Quantumplations

The purpose of this post is to review how the fields of an electrostatic dipole \(\boldsymbol{\pi}\) and magnetostatic dipole \(\boldsymbol{\mu}\) arise. For the electrostatic dipole, “fields” means the electrostatic potential \(\phi\) and by extension the electrostatic field \(\textbf E=-\partial\phi/\partial\textbf x\) while for the magnetostatic dipole, “fields” means the magnetostatic potential \(\textbf A\) and by extension the magnetostatic field \(\textbf B=\partial/\partial\textbf x\times\textbf A\).

Because one is working in the regime of electrostatics, Coulomb’s law is valid:

\[\phi(\textbf x)=\frac{1}{4\pi\varepsilon_0}\iiint_{\textbf x’\in\textbf R^3}\frac{\rho(\textbf x’)}{|\textbf x-\textbf x’|}d^3\textbf x’\]

An electrostatic dipole consists of two stationary charges \(\pm q\) separated by \(\Delta\textbf x\); arbitrarily placing the charge \(-q\) at the origin \(\textbf x’=\textbf 0\) implies that \(\rho(\textbf x’)=q(\delta^3(\textbf x’+\Delta\textbf x)-\delta^3(\textbf x’))\). Substituting this into Coulomb’s law picks out:

\[\phi(\textbf x)=\frac{q}{4\pi\varepsilon_0}\left(\frac{1}{|\textbf x+\Delta\textbf x|}-\frac{1}{|\textbf x|}\right)\]

As \(\Delta\textbf x\to\textbf 0\), one has by definition the directional derivative of \(1/|\textbf x|\) along \(\Delta\textbf x\):

\[\frac{1}{|\textbf x+\Delta\textbf x|}-\frac{1}{|\textbf x|}\to\Delta\textbf x\cdot\frac{\partial}{\partial\textbf x}\frac{1}{|\textbf x|}=-\Delta\textbf x\cdot\hat{\textbf x}/|\textbf x|^2\]

Using the definition \(\boldsymbol{\pi}:=-q\Delta\textbf x\) of the electrostatic dipole in this case, the electrostatic potential reduces to:

\[\phi(\textbf x)=\frac{\boldsymbol{\pi}\cdot\hat{\textbf x}}{4\pi\varepsilon_0|\textbf x|^2}\]

The electrostatic field then follows:

\[\textbf E(\textbf x)=-\frac{1}{4\pi\varepsilon_0}\frac{\partial}{\partial\textbf x}\frac{\boldsymbol{\pi}\cdot\textbf x}{|\textbf x|^3}=\frac{3(\boldsymbol{\pi}\cdot\hat{\textbf x})\hat{\textbf x}-\boldsymbol{\pi}}{4\pi\varepsilon_0|\textbf x|^3}\]

Meanwhile, because one is working in the regime of magnetostatics, the Biot-Savart law (in the Coulomb gauge) is valid:

\[\textbf A(\textbf x)=\frac{\mu_0}{4\pi}\iiint_{\textbf x’\in\textbf R^3}\frac{\textbf J(\textbf x’)}{|\textbf x-\textbf x’|}d^3\textbf x’\]

A magnetostatic dipole consists of a steady current loop \(I\) enclosing an area \(\textbf S\), with \(\textbf J(\textbf x’)d^3\textbf x’\mapsto Id\textbf x’\):

\[\textbf A(\textbf x)=\frac{\mu_0 I}{4\pi}\oint_{\textbf x’\in S^1}\frac{d\textbf x’}{|\textbf x-\textbf x’|}\]

As \(\textbf S\to\textbf 0\), one has (still looking for a simple way to see this):

\[\oint_{\textbf x’\in S^1}\frac{d\textbf x’}{|\textbf x-\textbf x’|}\to\frac{\partial}{\partial\textbf x}\frac{1}{|\textbf x|}\times\textbf S\]

Using the definition \(\boldsymbol{\mu}:=I\textbf S\) of the magnetostatic dipole, the magnetostatic vector potential reduces to:

\[\textbf A(\textbf x)=\frac{\mu_0\boldsymbol{\mu}\times\hat{\textbf x}}{4\pi|\textbf x|^2}\]

From which the magnetostatic field is:

\[\textbf B(\textbf x)=\frac{\mu_0(3(\boldsymbol{\mu}\cdot\hat{\textbf x})\hat{\textbf x}-\boldsymbol{\mu})}{4\pi|\textbf x|^3}\]

Given a flow field \(\textbf v(\textbf x,t)\), the vorticity \(\boldsymbol{\omega}\) of \(\textbf v\) is defined by taking its curl \(\boldsymbol{\omega}:=\frac{\partial}{\partial\textbf x}\times\textbf v\). For a flow field rotating rigidly with angular velocity vector \(\boldsymbol{\omega}_0\) so that \(\textbf v=\boldsymbol{\omega}_0\times\textbf x\). The vorticity associated with this purely rotational flow is:

\[\boldsymbol{\omega}=\frac{\partial}{\partial\textbf x}\times(\boldsymbol{\omega}_0\times\textbf x)=\boldsymbol{\omega}_0\left(\frac{\partial}{\partial\textbf x}\cdot\textbf x\right)+\left(\textbf x\cdot\frac{\partial}{\partial\textbf x}\right)\boldsymbol{\omega}_0-\textbf x\left(\frac{\partial}{\partial\textbf x}\cdot\boldsymbol{\omega}_0\right)-\left(\boldsymbol{\omega}_0\cdot\frac{\partial}{\partial\textbf x}\right)\textbf x=2\boldsymbol{\omega}_0\]

Thus, \(\boldsymbol{\omega}=2\boldsymbol{\omega}_0\). In other words, it is possible to rewrite the original flow field as \(\textbf v=\boldsymbol{\omega}_0\times\textbf x=\frac{1}{2}\boldsymbol{\omega}\times\textbf x=\frac{1}{2}\left(\frac{\partial}{\partial\textbf x}\times\textbf v\right)\times\textbf x\) (cf. Lamb vector?). Can this also be gotten from the vorticity equation?

Often in biochemistry, if a single substrate \(\text S\) needs to be become a product \(\text P\) via a chemical reaction of the form \(\text S\to \text P\). Assuming this is a first-order elementary chemical reaction, it would merely have rate law \(\dot{[\text P]}=-\dot{[\text S]}=k[\text S]\) for some rate constant \(k>0\), and thus the usual exponential time evolution. However, nature has made use of enzymes \(\text E\), which are simply biological catalysts. As with any catalysts, the thermodynamics \(\Delta H, \Delta S, \Delta G\) are invariant because they are state functions, but the kinetics (and thus activation energy barrier \(\Delta E_a\)) are significantly reduced by providing an alternative reaction pathway to the direct conversion \(\text S\to\text P\)). Specifically, the enzyme \(\text E\) first binds onto the substrate \(\text S\) via an elementary chemical reaction of the form \(\text E+\text S\to\text{ES}\), forming an enzyme-substrate complex \(\text{ES}\). The enzyme \(\text E\) then desorbs from the enzyme-substrate complex \(\text{ES}\) to yield the desired product \(\text P\) and reforming the enzyme \(\text{E}\) again via a second elementary reaction of the form \(\text{ES}\to\text{E}+\text{P}\), thus being involved in but not consumed by the overall chemical reaction \(\text{S}\to\text P\), another defining property of catalysts. By applying the steady state approximation to the only reaction intermediate there is, namely the enzyme-substrate complex \(\text{ES}\), one has that \(\dot{[\text{ES}]}=0\). This eventually yields the Michaelis-Menten equation for the velocity \(\dot{[\text P]}\) enzyme \(\text E\)-catalyzed reactions on a single substrate \(\text S\) to form a product \(\text P\):

\[\dot{[P]}=\frac{k_{\text{cat}}[\text E]_0[\text S]}{[S]+K_M}\]

where \([\text E]_0\) is the initial enzyme concentration at time \(t=0\), and \(K_M\) is the Michaelis constant. There is also a somewhat misleading graph of \(\dot{[P]}\) as a function of \([\text S]\) often shown, where in the limit \([\text S]\to\infty\) of an infinite substrate concentration (i.e. near the start of the reaction), the velocity of product formation is at a global maximum \(\dot{[P]}^*=k_{\text{cat}}[\text E]_0\).

The free, dispersionless wave equation on \((ct,\textbf x)\in\textbf R\times \textbf R^d\) is:

\[\frac{\partial^2\psi}{\partial (ct)^2}-\frac{\partial^2\psi}{\partial|\textbf x|^2}=0\]

or more simply as \(☐^2\psi=0\), where the d’Alembert operator is defined by \(☐^2:=\frac{\partial^2}{\partial (ct)^2}-\frac{\partial^2}{\partial|\textbf x|^2}=\partial^{\mu}\partial_{\nu}=\eta^{\mu\nu}\partial_{\mu}\partial_{\nu}\) with the usual metric \(\eta=\text{diag}(1,-1,-1,-1)\) defining the hyperbolic geometry of Minkowski spacetime. In \(d=1\) spatial dimension, d’Alembert discovered an \(\text{SO}(2)\) linear transformation on \(1+1\)-dimensional spacetime \(\textbf R\times\textbf R\) which substantially simplified the form of the d’Alembert operator \(☐^2\). Specifically, consider the spacetime coordinates \((u,v)\) rotated \(45^{\circ}\) counterclockwise from \((ct, x)\) via:

\[\begin{pmatrix}u\\v\end{pmatrix}=\frac{1}{\sqrt{2}}\begin{pmatrix}1&-1\\1&1\end{pmatrix}\begin{pmatrix} ct\\ x\end{pmatrix}\]

Then the claim is that \(☐^2=2\frac{\partial^2}{\partial v\partial u}\) is a composition of directional derivatives along the orthogonal “light-like” worldlines in spacetime. This follows from two observations. First, notice that the \(1+1\)-dimensional d’Alembert operator \(☐^2\) is separable via a difference of squares into two transport operators \(☐^2=\left(\frac{\partial}{\partial(ct)}+\frac{\partial}{\partial x}\right)\left(\frac{\partial}{\partial(ct)}-\frac{\partial}{\partial x}\right)\) (this trick only works in \(d=1\) spatial dimension!). Then, it suffices to show that \(\frac{\partial}{\partial v}\) corresponds to the first factor and likewise that \(\frac{\partial}{\partial u}\) corresponds to the second factor. The chain rule asserts that partial derivative operators transform under the transpose of the Jacobian:

\[\begin{pmatrix}\partial/\partial u\\ \partial/\partial v\end{pmatrix}=\left(\frac{\partial(ct,x)}{\partial(u,v)}\right)^T\begin{pmatrix} \partial/\partial(ct)\\ \partial/\partial x\end{pmatrix}\]

On the other hand, it is clear that the inverse Jacobian \(\frac{\partial(u,v)}{\partial(ct,x)}=\left(\frac{\partial(ct,x)}{\partial(u,v)}\right)^{-1}\) is given by (as it must, being the linearization of a map!):

\[\frac{\partial(u,v)}{\partial(ct,x)}=\frac{1}{\sqrt{2}}\begin{pmatrix}1&-1\\1&1\end{pmatrix}\]

So because \(\frac{\partial(u,v)}{\partial(ct,x)}\in\text{SO}(2)\triangleleft\text{O}(2)\), inverting it is precisely the same as transposing it. The involutory nature of transposition then yields the final result:

\[\begin{pmatrix}\partial/\partial u\\ \partial/\partial v\end{pmatrix}=\frac{1}{\sqrt{2}}\begin{pmatrix}1&-1\\1&1\end{pmatrix}\begin{pmatrix} \partial/\partial(ct)\\ \partial/\partial x\end{pmatrix}\]

consistent with the usual formulas for directional derivatives. From here, it is straightforward to directly integrate the free, dispersionless wave equation to see that the most general free, dispersionless wave \(\psi\in\ker(☐^2)\) in \(d=1\) spatial dimension is the superposition of a left travelling wave \(\psi_{\leftarrow}(x+ct)\) and a right travelling wave \(\psi_{\rightarrow}(x-ct)\), i.e. \(\psi(ct,x)=\psi_{\leftarrow}(x+ct)+\psi_{\rightarrow}(x-ct)\).

Problem #\(1\): Find the mistake in this derivation of the Green’s function for the inhomogeneous Helmholtz equation subject to the boundary condition of an outgoing wave:

Solution #\(1\): First of all, the correct answer should be \(G(r)=\frac{e^{ikr}}{4\pi r}\), not this standing wave with a \(\cos(kr)=(e^{ikr}+e^{-ikr})/2\) that superposes both an ingoing \(e^{-ikr}\) and outgoing wave \(e^{ikr}\). Reasoning that \(\cos(kr)\in\textbf R\) because it came after hitting with the \(\Im\) function, one might be led (correctly as it turns out) to the conclusion that this step:

\[\int_{-\infty}^{\infty}\frac{k’^2\text{sinc}(k’r)}{k^2-k’^2}\neq\frac{1}{r}\Im\int_{-\infty}^{\infty}dk’\frac{k’e^{ik’r}}{(k-k’)(k+k’)}\]

was the seemingly innocuous but incorrect step. Basically, although it’s certainly true that \(\sin(k’r)=\Im e^{ik’r}\), the difficulty is that in order to be able to pull the \(\Im\) out of the integral, one has to be sure that the prefactor \(\frac{k’^2}{k^2-k’^2}\) is real. But in fact, despite looking very real, it should really have an \(i\varepsilon\) in the denominator (or sometimes just written as \(i0\) to reflect taking the limit \(\varepsilon\to 0\)). It is this slight shifting of the poles at \(k’=\pm k\) off the real axis to \(k’=\pm k\pm’i\varepsilon\) that makes this naive manipulation invalid!

