4.5: Exercises

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Exercises

Exercise \(\PageIndex{1}\)

Consider a system of two identical particles. Each single-particle Hilbert space \(\mathscr{H}^{(1)}\) is spanned by a basis \(\{|\mu_i\}\). The exchange operator is defined on \(\mathscr{H}^{(2)} = \mathscr{H}^{(1)} \otimes \mathscr{H}^{(1)}\) by

\[P \Big (\sum_{ij} \psi_{ij} |\mu_i\rangle|\mu_j\rangle \Big) \;\equiv\; \sum_{ij} \psi_{ij} |\mu_j\rangle|\mu_i\rangle.\]

Prove that \(\hat{P}\) is linear, unitary, and Hermitian. Moreover, prove that the operation is basis-independent: i.e., given any other basis \(\{\nu_j\}\) that spans \(\mathscr{H}^{(1)}\),

\[P \Big (\sum_{ij} \varphi_{ij} |\nu_i\rangle|\nu_j\rangle \Big) \;=\; \sum_{ij} \varphi_{ij} |\nu_j\rangle|\nu_i\rangle.\]

Exercise \(\PageIndex{2}\)

Prove that the exchange operator commutes with the Hamiltonian

\[\hat{H} = - \frac{\hbar^2}{2m_e} \Big(\nabla_1^2 + \nabla^2_2\Big) + \frac{e^2}{4\pi\varepsilon_0|\mathbf{r}_1 - \mathbf{r}_2|}.\]

Exercise \(\PageIndex{3}\)

An \(N\)-boson state can be written as

\[|\phi_1,\phi_2,\dots,\phi_N\rangle = \mathcal{N} \sum_p \Big(|\phi_{p(1)}\rangle |\phi_{p(2)}\rangle |\phi_{p(3)}\rangle \cdots |\phi_{p(N)}\rangle\Big).\]

Prove that the normalization constant is

\[\mathcal{N} = \sqrt{\frac{1}{N!\prod_\mu n_\mu!}},\]

where \(n_\mu\) denotes the number of particles occupying the single-particle state \(\mu\).

Exercise \(\PageIndex{4}\)

\(\mathscr{H}_{S}^{(N)}\) and \(\mathscr{H}_{A}^{(N)}\) denote the Hilbert spaces of \(N\)-particle states that are totally symmetric and totally antisymmetric under exchange, respectively. Prove that

\[\begin{align}\begin{aligned} \mathrm{dim}\left(\mathscr{H}_{S}^{(N)}\right) &= \frac{(d+N-1)!}{N!(d-1)!}, \\ \mathrm{dim}\left(\mathscr{H}_{A}^{(N)}\right) &= \frac{d!}{N!(d-N)!}. \end{aligned}\end{align}\]

Exercise \(\PageIndex{5}\)

Prove that for boson creation and annihilation operators, \([\hat{a}_\mu,\hat{a}_\nu] = [\hat{a}_\mu^\dagger,\hat{a}_\nu^\dagger] = 0\).

Exercise \(\PageIndex{6}\)

Let \(\hat{A}_1\) be an observable (Hermitian operator) for single-particle states. Given a single-particle basis \(\{|\varphi_1\rangle,|\varphi_2\rangle,\dots\}\), define the bosonic multi-particle observable

\[\hat{A} = \sum_{\mu\nu} \,a^\dagger_\mu \; \langle\varphi_\mu|\hat{A}_1|\varphi_\nu\rangle \; a_\nu,\]

where \(a_\mu^\dagger\) and \(a_\mu\) are creation and annihilation operators satisfying the usual bosonic commutation relations, \([a_\mu,a_\nu] = 0\) and \([a_\mu,a_\nu^\dagger] = \delta_{\mu\nu}\). Prove that \(\hat{A}\) commutes with the total number operator:

\[\Big[\hat{A}, \sum_\mu a^\dagger_\mu a_\mu \Big] = 0.\]

Next, repeat the proof for a fermionic multi-particle observable

\[\hat{A} = \sum_{\mu\nu} \,c^\dagger_\mu \; \langle\varphi_\mu|\hat{A}_1|\varphi_\nu\rangle \; c_\nu,\]

where \(c_\mu^\dagger\) and \(c_\mu\) are creation and annihilation operators satisfying the fermionic anticommutation relations, \(\{c_\mu,c_\nu\} = 0\) and \(\{c_\mu,c_\nu^\dagger\} = \delta_{\mu\nu}\). In this case, prove that

\[\Big[\hat{A}, \sum_\mu c^\dagger_\mu c_\mu \Big] = 0.\]