation length, this is trivial:

\[\xi=\frac{\sqrt{\gamma}}{\mu}\sim\mu^{-1}=|T-T_c|^{-1/2}:=|T-T_c|^{-\nu}\]

so \(\nu=1/2\). In particular, as \(T\to T_c\), the correlation length diverges \(\xi\to\infty\). This also means that at the critical point, the only relevant regime in the Ornstein-Zernicke \(2\)-point correlator is \(r\ll\xi\to\infty\), so:

\[\langle\delta\phi(\textbf x)\delta\phi(\textbf x’)\rangle\sim\frac{1}{r^{d-2}}:=\frac{1}{r^{d-2+\eta}}\]

So \(\eta=0\).

Problem #\(24\): Using the Ginzburg criterion, rationalize why the upper critical dimension \(d_c=4\) for the Ising model.

Solution #\(24\): Conceptually, the Ginzburg criterion is a common sense idea. It says that mean-field theory has sensible things to say about the critical point \(T\to T_c\) iff:

\[\text{fluctuations about mean field}\ll\text{mean field itself}\]

or, mathematically (integrating only within a \(\xi\)-ball due to the exponentially-decaying Ornstein-Zernicke correlation for \(|\textbf x|>\xi\)):

\[\int_{|\textbf x|\leq\xi}d^d\textbf x\langle\delta\phi(\textbf x)\delta\phi(\textbf 0)\rangle\ll\int_{|\textbf x|\leq\xi}d^d\textbf x\phi_0^2\]

\[\frac{\xi^2}{\beta\gamma}\ll\xi^d\phi_0^2\]

\[\frac{\xi^{2-d}}{\phi_0^2}\ll\beta\gamma\]

\[\frac{|T-T_c|^{\nu(d-2)}}{|T-T_c|^{2\beta}}\ll\beta\gamma\]

But at the critical point \(\beta\gamma\to\beta_c\gamma_c\) which is just finite number…so in order for the left side of the equality to also remain bounded as \(T\to T_c\), require:

\[\nu(d-2)\geq 2\beta\]

Using the mean-field exponents \(\beta=\nu=1/2\), this amounts to constraining \(d\geq 4=d_c\). This may seem a bit sketchy given that the mean-field exponents for \(\beta,\nu\) were used in the Ginzburg criterion to show that mean-field theory is consistent. Rather, it just means that MFT is self-consistent. On the other hand, for \(d<4\), MFT literally predicts its own demise!

Problem #\(1\): Define the Poincaré group.

Solution #\(1\): In words, the Poincaré group is the isometry group of Minkowski spacetime \(\textbf R^{1,3}\). Mathematically, it is the semidirect product \(\textbf R^{1,3}⋊O(1,3)\) of the normal subgroup \(\textbf R^{1,3}\) of spacetime translations with the Lorentz subgroup \(O(1,3)\) of rotations, Lorentz boosts, parity, and time reversal:

The reason the Poincaré group is a semidirect product rather than simply a direct product \(\textbf R^{1,3}\times O(1,3)\) is that spacetime translations and Lorentz transformations talk to each other via the Poincaré group’s composition rule:

\[(\Delta X_2,\Lambda_2)\cdot(\Delta X_1,\Lambda_1):=(\Delta X_2+\Lambda_2\Delta X_1,\Lambda_2\Lambda_1)\]

Problem #\(2\): With the exception of parity and time reversal, all symmetries of the Poincaré group can be implemented continuously:

where the green box of proper, orthochronous Poincaré transformations is the connected component of the identity \(1\) and the only one of the \(4\) connected components which by itself comprises a subgroup of the Poincaré group. Among the \({4}\choose{2}\)\(=6\) pairs of connected components one can join together, the \(2\) pairs:

\[\textbf R^{1,3}⋊SO(1,3)=\textbf R^{1,3}⋊SO^+(1,3)\cup\textbf R^{1,3}⋊SO^-(1,3)\]

and

\[\textbf R^{1,3}⋊O^+(1,3)=\textbf R^{1,3}⋊SO(1,3)\cup\textbf R^{1,3}⋊\not{S}O(1,3)\]

are the only subgroups of the Poincaré group \(\textbf R^{1,3}⋊O(1,3)\). Anyways, this digression is just to say that, modulo the Klein \(4\)-group quotient structure \(\{1,\Pi,\Theta,\Pi\Theta\}\), the proper, orthochronous Poincaré group \(\textbf R^{1,3}⋊SO^+(1,3)\) is a Lie group and hence can be studied via its Lie algebra. Classify fully the structure of this so-called Poincaré algebra.

Solution #\(2\): Consider spacetime translations \(\Delta X\in\textbf R^{1,3}\) first. It is clear that:

\[e^{-\Delta X\cdot\frac{\partial}{\partial X}}X=X-\Delta X\]

where \(\Delta X\cdot\frac{\partial}{\partial X}=\Delta X^{\mu}\partial_{\mu}\) is a Lorentz scalar. Although there’s nothing quantum mechanical about what is being done here, for sake of comparison one can artificially introduce \(\hbar\) and define the generator of spacetime translations \(P:=-i\hbar\frac{\partial}{\partial X}\Leftrightarrow P_{\mu}=-i\hbar\partial_{\mu}\) so that a general spacetime translation would be given by the infinite-dimensional unitary representation \(e^{-i\Delta X\cdot P/\hbar}\). Since spacetime translations are abelian, the algebra is simply characterized by \([P_{\mu},P_{\nu}]=0\).

Considering proper, orthochronous Lorentz transformations \(\Lambda\in SO^+(1,3)\) next, recall that membership in \(O(1,3)\) means that:

\[\Lambda^T\eta\Lambda=\eta\Leftrightarrow(\Lambda^T)_{\mu}^{\space\space\nu}\eta_{\nu\rho}\Lambda^{\rho}_{\space\space\sigma}=\eta_{\mu\sigma}\]

So if one substitutes \(\Lambda=1+\omega\Leftrightarrow \Lambda^{\mu}_{\space\space\nu}=\delta^{\mu}_{\space\space\nu}+\omega^{\mu}_{\space\space\nu}\) for infinitesimal \(\omega\), one finds that any such generator \(\omega\in\frak{so}^+\)\((1,3)\) must be antisymmetric:

\[\omega^T\eta=-\eta\omega\Leftrightarrow\omega_{\mu\nu}=-\omega_{\nu\mu}\]

which is unsurprising considering \(SO^+(1,3)\) is similar to \(SO(4)\). Thus, one can adapt the obvious \(\frak{so}\)\((4)\) basis of \(6\) antisymmetric generators to the Lorentz Lie algebra \(\frak{so}\)\(^+(1,3)\). There are various notations with which one can express these \(6\) generators:

More Intuitive Notation: One has \(3\) generators \(\textbf J:=(J_1,J_2,J_3)\) for rotations and \(3\) generators \(\textbf K:=(K_1,K_2,K_3)\) for Lorentz boosts such that by definition a rotation through angular displacement \(\Delta\boldsymbol{\phi}\) is given by \(e^{-i\Delta\boldsymbol{\phi}\cdot\textbf J/\hbar}\) and a Lorentz boost through rapidity \(\Delta\boldsymbol{\varphi}\) is given by \(e^{-i\Delta\boldsymbol{\varphi}\cdot\textbf K/\hbar}\). One can kind of “cheat” using one’s prior knowledge about how macroscopic Lorentz transformation matrices \(\Lambda\) look in the case of rotations or Lorentz boosts about various axes, “infinitesimalize” them, and extract the corresponding generator. For example, for a rotation about the \(z\)-axis, as \(\Delta\phi\to 0\) one expects to \(\mathcal O(\Delta\phi)\):

\[\begin{pmatrix}1&0&0&0\\0&\cos\Delta\phi&-\sin\Delta\phi&0\\0&\sin\Delta\phi&\cos\Delta\phi&0\\0&0&0&1\end{pmatrix}\approx\begin{pmatrix}1&0&0&0\\0&1&-\Delta\phi&0\\0&\Delta\phi&1&0\\0&0&0&1\end{pmatrix}=\begin{pmatrix}1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}-\frac{i\Delta\phi J_3}{\hbar}\]

from which one obtains:

\[J_3=i\hbar\begin{pmatrix}0&0&0&0\\0&0&-1&0\\0&1&0&0\\0&0&0&0\end{pmatrix}\]

and similarly for \(J_1,J_2\). Meanwhile, repeating the procedure for an infinitesimal Lorentz boost by \(\Delta\varphi\to 0\) along the \(x\)-axis:

\[\begin{pmatrix}\cosh\Delta\varphi&-\sinh\Delta\varphi&0&0\\-\sinh\Delta\varphi&\cosh\Delta\varphi&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}\approx \begin{pmatrix}1&-\Delta\varphi&0&0\\-\Delta\varphi&1&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}=\begin{pmatrix}1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}-\frac{i\Delta\varphi K_1}{\hbar}\]

So that the generator of Lorentz boosts along the \(x\)-axis is:

\[K_1=i\hbar\begin{pmatrix}0&-1&0&0\\-1&0&0&0\\0&0&0&0\\0&0&0&0\end{pmatrix}\]

and similarly for \(K_2,K_3\). In light of these matrix representations for the Lorentz Lie algebra generators, one can directly compute their commutation relations:

\[[J_i,J_j]=i\hbar\varepsilon_{ijk}J_k\]

\[[J_i,K_j]=i\hbar\varepsilon_{ijk}K_k\]

\[[K_i,K_j]=-i\hbar\varepsilon_{ijk}J_k\]

where the last commutator expresses (via the BCH formula) the counterintuitive phenomenon of Wigner rotation associated with non-collinear \(i\neq j\) Lorentz boosts.

Less Intuitive Notation: The cross product on \(\textbf R^3\) realizes a representation of the rotation algebra \(\frak{so}\)\((3)\) given by assigning each antisymmetric tensor \(\omega\in\frak{so}\)\((3)\) to the unique vector \(\boldsymbol{\omega}\in\textbf R^3\) such that \(\omega\textbf x=\boldsymbol{\omega}\times\textbf x\) for all \(\textbf x\in\textbf R^3\). More explicitly:

\[\omega=\begin{pmatrix}0&-\omega_3&\omega_2\\\omega_3&0&-\omega_1\\-\omega_2&\omega_1&0\end{pmatrix}\]

\[\boldsymbol{\omega}=\begin{pmatrix}\omega_1\\\omega_2\\\omega_3\end{pmatrix}\]

Or in indices, \(\omega_{ij}=-\varepsilon_{ijk}\omega_k\) or, inverting by contraction with a suitable Levi-Civita symbol, one has the equivalent form \(\omega_k=-\frac{1}{2}\varepsilon_{ijk}\omega_{ij}\). Here, the vector \(\boldsymbol{\omega}\) is arguably more intuitive than the tensor \(\omega\) even though both contain exactly the same information.

In precisely this same spirit, one can take the more intuitive “vector” generators \(\textbf J,\textbf K\) for rotations and Lorentz boosts and convert them into the equivalent but less intuitive form of a \(4\times 4\) matrix, each of whose \(16\) elements is itself a \(4\times 4\) matrix (one could even think of this as a \(16\times 16\) matrix if one so desired). That is, it is typical to define:

\[(\mathcal M^{\rho\sigma})^{\mu\nu}=\eta^{\rho\mu}\eta^{\sigma\nu}-\eta^{\sigma\mu}\eta^{\rho\nu}\]

where \(0\leq \mu,\nu,\rho,\sigma\leq 3\) are all spacetime indices (this is partly the reason for introducing this less intuitive representation of the Lorentz algebra because it puts space and time on more equal footing in this compact expression). This is related to the more intuitive representation above by:

\[i\hbar\mathcal M=\begin{pmatrix}0&-K_1&-K_2&-K_3\\K_1&0&J_3&-J_2\\K_2&-J_3&0&J_1\\K_3&J_2&-J_1&0\end{pmatrix}=\begin{pmatrix}0&-\textbf K^T\\\textbf K& -J\end{pmatrix}\]

where \(J=\textbf J\times\) as in the example with \(\omega=\boldsymbol{\omega}\times\) earlier, or in indices:

\[i\hbar\mathcal M^{ij}=\varepsilon_{ijk}J_k\Leftrightarrow J_i=\frac{i\hbar}{2}\varepsilon_{ijk}\mathcal M^{jk}\]

\[K_i=i\hbar\mathcal M_{i0}\]

One can then compute the Lorentz algebra in a single sweep:

Lowering the \(\nu\) index with the metric (it’s more useful to have it like this since Lorentz transformations have a conventionally NW-SE index structure \(\Lambda^{\mu}_{\space\space\nu}\) and one is anticipating generating these by exponentiating linear combinations of these generators):

\[(\mathcal M^{\rho\sigma})^{\mu}_{\space\space\nu}=\eta^{\rho\mu}\delta^{\sigma}_{\space\space\nu}-\eta^{\sigma\mu}\delta^{\rho}_{\space\space\nu}\]

so that an arbitrary Lorentz algebra element \(\omega\in\frak{so}^+\)\((1,3)\) (not necessarily infinitesimal anymore) is some linear combination of the \(6\) generators \(\mathcal M^{\rho\sigma}\in\frak{so}^+\)\((1,3)\) with real coefficients \(\Delta\Phi_{\rho\sigma}\in\textbf R\) quantifying the extent of a rotation or Lorentz boost (they are related to the earlier more intuitive representation by \(\Delta\phi_i=\) and \(\Delta\varphi_i=\)):

\[\omega=\frac{1}{2}\Delta\Phi_{\rho\sigma}\mathcal M^{\rho\sigma}\Leftrightarrow\omega^{\mu}_{\space\space\nu}=\Delta\Phi^{\mu}_{\space\space\nu}\]

where the factor of \(1/2\) is to compensate for double-counting (since both \(\Delta\Phi_{\sigma\rho}=-\Delta\Phi_{\rho\sigma}\) and \(\mathcal M^{\sigma\rho}=-\mathcal M^{\rho\sigma}\) are antisymmetric in their indices \(\rho,\sigma\)) (explicitly write out \(\Phi\) to have \(\textbf J,\textbf K\) entries inside).