Problem #\(2\): Fix the mistake by redoing the derivation correctly

Solution #\(2\):

Problem #\(3\): Hence write down immediately the Lippman-Schwinger equation in the position representation for scattering off an arbitrary potential \(V\).

Solution #\(3\): All you need to remember is:

\[\left(k^2+\biggr|\frac{\partial}{\partial\textbf x}\biggr|^2\right)\psi=-\frac{2m}{\hbar^2}V\psi\]

where the RHS is viewed as a \(\psi\)-dependent forcing function, so the solution of this inhomogeneous PDE is just the particular integral + complementary function:

\[\psi(\textbf x)=e^{i\textbf k\cdot\textbf x}+\left(G*-\frac{2m}{\hbar^2}V\psi\right)(\textbf x)\]

where the convolution is:

\[\left(G*-\frac{2m}{\hbar^2}V\psi\right)(\textbf x)=-\frac{m}{2\pi\hbar^2}\int d^3\textbf x’V(\textbf x’)\psi(\textbf x’)\frac{e^{ik|\textbf x-\textbf x’|}}{|\textbf x-\textbf x’|}\]

Problem #\(4\): By aligning the incident momentum \(\textbf k=k\hat{\textbf z}\) along the \(z\)-axis and working asymptotically \(|\textbf x-\textbf x’|\approx |\textbf x|-\hat{\textbf x}\cdot\textbf x’\) in the exponent of the Green’s function \(G(r)=e^{ikr}/4\pi r\) while directly approximating \(|\textbf x-\textbf x’|\approx |\textbf x|\), obtain the Fraunhofer-Lippman-Schwinger equation:

\[\psi(\textbf x)=e^{ikz}+f(\hat{\textbf x})\frac{e^{ikr}}{r}\]

where the scattering amplitude \(f(\hat{\textbf x})\) in direction \(\hat{\textbf x}\) is defined by:

\[f(\hat{\textbf x}):=-\frac{m}{2\pi\hbar^2}\int d^3\textbf x’V(\textbf x’)\psi(\textbf x’)e^{-ik\hat{\textbf x}\cdot\textbf x’}\]

Solution #\(4\): Trivial.

Problem #\(5\): What is the scattering amplitude \(f(\hat{\textbf x})\) in the \(1^{\text{st}}\) Born approximation?

Solution #\(5\): Basically, just replace \(\psi(\textbf x’)\mapsto e^{i\textbf k\cdot\textbf x’}\) in the integrand (as that’s the plane wave solution when \(V=0\)). Thus, one finds that, up to some constants, the scattering amplitude \(f(\hat{\textbf x})\) in a given direction \(\hat{\textbf x}\) is determined by the Fourier structure of the potential \(V\), i.e. what is the amplitude of the plane wave in \(V\) travelling in the direction of the momentum transfer \(\textbf k’-\textbf k\):

\[f(\hat{\textbf x})=-\frac{m}{2\pi\hbar^2}\hat V(\textbf k’-\textbf k)\]

where the \(\hat{\textbf x}\)-dependence in \(f(\hat{\textbf x})\) is all inside \(\textbf k’:=k\hat{\textbf x}\).

Problem #\(6\): What about the corresponding cross section \(\sigma\) in the \(1^{\text{st}}\) Born approximation.

Solution #\(6\): \[\sigma=\int d^2\hat{\textbf x}|f(\hat{\textbf x})|^2=\left(\frac{m}{2\pi\hbar^2}\right)^2\int d^2\textbf x’|\hat V(\textbf k’-\textbf k)|^2\]

which is sort of like a Plancherel-type integral.

Problem #\(7\): Apply the \(1^{\text{st}}\) Born approximation to calculate the Rutherford scattering amplitude \(f(\theta)\) in the cone \(\theta\) from a repulsive Coulomb potential.

Solution #\(7\): The Coulomb potential itself is not Fourier transformable, but the Yukawa potential \(V(r)=Ae^{-\kappa r}/r\) (which is like the Helmholtz Green’s function of a bound state?) has \(\hat V(k)=4\pi A/(k^2+\kappa^2)\) (no contour integration is even needed, this integral is very elementary). Then (nonrigorously) taking the limit \(\kappa\to 0\) (i.e. letting the range \(1/\kappa\to\infty\)) reproduces the Coulomb potential with supposed inverse-square Fourier transform \(\hat V(k)=4\pi A/k^2\). So the scattering amplitude in the \(1^{\text{st}}\) Born approximation is thus:

\[f(\hat{\textbf x})=-\frac{2mA}{\hbar^2|\textbf k’-\textbf k|^2}=\frac{A}{4E\sin^2\theta/2}\]

with \(E=\hbar^2k^2/2m\).

Problem: Show that the \(s\)-wave scattering amplitude \(f_s(k)\) is isotropic and related to the \(s\)-wave scattering length \(a_s\) of the isotropic potential \(V(r)\) by the Mobius transformation:

\[f_s(k)\approx-\frac{1}{ik+a^{-1}_s}=-a_s+ika_s^2+O(k^2)\to -a_s\text{ as }k\to 0\]

and hence as \(k\to 0\) the asymptotic wavefunction approaches \(\psi(r)\to 1-\frac{a_s}{r}\).

Solution: The partial wave expansion of the \(z\)-axisymmetric scattering amplitude \(f(\theta)\) in an isotropic potential \(V(r)\) receives contributions from all angular momentum sectors:

\[f(\theta)=\sum_{\ell=0}^{\infty}f_{\ell}(k)P_{\ell}(\cos\theta)\]

with “Legendre coefficients”:

\[f_{\ell}(k)=\frac{2\ell+1}{k}e^{i\delta_{\ell}(k)}\sin\delta_{\ell}(k)\]

For low-energy \(k\to 0\) scattering, the \(\ell=0\) \(s\)-wave contribution dominates so that the scattering amplitude \(f(\theta)\approx f_0(k)P_0(\cos\theta)=f_0(k)=f_s(k)\) is not even a function of \(\theta\) anymore, i.e. it is isotropic!

\[f_s(k)=\frac{e^{i\delta_0(k)}\sin\delta_0(k)}{k}\]

In particular, in the low-energy limit \(k\to 0\) one has \(\delta_0(k)\to 0\) linearly as \(\delta_0(k)=-ka_s+O(k^2)\), so expanding \(\sin\delta_0(k)\approx\delta_o(k)\) and doing the slightly dodgy manipulation of writing \(e^{i\delta_0(k)}=\frac{1}{e^{-i\delta_0(k)}}=\frac{1}{1-i\delta_0(k)}\) one obtains the desired results. The asymptotic wavefunction is:

\[\psi(r,\theta,\phi)=e^{ikz}+f(k,\theta,\phi)\frac{e^{ikr}}{r}\to 1-\frac{a_s}{r}\]

upon setting \(k=0\).

The purpose of this post is to review the classical theory of rigid body dynamics by working through a few illustrative problems in that regard.

Problem #\(1\): What defines a rigid body? What is an immediate corollary of this?

Solution #\(1\): A rigid body obeys \(|\textbf x_i-\textbf x_j|=\text{const}\) for all pairs of points \(i,j\) comprising the rigid body. As an immediate corollary, it follows that:

\[\frac{d}{dt}|\textbf x_i-\textbf x_j|^2=0\]

Writing \(|\textbf x_i-\textbf x_j|^2=(\textbf x_i-\textbf x_j)\cdot (\textbf x_i-\textbf x_j)\) and using the product rule leads one to an orthogonality criterion between the relative positions and velocities in a rigid body for all time:

\[(\textbf x_i-\textbf x_j)\cdot(\dot{\textbf x}_i-\dot{\textbf x}_j)=0\]

Mathematically, this suggests that there should exist a single, universal vector \(\boldsymbol{\omega}\) called the angular velocity vector of the rigid body such that for all pairs \(i,j\) in the rigid body, one has:

\[\dot{\textbf x}_i-\dot{\textbf x}_j=\boldsymbol{\omega}\times(\textbf x_i-\textbf x_j)\]

Existence is always guaranteed, but one can find loopholes around uniqueness, e.g. the following rigid body consisting of a bunch of collinear point masses moving together with identical velocities \(\dot{\textbf x}_i=\dot{\textbf x}_j\) would admit any angular velocity \(\boldsymbol{\omega}\) along the line of their collinearity.

But in general, such situations never arise and so \(\boldsymbol{\omega}\) is unique at all times \(t\) (would be nice to obtain a rigorous proof of this intuition).

Problem #\(2\): How is the angular momentum \(\textbf L_{\textbf X}\) about a point \(\textbf X\) defined? What would it mean for \(\textbf X\) to be “fixed in some body frame” of a rigid body? Show that if \(\textbf X\) is fixed in some body frame, then the angular momentum \(\textbf L_{\textbf X}\) of the rigid body about \(\textbf X\) is:

\[\textbf L_{\textbf X}=\mathcal I_{\textbf X}\boldsymbol{\omega}\]

where the inertia tensor \(\mathcal I_{\textbf X}\) of the rigid body with respect to \(\textbf X\) is:

\[\mathcal I_{\textbf X}=\iiint_{\textbf x\in\text{rigid body}} d^3\textbf x\rho(\textbf x)(|\textbf x-\textbf X|^21-(\textbf x-\textbf X)^{\otimes 2})\]

i.e. inner product minus outer product. Note that \(\mathcal I_{\textbf X}^T=\mathcal I_{\textbf X}\) is clearly symmetric and positive semi-definite, so has \(3\) non-negative real eigenvalues \(I^1_{\textbf X},I^2_{\textbf X},I^3_{\textbf X}\) (for a given \(\textbf X\)) and orthogonal eigenspaces defining the so-called principal axes of the rigid body (with respect to \(\textbf X\)).

Solution #\(2\): The most useful definition in this context is:

\[\textbf L_{\textbf X}:=\sum_im_i(\textbf x_i-\textbf X)\times(\dot{\textbf x}_i-\dot{\textbf X})\]

And importantly, not to be confused with:

\[\tilde{\textbf L}_{\textbf X}:=\sum_i (\textbf x_i-\textbf X)\times \textbf p_i=\sum_i m_i(\textbf x_i-\textbf X)\times\dot{\textbf x}_i\]

The point \(\textbf X\) is said to be fixed in some body frame of a rigid body iff for all \(\textbf x_i\) comprising the rigid body, one has for all time \(|\textbf x_i-\textbf X|=\text{const}\); thus, if one likes one can think of \(\textbf X\) as just another mass element comprising the rigid body, except that it is massless.

In this case, it is clearly more desirable to work with the “fully relative” angular momentum \(\textbf L_{\textbf X}\) about \(\textbf X\) rather than the “partially relative” angular momentum \(\tilde{\textbf L}_{\textbf X}\) about \(\textbf X\) as it’s only in the former case that one can make use of the result \(\dot{\textbf x}_i-\dot{\textbf X}=\boldsymbol{\omega}\times(\textbf x_i-\textbf X)\) from Problem #\(1\) when \(\textbf X\) is fixed in some body frame:

\[\textbf L_{\textbf X}=\sum_i m_i(\textbf x_i-\textbf X)\times(\boldsymbol{\omega}\times(\textbf x_i-\textbf X))=\sum_im_i\left(|\textbf x_i-\textbf X|^2\boldsymbol{\omega}-((\textbf x_i-\textbf X)\cdot\boldsymbol{\omega})(\textbf x_i-\textbf X)\right)\]

where \(((\textbf x_i-\textbf X)\cdot\boldsymbol{\omega})(\textbf x_i-\textbf X)=(\textbf x_i-\textbf X)((\textbf x_i-\textbf X)\cdot\boldsymbol{\omega})=(\textbf x_i-\textbf X)((\textbf x_i-\textbf X)^T\boldsymbol{\omega})=((\textbf x_i-\textbf X)\textbf (\textbf x_i-\textbf X)^T)\boldsymbol{\omega}=(\textbf x_i-\textbf X)^{\otimes 2}\boldsymbol{\omega}\), so the result follows.

Problem #\(3\): Repeat Problem #\(2\) but for the kinetic energy \(T_{\textbf X}\) of a rigid body about a point \(\textbf X\) fixed in some body frame to show that:

\[T_{\textbf X}=\frac{1}{2}\boldsymbol{\omega}\cdot\textbf L_{\textbf X}\]

Solution #\(3\): As always, it is essential to think in a clear-headed way about how various quantities are defined in the first place. In this case, there is no ambiguity (unlike the ambiguity between \(\textbf L_{\textbf X}\) and \(\tilde{\textbf L}_{\textbf X}\)), and one simply has:

\[T_{\textbf X}:=\frac{1}{2}\sum_i m_i|\dot{\textbf x}_i-\dot{\textbf X}|^2\]

In this case, \(|\dot{\textbf x}_i-\dot{\textbf X}|^2=|\boldsymbol{\omega}\times(\textbf x_i-\textbf X)|^2=|\boldsymbol{\omega}|^2|\textbf x_i-\textbf X|^2-(\boldsymbol{\omega}\cdot(\textbf x_i-\textbf X))^2=\boldsymbol{\omega}^T|\textbf x_i-\textbf X|^2\boldsymbol{\omega}-\boldsymbol{\omega}^T(\boldsymbol{\omega}\cdot(\textbf x_i-\textbf X))(\textbf x_i-\textbf X)\) and the rest is straightforward.

Problem #\(4\): In any frame, the decompositions:

\[\textbf L=M\textbf X\times\dot{\textbf X}+\textbf L_{\textbf X}\]

and

\[T=\frac{1}{2}M|\dot{\textbf X}|^2+T_{\textbf X}\]

are true only for what point \(\textbf X\)? (note these decompositions are true not just for rigid bodies).

