Problem: Consider a system of \(N=2\) identical non-interacting spin \(s=1/2\) fermions in an infinite potential well of width \(L\) (nodes at \(x=0,L\)). Write down the general \(2\)-body wavefunction \(\Psi(1,2)\) for the system’s ground state, \(1^{\text{st}}\) excited state, and \(2^{\text{nd}}\) excited state by calculating Slater determinants of the single-fermion spin-orbitals.
Solution: Use the notation \(\chi_{n,m_s}\) for the spin-orbital (note the word “orbital” here isn’t really meant in the sense of e.g. orbital angular momentum \(\textbf L\), but rather in the chemist sense of “atomic orbital” though the \(2\) notions aren’t entirely disjoint):
\[\chi_{n,m_s}=|\psi_n\rangle\otimes|m_s\rangle\]
where \(\psi_n(x)=\sqrt{\frac{2}{L}}\sin\frac{n\pi x}{L}\) is the position space wavefunction of the \(n^{\text{th}}\) excited single-particle state. Use the notation \(\chi_{n,m_s}(i)\) to refer to the spin-orbital for the (arbitrarily labelled) \(i^{\text{th}}\) fermion (in this case \(i=1,2\)), for example:
\[\chi_{7,\uparrow}(2)\cong \sqrt{\frac{2}{L}}\sin\frac{7\pi x_2}{L}|\uparrow\rangle_2\]
In this case, a general Slater determinant of spin-orbitals is of the form:
\[\Psi(1,2)=\frac{1}{\sqrt{2!}}\det\begin{pmatrix}\chi_{n_1,m^{(1)}_s}(1)&\chi_{n_2,m^{(2)}_s}(1)\\\chi_{n_1,m^{(1)}_s}(2)&\chi_{n_2,m^{(2)}_s}(2)\end{pmatrix}\]
Now consider the ground state. Because the fermions are non-interacting, this amounts to the requirement \(n_1=n_2=1\), thus fixing the spatial part of the allowed spin-orbitals. In particular, since the spatial parts are identical, the spin parts cannot be \(m^{(1)}_s\neq m^{(2)}_s\) otherwise \(\Psi(1,2)=0\); this is just the Pauli exclusion principle. Arbitrarily letting \(m^{(1)}_s=-m^{(2)}_s=1/2\), it follows that the ground state manifold is one-dimensional, and spanned by the ground state:
\[\Psi(1,2)=\frac{1}{\sqrt{2!}}\det\begin{pmatrix}\chi_{1,\uparrow}(1)&\chi_{1,\downarrow}(1)\\\chi_{1,\uparrow}(2)&\chi_{1,\downarrow}(2)\end{pmatrix}\cong\frac{2}{L}\sin\frac{\pi x_1}{L}\sin\frac{\pi x_2}{L}\frac{|\uparrow\rangle_1|\downarrow\rangle_2-|\downarrow\rangle_1|\uparrow\rangle_2}{\sqrt{2}}\]
So the spatial part of the \(2\)-body wavefunction is clearly \(1\leftrightarrow 2\) symmetric while the singlet spin part is \(1\leftrightarrow 2\) antisymmetric, ensuring total antisymmetry \(\Psi(2,1)=-\Psi(1,2)\).
Meanwhile, for the system’s \(1^{\text{st}}\) excited state:
Problem: Briefly describe how the results above would be affected if instead it were \(N=2\) identical non-interacting spin \(s=1\) bosons.
Solution: Instead of Slater determinants, one would use “Slater permanents”, or just “permanents” for short. These are basically calculated in the same way as a determinant except all minus signs become plus signs. This implicitly enforces total symmetry \(\Psi(2,1)=\Psi(1,2)\) of the wavefunction. In addition, because \(2s+1=3\) now, the degeneracies of each manifold would be enhanced.