Finally, spacetime translations talk with Lorentz transformations (recall the proper, orthochronous Poincaré group is given by a semi-direct product \(\textbf R^{1,3}⋊SO^+(1,3)\)), so to fully specify the Poincaré algebra, one also has to figure out how the generators \(P:=(H/c,\textbf P)\) of \(\textbf R^{1,3}\) talk with the generators \(\textbf J,\textbf K\) of \(\frak{so}^+\)\((1,3)\), not just how they talk within their own Lie subalgebras.

It turns out one can concisely summarize all the remaining commutation relations within the Poincaré algebra in the more intuitive form:

\[[H,\textbf J]=\textbf 0\Leftrightarrow [H,J_i]=0\]

\[[H,\textbf K]=i\hbar c\textbf P\Leftrightarrow [H,K_i]=i\hbar cP_i\]

\[\textbf P\times\textbf J=i\hbar\textbf P\Leftrightarrow [P_i,J_j]=i\hbar\varepsilon_{ijk}P_k\]

\[[\textbf P,\textbf K]_{\otimes}=i\hbar\frac{H}{c}1\Leftrightarrow [P_i,K_j]=i\hbar\frac{H}{c}\delta_{ij}\]

or the less intuitive but more compact/Lorentz invariant form:

\[[\mathcal M^{\rho\sigma},P^{\mu}]=\eta^{\rho\mu}P^{\sigma}-\eta^{\sigma\mu}P^{\rho}\]

Problem #\(3\): Motivate the general definition of a Clifford algebra in mathematics, state the anticommutation relations of the specific Clifford algebra \(\text{Cl}_{1,3}(\textbf R)\), and state the chiral/Weyl representation of \(\text{Cl}_{1,3}(\textbf R)\).

Solution #\(3\): Given a vector space \(V\) over a field \(F\), a quadratic form \(Q:V\to F\) is any function with \(2\) properties:

1. \(Q(\lambda\textbf v)=\lambda^2Q(\textbf v)\) for all vectors \(\textbf v\in V\) and scalars \(\lambda\in F\).
2. The function \(\langle\space|\space\rangle:V\times V\to F\) defined by \(\langle\textbf v|\textbf w\rangle:=\frac{1}{2}(Q(\textbf v+\textbf w)-Q(\textbf v)-Q(\textbf w))\) is a bilinear form (called the polarization of \(Q\)).

Note that \(Q\) and \(\langle\space|\space\rangle\) contain exactly the same information, since one can also invert \(Q(\textbf v)=\langle\textbf v|\textbf v\rangle\). A general quadratic form on Euclidean space \(\textbf R^n\) is of the form \(Q(\textbf v):=\textbf v^Tg\textbf v\) (where without loss of generality one can take \(g^T=g\) to be symmetric) whose associated symmetric bilinear form is \(\langle\textbf v|\textbf w\rangle=(\textbf v^Tg\textbf w+\textbf w^Tg\textbf v)/2\). In particular, when \(g=1\) is the standard metric on Euclidean space, then \(\langle\textbf v|\textbf w\rangle=\textbf v\cdot\textbf w\) coincides with the usual dot product.

Given a quadratic space \((V,F,Q)\), one can construct from this the Clifford algebra \(\text{Cl}(V,F,Q)\) by starting with the tensor algebra \(\oplus_{k=0}^{\infty}V^{\otimes k}\) of \(V\) and quotienting it by the relation \(\textbf v^2:=\textbf v\otimes\textbf v=Q(\textbf v)1_V\) for all \(\textbf v\in V\). From this, it follows as a corollary that more generally, for any two vectors \(\textbf v,\textbf w\in V\):

\[\{\textbf v,\textbf w\}=\textbf v\textbf w+\textbf w\textbf v:=\textbf v\otimes\textbf w+\textbf w\otimes\textbf v=2\langle\textbf v|\textbf w\rangle 1_V\]

where in physics contexts the tensor product \(\textbf v\otimes\textbf w\) is often abbreviated to just \(\textbf v\textbf w\) and called the geometric product. Furthermore, since this anticommutator \(\{\textbf v,\textbf w\}\) is bilinear, it may be completely specified by finding a basis \(\textbf e_i\) of \(V\) and simply specifying the value of \(\{\textbf e_i,\textbf e_j\}\) for all pairs of basis vectors \(\textbf e_i,\textbf e_j\). Returning to the example of \(\textbf R^n\) with the standard metric, if one chooses an orthonormal basis \(\hat{\textbf e}_i\cdot\hat{\textbf e}_j=\delta_{ij}\), then the Clifford algebra is simply characterized by \(\{\hat{\textbf e}_i,\hat{\textbf e}_j\}=2\delta_{ij}1\). More explicitly, the \(\hat{\textbf e}_i\) all anticommute with each other and each squares to the identity \(\hat{\textbf e}^2_i=1\).

In \(\textbf R^2\), picking an orthonormal basis \(\hat{\textbf e}_1,\hat{\textbf e}_2\), although a general element of the Clifford algebra \(\text{Cl}_2(\textbf R)\) is formally:

\[a1+(b\hat{\textbf e}_1+c\hat{\textbf e}_2)+(d\hat{\textbf e}_1\hat{\textbf e}_1+e\hat{\textbf e}_1\hat{\textbf e}_2+f\hat{\textbf e}_2\hat{\textbf e}_1+g\hat{\textbf e}_2\hat{\textbf e}_2)+…\]

The anticommutation relations of the Clifford algebra allow one to substantially reduce the number of degrees of freedom e.g. \(\hat{\textbf e}_1\hat{\textbf e}_1=\hat{\textbf e}_2\hat{\textbf e}_2=1\) and \(\hat{\textbf e}_1\hat{\textbf e}_2=-\hat{\textbf e}_2\hat{\textbf e}_1\)). All higher-order multivectors also reduce to either a scalar, a vector, or a bivector, e.g. \(\hat{\textbf e}_1\hat{\textbf e}_2\hat{\textbf e}_1=-\hat{\textbf e}_1\hat{\textbf e}_1\hat{\textbf e}_2=-\hat{\textbf e}_2\). So in fact, a general element of the Clifford algebra \(\text{Cl}_2(\textbf R)\) is simply spanned by \(4\) generators:

\[a1+b\hat{\textbf e}_1+c\hat{\textbf e}_2+d\hat{\textbf e}_1\hat{\textbf e}_2\]

While the scalars and vectors \(1^2=\hat{\textbf e}_1^2=\hat{\textbf e}_2^2=1\) all square to the identity, the bivector \((\hat{\textbf e}_1\hat{\textbf e}_2)^2=-1\) squares to minus the identity! This is reminiscent of the identity \(i^2=-1\), where \(i=\sqrt{-1}\) is the imaginary unit of \(\textbf C\).

In \(\textbf R^3\), now with an orthonormal basis \(\hat{\textbf e}_1,\hat{\textbf e}_2,\hat{\textbf e}_3\), a general element of the Clifford algebra \(\text{Cl}_3(\textbf R)\) is parameterized by \(8\) real degrees of freedom (extrapolating, it should be clear by Pascal’s triangle that \(\dim\text{Cl}_{n}(\textbf R)=2^n\)):

\[a+b\hat{\textbf e}_1+c\hat{\textbf e}_2+d\hat{\textbf e}_3+e\hat{\textbf e}_1\hat{\textbf e}_2+f\hat{\textbf e}_2\hat{\textbf e}_3+g\hat{\textbf e}_1\hat{\textbf e}_3+h\hat{\textbf e}_1\hat{\textbf e}_2\hat{\textbf e}_3\]

note that \(\text{Cl}_3(\textbf R)\) can be explicitly represented by the \(3\) Pauli matrices \(\{\sigma_i,\sigma_j\}=2\delta_{ij}1\). In addition, the bivectors \((\hat{\textbf e}_1\hat{\textbf e}_2)^2=(\hat{\textbf e}_2\hat{\textbf e}_3)^2=(\hat{\textbf e}_1\hat{\textbf e}_3)^2=-1\) all square to minus the identity while \(\hat{\textbf e}_1\hat{\textbf e}_2\hat{\textbf e}_2\hat{\textbf e}_3\hat{\textbf e}_3\hat{\textbf e}_1=-1\), reminiscent of the defining relations \(i^2=j^2=k^2=ijk=-1\) of the quaternions \(\textbf H\) (H for “Hamilton”).

In special relativity, Minkowski spacetime \(\textbf R^{1,3}\) is a real vector space that naturally comes with a quadratic form \(Q(X):=X^T\eta X\) induced by the Minkowski metric \(g=\eta=\text{diag}(1,-1,-1,-1)\). Its associated \(16\)-dimensional Clifford algebra \(\text{Cl}_{1,3}(\textbf R)\) is therefore subject to the anticommutation relations:

\[\{\gamma^{\mu},\gamma^{\nu}\}=2\eta^{\mu\nu}1\]

for which one representation (and it turns out the only \(4\)-dimensional irreducible representation up to unitary similarity) is the chiral/Weyl representation:

\[\gamma^{\mu}=\begin{pmatrix}0&\sigma^{\mu}\\\sigma_{\mu}&0\end{pmatrix}\]

where \(\sigma^{\mu}=(1,\boldsymbol{\sigma})\) and \(\sigma_{\mu}=(1,-\boldsymbol{\sigma})\).

Problem #\(4\):

Solution #\(4\):

Problem #\(5\):

Solution #\(5\):

Problem #\(6\):

Solution #\(6\):

Problem #\(7\): Motivate (from a historical perspective) the classical Lagrangian density \(\mathcal L\) for the Dirac bispinor field \(\psi:\textbf R^{1,3}\to\textbf C^4\), and hence obtain the on-shell Dirac equation. Write down the general solution to the Dirac equation (given that it’s linear!).

Solution #\(7\): Recall that for a non-relativistic free particle, the on-shell dispersion relation \(H=\textbf P^2/2m\) is first order in the energy \(H\), whereas for a relativistic free particle it is second order \(H^2=c^2\textbf P^2+m^2c^4\). The former quantizes to the Schrodinger equation \((\partial_i\partial_i+2ik_c\partial_0)\psi=0\), while the latter quantizes to the Klein-Gordon equation \((\partial^{\mu}\partial_{\mu}+k_c^2)\phi=0\). At first glance, one might think that \(\phi(X)\) would, like the wavefunction \(\psi(X)\), also admit a probabilistic interpretation given by the Born rule, but historically there were \(2\) reasons why this was a bit suspicious.

Objection #\(1\): In non-relativistic QM one would only have needed to specify \(|\psi(t=0)\rangle\) to get the entire future time evolution of \(|\psi(t)\rangle\) because the Schrodinger equation is first-order in time \(t\). So it seems a bit weird that, just to be relativistically compatible, one would also need to specify the initial velocity \(\partial_0|\psi(t=0)\rangle\).

Objection #\(2\): On the same note of being \(2\)-nd order in time, the Klein-Gordon equation has \(2\) linearly independent solutions:

\[\phi(X)\sim\int d^3\textbf k(a_{\textbf k}e^{i(\textbf k\cdot\textbf x-\omega_{\textbf k}t)}+a_{\textbf k}^{\dagger}e^{-i(\textbf k\cdot\textbf x-\omega_{\textbf k}t)}\]

(this should be contrasted with the Schrodinger equation which would have only had a single linearly independent \(e^{i(\textbf k\cdot\textbf x-\omega_{\textbf k}t)}\) where the energy is always positive b/c it’s just kinetic energy, so get a positive semi-definite probability density). In particular, the conserved current is \(J^{\mu}=\) and the probability density \(J^0=\phi^2\) can be negative?

An obvious way to get around Objection #\(1\) is to instead Fourier transform the square root of the earlier relativistic dispersion relation \(H=\pm\sqrt{c^2\textbf P^2+m^2c^4}\)…indeed, in some sense this is what Dirac did. But notice the \(\pm\) signs!

Ironically though, in the modern understanding of QFT, despite the whole motivation being to somehow interpret \(\phi\) probabilistically, it actually does not admit such an interpretation. Similar to Bohr’s derivation of the gross structure of the hydrogenic atom, the method is not fully sound in its foundations but the answer it gives turns out to be correct.