Solution #\(4\): Only when \(\textbf X\) coincides with the center of mass of the system as this ensures that cross-terms such as \(\sum_im_i(\textbf x_i-\textbf X)=\sum_im_i(\dot{\textbf x}_i-\dot{\textbf X})=\textbf 0\) vanish.

Problem #\(5\): Define the external torque \(\boldsymbol{\tau}_{\textbf X}^{\text{ext}}\) about \(\textbf X\) and, using this definition, check under \(2\) assumptions (which should be stated) that \(\dot{\textbf L}_{\textbf X}=\boldsymbol{\tau}_{\textbf X}^{\text{ext}}\) (again, this result is true not just for rigid bodies).

Solution #\(5\): The sensible definition in this case is:

\[\boldsymbol{\tau}_{\textbf X}^{\text{ext}}:=\sum_i(\textbf x_i-\textbf X)\times\textbf F_i^{\text{ext}}=\boldsymbol{\tau}^{\text{ext}}-\textbf X\times\textbf F_{\text{ext}}\]

Now the first assumption is that \(\textbf X\) coincides with the center of mass of the system. This allows the decomposition of \(\textbf L\) in Problem #\(4\) to be used, in particular:

\[\dot{\textbf L}_{\textbf X}=\dot{\textbf L}-M\textbf X\times\ddot{\textbf X}\]

Now comes the second assumption, namely that one is working in an inertial frame. This allows one to replace \(\dot{\textbf L}=\boldsymbol{\tau}^{\text{ext}}\) and \(M\ddot{\textbf X}=\textbf F^{\text{ext}}\). The claim follows.

Problem #\(5\): Synthesizing the results of the previous few problems, derive Euler’s equations for rigid body rotation about the center of mass \(\textbf X\) in an inertial frame:

\[\mathcal I_{\textbf X}\dot{\boldsymbol{\omega}}+\boldsymbol{\omega}\times\textbf L_{\textbf X}=\boldsymbol{\tau}^{\text{ext}}_{\textbf X}\]

Solution #\(5\): The simple but important observation to make is that the center of mass \(\textbf X\) is also always fixed in any body frame (rigorous proof?). So conceptually, Euler’s equations are simply obtained by applying the product rule to:

\[\dot{\textbf L}_{\textbf X}=\frac{d}{dt}(\mathcal I_{\textbf X}\boldsymbol{\omega})=\dot{\mathcal I}_{\textbf X}\boldsymbol{\omega}+\mathcal I_{\textbf X}\dot{\boldsymbol{\omega}}\]

except that it isn’t so obvious how to simplify \(\dot{\mathcal I}_{\textbf X}\) from its definition in Problem #\(2\). So instead, it will be convenient to choose a specific basis to carry out the computation in, and then revert back to the general vector notation. Specifically, the algebra is dramatically simplified in the principal body frame \((\hat{\textbf e}_1,\hat{\textbf e}_2,\hat{\textbf e}_3)_{\textbf X}\) with respect to which \(\mathcal I_{\textbf X}\) about the center of mass \(\textbf X\) is diagonal \(\mathcal I_{\textbf X}=\sum_{i=1}^3I_{\textbf X}^i\hat{\textbf e}_i^{\otimes 2}\) and moreover the principal moments of inertia \(I^1_{\textbf X},I^2_{\textbf X},I^3_{\textbf X}\) are \(t\)-independent. In this \(\mathcal I_{\textbf X}\)-eigenbasis, one can express \(\boldsymbol{\omega}=\sum_{i=1}^3\omega_i\hat{\textbf e}_i\) and hence \(\textbf L_{\textbf X}=\sum_{i=1}^3I^i_{\textbf X}\omega_i\hat{\textbf e}_i\), so applying the product rule gives:

\[\dot{\boldsymbol{\omega}}=\sum_{i=1}^3(\dot{\omega}_i\hat{\textbf e}_i+\omega_i\boldsymbol{\omega}\times\hat{\textbf e}_i)=\sum_{i=1}^3\dot{\omega}_i\hat{\textbf e}_i+\boldsymbol{\omega}\times\boldsymbol{\omega}=\sum_{i=1}^3\dot{\omega}_i\hat{\textbf e}_i\]

\[\dot{\textbf L}_{\textbf X}=\sum_{i=1}^3(I^i_{\textbf X}\dot{\omega}_i\hat{\textbf e}_i+I^i_{\textbf X}\omega_i\boldsymbol{\omega}\times\hat{\textbf e}_i)=\mathcal I_{\textbf X}\dot{\boldsymbol{\omega}}+\boldsymbol{\omega}\times\textbf L_{\textbf X}\]

Problem #\(6\): What does it mean for a rigid body to be free? Verify explicitly that both the kinetic energy \(T_{\textbf X}\) and magnitude of the angular momentum \(|\textbf L_{\textbf X}|\) are conserved for a free rigid body.

Solution #\(6\): A rigid body is said to be free iff it experiences no net external torque \(\boldsymbol{\tau}^{\text{ext}}_{\textbf X}=\textbf 0\) about its center of mass \(\textbf X\) in an inertial frame so that \(\textbf L_{\textbf X}\) is immediately conserved (and thus so is \(|\textbf L_{\textbf X}|\)). Moreover, because the inertia tensor is symmetric:

\[\dot T_{\textbf X}=\frac{1}{2}\dot{\boldsymbol{\omega}}\cdot\textbf L_{\textbf X}=\frac{1}{2}\textbf L_{\textbf X}\cdot\dot{\boldsymbol{\omega}}=\frac{1}{2}(\mathcal I_{\textbf X}\boldsymbol{\omega})\cdot\dot{\boldsymbol{\omega}}=\frac{1}{2}(\mathcal I_{\textbf X}\boldsymbol{\omega})^T\dot{\boldsymbol{\omega}}=\frac{1}{2}\boldsymbol{\omega}^T\mathcal I_{\textbf X}\dot{\boldsymbol{\omega}}=0\]

vanishes by virtue of dotting Euler’s equations with \(\boldsymbol{\omega}\).

Problem #\(7\): Describe Poinsot’s construction for visualizing the rotational dynamics of an arbitrary free rigid body.

Solution #\(7\): The idea is to first go to the principal body frame where one has (simplifying the notation on the principal moments of inertia a bit from earlier):

\[T_{\textbf X}=\frac{1}{2}I_1\omega_1^2+\frac{1}{2}I_2\omega_2^2+\frac{1}{2}I_3\omega_3^2\]

\[|\textbf L_{\textbf X}|^2=I_1^2\omega_1^2+I_2^2\omega_2^2+I_3^2\omega_3^2\]

Thanks to Problem #\(6\), it follows that the angular velocity \(\boldsymbol{\omega}=\sum_{i=1}^3\omega_i\hat{\textbf e}_i\) is constrained to lie on a closed loop intersection of these two ellipsoids; it is common to visualize the first one (called the inertia ellipsoid) defined by the conserved kinetic energy \(T_{\textbf X}\) as embedded in the principal body frame of the rigid body so that the aforementioned closed loop on which \(\boldsymbol{\omega}\) travels is viewed as a trajectory (called the polhode) on the inertia ellipsoid.

So far this is just in the (principal) body frame, but in practice one is interested in how the rotation looks like in the inertial lab frame. For this, the key observation is that, reverting back to the vector notation, because \(T_{\textbf X}=\boldsymbol{\omega}\cdot\textbf L_{\textbf X}/2\) defines the equation of a plane in \(\boldsymbol{\omega}\)-space with normal vector \(\textbf L_{\textbf X}\), called the invariable plane; this also explains why one typically emphasizes the inertia ellipsoid over the “angular momentum ellipsoid”, since one thus has the nice interpretation that the inertia ellipsoid is tangent to the invariable plane at \(\boldsymbol{\omega}\), a fact which is confirmed by the gradient \(\partial T_{\textbf X}/\partial\boldsymbol{\omega}=\textbf L_{\textbf X}\) being orthogonal to contour surfaces. In the inertial lab frame, \(\boldsymbol{\omega}\) thus traces out a (not necessarily closed) trajectory called the herpolhode on the invariable plane as the inertia ellipsoid rolls without slipping on it (this is because, from the earlier rigid body kinematic formula \(\dot{\textbf x}_i-\dot{\textbf x}_j=\boldsymbol{\omega}\times(\textbf x_i-\textbf x_j)\), if one choose \(\textbf x_i\) to be the point of contact of the inertia ellipsoid with the invariable plane and \(\textbf x_j:=\textbf X\) the center of mass/center of the ellipsoid, then clearly \(\boldsymbol{\omega}\) is parallel to \(\textbf x_i-\textbf X\) and so the cross product vanishes, so \(\dot{\textbf x}_i=\dot{\textbf X}=\textbf 0\) which is the definition of rolling without slipping).

An important conceptual association to make:

\[\text{Polhode}\Leftrightarrow\text{Body Frame}\]

\[\text{Herpolhode}\Leftrightarrow\text{Lab Frame}\]

Another comment to make about Poinsot’s construction is that in some sense any free rigid body, no matter how jagged or complicated of an asteroid it is, can always be reduced to an ellipsoid with the same principal axes and moments of inertia; the \(2\) are kinematically indistinguishable but the latter has the advantage of being easier to interpret in terms of Poinsot’s construction. And clearly, understanding how the ellipsoid moves is equivalent to understanding how the principal axes of the original rigid body are rotating, which is all one could ask for to claim one has fully understood the dynamics of the rigid body!!! Systematically:

Step #\(1\): Mentally replace the rigid body by its inertia ellipsoid.

Step #\(2\): Figure out what the conserved \(\textbf L_{\textbf X}\) is; its magnitude determines the polhode on the inertia ellipsoid while its direction determines the invariable plane.

Step #\(3\): Keeping the invariable plane fixed in the lab frame, visualize the ellipsoid rolling without slipping on it along its polhode.

Step #\(4\): Now, optionally, get rid of the inertia ellipsoid as it was just a mental crutch and put the original rigid body back…but now it’s clear what it’s doing!

Here are some animations showing Poinsot’s construction for a particular free, asymmetric top with \(I_1\neq I_2\neq I_3\neq I_1\) varying the choice of \(\textbf L_{\textbf X}\), viewed in the lab frame

Problem #\(8\): State and explain intuitively the intermediate axis theorem (variously known as the tennis racket theorem or Dzhanibekov effect).

Solution #\(8\): For a free, asymmetric top (such as in the animations above), rotation about the principal axes \(\hat{\textbf e}_1\) and \(\hat{\textbf e}_3\) are stable, but rotation about \(\hat{\textbf e}_2\) is unstable. Although one can mathematically deduce it by linearizing, it also follows by inspecting the topology of the polhodes on the inertia ellipsoid for the asymmetric top (see also this nice qualitative explanation due to Terence Tao based on centrifugal force arguments in a rotating frame and video by Veritasium where it is explained that such a dramatic “flip” will not occur for the Earth by virtue of its non-rigidity which allows kinetic energy \(T_{\textbf X}\) to be dissipated internally as heat energy, thereby tending towards a minimal-energy state \(T_{\textbf X}=\sum_{i=1}^3L_i^2/2I_i\) of spinning about the principal axis with maximal moment of inertia).

Problem #\(9\): Integrate Euler’s equations exactly for a free, symmetric top with \(I_1=I_2\).

Solution #\(9\):

Problem #\(10\): Draw a diagram defining the Euler angles and rewrite Euler’s equations in terms of the Euler angles.

Solution #\(10\): The whole point of the Euler angles is to provide a language with which someone in the inertial lab frame can begin to talk about in a quantitative manner what the rigid body is doing (Poinsot’s construction with the herpolhode and all allows a good qualitative description but it doesn’t immediately offer anything quantitative).

Keep in mind there are many other (\(12\) to be precise) conventions for Euler angles. The \(SO(3)\) matrices should be interpreted not as active transformations of vectors in space, but as passive transformations of the basis which is why they look the way they do, for instance the last transformation says:

\[\begin{pmatrix}\hat{\textbf e}_1\\\hat{\textbf e}_2\\\hat{\textbf e}_3\end{pmatrix}=\begin{pmatrix}\cos\phi&\sin\phi&0\\-\sin\phi&\cos\phi&0\\0&0&1\end{pmatrix}\begin{pmatrix}\hat{\textbf e}_1^{\prime\prime}\\\hat{\textbf e}_2^{\prime\prime}\\\hat{\textbf e}_3^{\prime\prime}\end{pmatrix}\]

and composing together the other \(2\) rotations, one finds that the net Euler angle rotation from the lab frame to the body frame is:

\[\begin{pmatrix}\hat{\textbf e}_1\\\hat{\textbf e}_2\\\hat{\textbf e}_3\end{pmatrix}=\begin{pmatrix}
\cos\psi\cos\phi – \cos\theta\sin\phi\sin\psi &
\sin\phi\cos\psi + \cos\theta\sin\psi\cos\phi &
\sin\theta\sin\psi \\
-\cos\phi\sin\psi – \cos\theta\cos\psi\sin\phi &
-\sin\psi\sin\phi + \cos\theta\cos\psi\cos\phi &
\sin\theta\cos\psi \\
\sin\theta\sin\phi &
-\sin\theta\cos\phi &
\cos\theta
\end{pmatrix}\begin{pmatrix}\tilde{\hat{\textbf e}}_1\\\tilde{\hat{\textbf e}}_2\\\tilde{\hat{\textbf e}}_3\end{pmatrix}\]

with the inverse of course given by transposition:

\[\begin{pmatrix}\tilde{\hat{\textbf e}}_1\\\tilde{\hat{\textbf e}}_2\\\tilde{\hat{\textbf e}}_3\end{pmatrix}=\begin{pmatrix}
\cos\psi\cos\phi – \cos\theta\sin\phi\sin\psi &
-\cos\phi\sin\psi – \cos\theta\cos\psi\sin\phi &
\sin\theta\sin\phi \\
\sin\phi\cos\psi + \cos\theta\sin\psi\cos\phi &
-\sin\psi\sin\phi + \cos\theta\cos\psi\cos\phi &
-\sin\theta\cos\phi \\
\sin\theta\sin\psi &
\sin\theta\cos\psi &
\cos\theta
\end{pmatrix}\begin{pmatrix}\hat{\textbf e}_1\\\hat{\textbf e}_2\\\hat{\textbf e}_3\end{pmatrix}\]

Problem #\(11\): In order to check that one has truly understood the Euler angles, explain what coefficient should replace each of the question marks in the linear combination below:

\[\boldsymbol{\omega}=?\tilde{\hat{\textbf e}}_3+?\hat{\textbf e}_1’+?\hat{\textbf e}_3^{\prime\prime}\]

Solution #\(11\):

\[\boldsymbol{\omega}=\dot\phi\tilde{\hat{\textbf e}}_3+\dot\theta\hat{\textbf e}_1’+\dot\psi\hat{\textbf e}_3^{\prime\prime}\]

Problem #\(12\): Hence, how are the principal body frame angular velocities \((\omega_1,\omega_2,\omega_3)\) related to the Euler angles \((\phi,\theta,\psi)\)?

