Past exam of the quantum Condensed Matter Field Theory course of the University of Cambridge / 2015 / Paper 1 / 1 / d

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- Solution d

Solution

For a bulk bond,

K_{n + 1} = K_{n} (1 - 2 n_{n})

and

c_{n} (1 - 2 n_{n}) = - c_{n}

. Reordering the remaining different-site operators then gives

S_{n}^{+} S_{n + 1}^{-} = c_{n + 1}^{†} c_{n}, S_{n}^{-} S_{n + 1}^{+} = c_{n}^{†} c_{n + 1} (n < N) .

The periodic spin boundary condition requires separate attention. Set

P = \prod_{n = 1}^{N} (1 - 2 n_{n}) = (- 1)^{N_{f}}

. The end bond becomes

S_{N}^{+} S_{1}^{-} = - P c_{1}^{†} c_{N}

, with the conjugate relation for its adjoint. Consequently the exact finite-ring Hamiltonian is

H = - \frac{h N}{2} + h n = 1 \sum N n_{n} - \frac{J}{2} n = 1 \sum N - 1 (c_{n}^{†} c_{n + 1} + c_{n + 1}^{†} c_{n}) + \frac{J P}{2} (c_{N}^{†} c_{1} + c_{1}^{†} c_{N}) .

fermion parity is conserved, so diagonalize in a sector

P = \pm 1

. There the end bond can be written like every other bond provided

c_{N + 1} = - P c_{1}

. This parity-dependent boundary condition of the Jordan-Wigner chain is antiperiodic for even fermion number and periodic for odd fermion number. Simply assuming periodic fermions in both sectors changes the finite-size spectrum.

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