However, the Klein-Gordon equation, being a second order differential equation, has \(2\) linearly independent solutions:

\[\phi(X)\sim\int d^3\textbf k(a_{\textbf k}e^{i(\textbf k\cdot\textbf x-\omega_{\textbf k}t)}+a_{\textbf k}^{\dagger}e^{-i(\textbf k\cdot\textbf x-\omega_{\textbf k}t)}\]

Recalling that the

\[i\hbar\frac{\partial|\psi\rangle}{\partial t}=H|\psi\rangle\]

The Dirac Lagrangian \(\mathcal L=\mathcal L(\psi,\bar{\psi})\) is:

\[\mathcal L=\bar{\psi}(i\hbar\displaystyle{\not\!\partial}-mc1)\psi\]

So by varying the action with respect to the Dirac adjoint bispinor field \(\bar{\psi}:=\psi^{\dagger}\gamma^0\), one obtains the Dirac equation:

\[(i\hbar\displaystyle{\not\!\partial}-mc1)\psi=0\]

Solution #\(7\):

Problem #\(8\):

Solution #\(8\):

Problem #\(9\):

Solution #\(9\):

Problem #\(10\):

Solution #\(10\):

Problem #\(11\): Verify that, in spite of the anticommutation relations that were imposed in canonical quantization of the Dirac free field theory, the resulting creation and annihilation operators have the expected commutation relations with the Hamiltonian \(:H:\):

\[[:H:,(b_{\textbf k}^{m_s})^{\dagger}]=\hbar\omega_{\textbf k}(b_{\textbf k}^{m_s})^{\dagger}\]

\[[:H:,b_{\textbf k}^{m_s}]=-\hbar\omega_{\textbf k}b_{\textbf k}^{m_s}\]

\[[:H:,(c_{\textbf k}^{m_s})^{\dagger}]=\hbar\omega_{\textbf k}(c_{\textbf k}^{m_s})^{\dagger}\]

\[[:H:,c_{\textbf k}^{m_s}]=-\hbar\omega_{\textbf k}c_{\textbf k}^{m_s}\]

Solution #\(11\):

These commutation relations assert that the spectrum of Dirac free FT are given by Fock states describing particles and antiparticles carrying arbitrary momentum \(\textbf k\in\textbf R^3\) and spin angular momentum \(m_s\in\{\pm 1/2\}\) that emerge from excitations of the vacuum.

Problem #\(1\): Write down a general \(\phi\)-dependent perturbation to the Klein-Gordon Lagrangian density \(\mathcal L\) for a real scalar field \(\phi\), and explain why in practice only the first \(2\) terms of such a perturbation need to be considered.

Solution #\(1\): Assuming the potential \(V(\phi)\) is analytic in \(\phi\):

\[\mathcal L=\frac{1}{2}\partial^{\mu}\phi\partial_{\mu}\phi-\frac{1}{2}k_c^2\phi^2-\sum_{n\geq 3}\frac{\lambda_n}{n!}\phi^n\]

where the coupling constants \(\lambda_0=\lambda_1=0\), \(\lambda_2=k_c^2\), etc. are not to be confused with the Compton wavelength \(\lambda:=h/mc\). Because the action \(S=\int d^4|X\rangle\mathcal L\) has dimensions of angular momentum \([S]=[\hbar]\) and \([d^4|X\rangle]=[\lambda]^4\), it follows that \([\mathcal L]=[\hbar]/[\lambda]^4=[\hbar^{-3}(mc)^4]\) and thus \([\phi]=[\hbar^{-1/2}mc]\), so:

\[[\lambda_n\phi^n]=[\mathcal L]\Rightarrow [\lambda_n]=[\hbar^{(n-6)/2}(mc)^{4-n}]\]

So on dimensional analysis grounds, the \(n\)-th order coupling constant \(\lambda_n\) does not by itself give a meaningful assessment as to the relevance of the corresponding \(\phi^n\) perturbation, rather it is the dimensionless parameter \(\hbar^{(6-n)/2}(E/c)^{n-4}\lambda_n\) (where \([E]=[mc^2]\) is the relevant energy scale of the process/physics at work) that really matters, i.e. the \(n\)-th order \(\phi^n\) perturbation is small iff \(\hbar^{(6-n)/2}(E/c)^{n-4}\lambda_n\ll 1\).

By graphing \(\hbar^{(6-n)/2}(E/c)^{n-4}\lambda_n\) as a function of \(E\) and given the non-negotiable fact that one is typically interested in “low-\(E\) physics”, it follows that all the \(\phi^n\) couplings for \(n\geq 5\) are irrelevant at these low energy scales, and instead only the marginal \(\phi^4\) quartic coupling and relevant \(\phi^3\) cubic coupling need to be considered. This analysis is quite deep, and goes to show the power of dimensional analysis.

Problem #\(2\): What does it mean for an interacting QFT to be weakly coupled?

Solution #\(2\): It means that one can legitimately treat higher-order interaction terms as small perturbations of a free QFT, i.e. that perturbation theory is accurate (otherwise it would a strongly coupled QFT which is a whole different beast…)

Problem #\(3\): Derive the Dyson series solution to the linear dynamical system \(\dot{\textbf x}(t)=A(t)\textbf x(t)\) and show that it can be expressed as a time-ordered exponential. Show explicitly that it satisfies the ODE.

Solution #\(3\): First, “normalize” away the uninteresting initial condition \(\textbf x(0)\) by writing \(\textbf x(t)=U(t)\textbf x(0)\) so that \(\dot U(t)=A(t)U(t)\) with \(U(0)=1\). Then integrating both sides:

\[U(t)=1+\int_0^tdt’A(t’)U(t’)\]

And recursively substitute ad infinitum:

\[=1+\int_0^tdt’A(t’)\left(1+\int_0^{t’}dt^{\prime\prime}A(t^{\prime\prime})U(t^{\prime\prime})\right)=1+\int_0^tdt’A(t’)+\int_0^tdt’\int_0^{t’}dt^{\prime\prime}A(t’)A(t^{\prime\prime})+…\]

Recall that the definition of the time-ordering superoperator \(\mathcal T\) applied to a string of operators \(A_1(t_1)A_2(t_2)…A_n(t_n)\) is to reorder them so that the earliest time operators also act first (i.e. are on the right). For instance, for a string of \(2\) operators:

\[\mathcal T[A_1(t_1)A_2(t_2)]=A(t_1)A(t_2)\Theta(t_1-t_2)+A(t_2)A(t_1)\Theta(t_2-t_1)\]

or for a string of \(3\) operators, there will be \(3!=6\) terms:

\[\mathcal T[A_1(t_1)A_2(t_2)A_3(t_3)]=A_1(t_1)A_2(t_2)A_3(t_3)\Theta(t_1-t_2)\Theta(t_2-t_3)+A_1(t_1)A_3(t_3)A_2(t_2)\Theta(t_1-t_3)\Theta(t_3-t_2)+…\]

In particular, the Dyson series can be equivalently expressed as the time-ordering superoperator \(\mathcal T\) applied to the strings of operators obtained once one Taylor expands the naive exponential solution:

\[e^{\int_0^tA(t’)dt’}:=1+\int_0^tA(t’)dt’+\frac{1}{2}\int_0^tdt’\int_0^tdt^{\prime\prime}A(t’)A(t^{\prime\prime})+\frac{1}{6}\int_0^tdt’\int_0^tdt^{\prime\prime}\int_0^tdt^{\prime\prime\prime}A(t’)A(t^{\prime\prime})A(t^{\prime\prime\prime})+…\]

Applying \(\mathcal T\) (which is linear):

\[\mathcal T[e^{\int_0^tA(t’)dt’}]=1+\int_0^tA(t’)dt’+\frac{1}{2}\int_0^tdt’\int_0^tdt^{\prime\prime}\biggr(A(t’)A(t^{\prime\prime})\Theta(t’-t^{\prime\prime})+A(t^{\prime\prime})A(t’)\Theta(t^{\prime\prime}-t’)\biggr)+…\]

\[=1+\int_0^tA(t’)dt’+\frac{1}{2}\int_0^tdt’\int_0^{t’}dt^{\prime\prime}A(t’)A(t^{\prime\prime})+\frac{1}{2}\int_0^tdt’\int_{t’}^{t}dt^{\prime\prime}A(t’)A(t^{\prime\prime})+…\]

Among the \(2\) quadratic terms, the first one looks like the one in the Dyson series, but how about the second one? In fact it is equal to the first term; to see this draw a picture to facilitate interchanging the order of integration:

so it becomes clear that \(\int_0^t dt’\int_{t’}^{t}dt^{\prime\prime}=\int_0^tdt^{\prime\prime}\int_0^{t^{\prime\prime}}dt’\). Finally, if one wishes one can interchange the dummy variables \(t’\leftrightarrow t^{\prime\prime}\) to make the two double integrals look utterly identical. As one may anticipate, it turns out that all \(3!=6\) triple integrals at cubic order also coincide, precisely cancelling the \(1/3!=1/6\) prefactor in the Taylor expansion, etc. so that the Dyson series can indeed be written as a time-ordered exponential.

It is also satisfying to explicitly check that the Dyson series for \(U(t)\) satisfies the ODE:

\[\dot U(t)=\frac{d}{dt}\biggr(1+\int_0^tdt’A(t’)+\int_0^tdt’\int_0^{t’}dt^{\prime\prime}A(t’)A(t^{\prime\prime})+…\biggr)=A(t)+A(t)\int_0^t dt’A(t’)+…=A(t)U(t)\]

Or perhaps a more “slick” derivation is to appeal to the time-ordered exponential form of the Dyson series:

\[\dot U(t)=\frac{d}{dt}\mathcal T[e^{\int_0^tA(t’)dt’}]=\mathcal T\left[\frac{d}{dt}e^{\int_0^tA(t’)dt’}\right]\]

and then recognize that within \(\mathcal T\) all operators commute (e.g. for a string of \(2\) operators it is clear that \(\mathcal T[A_1(t_1)A_2(t_2)]=\mathcal T[A_2(t_2)A_1(t_1)]\)) so one can just differentiate naively:

\[=\mathcal T[A(t)e^{\int_0^tA(t’)dt’}]=A(t)\mathcal T[e^{\int_0^tA(t’)dt’}]\]

where the last step follows because \(t\) is the latest time so it can be “factored out” of \(\mathcal T\) to the left.

Problem #\(4\): Write down the Lagrangian density \(\mathcal L\) for scalar Yukawa QFT. What is the condition for this interacting QFT to be weakly coupled? By inspecting \(\mathcal L\), what charges remain conserved despite the interactions? Obtain the scalar Yukawa Hamiltonian density \(\mathcal H\).

Solution #\(4\): Given a real scalar field \(\phi\) and a complex scalar field \(\psi\):

\[\mathcal L=\frac{1}{2}\partial^{\mu}\phi\partial_{\mu}\phi+\partial^{\mu}\bar{\psi}\partial_{\mu}\psi-\frac{1}{2}k_{\phi}^2\phi^2-k_{\psi}^2\bar{\psi}\psi-g\phi\bar{\psi}\psi\]

where the relevant cubic coupling interaction \(g\phi\bar{\psi}\psi\) will be a small perturbation to the free QFT (so that this interacting QFT is weakly coupled) iff the coupling constant \(g=\lambda_3/6\) satisfies (using the results of Solution #\(1\)):

\[\sqrt{\hbar}g\ll k_{\phi},k_{\psi}\]

Just as for the free theory, since one still has the \(U(1)\) global internal symmetry \(\psi\mapsto e^{i\alpha}\psi\) in spite of the cubic coupling term, one continues to have a conserved charge \(Q=N_{\psi}-N_{\bar{\psi}}\).

Performing the relevant Legendre transforms:

\[\partial_0\phi\mapsto\pi_{\phi}=\partial_0\phi\]

\[\partial_0\psi\mapsto\pi_{\psi}=\partial_0\bar{\psi}\]

\[\partial_0\bar{\psi}\mapsto\pi_{\bar{\psi}}=\partial_0\psi\]

One obtains the Hamiltonian density \(\mathcal H=\mathcal H_{\text{free}}+\mathcal H_{\text{int}}\) for scalar Yukawa QFT, where:

\[\mathcal H_{\text{free}}=\frac{\pi^2_{\phi}}{2}+\frac{1}{2}\biggr|\frac{\partial\phi}{\partial\textbf x}\biggr|^2+\pi_{\bar{\psi}}\pi_{\psi}+\frac{\partial\bar{\psi}}{\partial\textbf x}\cdot\frac{\partial\psi}{\partial\textbf x}+\frac{1}{2}k^2_{\phi}\phi^2+k^2_{\psi}\bar{\psi}\psi\]

and:

\[\mathcal H_{\text{int}}=g\phi\bar{\psi}\psi\]

Problem #\(5\): Within scalar Yukawa QFT, explain why the scattering amplitude for the process \(\phi\to\bar{\psi}\psi\) is non-zero, and hence calculate it to order \(g\); what is an assumption implicit in this question?