Solution #\(12\): Basically, one just needs to express \(\tilde{\hat{\textbf e}}_3\) and \(\hat{\textbf e}’_1 \) in the principal body frame. Although one can basically see the answer in the bottom row of that massive matrix above, in practice to rederive it the efficient way is to just walk directly from start to destination, i.e. in the first transformation \(\tilde{\hat{\textbf e}}_3=\hat{\textbf e}_3’\), in the next transformation \(\hat{\textbf e}_3’=\sin\theta\hat{\textbf e}_2^{\prime\prime}+\cos\theta\hat{\textbf e}_3^{\prime\prime}\) and in the last transformation \(\hat{\textbf e}_3^{\prime\prime}=\hat{\textbf e}_3\) while \(\hat{\textbf e}_2^{\prime\prime}=\sin\psi\hat{\textbf e}_1+\cos\psi\hat{\textbf e}_2\) so overall:

\[\tilde{\hat{\textbf e}}_3=\sin\theta\sin\psi\hat{\textbf e}_1+\sin\theta\cos\psi\hat{\textbf e}_2+\cos\theta\hat{\textbf e}_3\]

as expected. Similarly, \(\hat{\textbf e}_1’=\hat{\textbf e}_1^{\prime\prime}=\cos\psi\hat{\textbf e}_1-\sin\psi\hat{\textbf e}_2\) and trivially \(\hat{\textbf e}_3^{\prime\prime}=\hat{\textbf e}_3\). Assembling everything, one obtains:

\[\omega_1=\dot{\phi}\sin\theta\sin\psi+\dot\theta\cos\psi\]

\[\omega_2=\dot\phi\sin\theta\cos\psi-\dot\theta\sin\psi\]

\[\omega_3=\dot\phi\cos\theta+\dot\psi\]

(and note one can of course also express the lab-frame components of \(\boldsymbol{\omega}\) in terms of the Euler angles \((\phi,\theta,\psi)\)).

Problem #\(13\): Armed with the Euler angle expressions of the angular velocity components, show for the free symmetric top (in the suggested order):

\[\dot{\theta}=0\]

\[\dot{\phi}=\frac{I_3\omega_3}{I_1\cos\theta}\]

\[\dot{\psi}=\left(1-\frac{I_3}{I_1}\right)\omega_3\]

Solution #\(14\): Orienting the lab frame axes so that \(\textbf L\) lies along \(\tilde{\hat{\textbf e}}_3\), then by definition of the Euler angle \(\theta\) one has:

\[\textbf L\cdot\hat{\textbf e}_3=|\textbf L|\cos\theta=I_3\omega_3\]

But both \(\textbf L\) and \(\omega_3\) are conserved for the free symmetric top, so \(\theta\) must also be conserved, i.e. \(\dot\theta=0\). This leads to the simplifications:

\[\omega_1=\dot{\phi}\sin\theta\sin\psi\]

\[\omega_2=\dot\phi\sin\theta\cos\psi\]

So that:

\[|\textbf L|^2=I_1^2(\omega_1^2+\omega_2^2)+I_3^2\omega_3^2=I_1^2\dot{\phi}^2\sin^2\theta+I_3^2\omega_3^2=\frac{I_3^2\omega_3^2}{\cos^2\theta}\]

Rearranging leads to the desired \(\dot{\phi}=I_3\omega_3/I_1\cos\theta\). Finally, substituting this result into the remaining equation:

\[\omega_3=\dot{\phi}\cos\theta+\dot{\psi}\]

and isolating for \(\dot{\psi}=\left(1-\frac{I_3}{I_1}\right)\omega_3\) completes the proof.

Aside: one could also have started with the kinetic Lagrangian \(L=\frac{1}{2}I_1\dot{\phi}^2\sin^2\theta+\frac{1}{2}I_3(\dot{\phi}\cos\theta+\dot\psi)^2\), extracting the conserved momenta \(\partial L/\partial\dot{\phi},\partial L/\partial\dot{\psi}=I_3\omega_3\) and using the equation of motion \(\partial L/\partial\theta=0\) to quickly also obtain the results above.

Aside: a famous example of the free symmetric top considered by Richard Feynman is the case of a lamina (e.g. a plate) with \(I_1=I_2=I_3/2\) by the perpendicular axis theorem. Then in the lab frame, the precession/wobble rate \(\dot\phi\) and the spin rate \(\dot{\psi}\) are related by:

\[\dot{\phi}=-\frac{2\dot{\psi}}{\cos\theta}\]

In particular, when the nutation angle is very slight \(\theta\ll 1\), it follows that:

\[\dot{\phi}\approx -2\dot{\psi}\]

so the lamina wobbles twice as fast as it spins, for small inclinations.

Problem #\(14\): The previous problems have all dealt with free rigid bodies. However, one can also consider a rigid body which is not free, being subject to an external torque \(\boldsymbol{\tau}_{\textbf X}^{\text{ext}}\). As an example, instead of a free symmetric top, consider a heavy symmetric top (e.g. a gyroscope) acted upon by a gravitational torque. Show that, just as there can be torque-free precession (Problem #\(9\)), there can also be torque-induced precession (in order to be able to do this problem, one absolutely requires the Euler angles in Solution #\(12\); thus, this just emphasizes again the point of Euler angles being to provide a language for discussing rigid body rotation in the lab frame).

Solution #\(14\):

The couple from gravity and the normal force is perfectly okay, but hard to express in the body frame? And also, I’m pretty sure the top will spin with its C.o.M fixed in the lab frame (rather than being pinned at the point of contact)? And as for the Lagrangian, just subtract \(-Mg\ell\cos\theta\), but \(\phi,\psi\) are still ignorable!

A ChatGPT thesaurus of all the synonyms of “propagator” across different disciplines:

Problem #\(1\): What is the retarded propagator of a quantum particle with Hamiltonian \(H\) in nonrelativistic quantum mechanics?

Solution #\(1\): At its heart, the “retarded” part should make one think of the causality factor \(\Theta(t-t’)\) while the “propagator” part should make one think of the time evolution operator \(\mathcal T\exp\left(-\frac{i}{\hbar}\int_{t’}^td\tilde t H(\tilde t)\right)\). So at its heart the retarded propagator is just the object:

\[\Theta(t-t’)\mathcal T\exp\left(-\frac{i}{\hbar}\int_{t’}^td\tilde t H\right)\]

that evolves states forward in time \(t\geq t’\). Indeed, the essence of this discussion holds for any kind of propagator in any context; they are basically just causal time evolution operators.

However, in the nonrelativistic (i.e. non-Lorentz invariant) quantum mechanics of a particle, one has access to its position observable \(\textbf X\) with position eigenstates \(|\textbf x\rangle\). So in this context, one may optionally (but conventionally) look at the matrix elements of the retarded propagator in the \(\textbf X\)-eigenbasis of that particle’s Hilbert space:

\[\Theta(t-t’)\langle\textbf x|\mathcal T\exp\left(-\frac{i}{\hbar}\int_{t’}^td\tilde t H\right)|\textbf x’\rangle\]

This now has the interpretation of the conditional probability amplitude for a measurement at time \(t\) of the particle’s position observable \(\textbf X\) to yield \(\textbf x\) given that at an earlier time \(t’\leq t\) the particle was prepared in a position eigenstate localized at \(\textbf x’\).

Problem #\(2\): Henceforth write \(U_H(t,t’):=\mathcal T\exp\left(-\frac{i}{\hbar}\int_{t’}^td\tilde t H\right)\) for the time evolution operator with respect to \(H\) so that \(i\hbar\dot U_H=HU_H\). Show that the retarded propagator is a retarded Green’s function for the Schrodinger operator \(\frac{\partial}{\partial t}-\frac{H}{i\hbar}\) in the sense that:

\[\langle\textbf x|\left(\frac{\partial}{\partial t}-\frac{H}{i\hbar}\right)\Theta(t-t’)U_H(t,t’)|\textbf x’\rangle=\delta(t-t’)\delta^3(\textbf x-\textbf x’)\]

Solution #\(2\): The key fact is that the derivative of the step function \(\dot{\Theta}=\delta\) is a Dirac delta, which one can heuristically justify on the grounds that \(\Theta(t)=\int_{-\infty}^tdx\delta(x)\). Thus, when the \(\partial/\partial t\) operator hits the \(\Theta(t-t’)\), it will ultimately give the requisite \(\delta(t-t’)\) expected of any retarded Green’s function. The spatial \(\delta^3(\textbf x-\textbf x’)=\langle\textbf x|\textbf x’\rangle\) then arises as an artifact of working in the \(\textbf X\)-eigenbasis.

Problem #\(3\): Although not as relevant for nonrelativistic QM, one can also consider the advanced propagator:

\[-\Theta(t’-t)\langle\textbf x|U_H(t,t’)|\textbf x’\rangle\]

and the Feynman propagator:

\[\Theta(t-t’)\langle\textbf x|U_H(t,t’)|\textbf x’\rangle+\Theta(t’-t)\langle\textbf x|U_H(t’,t)|\textbf x\rangle\]

Problem #\(3\): Hence, given any initial wavefunction \(\psi(\textbf x’,t’)\) at time \(t’\), how does the retarded propagator allow one to propagate this wavefunction forward in time to \(\psi(\textbf x,t)\) for \(t\geq t’\)?

Solution #\(3\): cf. Hugyen’s principle in wave optics:

\[\psi(\textbf x,t)=\int d^3\textbf x’\langle\textbf x|U_H(t,t’)|\textbf x’\rangle\psi(\textbf x’,t’)\]

Solution #\(2\): The nonrelativistic retarded propagator is intrinsic to the Hamiltonian \(H\) of the system, in particular being independent of whatever initial state \(|\psi(0)\rangle\) one prepares the particle in. More precisely, as the name “retarded propagator” suggests, its job is to propagate any initial state \(|\psi(0)\rangle\) forward in time by \(t>0\) (remember again that at its essence it’s just a causal time evolution operator!). That is, for \(t>0\) one can just set \(\Theta(t)=1\) and by insert a resolution of the identity:

\[|\psi(t)\rangle=e^{-iHt/\hbar}|\psi(0)\rangle\Leftrightarrow\langle\textbf x|\psi(t)\rangle=\int d^3\textbf x’\Delta_H(t,\textbf x-\textbf x’)\langle\textbf x’|\psi(0)\rangle\]

Emphasizing again, the \(\textbf x\) and \(\textbf x’\) are very much the unimportant labels in that equation; the causal time evolution of the state \(|\psi(0)\rangle\mapsto|\psi(t)\rangle\) is at the heart of the retarded propagator.

Problem #\(3\): Show that the nonrelativistic retarded propagator \(\Delta_H(t’,\textbf x’|t,\textbf x)\) is a retarded Green’s function for the Schrodinger operator:

\[\left(i\hbar\frac{\partial}{\partial t}-H\right)\Delta_H(t’,\textbf x’|t,\textbf x)=i\hbar\delta(t’-t)\delta^3(\tetbf x’-\textbf x)\]

Solution #\(3\):

Problem #\(4\): Show that the nonrelativistic propagator also admits the elegant path integral representation:

\[\Delta_H=\int\mathcal D\textbf xe^{iS[\textbf x]/\hbar}\]

Problem #\(5\): Compute the nonrelativistic propagator for a free particle.

Solution #\(5\): Inserting \(2\) resolutions of the identity:

\[\langle\textbf x|e^{-i|\textbf P|^2(t-t’)/2m\hbar}|\textbf x’\rangle=\int d^3\textbf p d^3\textbf p’\langle\textbf x|\textbf p\rangle\langle\textbf p|e^{-i|\textbf P|^2t/2m\hbar}|\textbf p’\rangle\langle\textbf p’|\textbf x’\rangle\]

and using the plane wave momentum eigenstate \(\langle\textbf x|\textbf p\rangle=e^{i\textbf p\cdot\textbf x/\hbar}/(2\pi\hbar)^{3/2}\), this can be shown to reduce to (after performing the relevant Gaussian integrals):

\[\langle\textbf x|e^{-i|\textbf P|^2(t-t’)/2m\hbar}|\textbf x’\rangle=\left(\frac{m}{2\pi i\hbar (t-t’)}\right)^{3/2}e^{-m|\textbf x-\textbf x’|^2/2i\hbar(t-t’)}\]

This result is often simplified by taking \(t’=0\) and \(\textbf x’=\textbf 0\):

\[\langle\textbf x|e^{-i|\textbf P|^2t/2m\hbar}|\textbf 0\rangle=\left(\frac{m}{2\pi i\hbar t}\right)^{3/2}e^{-m|\textbf x|^2/2i\hbar t}\]

Thus, roughly speaking, the propagator of a free particle at \(t=0\) is infinitely peaked at \(\textbf x=0\) but broadens over time \(t\) into an isotropic Gaussian with “imaginary” radial width \(\sigma_{|\textbf x|}=\sqrt{i\hbar t/m}\) exhibiting the usual \(\propto\sqrt{t}\) diffusion with “quantum diffusivity” \(D=i\hbar/m\).

