Past exam of the quantum Condensed Matter Field Theory course of the University of Cambridge / 2012 / Paper 1 / 1 / c

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- Solution c

Solution

With the specified Fourier transform convention, normal ordering gives

\sum_{n} (c_{n}^{†} c_{n + 1} - c_{n} c_{n + 1}^{†}) = 2 \sum_{k} cos k c_{k}^{†} c_{k},

where the possible constant

\sum_{k} e^{ik}

vanishes on the complete lattice momentum grid. Antisymmetry of opposite-momentum pairs similarly gives

\sum_{n} c_{n}^{†} c_{n + 1}^{†} = - i \sum_{k} sin k c_{k}^{†} c_{- k}^{†}, \sum_{n} c_{n} c_{n + 1} = i \sum_{k} sin k c_{k} c_{- k} .

Hence

H = - J g N + 2 J \sum_{k} (g - cos k) c_{k}^{†} c_{k} + i J \sum_{k} sin k (c_{k}^{†} c_{- k}^{†} + c_{k} c_{- k}) .

Use the Ising-chain Nambu spinor

Ψ_{k} = (c_{k}, - i c_{- k}^{†})^{T}

. Expanding its two diagonal terms uses

c_{- k} c_{- k}^{†} = 1 - c_{- k}^{†} c_{- k}

, retaining the constant. The result is

H = J k \sum Ψ_{k}^{†} [(g - cos k) σ_{z} - sin k σ_{x}] Ψ_{k} .

This is a full-zone sum. Its opposite-momentum components are redundant; treating each matrix eigenvalue as an independent physical excitation would miss the factor of two in the quasiparticle energy. Self-paired momenta have zero anomalous term and are ordinary number levels.

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