Solution #\(5\): The implicit assumption is that the initial and final scattering states \(|t=-\infty\rangle,|t=\infty\rangle\in\mathcal F\) are indeed Fock states of the free QFT, i.e. \(H_{\text{free}}=\int d^3\textbf x\mathcal H_{\text{free}}\)-eigenstates, specifically the Lorentz-normalized excitations of the vacuum:

\[|t=-\infty\rangle:=\sqrt{\frac{2}{\hbar}}\frac{\omega_{\textbf k_{\phi}}}{c}a_{\textbf k_{\phi}}^{\dagger}|0\rangle\]

\[|t=\infty\rangle:=\frac{2\omega_{\textbf k_{\bar{\psi}}}\omega_{\textbf k_{\psi}}}{\hbar c^2}b_{\textbf k_{\psi}}^{\dagger}c_{\textbf k_{\bar{\psi}}}^{\dagger}|0\rangle\]

After a long time, in the interaction picture the initial state \(|t=-\infty\rangle\) evolves to \(\mathcal T\exp\left(-\frac{i}{\hbar}\int_{-\infty}^{\infty}dtH_{\text{int}}\right)|t=-\infty\rangle\) (where \(H_{\text{int}}\) is the interaction Hamiltonian in the interaction picture). The overlap of this with a particular final state \(|t=\infty\rangle\) thus yields the probability amplitude of that scattering process:

\[\langle t=\infty|\mathcal T\exp\left(-\frac{i}{\hbar}\int_{-\infty}^{\infty}dtH_{\text{int}}\right)|t=-\infty\rangle\]

where this is expected to be non-zero because \(Q=0\) is conserved in this process. If one is only interested in computing this scattering amplitude to order \(g\), then as usual one Taylor expands the time-ordered exponential to first order:

\[\approx \langle t=\infty|t=-\infty\rangle-\frac{i}{\hbar}\langle t=\infty|\int_{-\infty}^{\infty}dt H_{\text{int}}|t=-\infty\rangle\]

In this case, the initial and final states are orthogonal \(\langle t=\infty|t=-\infty\rangle=0\) which follows intuitively because a \(\phi\)-particle is not the same as a \(\bar{\psi}\psi\)-particle antiparticle pair, or mathematically the creation operators of the different particles all commute so either \(a_{\textbf k_{\phi}}^{\dagger}\) annihilates the bra \(\langle 0|\) or \(b_{\textbf k_{\psi}},c_{\textbf k_{\bar{\psi}}}\) annihilate the ket \(|0\rangle\).

The \(\mathcal O(g)\) matrix element term simplifies to:

\[-\frac{i}{\hbar c}\int d^4|X\rangle\langle t=\infty|\mathcal H_{\text{int}}|t=-\infty\rangle\sim-ig\int d^4|X\rangle\langle 0|b_{\textbf k_{\psi}}c_{\textbf k_{\bar{\psi}}}\phi\bar{\psi}\psi a_{\textbf k_{\phi}}^{\dagger}|0\rangle\]

At this point one has to pull out the Fourier normal mode plane wave expansions of \(\phi,\psi,\bar{\psi}\) (with time dependence since operators evolve in the interaction picture as if they were in the Heisenberg picture of the free QFT), annihilate as much as you can and otherwise use commutation relations to pick up delta functions where needed to do the \(13\)-dimensional integral. When the dust settles, one finds that the scattering amplitude for the process goes like \(-ig\delta^4|K_{\psi}+K_{\bar{\psi}}-K_{\phi}\rangle\) so it’s zero unless \(4\)-momentum is conserved. As a corollary, in the ZMF (rest frame) of the \(\phi\)-particle, one can check that this implies \(m_{\phi}\geq 2m_{\psi}\).

Problem #\(6\): Verify Wick’s theorem for the case of \(3\) scalar fields:

\[\mathcal T\phi_{|X_1\rangle}\phi_{|X_2\rangle}\phi_{|X_3\rangle}=:\phi_{|X_1\rangle}\phi_{|X_2\rangle}\phi_{|X_3\rangle}:+\Delta^F_{|X_2-X_3\rangle}\phi_{|X_1\rangle}+\Delta^F_{|X_3-X_1\rangle}\phi_{|X_2\rangle}+\Delta^F_{|X_1-X_2\rangle}\phi_{|X_3\rangle}\]

Solution #\(6\):

Problem #\(7\): Within weakly coupled scalar Yukawa QFT, compute the scattering amplitude for the process \(\bar{\psi}\psi\to\bar{\psi}\psi\) at order \(\mathcal O(g^2)\).

Solution #\(7\): At order \(\mathcal O(g^2)\), the scattering amplitude for this interaction from the initial state \(|t=-\infty\rangle\sim b_{\textbf k_1}^{\dagger}b_{\textbf k_2}^{\dagger}|0\rangle\) to the final state \(|t=\infty\rangle=\sim b_{\textbf k’_1}^{\dagger}b_{\textbf k’_2}^{\dagger}|0\rangle\) is given by:

\[\langle t=-\infty|S-1|t=\infty\rangle\sim i\mathcal A_{\bar{\psi}\psi\to\bar{\psi}\psi}\delta^4|K_{\psi}+K_{\bar{\psi}}-K_{\phi}\rangle\]

where to order \(\mathcal O(g^2)\), the “amplitude” \(\mathcal A_{\bar{\psi}\psi\to\bar{\psi}\psi}\) is given by the sum of the following two tree-level Feynman diagrams, namely a \(t\)-channel and a \(u\)-channel:

Problem #\(8\): Repeat Problem #\(7\) for the case of scalar Yukawa scattering of mesons \(\phi\phi\to\phi\phi\) at order \(\mathcal O(g^4)\).

Solution #\(8\): Given that the interaction Hamiltonian density for scalar Yukawa theory is cubic \(\mathcal H_{\text{int}}=g\phi\bar{\psi}\psi\), it follows that any Feynman diagram must be a cubic graph with exactly one \(\phi\)-edge, one \(\psi\)-edge, and one \(\bar{\psi}\)-edge at each vertex of the directed graph. By initially drawing the following:

and pondering for a bit, it is intuitively plausible that the simplest way to satisfy the above constraint is for the \(\phi\)-mesons to exchange virtual \(\psi/\bar{\psi}\) nucleons/antinucleons via the following one-loop (no longer tree-level) Feynman diagram:

and so because there are \(4\) vertices, it follows that the \(\phi\phi\to\phi\phi\) interaction scattering amplitude is \(\mathcal O(g^4)\). The Feynman rules assert that to \(\mathcal O(g^4)\) it is given by (notation here is a bit loose, in particular \(|\phi\phi\rangle^{\dagger}\neq\langle\phi\phi|\)):

\[\langle\phi\phi|S-1|\phi\phi\rangle\sim i\mathcal A_{\phi\phi\to\phi\phi}\delta^4(K’_1+K’_2-K_1-K_2)\]

where \(\mathcal A_{\phi\phi\to\phi\phi}\) is given by the one-loop integral over \(4\)-momenta:

Problem #\(10\): Define the Mandelstam variables \(s,t,u\in\textbf R\) by drawing suitable \(s\)-channel, \(t\)-channel and \(u\)-channel Feynman diagrams for interactions between two particles (of arbitrary type).

Solution #\(10\):

The purpose of this post is to explain what the toric code is, and its potential use as a fault-tolerant quantum error correcting stabilizer surface code for topological quantum computing.

To begin, consider an \(N\times N\) square lattice \(\Lambda\) with a qubit placed at each edge as follows:

The vertices of the lattice \(\Lambda\) are called stars \(S\) while its faces are called plaquettes \(P\) (which can also be viewed as the vertices of the dual lattice \(\Lambda^*\), etc.). Since \(\Lambda\) is an \(N\times N\) square lattice, one might initially think that there are \(2N(N+1)\) qubits; however the catch is that opposite ends of the lattice are identified with each other, so that there are in fact only \(2N^2\) qubits. More importantly, this imposition of periodic boundary conditions endows \(\Lambda\cong S^1\times S^1\) with the topology of a torus, hence the name “toric code”.

A priori, the state space \(\mathcal H\) of \(2N^2\) qubits has dimension \(\text{dim}\mathcal H=2^{2N^2}\) (though if some/all of these qubits are identical then it may be less than this). Using the notation \(\sigma_i^x\) as a shorthand for the operator \(1_1\otimes…\otimes 1_{i-1}\otimes\sigma^x\otimes 1_{i+1}\otimes…\otimes 1_{2N^2}\) acting on \(\mathcal H\) and similarly for \(\sigma_i^y\) and \(\sigma_i^z\), one can define star and plaquette four-body interaction operators \(A_S,B_P\) associated respectively to stars \(S\) or plaquettes \(P\) in \(\Lambda\) by:

\[A_S:=\prod_{i\in S}\sigma_i^z\]

\[B_P:=\prod_{i\in P}\sigma_i^x\]

where the notation “\(i\in S\)” denotes the set of \(4\) qubits nearest to a given star \(S\), and similarly for “\(i\in P\)”; the notation \(\prod_{i\in S}\) or \(\prod_{i\in P}\) does not suffer from any operator ordering ambiguity because Pauli operators on distinct qubits (regardless of their \(x,y,z\) nature) trivially commute.

For any two stars \(S,S’\), and any two plaquettes \(P,P’\), one can check (without any motivation yet at this point) that:

\[[A_S,A_{S’}]=[B_{P},B_{P’}]=[A_S,B_P]=0\]

Specifically, if the stars or plaquettes are well-separated from each other then these are always trivially true so it suffices to just check the “edge cases” so to speak when the stars or plaquettes are close enough to share qubits. Then, for each such shared qubit, one just has to apply the anticommutation relation \(\{\sigma^{\alpha},\sigma^{\beta}\}=2\delta^{\alpha\beta}1\). Notably, the commutation relation \([A_S,B_P]=0\) relies on the fundamental fact that a star \(S\) and plaquette \(P\) which are adjacent to each other always share only \(2\) qubits, and \(2\) is even so \((-1)^2=1\).

Finally, suppose one thinks of the lattice of qubits as a spin lattice such that the spins interact via the following Hamiltonian:

\[H=-\sum_{S\in\Lambda}A_S-\sum_{P\in\Lambda^*}B_P\]

From the above discussion, it follows that for any star \(S\) or plaquette \(P\), one has:

\[[H,A_S]=[H,B_P]=0\]

There are \(N^2\) stars and \(N^2\) plaquettes, so this would suggest that one is free to specify the however, using a “Stokes theorem” type of argument (basically a more refined version of earlier arguments for showing \([A_S,A_{S’}]=[B_P,B_{P’}]=0\)), one can convince oneself inductively (i.e. playing around with small \(N=1,2,…\) etc.) that it is precisely thanks to the toric topology of \(\Lambda\) that one has the constraints:

\[\prod_{S\in\Lambda}A_S=\prod_{P\in\Lambda^*}B_P=1\]

(not to be confused with the quantities in the Hamiltonian \(H\) which involve sums \(\sum\) rather than products \(\prod\)).

To emphasize again, this is why a torus is desirable…4-fold degeneracy in the ground state manifold of \(H\).

The purpose of this post is to explain why, experimentally, one only observes \(4\) electric dipole transitions in the IR spectrum of buckminsterfullerene, also known as \(\text C_{60}\) or informally as the buckyball:

Buckyball Basics

The simplest conceptual way to construct a buckyball is to start with a regular icosahedron:

which has \(F=20\) equilateral triangular faces, \(V=12\) vertices, and \(E=30\) edges; notice this obeys Euler’s formula \(F+V=E+2\). Then, by simply “shaving off” each of the \(V=12\) vertices; it is clear from the picture these \(12\) vertices would transform into \(12\) pentagonal faces, each surrounded by \(5\) hexagonal faces so that no \(2\) pentagonal faces share an adjacent edge, yielding a buckyball topology (and geometrically, all \(\text C-\text C\) covalent bonds should be of equal length). This shaving process increases the number of faces to \(F’=32\) (of which \(12\) are pentagonal and \(20\) are hexagonal), the number of vertices to \(V’=60\) (i.e. just the number of carbon atoms), and the number of edges to \(E’=90\) such that Euler’s formula is maintained \(F’+V’=E’+2\).

IR Spectroscopy Selection Rules

Recall that within the Born-Oppenheimer approximation, the \(N\) nuclei of some molecule are “clamped” at positions \(\textbf X:=(\textbf X_1,…,\textbf X_N)\) and are regarded as moving in the effective potential \(V_{\text{eff}}(\textbf X)\) due to the electrons \(e^-\), so that the molecular Hamiltonian is:

\[H=T_{\text n}+V_{\text{eff}}(\textbf X)\]

If in addition one approximates \(V_{\text{eff}}(\textbf X)\) by a harmonic potential about the (stable) equilibrium configuration \(\textbf X_0\) of the nuclei:

\[V_{\text{eff}}(\textbf X)\approx\frac{1}{2}(\textbf X-\textbf X_0)^T\left(\frac{\partial^2 V_{\text{eff}}}{\partial\textbf X^2}\right)_{\textbf X_0}(\textbf X-\textbf X_0)\]

then, upon diagonalizing the Hessian \(\left(\frac{\partial^2 V_{\text{eff}}}{\partial\textbf X^2}\right)_{\textbf X_0}\) into the orthonormal eigenbasis of its normal modes, one obtains \(3N\) decoupled simple harmonic oscillators so the spectrum of \(H\) is just that of an anisotropic harmonic oscillator in \(\textbf R^{3N}\), at least within all the assumptions made so far (e.g. ignoring anharmonicity of \(V_{\text{eff}}(\textbf X)\)). Each Fock eigenstate \(|n_1,n_2,…,n_{3N}\rangle\) of \(H\) (thought of as a vibrational eigenstate because one can view \(H\) as a Hamiltonian governing vibrations/SHM of the nuclei about equilibrium) is thus a product \(|n_1,n_2,…,n_{3N}\rangle=\otimes_{i=1}^{3N}|n_i\rangle\) of suitable \(1\)D quantum harmonic oscillator wavefunctions.