Problem: Using Fourier transforms, calculate the Green’s function for the operator \(i\hbar\frac{\partial}{\partial t}-\frac{|\textbf P|^2}{2m}\) and by taking an inverse Fourier transform, show explicitly that it agrees, up to a factor of \(i\hbar\) with the propagator above.

Solution: The Green’s function in \(\textbf k,\omega\)-space is a simple Mobius transformation:

\[G_0(\textbf k,\omega)=\frac{1}{\hbar\omega-\frac{\hbar^2|\textbf k|^2}{2m}}\]

Choosing the retarded Green’s function:

\[G_0^+(\textbf x,t)=\int\frac{d^3\textbf k}{(2\pi)^3}\int\frac{d\omega}{2\pi}\frac{e^{i(\textbf k\cdot\textbf x-\omega t)}}{\hbar\omega-\frac{\hbar^2|\textbf k|^2}{2m}+i0^+}\]

So the purpose of the pole at \(\omega=\omega_k=\hbar k^2/2m\) is to ensure that when using Cauchy’s integral formula in this inverse FT, the particle is put on-shell with that dispersion. Doing the Gaussian integrals then reproduces \(1/i\hbar\) times the propagator.

Problem #\(6\): Repeat Problem #\(5\) but for the \(1\)-dimensional quantum harmonic oscillator of trapping frequency \(\omega\). Hence, deduce the identity of Hermite polynomials:

\[\sum\]

Solution #\(6\):

Newton’s second law in its most basic form states that for a single point mass, \(\dot{\textbf p}=\textbf F\) in any inertial frame. Combining this with Newton’s third law (antisymmetry of forces \(\textbf F_{i\to j}=-\textbf F_{j\to i}\), similar to many other antisymmetric phenomena in physics such as relative velocities \(\textbf v_{ij}=-\textbf v_{ji}\)) gives the most useful system formulation of Newton’s second law \(\dot{\textbf P}=\textbf F^{\text{ext}}\) in any inertial frame where \(\textbf P:=\sum_{i=1}^N\textbf p_i\) is the total momentum of a system of \(N\) point masses.

For the single-particle form \(\dot{\textbf p}=\textbf F\) of Newton’s second law, crossing both sides with the particle’s position \(\textbf x\) (relative to any origin, stationary, moving, accelerating, etc. as seen in the inertial frame) gives \(\dot{\textbf L}=\boldsymbol{\tau}\) in any inertial frame (this is thanks to a combination of two facts, namely: the cross product is a derivation, and it is non-degenerate, I wonder if there is a theory which generalizes this observation?) a rotational analog of Newton’s second law. Thus, naturally one might hope that an analog \(\textbf L=\boldsymbol{\tau}^{\text{ext}}\) should hold at the system level too, with \(\textbf L=\sum_{i=1}^N\textbf L_i\) and \(\boldsymbol{\tau}^{\text{ext}}=\sum_{i=1}^N\boldsymbol{\tau}^{\text{ext}}_i\). This identity is indeed true provided that the angular momenta \(\textbf L_i\) and external torques \(\boldsymbol{\tau}^{\text{ext}}_i\) are all measured in an inertial frame relative to a point \(\tilde{\textbf X}\) moving parallel at all times to the center of mass \(\textbf X\) of the system. The proof is a direct computation:

$$\dot{\textbf L}=\sum_{i=1}^N\dot{\textbf L}_i=\sum_{i=1}^N(\dot{\textbf x}_i-\dot{\tilde{\textbf X}})\times\textbf p_i+(\textbf x_i-\tilde{\textbf X})\times\textbf F_i$$

Now, the idea is to decompose the forces on each particle as \(\textbf F_i=\textbf F^{\text{ext}}_i+\sum_{j=1}^N\textbf F_{j\to i}\) with the non-degenerate \(\textbf F_{i\to i}=\textbf 0\). This finally gives:

$$\dot{\textbf L}=-\dot{\tilde{\textbf X}}\times\textbf P+\boldsymbol{\tau}^{\text{int}}+\boldsymbol{\tau}^{\text{ext}}$$

where \(\boldsymbol{\tau}^{\text{int}}=\sum_{1\leq i<j\leq N}(\textbf x_i-\textbf x_j)\times\textbf F_{j\to i}\) are \(N\choose 2\)\(=\frac{N(N-1)}{2}\) internal, reference-independent couples (and one would normally use one’s intuition to think about these rather than explicitly appealing to the formula) and \(\boldsymbol{\tau}^{\text{ext}}=\sum_{i=1}^N(\textbf x_i-\tilde{\textbf X})\times\textbf F^{\text{ext}}_i\). So it is clear that in order to have \(\dot{\textbf L}=\boldsymbol{\tau}^{\text{ext}}\) there are two obstacles in the way. Although actually, one of them is always zero, namely the internal couples \(\boldsymbol{\tau}^{\text{int}}=\textbf 0\). This follows from Galilean covariance (rather than from God, as my high school physics teacher told me); forces at a distance like gravity, electrostatics, masses pulled by springs or tension forces through ropes all have to be central (and thus conservative) so the cross product vanishes. On the other hand, the only time one can achieve an azimuthal/tangential/angular component of force is with friction, but then friction is not a force at a distance, but is rather a contact force and so the relative position \(\textbf x_i-\textbf x_j=\textbf 0\) at the point of transmission (the normal force is thus also zero because it simultaneously falls into both of those categories). I suspect any force of only position has to be like this, velocity-dependent forces like the magnetic force may be more subtle depending on Newtonian vs. relativistic frameworks, etc. To zero the remaining obstacle, since \(\textbf P=M\dot{\textbf X}\), it is both sufficient and necessary to have \(\tilde{\textbf X}||\textbf X\) at all times as claimed (e.g. a common, but certainly not the only choice is \(\tilde{\textbf X}:=\textbf X\)).

For instance, consider a hollow cylinder rolling without slipping down a ramp (couched in some inertial frame obviously). Take \(\tilde{\textbf X}\) to be the point of contact of the wheel with the ramp (clearly not associated with any physical particle). However, because \(\tilde{\textbf X}\) clearly moves parallel to \(\textbf X\) (indeed, it moves with the exact same velocity, just displaced from it), we know that \(\dot{\textbf L}=\boldsymbol{\tau}^{\text{ext}}\) relative to \(\tilde{\textbf X}\). Since this particular system of particles is in fact a rigid body, it is convenient to work with their common \(\omega\) and so \(L=I\omega\) (this step only works because of rolling without slipping) where by the parallel axis theorem \(I=2MR^2\). Now, one convenience of choosing \(\tilde{\textbf X}\) as the point of contact is that all contact forces acting at \(\tilde{\textbf X}\) by definition exert no torque about \(\tilde{\textbf X}\). So the only one is gravity. This yields \(\dot{\omega}=\frac{g\sin(\theta)}{2R}\). Since \(\omega\) is the same about any point (still don’t have very good intuition for this fact), one can do the derivation choosing \(\tilde{\textbf X}:=\textbf X\) instead, but it will be more lengthy since one has to more explicitly invoke the rolling without slipping constraint and Newton’s second law in its translation form to eliminate the static friction enforcing the rolling without slipping. The result is the same however, giving credence to the method.

My first year Cambridge physics supervisor also showed me a cute way to think about the situation which is mathematically equivalent to the above but conceptually somewhat different. The thing is that Newton’s second law refers only to a vector addition of force vectors, with no regards for where the point of application of those forces is (clearly if one is only interested in the center of mass motion, it doesn’t matter!). The torques are the devices that are sensitive to the point of application. So the idea is to basically draw the free body diagram of a system. Then, any forces that are not acting at the center of mass, draw a fictitious copy of the force acting at the center of mass but then to balance this out draw its negation also acting at the center of mass (if it acted elsewhere it wouldn’t affect the CoM motion but definitely the rotational dynamics!). Do this for all forces not acting at the CoM. The final result is that the torque about the CoM is completely comprised of reference-independent couples which dramatically clarifies the situation. Meanwhile, there is a unbalanced force left acting on the center of mass, which, intuitively, dictates its motion by Newton’s second law (actually, all this discussion holds equally well with CoM replaced by \(\tilde{\textbf X}\)). Hopefully this discussion will help me go back to learn rigid body dynamics more carefully and understand better.

The purpose of this post is to appreciate that the familiar idea of non-interacting particles from e.g. statistical mechanics manifests in the context of QFT as a free quantum field theory.

Problem #\(1\): What defines a free classical field theory? Give an example.

Solution #\(1\): A classical field theory is said to be free iff its Lagrangian density \(\mathcal L\), or equivalently Hamiltonian density \(\mathcal H\), is quadratic in all fields \(\phi_i\). Why? Because the corresponding Euler-Lagrange equations of motion for the fields \(\phi_i\) would then be linear in the \(\phi_i\). An example of a free classical field theory is the classical Klein-Gordon field theory which governs just a single, relativistic, real scalar field \(\phi(X)\):

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

\[\mathcal H=\frac{c^2\pi^2}{2}+\frac{1}{2}\biggr|\frac{\partial\phi}{\partial\textbf x}\biggr|^2+\frac{1}{2}k_c^2\phi^2\]

\[\partial^{\mu}\partial_{\mu}\phi+k_c^2\phi=0\]

(aside: Klein-Gordon field theory is “relativistic” in the sense of being Lorentz-invariant as manifest from the Lagrangian density \(\mathcal L\). It is “real” \(\phi(X)\in\textbf R\) because, when canonically quantized, it turns out to only give rise to neutral particles. It is clearly free. And it is a scalar field (as opposed to a vector field like in E&M) because, again when canonically quantized from a CFT to a QFT, turns out to also only describe spinless \(s=0\) bosons). Think Higgs boson for instance.

Problem #\(2\): In the context of the quantum harmonic oscillator from non-relativistic particle mechanics, the operators \(a,a^{\dagger}\) are typically called lowering and raising operators respectively. In moving from quantum particles to quantum fields however, the key conceptual leap that the number of particles is not conserved gives rise to an entirely new perspective on what \(a,a^{\dagger}\) are so that in QFT they are called annihilation and creation operators.

But, first, just using “classical cheating”, define \(a\) and \(a^{\dagger}\) and compute their commutator \([a,a^{\dagger}]\). Also define the number operator \(N\) and compute \([N,a]\) and \([N,a^{\dagger}]\).

Solution #\(2\): By inspecting the classical Hamiltonian:

\[H(x,p)=\frac{p^2}{2m}+\frac{m\omega^2x^2}{2}\]

One can construct a characteristic quantum oscillator length \(x_0\) and oscillator momentum \(p_0\) through the virial relations:

\[\frac{\hbar\omega}{2}=\frac{p_0^2}{2m}=\frac{m\omega^2x_0^2}{2}\]

So that:

\[H=\frac{\hbar\omega}{2}\left(\left(\frac{x}{x_0}\right)^2+\left(\frac{p}{p_0}\right)^2\right)\]

This sum of squares suggests the classical factorization:

\[H=\hbar\omega N\]

where \(N:=aa^{\dagger}\) and the isomorphism \(\cong\) between phase space \(\textbf R^2\) and the complex plane \(\textbf C\) is exploited:

Since canonical quantization is a Lie algebra representation up to quadratic polynomials on phase space (which applies to everything here), it follows that all Poisson bracket computations must agree with the corresponding commutators. In particular:

\[\{a,a^{\dagger}\}=\det\begin{pmatrix}\partial a/\partial x&\partial a^{\dagger}/\partial x \\ \partial a/\partial p & \partial a^{\dagger}/\partial p\end{pmatrix}=-\frac{i}{x_0p_0}=\frac{1}{i\hbar}\]

which could also be obtained just from drawing pictures like:

This “proves” that \([a,a^{\dagger}]=i\hbar\{a,a^{\dagger}\}=1\). The other commutators can be quickly obtained using the “product rule” for Poisson brackets:

\[\{N,a\}=\{aa^{\dagger},a\}=a\{a^{\dagger},a\}+\{a,a\}a^{\dagger}=-\frac{a}{i\hbar}\Rightarrow [N,a]=-a\]

\[\{N,a^{\dagger}\}=\{N,a\}^{\dagger}=\frac{a^{\dagger}}{i\hbar}\Rightarrow [N,a^{\dagger}]=a^{\dagger}\]

(note that although \([F,G]^{\dagger}=[G^{\dagger},F^{\dagger}]\) for commutators, for Poisson brackets it’s just \(\{f,g\}^{\dagger}=\{f^{\dagger},g^{\dagger}\}\)).

