 Solution

ID: past-exam-of-the-quantum-matter-course-of-the-university-of-oxford/2025/1/b/solution

e^{ik M} = 1

, so there are exactly

M

k = \frac{2 π l}{M}, l = 0, \dots, M - 1 (mod 2 π) .

c_{σ, j}^{†} = \frac{1}{M} \sum_{k} e^{- ikj} c_{σ, k}^{†}, c_{σ, j} = \frac{1}{M} \sum_{k} e^{ikj} c_{σ, k} .

ϵ_{k} = - 2 t cos k

. For the interaction, first use the canonical anticommutation relations at distinct sites to write

n_{σ, j} n_{σ, j + 1} = c_{σ, j}^{†} c_{σ, j + 1}^{†} c_{σ, j + 1} c_{σ, j}

j

H = σ, k \sum (- 2 t cos k) c_{σ, k}^{†} c_{σ, k} + \frac{V}{M} σ, k, p, q \sum e^{i q} c_{σ, k + q}^{†} c_{σ, p - q}^{†} c_{σ, p} c_{σ, k} .

Every wavevector is reduced modulo

2 π

. The phase

e^{i q}

comes from the displacement of the second site by one lattice spacing. Equivalently replace

e^{i q}

cos q

: exchanging the two incoming and the two outgoing fermions changes

q

- q

with two cancelling fermionic signs. Thus the usual interaction vertex is

2 V cos q

accompanied by the factor

1/ (2 M)

 New to topics? Read the docs here!