IR radiation is just like any other electromagnetic radiation in that (within the dipole approximation, which is fair for long-wavelength IR radiation despite the larger size of molecules) it stimulates electric dipole transitions via the time-dependent perturbation \(\Delta H=\Delta H(t)\) to the molecular Hamiltonian \(H\):

\[\Delta H=-\boldsymbol{\pi}\cdot\textbf E_0\cos \omega_{\text{IR}} t\]

where the electric dipole moment \(\boldsymbol{\pi}=\boldsymbol{\pi}(\textbf X,\textbf x)\) of the molecule is:

\[\boldsymbol{\pi}:=e\sum_{\text{nuclei }i}Z_i\textbf X_i-e\sum_{\text{electrons }i}\textbf x_i\]

and the molecule is assumed to be neutral \(\sum_{\text{nuclei }i}Z_i=\sum_{\text{electrons }i}\). By Fermi’s golden rule, the molecular transition rate between two distinct vibrational \(H\)-eigenstates \(|n_1,…,n_{3N}\rangle\to|n’_1,…,n’_{3N}\rangle\) is proportional to the mod-square of the matrix element of the (time-independent amplitude of the) perturbation:

\[|\langle n’_1,…,n’_{3N}|\boldsymbol{\pi}\cdot\textbf E_0|n\rangle|^2\]

However, most IR spectroscopy experiments (e.g. IR laser sources in FTIR spectrometers) use unpolarized IR radiation, so this means one should really replace by an isotropic averaging factor of \(1/3\) and forget about (as far as selection rules are concerned) a factor of \(|\textbf E_0|^2\), so just focus on:

\[|\langle n’_1,…,n’_{3N}|\boldsymbol{\pi}|n\rangle|^2\]

Similar to what was done earlier for the effective potential \(V_{\text{eff}}(\textbf X)\), one can also Taylor expand the dipole moment operator \(\boldsymbol{\pi}\) within the configuration space \(\textbf X\) of the nuclei about the equilibrium configuration \(\textbf X_0\):

\[\boldsymbol{\pi}\approx\boldsymbol{\pi}(\textbf X_0)+\left(\frac{\partial\boldsymbol{\pi}}{\partial\textbf X}\right)_{\textbf X_0}(\textbf X-\textbf X_0)+…\]

Sandwiching this in the matrix element, the constant term \(\boldsymbol{\pi}(\textbf X_0)\) (which represents the possibility of a permanent electric dipole like in a water molecule) vanishes by orthogonality of the distinct vibrational \(H\)-eigenstates, so the quantity of interest is just:

\[\left|\left(\frac{\partial\boldsymbol{\pi}}{\partial\textbf X}\right)_{\textbf X_0}\langle n’_1,…,n’_{3N}|\textbf X-\textbf X_0|n_1,…,n_{3N}\rangle\right|^2\]

This immediately leads to a “gross selection rule” for two vibrational \(H\)-eigenstates to be coupled by the IR perturbation \(\Delta H\), namely that the Jacobian \(\left(\frac{\partial\boldsymbol{\pi}}{\partial\textbf X}\right)_{\textbf X_0}\) must be non-vanishing; physically, this means that only the normal modes of the nuclear vibration that experience a change in \(\boldsymbol{\pi}\) when the nuclei are slightly displaced from \(\textbf X_0\) will be IR-active.

…main point of this section is to explain that IR transitions are low-energy excitations require a change in electric dipole moment b/w initial and final states, which, factoring out the charge e, means the matrix element of the position observable b/w two states must be non-zero by Fermi’s golden rule. There are actually \(2\) selection rules for IR spectroscopy; the gross selection rule is that \(\partial\boldsymbol{\pi}/\partial\) must be zero.

Character Tables of Finite Group Representations

As a warmup, consider the symmetric group \(S_3=\{1, (12), (13), (23), (123), (132)\}\) of order \(|S_3|=3!=6\). Then \(S_3\) can be partitioned into \(3\) conjugacy classes, namely \(S_3=\{1\}\cup\{(12), (13), (23)\}\cup\{(123), (132)\}\). Each \(S_3\) conjugacy class maps onto an \(S_3\)-irrep, so there will also be three \(S_3\)-irreps. One of them is always just the trivial irrep whereby the permutations do nothing to all vectors. For symmetric groups, there is also always the sign irrep \(1,(123),(132)\mapsto 1\) and \((12),(13),(23)\mapsto -1\) such that odd permutations do nothing to all vectors but even \(A_3\)-permutations flip vectors across the origin. Finally, there is the standard \(S_3\)-irrep that one might intuitively think about as acting on the vertices of an equilateral triangle which rigidly drags the whole Cartesian plane \(\textbf R^2\) along with it. Note the dimensions of these \(S_3\)-irreps are the only ones that could have been compatible with its order \(6=1^2+1^2+2^2\).

Each \(S_3\)-irrep is associated with its own character class function which can be evaluated on an arbitrary representative of each conjugacy class. For instance, the trivial irrep is \(1\)-dimensional and its character always evaluates to \(1\) on all conjugacy classes. The sign irrep is also \(1\)-dimensional but its character evaluated on each conjugacy class is instead the sign of the permutations in that conjugacy class. Finally, noting that \(\cos(2\pi/3)=-1/2\) and that the trace of a rotation in \(\textbf R^2\) by angle \(\theta\) is \(2\cos\theta\), the following character table for \(S_3\) may be obtained:

So far this example has been fairly abstract. To bridge this “abstract nonsense” with chemistry, consider the specific example of the ammonia molecule \(\text{NH}_3\):

In this case, rather than pulling a random group (like \(S_3\)) out of thin air, here one can use the ammonia molecule to motivate studying the specific group of actions on \(\textbf R^3\) that leave it looking like nothing happened. In chemist jargon, one is interested in the point group of the ammonia molecule, the adjective “point” implying that any such action on \(\textbf R^3\) must have at least one fixed point in space (commonly chosen to be the origin) so that e.g. translations of the ammonia molecule are disregarded (cf. the distinction in special relativity between the Poincare group and its Lorentz subgroup). Note that this is indeed a group thanks to its very definition (if one action leaves ammonia looking like nothing happened, and a second action also leaves ammonia looking like nothing happened, then composing them will also leave ammonia looking like nothing happened).

So what is the point group of the ammonia molecule? For this, there isn’t really any super rigorous/systematic way to do it, one just has to stare hard at the molecule and think…

Clearly, one way to leave ammonia looking like nothing happened is to literally do nothing. There is also manifestly \(C_3\) rotational symmetry (by \(120^{\circ}\) or \(240^{\circ}\)). Finally, there are \(3\) reflection symmetries about “vertical” mirror planes. As a mathematician, it is thus easy to recognize that the point group of the ammonia molecule is just the dihedral group \(D_3\) (which happens to be isomorphic to \(S_3\)). However, as a chemist one would instead refer to this as the “\(C_{3v}\) point group”, where the \(C\) and the \(3\) subscript are meant to emphasize the \(C_3\) subgroup of rotational symmetries mentioned above, while the “\(v\)” subscript is meant to emphasize that the mirror planes are “vertical”. Strictly speaking, one should also check that there are no horizontal mirror planes, inversion centers, or improper rotation axes in order to be able to confidently assert that the point group of ammonia really is \(C_{3v}\) rather than \(C_{3v}\) merely being a subgroup of an actually larger point group.

The isomorphism \(C_{3v}\cong D_3\cong S_3\) means that by happy chance the character table for the ammonia point group \(C_{3v}\) is already known. For instance, the transpositions \((12),(23),(13)\) in \(S_3\) map onto vertical mirror plane reflections in \(C_{3v}\), while the \(3\)-cycles \((123),(132)\) are simply \(C_3\) rotations. Indeed, the \(3\)-dimensional \(C_{3v}\)-representation on the ammonia molecule itself embedded rigidly in \(\textbf R^3\) is reducible into a direct sum of the trivial irrep acting on the \(1\)-dimensional \(z\)-axis and the standard irrep acting on the \(2\)-dimensional \(xy\)-plane. It is also worth verifying that the columns of the character table satisfy orthonormality \(\sum_{\text{ccl}\text{ of }C_{3v}}|\text{ccl}|\chi_{\phi}(\text{ccl})\chi_{\phi’}(\text{ccl})=|C_{3v}|\delta_{\phi\cong\phi’}\) for any \(2\) \(C_{3v}\)-irreps \(\phi,\phi’\).

One can also consider how various scalar-valued and vector-valued polynomials on \(\textbf R^3\) transform passively under the \(C_{3v}\) point group. For instance, any function \(f=f(\rho)\) of the cylindrical coordinate \(\rho:=\sqrt{x^2+y^2}\) only, such as \(f(\rho)=\rho^2\), will transform under the trivial \(C_{3v}\)-irrep.

Making the reasonable assumption that all \(60\) carbon atoms are specifically of the \(^{12}\text C\) isotope, and are therefore identical bosons, so something about the permutation group \(S_{60}\)? (formalize the earlier comment about “looking the same” via the idea of identical quantum particles/ bosons and fermions where in that case the vectors are not in \(\textbf R^3\) but state vectors in the symmetrized tensor product Hilbert space of identical bosons)

The purpose of this post is to explain several techniques for laser cooling and laser trapping of atoms. In order to better emphasize key conceptual points, it will take the approach of posing problems, followed immediately by their solutions.

Problem #\(1\): Recall that any locally conserved field is associated to a continuity equation \(\dot{\rho}+\partial_{\textbf x}\cdot\textbf J=0\). In particular, the local flow of the corresponding conserved quantity can always be interpreted in terms of a velocity field \(\textbf v\) (with SI units of meters/second) by enforcing that \(\textbf J=\rho\textbf v\). Interpret this in the context of fluid mechanics and electromagnetism.

Solution #\(1\): In fluid mechanics, mass is locally conserved, so \(\rho\) could represent mass density while \(\textbf J\) represents mass flux. Momentum is also locally conserved provided there are no external body forces acting on the fluid, so in that case \(\rho\) would be a vector field representing momentum density while \(\textbf J\) would be the stress tensor field representing momentum transport (this is just the Navier-Stokes equations). Similar comments can be made about energy, angular momentum, etc.

In electromagnetism, electric charge is locally conserved, so in that case \(\rho\) represents charge density while \(\textbf J\) represents an electric current. Similarly, the total energy stored in both electromagnetic fields and electric charges is locally conserved. In that case, the continuity equation in linear dielectrics takes the form:

\[\frac{\partial\mathcal E}{\partial t}+\textbf J_f\cdot\textbf E+\frac{\partial}{\partial\textbf x}\cdot\textbf S=0\]

where the electromagnetic field energy density is \(\mathcal E:=(\textbf D\cdot\textbf E+\textbf H\cdot\textbf B)/2\) and the Poynting vector \(\textbf S:=\textbf E\times\textbf H\). If an electromagnetic wave propagates with group velocity \(|\textbf v|\leq c\) in some linear dielectric, it follows that \(\textbf S=\mathcal E\textbf v\) (similar comments apply to momentum density; indeed they are unified in the Maxwell stress tensor).

Problem #\(2\): Within the framework of classical electromagnetism, what is the formula for the radiation pressure \(p_{\gamma}\) exerted at normal incidence to an absorbing surface \(\hat{\textbf n}\) with reflection coefficient \(R\)?

Solution #\(2\): Just as \(\textbf S=\mathcal E\textbf v\) (from Solution #\(1\)), so the radiation pressure is a certain projection of an isotropic current for the momentum density \(\boldsymbol{\mathcal P}\):

\[p_{\gamma}=(1+R)(\boldsymbol{\mathcal P}\cdot\hat{\textbf n})(\textbf v\cdot\hat{\textbf n})=(1+R)|\boldsymbol{\mathcal P}||\textbf v|\cos^2\theta\]

The photon dispersion relation \(\mathcal E=|\boldsymbol{\mathcal P}||\textbf v|\) in a linear dielectric then implies that the radiation pressure \(p_{\gamma}=(1+R)\mathcal E\cos^2\theta\) is given by a Malusian proportionality with the energy density stored in the EM field (aside: how is this related to the equation of state \(p=\mathcal E/3\) for a photon gas?).

Problem #\(3\): An atom moving at (nonrelativistic) speed \(v\) (in the lab frame) and meets a counter-propagating laser beam of incident intensity \(I\) and frequency \(\omega\) (both in the lab frame, where of course it is assumed that the laser source is also at rest in the lab frame). Estimate the velocity-dependent scattering force \(F_{\gamma}(v)\) exerted on the atom and its maximum possible value \(F_{\gamma}^*> F_{\gamma}(v)\).

Solution #\(3\): Heuristically, the scattering force might also be called the radiation force as it’s roughly just the radiation pressure \(p_{\gamma}\) exerted on the cross-section \(\sigma(\omega_D)\) of the atom (evaluated at the Doppler-shifted frequency \(\omega_D:=\omega\sqrt{(1+\beta)/(1-\beta)}\approx \omega+kv\)):

\[F_{\gamma}(v)\approx p_{\gamma}\sigma(\omega_D)=\frac{I}{c}\frac{\sigma_{01}}{1+(2\delta_D/\Gamma)^2}\]

where in the last expression it is assumed that the Doppler detuning \(\delta_D=\omega_D-\omega_{01}\) is small so in particular both \(I\) and \(\sigma_{01}\) are taken to be independent of \(\omega_D\). This is from the perspective of stimulated absorption. On the other hand, because all the analysis is implicitly in the steady state, one can also view this from the perspective of spontaneous emission:

\[F_{\gamma}(v)=(\hbar k)(\Gamma\rho_{11})=\frac{\Gamma}{2}\frac{s}{1+s+(2\delta_D/\Gamma)^2}\]

Note that these two formulas for \(F_{\gamma}(v)\) are not exactly equal, but approximately so at low irradiance \(s\ll 1\) (how to rationalize why?). Meanwhile, on resonance \(\delta_D=0\), in the high-irradiance \(s\to\infty\) limit, the theoretical max scattering force achievable is \(F_{\gamma}^*=\hbar k\Gamma/2\).

