Phys 712 — Spring 2015 — Assignment 4

Quantum particles are either bosonic or fermionic in character and can be described by so-called “second-quantized” operators (\(a\), \(a^\dagger\) and \(c\), \(c^\dagger\)) that obey the appropriate commutator or anticommutator algebra. In the case of fermions, the anticommutation relation implies that \((c^\dagger)^2 = 0\), which is to say that the amplitude for creating two identical fermions is exactly zero. This is the famous Pauli exclusion principle. Hence, there are only two possible fermionic states, \(\lvert 0 \rangle\) and \(\lvert 1 \rangle\), which are labelled according to the eigenvalues of the number operator \(c^\dagger c\). Bosons, on the other hand, follow the algebra familiar from the quantum harmonic oscillator. The occupancy in each level is unbounded, and the \(a^\dagger a\) eigenstates allow for all possible integer values of the occupation: \(\lvert 0 \rangle\), \(\lvert 1 \rangle\), \(\lvert 2 \rangle\), \(\lvert 3 \rangle\), ….

  1. Prove the “contraction” formula

    \[\langle \text{vac} | g_{\alpha_1}g_{\alpha_2}\cdots g_{\alpha_N} g_{\beta_N}^\dagger \cdots g_{\beta_2}^\dagger g_{\beta_1}^\dagger | \text{vac} \rangle = \sum_{\sigma} (\pm 1)^\sigma \prod_{k=1}^N \delta_{\alpha_k,\beta_{\sigma(k)}}.\]

    Here, the sum ranges over all permutations of \(N\) elements. (E.g., \(\sigma = (1\ 2\ 4)(3)\) is one of the \(4!=24\) possible permutatons when \(N=4\); it behaves as \(\sigma(1) = 2\), \(\sigma(2) = 4\), \(\sigma(3) = 3\), and \(\sigma(4) = 1\).) For bosons (\(g = a\)), there is no alternating sign: \((+1)^\sigma = 1\). But for fermions (\(g=c\)), \((-1)^\sigma = 1\) when \(\sigma\) is an even permutation and \((-1)^\sigma = -1\) when it’s odd. Hint: check the \(N=1\) and \(N=2\) cases explicitly and proceed by recursion.

  2. Use a Taylor series expansion of the exponential to prove that

    \[e^{-A c^\dagger c} = 1 + \bigl( e^{-A} - 1 \bigr) c^\dagger c\]

    and that consequently

    \[e^{-A c^\dagger c}c^\dagger = e^{-A} c^\dagger.\]

    Explain why this expression does not also hold for bosons.

  3. Consider a system of \(N\) noninteracting (spinless) fermions on a lattice of \(M\ge N\) sites. (Here, we’re working in the microcanonical ensemble, in which energy \(E\) and particle number \(N\) are well defined.) The Hamiltonian includes only a hopping term:

    \[\hat{H} = - \sum_{ij} t_{ij} c_i^\dagger c_j.\]

    We introduce a set of basis functions \(\{ \phi_i^{(\alpha)} \}\)—these are orthonormal, so they satisfy

    \[\begin{split}\sum_{\alpha=1}^M{ \phi_i^{(\alpha)}}^*\phi^{(\alpha)}_j &= \delta_{ij}, \\ \sum_{i=1}^M{ \phi_i^{(\alpha)}}^*\phi^{(\beta)}_i &= \delta_{\alpha\beta} \end{split}\]

    —in order to define a unitary transformation

    \[c_i = \sum_{\alpha = 1}^M \phi_i^{(\alpha)}d_\alpha.\]

    In this new basis, the Hamiltonian has the form

    \[\begin{split} \hat{H} &= - \sum_{ij} t_{ij} \biggl( \sum_{\alpha = 1}^M \phi_i^{(\alpha)} d_\alpha \biggr)^\dagger \biggl( \sum_{\beta = 1}^M \phi_j^{(\beta)}d_\beta \biggr)\\ &= \sum_{\alpha\beta} \underbrace{\biggl(-\sum_{ij} {\phi_i^{(\alpha)}}^* t_{ij} \phi^{(\beta)}_j\biggr)}_{\equiv \epsilon_{\alpha} \delta_{\alpha \beta}} d_\alpha^\dagger d_{\beta} = \sum_\alpha \epsilon_\alpha d^\dagger_\alpha d_\alpha. \end{split}\]

    Here, we’ve assumed that the basis has been specifically chosen to diagonalize the hopping matrix. Let’s also assume that the energy eigenstates are labelled in such a way that \(\epsilon_1 \le \epsilon_2 \le \cdots \le \epsilon_M\).

    Show that a state of the form

    \[d^\dagger_{\alpha_N} \cdots d^\dagger_{\alpha_2} d^\dagger_{\alpha_1} \lvert \text{vac} \rangle\]

    with \(\alpha_1 \neq \alpha_2 \neq \cdots \alpha_N\) is a number eigenstate with eigenvalue \(N\) and an energy eigenstate with eigenvalue \(\epsilon_{\alpha_1} + \epsilon_{\alpha_2} + \cdots + \epsilon_{\alpha_N}\).

  4. It follows that the \(N\)-particle ground state, the state with lowest energy \(E_0 = \epsilon_1 + \epsilon_2 + \cdots \epsilon_N\), is

    \[| \psi_0 \rangle = d_N^\dagger \cdots d_2^\dagger d_1^\dagger \lvert \text{vac} \rangle.\]

    (a) Show that its time evolution is governed by

    \[e^{-i\hat{H}t/\hbar}| \psi_0 \rangle = e^{-i(\sum_{\alpha=1}^M d^\dagger_\alpha d_\alpha)t/\hbar}d_N^\dagger \cdots d_2^\dagger d_1^\dagger \lvert \text{vac} \rangle = e^{-iE_0t/\hbar}| \psi_0 \rangle.\]

    (b) Define the state \(\lvert j_1 \cdots j_N \rangle = c_{j_N}^\dagger \cdots c_{j_2}^\dagger c_{j_1}^\dagger \lvert \text{vac} \rangle\) that describes \(N\) fermions at positions \(j_1, \ldots, j_N\). Show that the properly normalized ground-state wave function is the Slater determinant

    \[\Psi_{j_1\cdots j_N} = \frac{1}{\sqrt{N!}} \langle j_1\cdots j_N | \psi_0 \rangle = \frac{1}{\sqrt{N!}} \begin{vmatrix} \phi^{(1)}_{j_1} & \phi^{(1)}_{j_2} & \cdots & \phi^{(1)}_{j_N} \\ \phi^{(2)}_{j_1} & \phi^{(2)}_{j_2} & \cdots & \phi^{(2)}_{j_N} \\ \vdots & \vdots & & \vdots \\ \phi^{(N)}_{j_1} & \phi^{(N)}_{j_N} & \cdots & \phi^{(N)}_{j_N} \end{vmatrix}.\]

  5. Show that a noninteracting gas of \(N\) bosons has a normalized ground state

    \[\lvert \psi_0 \rangle = \frac{1}{\sqrt{N!}}(b_1^\dagger)^N \lvert \text{vac} \rangle\]

    with energy \(N\epsilon_1\) and a wave function

\[\Psi_{j_1\cdots j_N} = \phi^{(1)}_{j_1}\phi^{(1)}_{j_2}\cdots \phi^{(1)}_{j_N}.\]