Problem #\(3\): A \(1\)D string has Hamiltonian density:

\[\mathcal H=\sum_{n=1}^{\infty}\left(\frac{p_n^2}{2\mu}+\frac{1}{2}\mu\omega_n^2x_n^2\right)\]

Unlike the previous problem where there was just \(1\) harmonic oscillator, here there are an infinite number of harmonic oscillators, one for each normal mode \(n=1,2,3,…\) (however, here it is only countably infinite; in QFT it will often be uncountably infinite!). Thus, each one gets its own pair of creation and annihilation operators \(a_n,a_n^{\dagger}\) defined analogously as in Problem #\(2\) (but where each oscillator now also has its own characteristic length \(x_0^{(n)}=\sqrt{\hbar/\mu\omega_n}\) and characteristic momentum \(p_0^{(n)}=\sqrt{\hbar\mu\omega_n}\)). Check that for distinct oscillators \(n\neq m\), these operators don’t talk to each other in the sense that \([a_n,a_m]=[a_n^{\dagger},a_m^{\dagger}]=0\) and \([a_n,a_m^{\dagger}]=\delta_{nm}\). Show that the Hamiltonian density can be rewritten:

\[\mathcal H=\frac{\hbar}{2}\sum_{n=1}^{\infty}\omega_n\{a_n,a_n^{\dagger}\}\]

Assuming there exists a single universal ground state \(|0\rangle\) annihilated by \(a_n|0\rangle=0\) for all \(n=1,2,3,…\), show that, after discarding the vacuum energy, \(\mathcal H\mapsto :\mathcal H:\) takes the normally ordered form:

\[:\mathcal H:=\hbar\sum_{n=1}^{\infty}\omega_na_n^{\dagger}a_n\]

And also show that \([:\mathcal H:,a_n^{\dagger}]=\hbar\omega_na_n^{\dagger}\). Hence calculate the energy of the excited Fock state:

\[|N_1,N_2,…,N _n\rangle=(a_1^{\dagger})^{N_1}(a_2^{\dagger})^{N_2}…(a_n^{\dagger})^{N_n}|0\rangle\]

Solution #\(3\):

Problem #\(4\): What is the most general real scalar field solution of the classical Klein-Gordon equation?

Solution #\(4\): Since it’s a linear PDE, it makes sense to change to the Fourier basis for the real scalar field:

\[\phi(\textbf x,t)=\frac{1}{(2\pi)^{3/2}}\int\hat{\phi}(\textbf k,t)e^{i\textbf k\cdot\textbf x}d^3\textbf k\]

Substituting this as an ansatz into the Klein-Gordon equation yields, for each \(\textbf k\in\textbf R^3\), the equation of motion of a simple harmonic oscillator for \(\hat{\phi}(\textbf k,t)\):

\[\frac{\partial^2\hat{\phi}}{\partial t^2}+\omega_{\textbf k}^2\hat{\phi}=0\]

where the relativistic dispersion relation for the oscillator frequency is:

\[\omega_{\textbf k}=c\sqrt{|\textbf k|^2+k_c^2}\]

(defined here as the positive square root). So the most general solution in \(\textbf k\)-space is parameterized by two arbitrary scalar fields \(a_{\textbf k},b_{\textbf k}\):

\[\hat{\phi}(\textbf k,t)=a_{\textbf k}e^{-i\omega_{\textbf k}t}+b_{\textbf k}e^{i\omega_{\textbf k}t}\]

leading to the corresponding general solution in \(\textbf x\)-space:

\[\phi(\textbf x,t)=\frac{1}{(2\pi)^{3/2}}\int\left(a_{\textbf k}e^{i(\textbf k\cdot\textbf x-\omega_{\textbf k}t)}+b_{\textbf k}e^{i(\textbf k\cdot\textbf x+\omega_{\textbf k}t)}\right)d^3\textbf k\]

However, demanding that \(\phi(\textbf x,t)\) be a real scalar field so that \(\phi^{\dagger}(\textbf x,t)=\phi(\textbf x,t)\) (which is equivalent to demanding that its Fourier transform be Hermitian \(\hat{\phi}^{\dagger}(\textbf k,t)=\hat{\phi}(-\textbf k,t)\)) forces \(b_{\textbf k}=a_{-\textbf k}^{\dagger}\). Thus actually one can simply write:

\[\phi(\textbf x,t)=\frac{1}{(2\pi)^{3/2}}\int\left(a_{\textbf k}e^{i(\textbf k\cdot\textbf x-\omega_{\textbf k}t)}+a_{\textbf k}^{\dagger}e^{-i(\textbf k\cdot\textbf x-\omega_{\textbf k}t)}\right)d^3\textbf k\]

Or, defining the four-wavevector \(|K\rangle:=(\omega_{\textbf k}/c, \textbf k)\), working with the standard four-position \(|X\rangle:=(ct,\textbf x)\), and using the time-like convention for the Minkowski metric \(\eta=\text{diag}(1,-1,-1,-1)\):

\[\phi(X)=\frac{1}{(2\pi)^{3/2}}\int\left(a_{\textbf k}e^{-i\langle K|X\rangle}+a_{\textbf k}^{\dagger}e^{i\langle K|X\rangle}\right)d^3\textbf k\]

Note here that one is still working entirely within the realm of classical field theory (not QFT yet!) so the daggers \(\dagger\) are nothing more than complex conjugation, and in particular \(\phi(X)\) is a (real) scalar field rather than a (Hermitian) operator field. This is similarly the case with \(a_{\textbf k},a_{\textbf k}^{\dagger}\) being (complex) scalar fields at this stage rather than operator fields. Finally, it is worth noting that the conjugate real scalar momentum field is:

\[\pi(X):=\frac{1}{c^2}\frac{\partial\phi}{\partial t}=\frac{-i}{(2\pi)^{3/2}c^2}\int\omega_{\textbf k}\left(a_{\textbf k}e^{-i\langle K|X\rangle}-a_{\textbf k}^{\dagger}e^{i\langle K|X\rangle}\right)d^3\textbf k\]

Problem #\(5\): As warmup for the next problem, evaluate the Fourier transform of the plane wave with wavevector \(\textbf k_0\):

\[\frac{1}{(2\pi)^{3/2}}\int e^{i(\textbf k_0\cdot\textbf x-\omega t)}e^{-i\textbf k\cdot\textbf x}d^3\textbf x\]

Solution #\(5\): Under this unitary convention for the Fourier transform, it is:

\[\frac{1}{(2\pi)^{3/2}}\int e^{i(\textbf k_0\cdot\textbf x-\omega t)}e^{-i\textbf k\cdot\textbf x}d^3\textbf x=(2\pi)^{3/2}\delta^3(\textbf k-\textbf k_0)e^{-i\omega t}\]

As one may check by computing the “inverse Fourier transform” and using the non-negotiable, defining sifting property of the \(\delta\)-distribution:

\[\frac{1}{(2\pi)^{3/2}}\int (2\pi)^{3/2}\delta^3(\textbf k-\textbf k_0)e^{-i\omega t}e^{i\textbf k\cdot\textbf x}d^3\textbf k=e^{i(\textbf k_0\cdot\textbf x-\omega t)}\]

Problem #\(6\): Explain what it means to take the classical Klein-Gordon theory governing the real scalar fields \(\phi(X),\pi(X)\) and canonically quantize them in the Heisenberg picture to Hermitian operator fields, thereby constructing a free quantum Klein-Gordon field theory. What changes in the Schrodinger picture?

Solution #\(6\): On paper, the equations are invariant. Canonical quantization is really more of a conceptual leap, in that the fundamental “data type” of the objects in the equations are what changes. Specifically, one now promotes \(\phi(X),\pi(X)\) to operator fields, i.e. a function associating an operator to each event \(|X\rangle\) in Minkowski spacetime. It follows that the only way the classical Klein-Gordon field decompositions in the Fourier basis of plane waves can be compatible with canonical quantization is iff, in addition to promoting \(\phi(X),\pi(X)\) to operator fields, one also/simultaneously promotes the “Fourier coefficients” \(a_{\textbf k},a_{\textbf k}^{\dagger}\) to operators.

Furthermore, it turns out one should postulate equal-\(t\) commutation relations between these operators in the Heisenberg picture:

\[[\phi(\textbf x,t),\phi(\textbf x’,t)]=[\pi(\textbf x,t),\pi(\textbf x’,t)]=0\]

\[[\phi(\textbf x,t),\pi(\textbf x’,t)]=i\hbar\delta^3(\textbf x-\textbf x’)\]

though one can show through straightforward computations in the spirit of Problem #\(5\) that these are logically equivalent to the following commutation relations on the Fourier coefficients (cf. Problem #\(3\)):

\[[a_{\textbf k},a_{\textbf k’}]=[a_{\textbf k}^{\dagger},a_{\textbf k’}^{\dagger}]=0\]

\[[a_{\textbf k},a_{\textbf k’}^{\dagger}]=\frac{\hbar c^2}{2\omega_{\textbf k}}\delta^3(\textbf k-\textbf k’)\]

(note other normalization conventions also exist). For example:

\[\hat{\pi}(\textbf k,t)=\frac{1}{(2\pi)^{3/2}}\int d^3\textbf x e^{-i\textbf k\cdot\textbf x}\frac{-i}{(2\pi)^{3/2}c^2}\int\omega_{\textbf k’}\left(a_{\textbf k’}e^{-i\langle K’|X\rangle}-a_{\textbf k’}^{\dagger}e^{i\langle K’|X\rangle}\right)d^3\textbf k’\]

\[=-\frac{i}{(2\pi)^3c^2}\int d^3\textbf k’ d^3\textbf x\omega_{\textbf k’}\left(a_{\textbf k’}e^{i(\textbf k’-\textbf k)\cdot\textbf x}e^{-i\omega_{\textbf k’}t}-a_{\textbf k’}^{\dagger}e^{-i(\textbf k’+\textbf k)\cdot\textbf x}e^{i\omega_{\textbf k’}t}\right)\]

\[=-\frac{i}{(2\pi)^3c^2}\int d^3\textbf k’\omega_{\textbf k’}\left(a_{\textbf k’}(2\pi)^3\delta^3(\textbf k’-\textbf k)e^{-i\omega_{\textbf k’}t}-a_{\textbf k’}^{\dagger}(2\pi)^3\delta^3(\textbf k+\textbf k’)e^{i\omega_{\textbf k’}t}\right)\]

\[=-\frac{i\omega_{\textbf k}}{c^2}(a_{\textbf k}e^{-i\omega_{\textbf k}t}-a^{\dagger}_{-\textbf k}e^{i\omega_{\textbf k}t})\]

where the isotropy of the dispersion relation \(\omega_{-\textbf k}=\omega_{\textbf k}=\omega_{|\textbf k|}\) has been used. The expression for \(\hat{\phi}(\textbf k,t)\) was already given above, but can also of course be recovered by identical manipulations. So, in analogy with the classical case in Problem #\(3\), one has:

\[a_{\textbf k}=\frac{e^{i\omega_{\textbf k}t}}{2}\left(\hat{\phi}(\textbf k,t)+\frac{ic^2}{\omega_{\textbf k}}\hat{\pi}(\textbf k,t)\right)\]

\[a_{-\textbf k}^{\dagger}=\frac{e^{-i\omega_{\textbf k}t}}{2}\left(\hat{\phi}(\textbf k,t)-\frac{ic^2}{\omega_{\textbf k}}\hat{\pi}(\textbf k,t)\right)\]

which are happily consistent with each other (note also that \(a_{\textbf k}\) is \(t\)-independent so all \(t\)-dependence on the right-hand side is spurious). One then has e.g.

\[[a_{\textbf k},a_{\textbf k’}^{\dagger}]=\frac{e^{i(\omega_{\textbf k}-\omega_{\textbf k’})t}}{4}\left([\hat{\phi}(\textbf k,t),\hat{\phi}(-\textbf k,t)]+\frac{ic^2}{\omega_{\textbf k}}[\hat{\pi}(\textbf k,t),\hat{\phi}(-\textbf k’,t)]-\frac{ic^2}{\omega_{\textbf k}}[\hat{\phi}(\textbf k,t),\hat{\pi}(-\textbf k’,t)]+\frac{c^4}{\omega_{\textbf k}\omega_{\textbf k’}}[\hat{\pi}(\textbf k,t),\hat{\pi}(-\textbf k’,t)]\right)\]

But the commutation relations postulated on \(\phi(\textbf x,t)\) and \(\pi(\textbf x,t)\) in real space directly imply their corresponding commutation relations in \(\textbf k\)-space, for instance:

\[[\hat{\phi}(\textbf k,t),\hat{\pi}(-\textbf k’,t)]=\biggr[\frac{1}{(2\pi)^{3/2}}\int d^3\textbf x e^{-i\textbf k\cdot\textbf x}\phi(\textbf x,t),\frac{1}{(2\pi)^{3/2}}\int d^3\textbf x’e^{i\textbf k’\cdot\textbf x’}\pi(\textbf x’,t)\biggr]\]

\[=\frac{1}{(2\pi)^3}\int d^3\textbf x d^3\textbf x’e^{i(\textbf k’\cdot\textbf x’-\textbf k\cdot\textbf x)}[\phi(\textbf x,t),\pi(\textbf x’,t)]\]

\[=\frac{i\hbar}{(2\pi)^3}\int d^3\textbf x d^3\textbf x’e^{i(\textbf k’\cdot\textbf x’-\textbf k\cdot\textbf x)}\delta^3(\textbf x-\textbf x’)\]

\[=\frac{i\hbar}{(2\pi)^3}\int d^3\textbf x e^{i(\textbf k’-\textbf k)\cdot\textbf x}=i\hbar\delta^3(\textbf k-\textbf k’)\]

From here the result follows. In a way, the fact that it is equal-\(t\) commutators of the operator fields which appear could be construed as motivation for imposing equal-\(t\) commutation relations in the first place. One can of course also easily show the converse, i.e. that imposing the commutation relations on \(a_{\textbf k},a_{\textbf k}^{\dagger}\) yield those of \(\phi\) and \(\pi\).

Problem #\(7\): Calculate the Hamiltonian \(H=\int d^3\textbf x\mathcal H\) for the free Klein-Gordon quantum field theory. In particular, comment on the infrared and ultraviolet divergences present and how one can regularize them.