Problem: Define the \(N\)-symmetrizer \(\mathcal S_N\) and the \(N\)-antisymmetrizer \(\mathcal A_N\) operators on the space \(\mathcal H^{\otimes N}\) of \(N\) identical particles (where \(\mathcal H\) is a single-particle state space).
Solution: The \(N\)-symmetrizer \(\mathcal S_N:\mathcal H^{\otimes N}\to S^N\mathcal H\) is defined by:
\[\mathcal S_N:=\frac{1}{\#S_N}\sum_{\sigma\in S_N}\sigma\]
And the \(N\)-antisymmetrizer \(\mathcal A_N:\mathcal H^{\otimes N}\to \bigwedge^N\mathcal H\) is defined by:
\[\mathcal A_N:=\frac{1}{\#S_N}\sum_{\sigma\in S_N}\text{sgn}(\sigma)\sigma=\frac{1}{\#S_N}\left(\sum_{\sigma\in A_N}\sigma-\sum_{\sigma\in S_N-A_N}\sigma\right)\]
note that \(\#S_N=N!\). Strictly speaking each permutation \(\sigma\in S_N\) is an abstract group element for which there is no notion of “group addition” (and it certainly doesn’t refer to compositions of permutations either), rather this is a faithful representation of \(S_N\) on \(\mathcal H^{\otimes N}\) (and so a more pedantic notation could be \(\hat{\sigma}\) for the operator associated to \(\sigma\)). For instance, if \(N=6\) and \(\sigma=(352)(14)\) in cycle notation, then:
\[\sigma\Psi(1,2,3,4,5,6)=\Psi(4,3,5,1,2,6)\]
Problem: What is the explicit connection between the Slater permanents/determinants and the symmetrizer/antisymmetrizer?
Solution: The idea is to symmetrize/antisymmetrize the Hartree product ansatz, essentially providing the bridge to Hartree-Fock. This immediately yields Slater permanents/determinants respectively (up to a normalization):
\[\sqrt{N!}\mathcal S_N\chi_1(1)\chi_2(2)…\chi_N(N)=\frac{1}{\sqrt{N!}}\text{perm}\begin{pmatrix}\chi_1(1)\end{pmatrix}\]
\[\sqrt{N!}\mathcal A_N\chi_1(1)\chi_2(2)…\chi_N(N)=\frac{1}{\sqrt{N!}}\text{det}\begin{pmatrix}\chi_1(1)\end{pmatrix}\]
Problem: Establish the following useful properties of \(\mathcal S_N\) and \(\mathcal A_N\):
i) \[\sigma\mathcal S_N=\mathcal S_N\sigma=\mathcal S_N\] and \[\sigma\mathcal A_N=\mathcal A_N\sigma=\text{sgn}(\sigma)\mathcal A_N\] for any \(\sigma\in S_N\).
ii) \[\mathcal S^{\dagger}_N=\mathcal S_N\] and \[\mathcal A^{\dagger}_N=\mathcal A_N\] (i.e. the \(N\)-symmetrizer and \(N\)-antisymmetrizer are both Hermitian observables).
iii) \(\mathcal S^2_N=\mathcal S_N\) and \(\mathcal A^2_N=\mathcal A_N\) and (i.e. \(\mathcal S_N\) and \(\mathcal A_N\) are orthogonal projectors, which makes sense because they project onto the symmetric subspace \(S^N\mathcal H\) and its orthogonal complement? \(\bigwedge^N\mathcal H\) (this also explains the factor of \(1/\#S_N\) in the definition as a “idempotence factor”?)
iv) \[[\mathcal S_N,H]=[\mathcal A_N,H]=0\] where \(H\) is any Hermitian observable.
Solution:
Problem: Show that for \(N=2\) identical particles, \(S_2+A_2=1\) partitions the space \(\mathcal H^{\otimes 2}\) but for \(N=3\) identical particles \(S_3+A_3\neq 1\).