Problem #\(4\): Atoms flying out of an oven at temperature \(T\) are collimated to form an atomic beam which meets a counterpropagating laser \(s,\omega\). Without any frequency compensation on the laser \(\omega\), show how to determine the terminal velocity \(v_{\infty}\) of the atoms.

Solution #\(4\): The idea is that when one is far detuned from \(\omega_D\), there is hardly any scattering force and so the atoms move at roughly constant velocity. It is only in a small window \(\omega_D\in[\omega_{01}-\Gamma/2,\omega_{01}+\Gamma/2]\) that there is any significant scattering force. In this near-resonance regime, one can therefore set up an equation of motion \(M\dot v=-F_{\gamma}(v)\) whose solution is:

\[(1+s)(v_{\infty}-v_0)+\frac{4}{3\Gamma k^2}(\delta_{D,\infty}^3-\delta_{D,0}^3)=-\frac{\hbar k\Gamma s}{2M}\Delta t\]

where the initial speed can be taken as the most probable one \(v_0=\sqrt{2k_BT/M}\), and the time \(\Delta t\) can be estimated by considering the situation where \(\omega+kv_0=\omega_{01}+\Gamma/2\) and \(\omega+kv_{\infty}=\omega_{01}-\Gamma/2\) (actually not sure about this).

Problem #\(5\): If the atomic beam emerging from the oven along with the counterpropagating laser are oriented along some “\(z\)-axis”, write down the magnetic field \(\mathbf B(z)=B(z)\hat{\textbf k}\) required to make a Zeeman slower.

Solution #\(5\): Assuming one is working at relatively low magnetic field strengths, the low-field basis which diagonalizes the perturbative hyperfine Hamiltonian \(\Delta H_{\text{HFS}}\propto\textbf I\cdot\mathbf J\) is \(|f,m_f\rangle\). In this basis, the expectation of the perturbative Zeeman Hamiltonian \(\Delta H_{\text{Zeeman}}\) is approximately:

\[\langle f,m_f|\Delta H_{\text{Zeeman}}|f,m_f\rangle=g_f\mu_BB(z)m_f\hbar\]

Thus, if one is exploiting an electric dipole transition from a “ground state” \(|fm_f\rangle\) to an “excited state” \(|f’m’_f\rangle\), then one requires the difference in the Zeeman shifts experienced by these \(2\) levels to precisely compensate for the Doppler detuning \(\delta_D=\omega+kv-\omega_{01}\):

\[(g_{f’}m’_f-g_fm_f)\frac{\mu_BB(z)}{\hbar}=\delta_D\]

In practice, one would like to achieve a constant deceleration \(a>0\) so that \(v^2=v_0^2-2az\). This requires a magnetic field from e.g. a tapered solenoid of the form:

\[B(z)=B_0\sqrt{1-\frac{z}{z_0}}+B_{\text{bias}}\]

where \(z_0:=v_0^2/2a\) is the stopping distance, \(B_0:=\frac{\hbar kv_0}{\mu_B(g_{f’}m’_f-g_fm_f)}\) can be interpreted as \(B_0=B(0)-B(z_0)\), and the bias field is \(B_{\text{bias}}=\frac{\hbar(\omega-\omega_{01})}{\mu_B(g_{f’}m’_f-g_fm_f)}\); or at least, if one wishes to completely stop the atoms at \(z=z_0\) (if not, then one may need to experimentally lower \(B_{\text{bias}}\) until one obtains a desired terminal velocity; in that case one should then discontinuously turn off the magnetic field \(B(z):=0\) for \(z>z_0\) so that the E\(1\) transition would again be substantially detuned from resonance and so experience a negligible scattering force \(F_{\gamma}\) after the atoms exit the Zeeman slower).

Problem #\(6\): For the specific case of say sodium atoms \(\text{Na}\) (analogous remarks apply to any alkali atom), explain why electric dipole transitions between the stretched states \(|fm_f\rangle=|22\rangle\) in \(3s_{1/2}\) and \(|f’m’_f\rangle=|33\rangle\) in \(3p_{3/2}\) are a good choice for laser cooling via the scattering force \(F_{\gamma}\)? In this case, what should be the corresponding value of \(B_0\) used in a Zeeman slower (the transition has \(\lambda_{01}=589\text{ nm}\))? What should be the polarization of the incident laser light?

Solution #\(6\): This E\(1\) transition is not only allowed but also forms a closed cycle thanks to the usual E\(1\) selection rules, ensuring no optical pumping into dark states that would prevent a given atom from experiencing any further scattering. One can check that \(B_0\approx 0.12\text{ T}=1200\text{ G}\) is reasonable to achieve in the lab. The incident photons should be \(\sigma^+\) circularly-polarized along the quantization axis \(\hat{\textbf k}\) defined by the \(\textbf B\)-field along which the atomic beam is also propagating.

Problem #\(7\): Show, stating any assumptions made, that the velocity-dependent optical molasses force \(F_m(v)\approx -\alpha v\) along an arbitrary axis looks like linear damping with some damping coefficient \(\alpha>0\) for an appropriately-chosen laser detuning \(\delta\). Hence, show that the atom’s kinetic energy \(T(t)=T(0)e^{-t/\tau}\) (supposedly) decays exponentially with time constant \(\tau=m/2\alpha\).

Solution #\(7\): Along any of the \(3\) Cartesian axies, the component of the optical molasses force \(F_m(v)\) along that axis is a superposition of two opposite (but not necessarily equal) scattering forces:

\[F_m(v)=-F_{\gamma}(v)+F_{\gamma}(-v)\approx -2v\frac{\partial F_{\gamma}}{\partial v}(v=0)\]

where the low-velocity assumption \(kv/\Gamma\ll 1\) has been made. This shows that the damping coefficient is:

\[\alpha=2\frac{\partial F_{\gamma}}{\partial v}(v=0)=-4\hbar k^2s\frac{2\delta/\Gamma}{(1+s+(2\delta/\Gamma)^2)^2}\]

So in order for \(\alpha>0\) to actually be damping, one requires \(\delta<0\) to be red-detuned as suggested in the picture. It turns out that in order to treat the two laser beams independently (as has been implicitly done above), they both have the same low \(s\ll 1\) so that neither one saturates the E\(1\) transition too much. In this simple model of optical molasses, one should therefore really take:

\[\alpha=2\frac{\partial F_{\gamma}}{\partial v}(v=0)=-4\hbar k^2s\frac{2\delta/\Gamma}{(1+(2\delta/\Gamma)^2)^2}\]

In some sense, the fact that it looks like linear damping near \(v=0\) is not really conceptually any different from the fact that e.g. systems looks like simple harmonic oscillators near stable equilibria; it’s just taking the \(1\)-st order term in a Taylor series in \(v\).

Kinetic energy is lost through the dissipative power of the molasses force:

\[\dot T=F_m(v)v=-\alpha v^2=-\frac{2\alpha}{m}T\Rightarrow T(t)=T(0)e^{-t/\tau}\]

where \(\tau=m/2\alpha\) is usually on the order of microseconds. This suggests that there is no cooling limit, i.e. that one can quickly cool down arbitrarily close to \(T\to 0\), but this turns out to be impossible, see Problem #\(9\).

Problem #\(8\): For a random walk \(\textbf X_1,\textbf X_2,…,\textbf X_N\) of \(N\) steps, each of identical length \(L=|\textbf X_1|=|\textbf X_2|=…=|\textbf X_N|\), what is the expectation of the total displacement squared \(\langle|\textbf X_1+\textbf X_2+…+\textbf X_N|^2\rangle\)?

Solution #\(8\): All the dot products vanish so:

\[\langle|\textbf X_1+\textbf X_2+…+\textbf X_N|^2\rangle=NL^2\]

Problem #\(9\): Conceptually, what is the issue with the approach taken in Solution #\(7\)? Upon amending it, show that there is actually a Doppler cooling limit, namely a minimum temperature \(T_D\) that can be achieved using the optical molasses technique, given by:

\[k_BT_D=\frac{\hbar\Gamma}{2}\]

Solution #\(9\): The issue with Solution #\(7\) is that, while it accounts for the momentum kick given to the atom during stimulated absorption of photons and also recognizes that spontaneous emission averages to no momentum kick, it neglects fluctuations in both the number of photons absorbed and in the number of photons scattered. Using the result of Problem #\(8\), except here conceptually it’s not a random walk in real space but rather in momentum space with \(\langle p_z^2\rangle(t)=2(\hbar k)^2(2\gamma)t\), the kinetic energy \(T_z\) of an atom in the optical molasses now evolves as (see Foote textbook for the detailed arguments, some sketchy assumptions involved too, and turns out ultimately wrong, see Problem #\(10\)):

\[\dot T_z=-\frac{2\alpha}{m} T_z+\left(1+3\times\frac{1}{3}\right)\frac{\hbar^2k^2}{2m}(2\gamma)\]

So in the steady state \(\dot T_z=0\) and using equipartition \(T_z=k_BT/2\), one finds that:

\[k_BT=-\frac{\hbar\Gamma}{4}\frac{1+(2\delta/\Gamma)^2}{2\delta/\Gamma}\]

is minimized for red-detuned \(\delta<0\) at \(2\delta/\Gamma=-1\), yielding the Doppler cooling limit \(k_BT_D=\hbar\Gamma/2\) as claimed.

Problem #\(10\): Explain why, even after accounting for Poissonian fluctuations in the stimulated absorption and spontaneous emission of photons, the Doppler cooling limit for the optical molasses technique above rests on a “spherical cows in vacuum” foundation.

Solution #\(10\): Simply put, the assumption of a \(2\)-level atom has been implicitly lurking in the discussion the whole time, but it turns out that if one leverages the existence of other sublevels, one can do even better than the Doppler cooling limit above would suggest. Note also that the optical molasses laser cooling technique only provides confinement in \(\textbf k\)-space, but not so much in \(\textbf x\)-space which is also relevant; in other words, laser trapping is to \(\textbf x\) what laser cooling is to \(\textbf k\).

Problem #\(11\): In light of Problem #\(10\), draw \(3\) pictures that summarize how a magneto-optical trap (MOT) works. In particular, it should be clear that the MOT could not work with just a \(2\)-level atom. Also write down explicitly a formula for the MOT force \(F_{\text{MOT}}(z,v)\).

Solution #\(11\):

The MOT force \(F_{\text{MOT}}(z,v)\) is conceptually identical to the optical molasses force \(F_m(v)\) only now the detuning includes not only the \(v\)-dependent Doppler shift but also the \(z\)-dependent Zeeman shift (more precisely, the formula \(F_{\text{MOT}}(z,v)\) that will be given below is the component of the net MOT force \(\textbf F_{\text{MOT}}(\textbf x,\textbf v)\) along the quantization \(z\)-axis of the quadrupolar anti-Helmholtz MOT \(\textbf B\)-field near the origin \(\textbf x\approx\textbf 0\) whose purpose is to generate a nearly uniform magnetic field gradient; a similar equation holds in the \(\rho\)-direction except one should note from \(\frac{\partial}{\partial\textbf x}\cdot\textbf B=0\) that \(\frac{\partial B_{\rho}}{\partial\rho}=-\frac{1}{2}\frac{\partial B_z}{\partial z}\)).

\[F_{\text{MOT}}(z,v)=F_{\gamma}\left(\delta=\omega-kv-\left(\omega_{01}+\frac{g’_f\mu_B}{\hbar}\frac{\partial B_z}{\partial z}z\right)\right)-F_{\gamma}\left(\delta=\omega+kv-\left(\omega_{01}-\frac{g’_f\mu_B}{\hbar}\frac{\partial B_z}{\partial z}z\right)\right)\]

Making the same low Doppler shift assumption \(kv\ll\Gamma\) but now also a low Zeeman shift assumption \(\frac{g’_f\mu_B}{\hbar}\frac{\partial B_z}{\partial z}z\ll\Gamma\), one can check that this reduces to the previous linear damping force of the optical molasses together with a Hookean spring force:

\[F_{\text{MOT}}(z,v)=-\alpha v-\beta z\]

where the spring constant is \(\beta=\frac{\alpha g’_f\mu_B}{\hbar k}\frac{\partial B_z}{\partial z}\) and the damping coefficient \(\alpha\) is as before in Problem #\(7\). Usually, the atom will be overdamped (as is desirable) \(\alpha^2>4\beta m\).

Problem #\(12\): State one advantage and one disadvantage of a MOT compared with just the optical molasses technique.

Solution #\(12\): The advantage of a MOT is that its capture velocity \(v_{c,\text{MOT}}\gg v_{c,m}\) is much greater than that of optical molasses so it is a better trapper, hence its name (indeed, one can load a MOT simply by firing a room-\(T\) vapor, or to get more atoms one can first Zeeman slow them). The disadvantage is that it is not as good of a laser cooler; optical molasses is able to reach much colder temperatures, surpassing the naive Doppler cooling limit \(k_BT_D=\hbar\Gamma/2\) mentioned earlier (this is because of various sub-Doppler cooling mechanisms, notably Sisyphus cooling, that are elaborated upon in the next problem; all that matters for now is that such sub-Doppler cooling mechanisms are effectively disabled when the Zeeman shift exceeds the light shift as in a MOT).