Solution #\(7\):

\[H=\frac{c^2}{2}\int d^3\textbf x\pi^2+\frac{1}{2}\int d^3\textbf x\biggr|\frac{\partial\phi}{\partial\textbf x}\biggr|^2+\frac{k_c^2}{2}\int d^3\textbf x\phi^2\]

As an example:

\[\int d^3\textbf x\pi^2=-\frac{1}{(2\pi)^3c^4}\int d^3\textbf kd^3\textbf k’d^3\textbf x\omega_{\textbf k}\omega_{\textbf k’}(a_{\textbf k}a_{\textbf k’}e^{i(\textbf k+\textbf k’)\cdot\textbf x}e^{-i(\omega_{\textbf k}+\omega_{\textbf k’})t}-a_{\textbf k}a^{\dagger}_{\textbf k’}e^{i(\textbf k-\textbf k’)\cdot\textbf x}e^{i(\omega_{\textbf k’}-\omega_{\textbf k})t}+\approx\text{h.c.})\]

\[=-\frac{1}{c^4}\int d^3\textbf kd^3\textbf k’\omega_{\textbf k}\omega_{\textbf k’}(a_{\textbf k}a_{\textbf k’}\delta^3(\textbf k+\textbf k’)e^{-i(\omega_{\textbf k}+\omega_{\textbf k’})t}-a_{\textbf k}a^{\dagger}_{\textbf k’}\delta^3(\textbf k-\textbf k’)e^{i(\omega_{\textbf k’}-\omega_{\textbf k})t}+\approx\text{h.c.})\]

\[=-\frac{1}{c^4}\int d^3\textbf k\omega_{\textbf k}^2(a_{\textbf k}a_{-\textbf k}e^{-2i\omega_{\textbf k}t}-a_{\textbf k}a^{\dagger}_{\textbf k}+\text{h.c.})\]

And similar manipulations give:

\[\int d^3\textbf x\biggr|\frac{\partial\phi}{\partial\textbf x}\biggr|^2=\int d^3\textbf k|\textbf k|^2(a_{\textbf k}a_{-\textbf k}e^{-2i\omega_{\textbf k}t}+a_{\textbf k}a^{\dagger}_{\textbf k}+\text{h.c.})\]

\[\int d^3\textbf x\phi^2=\int d^3\textbf k(a_{\textbf k}a_{-\textbf k}e^{-2i\omega_{\textbf k}t}+a_{\textbf k}a^{\dagger}_{\textbf k}+\text{h.c.})\]

Using the relativistic dispersion relation \(\omega_{\textbf k}^2/c^2=|\textbf k|^2+k_c^2\), the \(t\)-dependent oscillatory terms vanish and one finds that:

\[H=\frac{1}{c^2}\int d^3\textbf k\omega_{\textbf k}^2\{a_{\textbf k},a^{\dagger}_{\textbf k}\}\]

At this point, wishing to normally order \(H\mapsto :H:\), one has:

\[\{a_{\textbf k},a_{\textbf k}^{\dagger}\}=2a_{\textbf k}^{\dagger}a_{\textbf k}+\frac{\hbar c^2}{2\omega_{\textbf k}}\delta^3(\textbf 0)\]

This particularly infinity involving the \(\delta^3(\textbf 0)\) is an example of an infrared divergence; it can be dealt with using the identity:

\[\delta^3(\textbf 0)=\frac{1}{(2\pi)^3}\int d^3\textbf x=\frac{V}{(2\pi)^3}\]

where \(V\) is the volume of the system that one subsequently send to \(V\to\infty\), hence \(\delta^3(\textbf 0)=\infty_{\text{IR}}\). So actually:

\[H=\frac{1}{c^2}\int d^3\textbf k\omega_{\textbf k}^2\left(2a_{\textbf k}^{\dagger}a_{\textbf k}+\frac{V}{(2\pi)^3}\frac{\hbar c^2}{2\omega_{\textbf k}}\right)\]

However, even after this replacement the second term still goes like \(\sim\int d^3\textbf k\omega_{\textbf k}=\infty_{\text{UV}}\) which diverges when integrated over \(\textbf k\in\textbf R^3\); it is called an ultraviolet divergence. In this case, it turns out one can deal with it by just subtracting it off from \(H\) (note it can also be interpreted as the zero-point energy \(E_0\) of the vacuum state \(|0\rangle\)) to obtain:

\[:H:=\frac{2}{c^2}\int d^3\textbf k\omega_{\textbf k}^2a_{\textbf k}^{\dagger}a_{\textbf k}\]

though the Casimir force and cosmological constant are \(2\) cases where this cannot be done.

Pragmatically speaking, normal ordering allows one to commute creation and annihilation operators with impunity (no need to worry about IR or UV divergences), i.e. \(:[a_{\textbf k},a_{\textbf k’}^{\dagger}]:=0\).

Problem #\(8\): Show that \([:H:,a_{\textbf k}^{\dagger}]=\hbar\omega_{\textbf k}a_{\textbf k}^{\dagger}\) and \([:H:,a_{\textbf k}]=-\hbar\omega_{\textbf k}a_{\textbf k}\) are identical to the classical results in Problem #\(3\). Hence, explain how excited states of the free quantum field \(\phi\) can be interpreted as identical spinless bosons.

Solution #\(8\): As an example:

\[[:H:,a_{\textbf k}^{\dagger}]=\frac{2}{c^2}\int d^3\textbf k’\omega_{\textbf k’}^2[a_{\textbf k’}^{\dagger}a_{\textbf k’},a_{\textbf k}^{\dagger}]\]

\[=\frac{2}{c^2}\int d^3\textbf k’\omega_{\textbf k’}^2a_{\textbf k’}^{\dagger}\left(\frac{\hbar c^2}{2\omega_{\textbf k}}\right)\delta^3(\textbf k-\textbf k’)=\hbar\omega_{\textbf k}a_{\textbf k}\]

and taking its adjoint yields the other commutation relation. Hence, one can declare victory; the excited state \(a_{\textbf k}^{\dagger}|0\rangle\) may be checked to be an \(:H:\)-eigenstate:

\[:H:a_{\textbf k}^{\dagger}|0\rangle=(a_{\textbf k}:H:+\hbar\omega_{\textbf k}a_{\textbf k}^{\dagger})|0\rangle=E_{\textbf k}a_{\textbf k}^{\dagger}|0\rangle\]

with energy \(E_{\textbf k}:=\hbar\omega_{\textbf k}\) above the ground state energy \(E_{\textbf 0}=0\). This already provides some evidence for the particle interpretation

As further evidence, one can check that the excited state \(a_{\textbf k}^{\dagger}|0\rangle\) has momentum \(\textbf p_{\textbf k}=\hbar\textbf k\) just like for a particle. To see this, note that for Klein-Gordon theory the energy-momentum tensor is symmetric with:

\(T^{\mu\nu}=\frac{\partial\mathcal L}{\partial(\partial_{\mu}\phi)}\partial^{\nu}\phi-\delta^{\mu\nu}\mathcal L=\partial^{\mu}\phi\partial^{\nu}\phi-\delta^{\mu\nu}\mathcal L=\begin{pmatrix}\mathcal H & c\boldsymbol{\mathcal P}^T\\ c\boldsymbol{\mathcal P} & \left(\frac{\partial\phi}{\partial\textbf x}\right)^{\otimes 2}-\mathcal L1_3\end{pmatrix}\)

where the momentum density is \(\boldsymbol{\mathcal P}=-\frac{1}{c^2}\frac{\partial\phi}{\partial t}\frac{\partial\phi}{\partial\textbf x}\) from which one obtains the conserved momentum \(\textbf P:=\int d^3\textbf x\boldsymbol{\mathcal P}\):

\[\textbf P=-\int d^3\textbf x\pi\frac{\partial\phi}{\partial\textbf x}\]

\[=\frac{1}{c^2}\int d^3\textbf k\omega_{\textbf k}\textbf k(a_{\textbf k}a_{-\textbf k}e^{-2i\omega_{\textbf k}t}+a^{\dagger}_{\textbf k}a^{\dagger}_{-\textbf k}e^{2i\omega_{\textbf k}t}+a_{\textbf k}a^{\dagger}_{\textbf k}+a^{\dagger}_{\textbf k}a_{\textbf k})\]

but the \(t\)-dependent part together with the factor of \(\textbf k\) makes that whole part of the integral vanish by oddness, so:

\[\textbf P=\frac{1}{c^2}\int d^3\textbf k\omega_{\textbf k}\textbf k\{a_{\textbf k},a_{\textbf k}^{\dagger}\}\]

Canonically quantizing \(\textbf P\) to operator status in the Heisenberg picture and applying normal ordering, one thus has:

\[:\textbf P:=\frac{2}{c^2}\int d^3\textbf k\omega_{\textbf k}\textbf ka^{\dagger}_{\textbf k}a_{\textbf k}\]

so acting on the excited state \(a_{\textbf k}^{\dagger}|0\rangle\) yields:

\[:\textbf P:a_{\textbf k}^{\dagger}|0\rangle=\frac{2}{c^2}\int d^3\textbf k\omega_{\textbf k}\textbf ka^{\dagger}_{\textbf k}\left(a^{\dagger}_{\textbf k}a_{\textbf k}+\frac{\hbar c^2}{2\omega_{\textbf k}}\delta^3(\textbf k-\textbf k’)\right)|0\rangle=\hbar\textbf ka_{\textbf k}^{\dagger}|0\rangle\]

Finally, the total angular momentum of the classical field is:

\[\textbf J=\int d^3\textbf x(\textbf x\times\boldsymbol{\mathcal P})\]

which includes contributions from both orbital angular momentum and spin angular momentum \(\textbf J=\textbf L+\textbf S\). In particular, if one artificially fixes \(\textbf L=\textbf 0\) by considering a stationary particle \(|\textbf k=\textbf 0\rangle=a_{\textbf k=\textbf 0}^{\dagger}|0\rangle\), then this has:

\[\textbf J|\textbf k=\textbf 0\rangle=0\]

so the particles that arise as excitations of the free quantum field are spinless as mentioned earlier. Furthermore, given any multi-particle excited state:

\[|\textbf k_1,\textbf k_2,…\rangle=a_{\textbf k_1}^{\dagger}a_{\textbf k_2}^{\dagger}…|0\rangle\]

the exchange of any two particles \(\textbf k_1\leftrightarrow\textbf k_2\) yields a physically equivalent state (since all creation operators commute) so these are by definition identical bosons.

As an aside, this particle interpretation of the excited states is also what justifies calling the ground state \(|0\rangle\) as the “vacuum state” because classically a vacuum is thought of as containing no particles.

Problem #\(9\): What is the state space \(\mathcal F\) of the free Klein-Gordon quantum field theory?

Solution #\(9\): It is known as Fock space, and is just the direct sum of all \(N\)-particle sectors \(\mathcal F_N=\text{span}_{\textbf C}\{\prod_{i=1}^Na_{\textbf k_i}^{\dagger}|0\rangle:\textbf k_i\in\textbf R^3\}\):

\[\mathcal F:=\oplus_{N=0}^{\infty}\mathcal F_N\]

It is denoted by \(\mathcal F\) rather than \(\mathcal H\) to avoid confusion with the Hamiltonian density (cf. in statistical mechanics where, to specify the microstate of a Bose gas, one just has to specify Bose occupation numbers in each single-boson state; this is the Fock space at work).

Problem #\(10\): Define the number operator \(N\), explain what it does, and check that it commutes with \(:H:\); what is the implication of this for free quantum field theories?

Solution #\(10\): It takes the normally ordered form:

\[N:=\frac{2}{\hbar c^2}\int d^3\textbf k\omega_{\textbf k}a_{\textbf k}^{\dagger}a_{\textbf k}\]

And it is easy to convince oneself that \(N|\textbf k_1,\textbf k_2,…,\textbf k_n\rangle=n|\textbf k_1,\textbf k_2,…,\textbf k_n\rangle\) counts the number \(n\) of spinless bosons in a given multi-boson state \(|\textbf k_1,\textbf k_2,…,\textbf k_n\rangle\in \mathcal F_n\). Finally:

\[[:H:,N]=\frac{4}{\hbar c^4}\int d^3\textbf k d^3\textbf k’\omega_{\textbf k}^2\omega_{\textbf k’}[a_{\textbf k}^{\dagger}a_{\textbf k},a_{\textbf k’}^{\dagger}a_{\textbf k’}]\]

\[=\frac{2}{c^2}\int d^3\textbf k\omega_{\textbf k}^2(a_{\textbf k}^{\dagger}a_{\textbf k}-a_{\textbf k}^{\dagger}a_{\textbf k})=0\]

so the number of particles in free field theories are conserved (a property which is decisively not true of interacting field theories).

Problem #\(11\): Show that the neither the state \(a_{\textbf k}^{\dagger}|0\rangle\) nor \(\phi(\textbf x,t)|0\rangle\) are normalizable in \(\mathcal F\).

Solution #\(11\): Assuming the vacuum state is normalized \(\langle 0|0\rangle=1\), then for the single-boson excitation \(|\textbf k\rangle=a_{\textbf k}^{\dagger}|0\rangle\), one picks up the same IR divergence as before:

\[\langle\textbf k|\textbf k\rangle=\langle 0|a_{\textbf k}a_{\textbf k}^{\dagger}|0\rangle=\langle 0|a_{\textbf k}^{\dagger}a_{\textbf k}+\frac{\hbar c^2}{2\omega_{\textbf k}}\delta^3(\textbf 0)|0\rangle=\frac{\hbar c^2}{2\omega_{\textbf k}}\delta^3(\textbf 0)\]

Similarly, although \(\langle 0|\phi(\textbf x,t)|0\rangle=0\), the fluctuations of \(\phi\) at a fixed event \((\textbf x,t)\) (which is also the norm-squared of the state since \(\phi^{\dagger}=\phi\)):

\[\langle 0|\phi^2(\textbf x,t)|0\rangle=\frac{4\pi\hbar c}{2(2\pi)^3}\delta^3(\textbf 0)\int_0^{\infty}\frac{|\textbf k|^2}{\sqrt{|\textbf k|^2+k_c^2}}d|\textbf k|\]

suffers both IR and UV divergences.