Solution:
Problem: The purpose of the previous problems wasn’t so much to actually compute all those \(N\)-body wavefunctions \(\Psi(1,…,N)\), but rather to force one to compute them in order to realize how tedious and redundant the whole business is. That is, because the particles are identical, it is highly inefficient to be asking “which state is which particle in” since the notion of “which particle” is meaningless. Instead, common sense dictates that the more efficient question to ask is “how many particles are in each state?” as this doesn’t care about which particle is which. This conceptual simplification lies at the heart of the second quantization approach to the quantum mechanics of a many-body system; it is also called the occupation number representation.
The canonical example of this is the ideal Bose/Fermi gas which, although it can be described by linear combinations of \(N\)-body wavefunctions with \(N!\) terms, for \(N\sim 10^{23}\) this quickly becomes inconvenient, so in practice one sweeps the wavefunctions under the rug and just speaks about occupation numbers \(N_{\textbf k}\) of various \(\textbf k\)-states, aka the Bose-Einstein and Fermi-Dirac distributions.
A warning: second quantization is only really useful for systems of identical particles; if the \(N\) particles were all distinguishable then in principle one can still use the second quantization framework, but in that case it doesn’t offer any advantage over just plain wavefunction language (aka first quantization).
Now then, go back to the earlier problems and redo everything in second quantization/occupation number language.
Solution:
Problem: What is the name for a many-body quantum state \(|N_1,N_2,…\rangle\) written in the occupation number representation? What is the name of the space that such states live in?
Solution: Many-body quantum states in the occupation number representation are called Fock states, and live in Fock space \(\mathcal F:=\oplus_{N=0}^{\infty}\mathcal F_N\) which considers variable number of particles via the various sectors \(\mathcal F_N\).
Problem: How does one navigate the Fock space \(\mathcal F\).
Solution: Using creation and annihilation operators. To be precise, for a given basis of \(\mathcal H^{\otimes N}\), one associates a creation and annihilation operator to each basis state in the basis…
Problem: Derive the commutation relations for the bosonic creation/annihilation operators and similarly derive the anticommutation relations for the fermionic creation/annihilation operators, starting from … . This shows that commutation and anticommutation relations completely encode the permutation symmetries of the bosonic and fermionic states.
Problem: What does it mean for a linear operator \(H\) to be an \(N\)-body operator?
Solution: An operator \(H\) is said to be an \(N\)-body operator iff there exists a decomposition of \(H\) in the form:
\[H=\sum_i H_i\]
where each operator \(H_i\) acts only on \(N\) particles at a time, acting as the identity operator on all other particles. For instance, the kinetic energy operator for an arbitrary system of particles (whether identical or not) is a \(1\)-body operator, as are most external potentials. By contrast, common \(2\)-body operators include interaction potentials between particles.
Problem: Given a system of \(N\) identical particles (fermions or bosons), and a \(1\)-body operator \(H\), and a basis \(\{|i\rangle\}\) of the single-particle Hilbert space, explain why the second quantization functor acting on \(H\) is given by the “dictionary”:
\[|i\rangle\mapsto c^{\dagger}_{|i\rangle}|\space\rangle\]
\[H\mapsto\sum_{|i\rangle,|j\rangle}\langle i|H|j\rangle c^{\dagger}_{|i\rangle}c_{|j\rangle}\]
Solution: Because the matrix elements are preserved under this homomorphism, so the functor is sort of “unitary” in a way? More precisely, matrix elements between any \(2\) Fock states are unchanged.
Problem: Write the one-body Rabi drive operator \(V_{\text{Rabi}}=\frac{\hbar}{2}\tilde{\boldsymbol{\Omega}}\cdot\boldsymbol{\sigma}\) in \(2^{\text{nd}}\) quantization.
Solution: Since the states \(\{|\textbf k,\sigma\rangle\}\) are a basis for the single-particle Hilbert space:
Problem: Write the \(2\)-body scattering contact pseudopotential operator \(V:=g\delta^3(\textbf X-\textbf X’)\) in \(2^{\text{nd}}\) quantization.