Position-space particle-hole symmetry complex-conjugates plane waves, so the <Fourier transform> sends $k$ to $-k$. Fourier transforming the stated relation therefore gives
$$
H_{\rm BdG}(k)=-\Sigma_1H_{\rm BdG}^*(-k)\Sigma_1.
$$
If $H_{\rm BdG}(k)u_k=\varepsilon_k u_k$, complex conjugation and multiplication by $\Sigma_1$ yield
$$
H_{\rm BdG}(-k)(\Sigma_1u_k^*)
=-\varepsilon_k(\Sigma_1u_k^*).
$$
Thus the <Particle-hole symmetry of a Bogoliubov--de Gennes Hamiltonian> imposes
$$
\operatorname{spec}H_{\rm BdG}(-k)
=-\operatorname{spec}H_{\rm BdG}(k),
$$
or, band by band after a suitable relabelling, $\varepsilon_{-k}=-\varepsilon_k$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 342 / 3 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-342/3/b/solution