Problem #\(12\): Outline why \(\int\frac{d^3\textbf k}{\omega_{\textbf k}}\) is Lorentz invariant. Hence, define a relativistically “normalized” one-particle Fock state \(|\textbf k\rangle\).

Solution #\(12\): Clearly the measure \(d^4|K\rangle\) is Lorentz invariant (more precisely, \(SO(1,3)\)-invariant since \(|K’\rangle=\Lambda|K\rangle\) gives a Jacobian determinant \(\det\frac{\partial|K’\rangle}{\partial|K\rangle}=\det\Lambda=1\) for \(\Lambda\in SO(1,3)\)) and the inner product \(\langle K|K\rangle=\frac{\omega^2}{c^2}-|\textbf k|^2\) is also Lorentz invariant (this time \(O(1,3)\)-invariant), and finally the step function \(\Theta(\omega)\) is also Lorentz invariant (this time \(O^+(1,3)\)-invariant), so all in all, when the measure \(d^4|K\rangle\) is integrated on the mass shell \(\langle K|K\rangle=k_c^2\) and with the condition \(\omega>0\):

\[\int d^4|K\rangle\delta\left(\frac{\omega^2}{c^2}-|\textbf k|^2-k_c^2\right)\Theta(\omega)=\int\frac{cd^3\textbf k}{2\omega_{\textbf k}}\]

the resulting object must be Lorentz invariant (i.e. \(SO^+(1,3)\)-invariant). Note that the identity \(\delta^n(\textbf F(\textbf x))=\sum_{\textbf x_0\in\textbf R^n:\textbf F(\textbf x_0)=\textbf 0}\frac{\delta(\textbf x-\textbf x_0)}{|\det\frac{\partial\textbf F}{\partial\textbf x}(\textbf x_0)|}\) has been used.

Therefore, if one works with the following relativistic normalization for the Fock states:

\[|\textbf k\rangle:=\sqrt{\frac{2}{\hbar}}\frac{\omega_{\textbf k}}{c}a_{\textbf k}^{\dagger}|0\rangle\]

Then the inner product of two single-boson Fock states yields a Lorentz invariant delta:

\[\langle\textbf k’|\textbf k\rangle=\omega_{\textbf k}\delta^3(\textbf k’-\textbf k)\]

which is Lorentz invariant because \(1\) is a Lorentz scalar and for any \(\textbf k’\in\textbf R^3\):

\[\int\frac{d^3\textbf k}{\omega_{\textbf k}}\omega_{\textbf k}\delta^3(\textbf k-\textbf k’)=1\]

Problem #\(13\): Recall that when canonically quantizing the classical free Klein-Gordon field theory for a real scalar field in the Hamiltonian formulation, the canonical commutation relations in the Heisenberg picture were equal-\(t\) commutators. For instance, at an equal time \(t\) and for any two distinct locations in space \(\textbf x\neq\textbf x’\), the operators \(\phi(\textbf x,t),\phi(\textbf x’,t)\) associated to the space-like separated events \((ct,\textbf x),(ct,\textbf x’)\) were postulated to commute:

\[[\phi(\textbf x,t),\phi(\textbf x’,t)]=0\]

However, in order for the theory to be compatible with special relativity, and in particular causality, it must be the case that:

\[[\phi(X),\phi(X’)]=0\]

for any two space-like separated events \(|X-X’|^2<0\) outside each other’s light cones, not just for two simultaneous events. Despite not being manifest, show that this is indeed the case.

Solution #\(13\): By directly chugging in the Fourier mode expansion for \(\phi\), one finds that the commutator is simply proportional to the identity in this free QFT (cf. many other commutators like \([X,P_x]=i\hbar 1\) and \([a,a^{\dagger}]=1\) in quantum mechanics where this occurs, so commonly that it is sometimes called a “c-number function” where the “c” stands for “classical”):

\[[\phi(X),\phi(X’)]=\frac{\hbar c^2}{2(2\pi)^3}\int\frac{d^3\textbf k}{\omega_{\textbf k}}(e^{-i\langle K|X-X’\rangle}-e^{i\langle K|X-X’\rangle})\]

The fact that the integrand involves only the Lorentz invariant inner product \(\langle K|X-X’\rangle\) and that the measure \(\int d^3\textbf k/\omega_{\textbf k}\) is Lorentz invariant (thanks to Solution #\(12\)), it follows that the c-number commutator is also Lorentz invariant, i.e. for any \(\Lambda\in SO^+(1,3)\), \([\phi(\Lambda^{-1}X),\phi(\Lambda^{-1}X’)]=[\phi(X),\phi(X’)]\). Now, let \(|X-X’|^2<0\) be spacelike separated events. Then, and this is the key insight, there exists some Lorentz transformation \(\Lambda\in SO^+(1,3)\) such that \(\Lambda X\) and \(\Lambda X’\) become simultaneous events, so write \(\Lambda X-\Lambda X’=(0,\Delta\textbf x)\). But this then reproduces the originally postulated equal-\(t\) commutation relation:

\[[\phi(X),\phi(X’)]=[\phi(\Lambda X),\phi(\Lambda X’)]=\frac{\hbar c^2}{2(2\pi)^3}\int\frac{d^3\textbf k}{\omega_{\textbf k}}(e^{i\textbf k\cdot\Delta\textbf x}-e^{-i\textbf k\cdot\Delta\textbf x})=0\]

since the integrand \(\sim\sin(\textbf k\cdot\Delta\textbf x)\) is odd in \(\textbf k\). Thus, causality is preserved.

Problem #\(14\): Define the Wightman propagator of Klein-Gordon free QFT to be the two-point correlation function (change of notation here to emphasize that \(X\) is a mere label):

\[\Delta_W(X-X’):=\langle 0|\phi_X\phi_{X’}|0\rangle\]

and show that it is given by the Lorentz invariant integral:

\[\Delta_W(X-X’)=\frac{\hbar c^2}{2(2\pi)^3}\int\frac{d^3\textbf k}{\omega_{\textbf k}}e^{-i\langle K|X-X’\rangle}\]

Earlier in Solution #\(11\), it was seen that for \(X=X’\), the propagator \(\Delta_W(0)=\infty\) suffered both IR and UV divergences. By contrast, if \(X\neq X’\) but instead are space-like separated events \(X-X’\approx (0,\Delta\textbf x)\), show that:

\[|\Delta_W(X-X’)|\sim e^{-k_c|\Delta\textbf x|}\]

Solution #\(14\): As usual, just plug and chug the Fourier mode expansion for \(\phi(X)\) in the Heisenberg picture. Since the ket \(|0\rangle\) is annihilated by all \(a_{\textbf k’}\) while the bra \(\langle 0|\) is annihilated by all \(a_{\textbf k}^{\dagger}\) so the only surviving combination is of the form \(a_{\textbf k}a_{\textbf k’}^{\dagger}\) which can be commuted to produce one term that annihilates as above and a delta function that kills the \(d^3\textbf k’\) integral to obtain the desired result.

For the second part, evaluate the integral in \(\textbf k\)-space spherical coordinates:

\[\Delta_W(X-X’)=\frac{\hbar c}{2(2\pi)^2}\int_0^{\infty}d|\textbf k|\frac{|\textbf k|^2}{\sqrt{|\textbf k|^2+k_c^2}}\int_0^{\pi}d\theta e^{i|\textbf k||\Delta\textbf x|\cos\theta}\sin\theta\]

\[=\frac{\hbar c}{(2\pi|\Delta\textbf x|)^2i}\int_0^{\infty}\frac{\cos x}{\sqrt{x^2+(k_c|\Delta\textbf x|)^2}}dx\]

The integral cannot be evaluated analytically, however one can apply Laplace’s saddle-point approximation to it in the limit \(a:=k_c|\Delta\textbf x|\gg 1\) of far spacelike separation:

\[\int_0^{\infty}\frac{\cos x}{\sqrt{x^2+a^2}}dx=\text{Re}\int_0^{\infty}e^{ix-\frac{1}{2}\ln(x^2+a^2)}dx\]

which is stationary when \(x^2+ix+a^2=0\Rightarrow x\approx ia\), though one has to be a bit delicate about this, in particular \(x^2+a^2=-ix=a\neq 0\). One obtains the result:

\[|\Delta_W(X-X’)|\sim |\Delta\textbf x|^{-5/2}e^{-k_c|\Delta\textbf x|}\]

in particular, even for arbitrarily spacelike separated events \(X,X’\), the quantum field “leaks” outside of the light cone in an evanescent manner. Nevertheless, this does not violate causality which only requires that \(\langle 0|[\phi_X,\phi_{X’}]|0\rangle=\Delta_W(X-X’)-\Delta_W(X’-X)=0\) vanish for spacelike separated \(|X-X’|^2<0\).

Problem #\(15\): Give \(3\) logically equivalent definitions of the Feynman propagator \(\Delta_F(X-X’)\) (useful in interacting QFTs). Show that Lorentz invariance is manifest in all \(3\) cases.

Solution #\(15\): The simplest way to think about it is that it is the time-ordered Wightman propagator:

\[\Delta_F(X-X’):=\langle 0|\mathcal T\phi_X\phi_{X’}|0\rangle:=\Delta_W(X-X’)\Theta(X^0-X’^{0})+\Delta_W(X’-X)\Theta(X’^{0}-X^0)\]

(with the convention that \(\Theta(0)=1/2\) for simultaneous events). To check that \(\Delta_F(X-X’)\) is Lorentz invariant, first note that \(\Delta_W(X-X’)\) is Lorentz invariant for any pair \(X,X’\) of events. Then, if \(|X-X’|^2>0\) are timelike-separated, their time ordering \(\Theta(X^0-X’^{0})\) is Lorentz invariant so \(\Delta_F(X-X’)\) will be Lorentz invariant. On the other hand, if \(|X-X’|^2<0\) are spacelike separated, then their time ordering will in general not be Lorentz invariant, however in that case \(\Delta_W(X-X’)=\Delta_W(X’-X)\) as established in Solution #\(14\) so regardless of how their time ordering gets shuffled, the Feynman propagator \(\Delta_F(X-X’)\) takes the same value, hence it is Lorentz invariant.

On the other hand, the Feynman propagator is also, roughly speaking, given by the off-shell integral representation (containing manifestly Lorentz invariant objects like \(d^4|K\rangle\), Minkowski inner products, etc.):

\[\Delta_F(X-X’)=\frac{i\hbar c}{(2\pi)^4}\int d^4|K\rangle\frac{e^{-i\langle K|X-X’\rangle}}{\langle K|K\rangle-k_c^2}\]

A priori, the integral is over all four-vectors \(|K\rangle\) in Minkowski spacetime which runs into trouble at the on-shell poles \(\langle K|K\rangle=k_c^2\). Thus, to be more precise, one can first write \(d^4|K\rangle=d^3\textbf kd\omega/c\) and also emphasize this by replacing \(\langle K|K\rangle-k_c^2=(\omega+\omega_{\textbf k})(\omega-\omega_{\textbf k})/c^2\), rather than integrating over the real axis \(\omega\in\textbf R\), the idea is to complexify it \(\omega\in\textbf C\) so that it hops around the on-shell poles at \(\omega=\pm\omega_{\textbf k}\) in a very specific way that enforces the time-ordering property from earlier:

where the choice of which semicircular contour to close with is motivated by considering the time-ordering of \(X,X’\) and using Jordan’s lemma. Picking the appropriate residue in each case, one can thus verify this contour integral representation of the Feynman propagator, and indeed it is often also equivalently written without any complexification \(\omega\in\textbf R\) but instead introducing an \(i\varepsilon\)-prescription (with \(\varepsilon>0\) but infinitesimal in suitable units) to shift the poles slightly off the \(\Re\)-axis such that the pole originally at \(\omega=-\omega_{\textbf k}\) shifts slightly above the \(\Re\)-axis and vice versa for the pole at \(\omega=\omega_{\textbf k}\) to maintain the same topology as above:

\[\Delta_F(X-X’)=\frac{i\hbar c}{(2\pi)^4}\int d^4|K\rangle\frac{e^{-i\langle K|X-X’\rangle}}{\langle K|K\rangle-k_c^2+i\varepsilon}\]

(think of this as more of just a formality/notation to express a topological statement)

One last interpretation of the Feynman propagator \(\Delta_F(X-X’)\) is that, being suggestively written as a function of the argument \(X-X’\), it is essentially (irrespective of the choice of contour) a Green’s function for the Klein-Gordon operator on Minkowski spacetime:

\[(\partial^{\mu}\partial_{\mu}+k_c^2)\Delta_F(X-X’)=\frac{i\hbar c}{(2\pi)^4}\int \frac{d^4|K\rangle}{\langle K|K\rangle-k_c^2}(-\langle K|K\rangle+k_c^2)e^{-i\langle K|X-X’\rangle}\]

\[=-i\hbar c\delta^4(X-X’)\]

where note that \(\partial_{\mu}=\frac{\partial}{\partial X^{\mu}}\neq\frac{\partial}{\partial X’^{\mu}}\). This means that if one wished to solve an inhomogeneous Klein-Gordon equation with some forcing field \(f(X)\), one could just convolve \(f(X)\) with the Feynman propagator \(\Delta_F(X-X’)\) as a kernel (boundary conditions tied to a choice of contour for the Green’s function? Retarded vs. advanced Green’s functions…).