Problem #\(13\): In a MOT, where does the trapping come from?

Solution #\(13\): It does not come from the quadrupolar \(\textbf B\)-field, but rather

Problem #\(14\): Draw a picture to explain how Sisyphus cooling allows. What is the new cooling limit?

Problem #\(15\): Show that the magnetic field strength \(|\textbf B(\textbf x)|\) in free space can never have a strict local maximum provided \(\dot{\textbf E}=\textbf 0\) (this is one version of Earnshaw’s theorem).

Solution #\(15\): One has the usual identity:

\[\frac{\partial}{\partial\textbf x}\times\left(\frac{\partial}{\partial\textbf x}\times\textbf B\right)=\frac{\partial}{\partial\textbf x}\left(\frac{\partial}{\partial\textbf x}\cdot\textbf B\right)-\left|\frac{\partial}{\partial\textbf x}\right|^2\textbf B\]

But thanks to the assumptions of the problem the term on the left vanishes, and in addition to the absence of magnetic monopoles, one concludes that \(\left|\frac{\partial}{\partial\textbf x}\right|^2\textbf B=\textbf 0\).

Now, \(|\textbf B(\textbf x)|\) satisfies the claim if and only if \(|\textbf B(\textbf x)|^2\) satisfies it. In turn, one just has to check that the Hessian \(\frac{\partial^2|\textbf B|^2}{\partial\textbf x^2}\) does not have \(3\) negative eigenvalues anywhere. A sufficient (but not necessary) condition for this would be if \(\text{Tr}\frac{\partial^2|\textbf B|^2}{\partial\textbf x^2}\geq 0\) everywhere. But the trace of the Hessian is just the Laplacian of the original scalar field:

\[\text{Tr}\frac{\partial^2|\textbf B|^2}{\partial\textbf x^2}=\left|\frac{\partial}{\partial\textbf x}\right|^2|\textbf B|^2=2\left|\frac{\partial|\textbf B|}{\partial\textbf x}\right|^2+2|\textbf B|\left|\frac{\partial}{\partial\textbf x}\right|^2\textbf B=2\left|\frac{\partial|\textbf B|}{\partial\textbf x}\right|^2\geq 0\]

which completes the proof.

Problem #\(16\): What is the high-level mechanism of magnetic trapping? What are the \(3\) most common kinds of magnetic traps? (just draw a picture of each one with some annotations, no need for long explanations)

Solution #\(16\): Just as in the Stern-Gerlach experiment, magnetic dipoles in non-uniform magnetic fields experience a force that pushes them towards weaker or stronger magnetic field depending on their orientation relative to the magnetic field. The difference is that in the Stern-Gerlach experiment the atoms are flying in at hundreds of meters/second and getting deflected, way beyond capture velocity, whereas in magnetic trapping they have already been laser cooled. The diThe \(3\) most common magnetic traps are the quadrupole trap (made from a pair of anti-Helmholtz coils as in a MOT), a time-averaged orbiting potential (TOP) trap, and an Ioffe-Pritchard trap, the last of which is the most common and simply consists of \(4\) long cylindrical bars (which form a “linear quadrupole” in direct analogy with an electrostatic quadrupole) and \(2\) Helmholtz-like coils. Key words: radial/\(\rho\)-trapping and axial/\(z\)-trapping. Due to the result of Problem #\(14\), it follows that one can only trap atoms in low-field seeking states (such atoms are called low-field seekers).

Good overview here: https://arxiv.org/pdf/1310.6054

The purpose of this post is to review the theory behind several standard amplitude-splitting interferometers.

Michelson Interferometer

The simplest possible version of a Michelson interferometer is the following:

For now, consider for simplicity a monochromatic light source of frequency \(\omega=ck=nvk=n’v’k=n^{\prime\prime}v^{\prime\prime}k\). In “chronological order”, the phases accrued by \(2\) amplitude-split reflected and transmitted rays in their respective journeys from the light source to the photodiode are explicitly:

\[\phi_1=nkx_0+\frac{n’ka}{2}+\pi+\frac{n’ka}{2}+nkx_2+\pi+nkx_2+\frac{n’ka}{2}+\frac{n^{\prime\prime}kb}{\sqrt{1-n’^2/2{n^{\prime\prime}}^2}}+\frac{n’ka}{2}+nkx_3\]

\[\phi_2=nkx_0+\frac{n’ka}{2}+\frac{n^{\prime\prime}kb}{\sqrt{1-n’^2/2{n^{\prime\prime}}^2}}+\frac{n’ka}{2}+nkx_1+\pi+nkx_1+\frac{n’ka}{2}+\pi+\frac{n’ka}{2}+nkx_3\]

where it has been assumed that \(n'<n^{\prime\prime}\) and that \(n<n_{\text{mirrors}}\), though as far as their relative phase difference \(\Delta\phi\) is concerned, because \(\textbf R\) is an additive abelian group it will always just be:

\[\Delta\phi=|\phi_2-\phi_1|=2nk\Delta x\]

where the arm-length difference is \(\Delta x:=|x_2-x_1|\) (to get a \(50:50\) beam splitter there should also be some constraint on \(a,b,n,n’,n^{\prime\prime}\) coming from the Fresnel equations). If the photodiode has sampling period \(T_s>0\), then one expects that the signal it measures will be a time-averaged irradiance of the form (it is being assumed here that the beam splitter really is exactly \(50:50\)):

\[I\sim\langle|e^{-i\omega t}+e^{i\Delta\phi}e^{-i\omega t}|^2\rangle_{T_s}\sim 1+\cos\Delta\phi\sim\cos^2nk\Delta x\]

where \(\cos\Delta\phi\) is the interference term which is sensitive to the relative phase difference \(\Delta\phi\) between the two interfering waves of identical frequency \(\omega\).

On the other hand then, given two distinct frequencies \(\omega<\omega’\) separated by \(\Delta\omega:=\omega’-\omega>0\), one can repeat the story above, this time measuring an irradiance:

\[I\sim\langle|(1+e^{i\Delta\phi})e^{-i\omega t}+(1+e^{i\Delta\phi’})e^{-i\omega’ t}|^2\rangle_{T_s}\sim 1+\cos\frac{\Delta\phi+\Delta\phi’}{2}\cos\frac{\Delta\phi’-\Delta\phi}{2}\]

where the coupled interference cross-term:

\[\frac{\sin\Delta\omega T_s+\sin(\Delta\omega T_s-\Delta\phi’)+\sin(\Delta\omega T_s+\Delta\phi)+\sin(\Delta\omega T_s+\Delta\phi-\Delta\phi’)}{\Delta\omega T_s}\]

is only non-negligible in the limit \(\Delta\omega T_s\ll 1\) which is typically not satisfied, and so here is taken to vanish. The key takeaway from this is that, while any two waves will interfere, waves of distinct frequencies \(\omega’\neq\omega\) do not in general interfere in an easy-to-measure way, i.e. only interference between waves of the same frequency \(\omega\) tends to be measurable.

That being said, it is essential to stress that interference between waves of the same \(\omega\) is interference nonetheless, and the resultant interference pattern \(I(\Delta\phi,\Delta\phi’)\) is readily measured in experiments and still extremely useful. One example of this is in the case where \(\Delta\omega\) is small, such as between \(2\) atomic fine or hyperfine transitions. Then the interference pattern can be written as a function of the Michelson interferometer arm-length difference \(\Delta x\):

\[I(\Delta x)\sim 1+\cos n(k+k’)\Delta x\cos n(k’-k)\Delta x\]

This is analogous to the phenomenon of beats in the \(t\)-domain; here in a spatial domain \(\Delta x\), the low-frequency envelope \(\cos n(k’-k)\Delta x\) is modulated by high-frequency fringes \(\cos n(k+k’)\Delta x\). By finding a \(\Delta x=\Delta x_0\) where fringes disappear due to vanishing of the envelope \(\Delta k:=k’-k=\pi/2n\Delta x_0\), one can thereby obtain the frequency difference \(\Delta\omega=nv\Delta k\) assuming \(n\) is non-dispersive. The envelope is sometimes phrased in terms of the fringe visibility:

\[V(\Delta x):=\frac{\Delta I_{\text{envelope}}(\Delta x)}{\bar I(\Delta x)}=\cos n(k’-k)\Delta x\]

As its name suggests, when the fringes become invisible (i.e. fringe visibility becomes \(V(\Delta x_0)=0\)), then one again recovers the same frequency difference \(\Delta\omega\) as above.

Another point implicitly assumed above and worth stressing is that the relative phase difference \(\Delta\phi\) of the two interfering waves have to maintain temporal coherence \(\frac{\partial\Delta\phi}{dt}=0\) at the photodiode position.

More generally, assuming that waves of any two distinct frequencies, no matter how close, do not interfere, a highly polychromatic signal containing irradiance \(2\hat I(k)dk\) in the wavenumber interval \([k,k+dk]\) would be expected to exhibit a Michelson interference pattern of the form:

\[I(\Delta x)=2\int_0^{\infty}\hat I(k)(1+\cos nk\Delta x)dk\]

Or equivalently, because \(\hat I(\Delta x)\in\textbf R\) is expected to be real-valued for all \(\Delta x\), it follows that its Fourier transform \(\hat I(k)\) needs to be Hermitian \(\hat I(-k)=\hat I^{\dagger}(k)\). This means one can rewrite this as an inverse Fourier transform:

\[I(\Delta x)=I_0+\int_{-\infty}^{\infty}\hat I(k)e^{ink\Delta x}dk\]

with \(I_0=\int_{-\infty}^{\infty}\hat I(k)dk\) the background irradiance. The Fourier transform then provides the spectrum of the light source:

\[\hat I(k)=\int_{-\infty}^{\infty}(I(\Delta x)-I_0)e^{-ink\Delta x}dx\]

and is the basis of the Fourier transform infrared spectrometer (FTIR) where a Michelson interferometer is employed along with one mirror on a motorized translation stage to vary \(\Delta x\) and a photodiode to measure the corresponding total interference pattern \(I(\Delta x)\) from all the wavenumbers \(k\in(0,\infty)\).

Note in all of the above discussion it has been implicitly assumed that all the beams in the Michelson interferometer are perfectly collimated, etc. so that the photodiode observes an interference pattern \(I(\Delta x)\) which would vary with arm-length difference \(\Delta x\) but spatially across the photodiode surface would be uniform; in practice, due to misalignment (accidental or deliberate) or imperfect collimation because the light source is not a point but extended, etc. there is not only an interference pattern in \(\Delta x\)-space, but in real space \((x,y)\) across the photodiode surface too; this whole interference profile \(I(x,y,\Delta x)\) would then change as \(\Delta x\) were to evolve (this is best understood by putting one’s eye at the photodiode location and looking into the beam splitter; one would then see \(2\) virtual images of the light source behind a mirror sitting around \(\sim 2 x_2\) away, separated by \(\sim 2\Delta x\). Conceptually, one can then just forget about the whole Michelson interferometer setup and pretend as if one just had \(2\) coherent light sources interfering with each other on a distant screen. Using this perspective, it is clear that the spatial fringes one observes will be sections of hyperbolae.

Finally, it is worth mentioning that sometimes a minor variant of the Michelson interferometer (called a Twyman-Green interferometer) is used for better collimation of the incident light beam. In addition, if instead of a beam splitter one were to use a half-silvered mirror, then a compensator may also need to be added (this would compensate not only for the optical path length difference in monochromatic incident light but also for optical dispersion in the case of polychromatic incident light).

Mach-Zehnder Interferometer

The setup is the following:

• Kinda similar to Michelson except light never retravels its path.
• Also, instead of a single photodiode on which equal-frequency waves interfere, have two photodiodes to measure the intensities on each output port of the last beamsplitter.
• Idea is that for a \(50:50\) beamsplitter, a unitary (probability-conserving) to describe action on photon probability amplitudes for going down each arm of the Mach-Zehnder.

Fabry-Perot Interferometer

Miscellaneous: Thin Film Interferometer, Haidinger Fringes & Newton’s Rings

The purpose of this post is to review several mechanisms that broaden the shape of spectral lines, whether these be absorption or emission spectra (it seems that there are no such things as “narrowing mechanisms” for spectral lines). Roughly speaking, all spectral line broadening mechanisms can be broadly classified (no pun intended) into two categories, namely homogeneous broadening mechanisms and inhomogeneous broadening mechanisms. Examples of homogeneous broadening mechanisms include natural broadening, pressure broadening and power broadening, and all have in common the fact that they affect each atom indistinguishably, causing the spectral line shape to broaden into a Lorentzian. By contrast, examples of inhomogeneous broadening mechanisms include Doppler broadening.

Natural Broadening

This is also called lifetime broadening, because it has its origins in the finite lifetime \(\tau<\infty\) of an excited state \(|1\rangle\) relative to some ground state \(|0\rangle\) due to spontaneous emission. In the absence \(\Omega=0\) of any external driving, the optical Bloch equations are simple to solve analytically, and predict that \(\langle 1|\psi_I(t)\rangle=\langle 1|\psi_I(t)\rangle\).

The bandwidth of a broadened spectral line is, unsurprisingly, called its linewidth \(\Gamma\).

Pressure Broadening

This is also called collisional broadening and is homogeneous.

sum of independent random variables yields convolution.

Power Broadening

This is also homogeneous.

Doppler Broadening

Need to use that formula for converting a probability distribution of one random variable into another function of it with the derivative…(this is how you can experimentally verify the Maxwell distribution btw).

The purpose of this post is to explain how Boltzmann’s equation in kinetic theory arises.

Problem #\(1\): Write down Liouville’s equation from classical Hamiltonian mechanics governing the incompressible phase space flow (i.e. time evolution) of the joint probability density function \(\rho(\textbf x_1,…,\textbf x_N,\textbf p_1,…,\textbf p_N,t)\).

Solution #\(1\): Remembering the intuition that \(\{\space\space,H\}\) implements an advective derivative on the joint phase space:

\[\frac{D\rho}{Dt}=0\Rightarrow\frac{\partial\rho}{\partial t}+\{\rho,H\}=0\]

The idea is that \(\rho(\textbf x_1,…,\textbf x_N,\textbf p_1,…,\textbf p_N,t)d^3\textbf x_1…d^3\textbf x_Nd^3\textbf p_1…d^3\textbf p_N\) is the (purely due to classical ignorance) probability that the system of \(N\) identical particles is, at some time \(t\in\textbf R\), living within an infinitesimal volume \(d^3\textbf x_1…d^3\textbf x_Nd^3\textbf p_1…d^3\textbf p_N\) centered around the microstate \((\textbf x_1,…,\textbf x_N,\textbf p_1,…,\textbf p_N)\in\textbf R^{6N}\) in phase space.

Problem #\(2\): What is the marginal probability density \(\rho_1(\textbf x,\textbf p,t)\) of the any given particle being in state \((\textbf x,\textbf p)\in\textbf R^6\) at time \(t\in\textbf R\)? How is \(\rho_1\) related to the \(1\)-particle distribution function \(n_1(\textbf x,\textbf p,t)\)?

Solution #\(2\): One simply takes the joint distribution and integrates away the \(6(N-1)\) degrees of freedom of the other \(N-1\) particles:

\[\rho_1(\textbf x,\textbf p,t)=\int d^3\textbf x_2…d^3\textbf x_Nd^3\textbf p_2…d^3\textbf p_N\rho(\textbf x,…\textbf x_N,\textbf p,…,\textbf p_N,t)\]

Moreover, because all \(N\) particles are identical, the same marginal distribution \(\rho_1\) applies equally well to all the other \(N-1\) particles; thus \(\rho_1(\textbf x,\textbf p,t)=\rho_2(\textbf x,\textbf p,t)=…=\rho_N(\textbf x,\textbf p,t)\). The one-particle distribution function \(n_1(\textbf x,\textbf p,t)\) at \((\textbf x,\textbf p)\in\textbf R^6\) at time \(t\in\textbf R\) is therefore:

\[n_1(\textbf x,\textbf p,t)=\sum_{i=1}^N\rho_i(\textbf x,\textbf p,t)=N\rho_1(\textbf x,\textbf p,t)\]

Problem #\(3\): In terms of the \(1\)-particle distribution function \(n_1(\textbf x,\textbf p,t)\), write down formulas for the number density \(n(\textbf x,t)\) in configuration space. What about the momentum density and kinetic energy density (also in \(\textbf x\)-space)?

Solution #\(3\): If one further marginalizes \(n_1(\textbf x,\textbf p,t)\) over the momenta \(\textbf p\), one obtains the number density of particles purely in configuration space:

\[n(\textbf x,t)=\int d^3\textbf p n_1(\textbf x,\textbf p,t)\]

Similarly, the momentum density in phase space is \(\textbf pn_1(\textbf x,\textbf p,t)\) while the kinetic energy density in phase space is \(\textbf |\textbf p|^2n_1(\textbf x,\textbf p,t)/2m\), and both of these can be similarly marginalized over \(\textbf p\) to obtain corresponding density distributions in configuration space.

Problem #\(4\): What is the analog of Liouville’s equation for the \(1\)-particle distribution function \(n_1\)?

Solution #\(4\): For a generic Hamiltonian \(H\):

\[\frac{\partial n_1}{\partial t}(\textbf x,\textbf p,t)=N\int d^3\textbf x_2…d^3\textbf x_Nd^3\textbf p_2…d^3\textbf p_N\{H(\textbf x,…\textbf x_N,\textbf p,…,\textbf p_N,t),\rho(\textbf x,…\textbf x_N,\textbf p,…,\textbf p_N,t)\}\]

Problem #\(5\): To make further progress, it is necessary to actually specify some physics. The following typical dispersion relation for the Hamiltonian \(H\) is chosen (here the identification \(\textbf x_1:=\textbf x\) is being made):

\[H(\textbf x,…\textbf x_N,\textbf p,…,\textbf p_N,t)=\sum_{i=1}^N\left(\frac{|\textbf p_i|^2}{2m}+V_{\text{ext}}(\textbf x_i,t)\right)+\sum_{1\leq i<j\leq N}V_{\text{int}}(\textbf x_i-\textbf x_j)\]

Show that:

\[\frac{\partial n_1}{\partial t}(\textbf x,\textbf p,t)=\{H_1(\textbf x,\textbf p,t),n_1(\textbf x,\textbf p,t)\}+\left(\frac{\partial n_1}{\partial t}\right)_{\text{collision}}(\textbf x,\textbf p,t)\]

where \(H_1(\textbf x,\textbf p,t):=|\textbf p|^2/2m+V_{\text{ext}}(\textbf x,t)\) is the \(1\)-particle Hamiltonian which features in the so-called streaming term \(\{H_1,n_1\}\), and \(\left(\partial n_1/\partial t\right)_{\text{coll}}\) is called the collision integral and is given by:

\[\left(\frac{\partial n_1}{\partial t}\right)_{\text{coll}}(\textbf x,\textbf p,t)=\int d^3\textbf x_2d^3\textbf p_2\frac{\partial V_{\text{int}}(\textbf x-\textbf x_2)}{\partial\textbf x}\cdot\frac{\partial n_2}{\partial\textbf p}\]

where the \(2\)-particle distribution function \(n_2(\textbf x,\textbf x_2,\textbf p,\textbf p_2,t)\) is defined by:

\[n_2(\textbf x,\textbf x_2,\textbf p,\textbf p_2,t):=N(N-1)\int d^3\textbf x_3…d^3\textbf x_Nd^3\textbf p_3…d^3\textbf p_N\rho(\textbf x,…,\textbf x_N,\textbf p,…,\textbf p_N,t)\]

and has an interpretation entirely analogous to the one-particle distribution function \(n_1(\textbf x,\textbf p,t)\).

Solution #\(5\):

Problem #\(6\): Explain why only the streaming term \(\{H_1,n_1\}\) (and not the collision integral \((\partial n_1/\partial t)_{\text{coll}}\)) matters for the time evolution of the real space number density \(n(\textbf x,t)\). Is this also the case for the momentum density \(\textbf pn_1\) or the kinetic energy density \(|\textbf p|^2n_1/2m\)?

Solution #\(6\): Using \(n=\int d^3\textbf p n_1\):

\[\frac{\partial n}{\partial t}=\int d^3\textbf p\{H_1,n_1\}+\int d^3\textbf p\left(\frac{\partial n_1}{\partial t}\right)_{\text{coll}}\]

But the integral over the collision integral vanishes for the same kind of reasons as in Solution #\(5\):

(note that if instead one were interested in the time evolution of the momentum space number density \(n(\textbf p,t):=\int d^3\textbf xn_1(\textbf x,\textbf p,t)\) then the collision integral does matter. Similar remarks apply to the real space momentum density \(\int d^3\textbf p\textbf pn_1\) or the real space kinetic energy density \(\int d^3\textbf p|\textbf p|^2n_1/2m\)).

Problem #\(7\): Explain how the methods of Solution #\(5\) generalize to yield the BBGKY hierarchy of \(k=1,2,…,N\sim 10^{23}\) coupled PDEs.

Solution #\(7\): Using the same methods as Solution #\(5\), one can check that:

\[\frac{\partial n_k}{\partial t}=\{H_k,n_k\}+\sum_{i=1}^{k}\int d^3\textbf x_{k+1}d^3\textbf p_{k+1}\frac{\partial V_{\text{int}}(\textbf x_i-\textbf x_{k+1})}{\partial\textbf x_i}\cdot\frac{\partial n_{k+1}}{\partial\textbf p_i}\]

with \(k\)-particle distribution function:

\[n_k(\textbf x_1,…\textbf x_k,\textbf p_1,…,\textbf p_k,t)=\frac{N!}{(N-k)!}\int d^3\textbf x_{k+1}…d^3\textbf x_Nd^3\textbf p_{k+1}…d^3\textbf p_N n(\textbf x_1,…,\textbf x_N,\textbf p_1,…,\textbf p_N,t)\]

and \(k\)-particle Hamiltonian \(H_k\) including both \(V_{\text{ext}}\) and interactions \(V_{\text{int}}\) among the first \(k\) particles but ignores interactions with the other \(N-k\) particles:

\[H_k=\sum_{i=1}^{k}\left(\frac{|\textbf p_i|^2}{2m}+V_{\text{ext}}(\textbf x_i)\right)+\sum_{1\leq i<j\leq k}V_{\text{int}}(\textbf x_i-\textbf x_j)\]

The Boltzmann equation arises by truncating the BBGKY hierarchy. The idea is that there is a long time scale \(\tau\) (interscattering time) and a short time scale \(\tau_c\ll\tau\) (scattering time).

• The initial \(t=0\) Bose gas momentum space distribution \(n_{|k\rangle}(t=0)\) is sharply peaked at some \(k=k_0\), with the low-energy bosons of \(k<k_0\) in the so-called IR regime and the high-energy bosons of \(k>k_0\) in the UV regime.
• The strongly interacting regime corresponds to bosons with \(k\xi\ll 1\) with \(\xi\propto a^{-1/2}\) the healing length, whereas the weakly interacting regime is \(k\xi\gg 1\).
• In the UV regime, if it starts weakly interacting then it will remain like that because \(k\) just grows, pushing further into weakly interacting.
• In the IR regime, there is a transition from strongly to weakly interacting, so more interesting to study.
• Gevorg’s speed limit paper showed that the coherent coarsening dynamics as quantified by the speed limit \(3\hbar/m\) is, after an initial transient, independent of the (dimensionless) interaction strength \(1/(k\xi)\).
• WWT applies to the weakly interacting \(k\xi\gg 1\) regime.
• # the wiggles at the end are due to diffraction off the BEC, not the sinc momentum space contribution from the BEC which is approximately (need to crop them from the data)
•         # a homogeneous top hat in real space.
•         # Each experimental cycle lasts about 30 seconds, the first 25 seconds is just the standard steps (MOT, evaporation, Sisyphus cooling, molasses, etc.), the last 5 seconds
•         # only like a few seconds is actually the physics.
•         # Right now, Gevorg & Simon decided to have 15 increasing TOFs, in order to probe the momentum space distribution n_k of the BEC.
•         # The idea is that you measure a 2D momentum space distribution of the BEC along your line of sight, then inverse Abel transform (involves a derivative)
•         # it to get the 3D momentum space distribution. But also, the only part you can reliably inverse Abel transform is the part that is not diffracted
•         # off the BEC and doesn’t saturate the optical density OD at 3, so can only reliably measure in sort of the “outskirt” regions of the cylinder so to speak
•         # where the BEC is not too dense (i.e. OD < 3). Also this is why it’s hard to measure the low-k part of the momentum space distribution, b/c the BEC is so dense
•         # there (OD > 3) that it saturates the imaging system.
•         # Numerical differentiation is much less robust than numerical integration (Gevorg gave example of a monotonic function like exp(x)),
•         # so it’s much harder to get the 3D momentum space distribution. In the case of the inverse Abel transform, you have to differentiate by subtracting the
•         # of neighbouring pixels.
•         # Need to be on the Cambridge VPN to “SSH” into the VNC viewer to see Analysis GpUI (imaging computer), Cicero, Origin, or any of the office computers
•         # each of the lab computers has a IP address, and you can only access them from the Cambridge network. Also a remote Toptica software for
•         # relocking the laser if it drifts/unlocks overnight (somehow not so easy to just automate this b/c need to manually play with the current and piezo voltage in the AOMs).
•         # Ground state wavefunction of BEC in k-space is not exactly a sinc, rather a Bessel b/c it’s a cylinder.
•         # one-body loss, evaporative loss, usually temperatures too high cf. box trap depth, in that case lose a lot of energy but not many particles
•         # b/c only particels you lose are the high-energy ones, so you lose a lot of energy but not many particles
•         # another worry is counting worry, the n_rad_k distributions might overestimate or underestimate the number of particles
•         # e.g. the 350000 atoms seems too high…truncate integrals as well to avoid the noisy region at high-k.
•         # for n_k use log-log, for k**2*n_k or k**4*n_k use lin-lin maybe? or log-lin… advantage of lin-lin is that area you visually see is proportional to the
•         # integral of the function and for k**2 and k**4 this is nice to see.
• #try playing around with some different definition of “speed of thermalization”, see if they give a monotonic thing or not
• # also compare with the GPE & WWT theory of the paper (which plots k**2*n_k) see if it matches…
• # change energy to temperature scale (nK), also get E/N for each set