Let us consider a system of a 1D edge of a 2D topological insulator in proximity to an s-wave superconductor. The system is described by the Hamiltonian: $$ H =\frac{1}{2} \int \mathrm{d}x \ \Psi^{\dagger}(x) \mathcal{H}(x) \Psi(x)$$ with single particle Hamiltonian $$ \mathcal{H}(x) = \begin{pmatrix} -i\hbar v\partial_{x} & 0 & 0 & \Delta \\ 0 & i\hbar v\partial_{x} & -\Delta & 0 \\ 0 & -\Delta & -i\hbar v\partial_{x} & 0 \\ \Delta & 0 & 0 & i\hbar v\partial_{x} \end{pmatrix} $$ and the four-component spinor $$ \Psi(x) = \begin{pmatrix} \Psi_{\uparrow}(x) \ e^{i\Phi/2} \\ \Psi_{\downarrow}(x) \ e^{i\Phi/2} \\ \Psi^{\dagger}_{\uparrow}(x) \ e^{-i\Phi/2} \\ \Psi^{\dagger}_{\downarrow}(x) \ e^{-i\Phi/2} \end{pmatrix}. $$ Here $v$ is the Fermi velocity, $\Delta>0$ is the superconducting gap and $\Phi$ is the superconducting phase. I have removed the superconducting phase from the single-particle Hamiltonian and I have reinstalled it in the definition of my electron spinor (by a suitable unitary transformation). This reflects the fact that the absolute superconducting phase is not measurable when we consider just a single superconductor.

I am closely following http://arxiv.org/abs/0912.2157 and I am introducing the operation of time-reversal symmetry for the single particle Hamiltonian $\mathcal{H}(x)$. This is done by defining the matrix $$ U_{T}=\begin{pmatrix} 0 & 1 & 0 & 0 \\ -1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & -1 & 0 \\ \end{pmatrix} $$ We then observe that $$U_{T}^{\dagger}\mathcal{H}^{*}(x)U_{T}=\mathcal{H}(x)$$ and so the system is expected to be time-reversal symmetric. Now I am moving to the second-quantized picture by declaring that the operation of time-reversal symmetry $\mathcal{T}$ acts on my operators as $$ \mathcal{T}\Psi(x)\mathcal{T}^{-1}=U_{T}\Psi(x) $$ So for example $$ \mathcal{T}\Psi_{\uparrow}(x)e^{i\Phi/2}\mathcal{T}^{-1}=\Psi_{\downarrow}(x)e^{i\Phi/2} $$ This transformation law (although is closely follows http://arxiv.org/abs/0912.2157 page 7) looks strange to me, because due to the anti unitary of time reversal symmetry, i.e. $\mathcal{T}i\mathcal{T}^{-1}=-i$, I would have expected the transformation law $$ \mathcal{T}\Psi_{\uparrow}(x)e^{i\Phi/2}\mathcal{T}^{-1}=\Psi_{\downarrow}(x)e^{-i\Phi/2} $$ When I am setting $\Phi=0$ or $\Phi=\pi$ the problem of course disappears. But since the global superconducting phase is not physical in this case the system should be time-reversal symmetric independent of the choice of $\Phi$.

Can someone resolve my confusion?

  • $\begingroup$ I do not know where you get $T\Psi T^{-1}=U_{T}\Psi$ but it's wrong, the definition of time-reversal symmetry is $T\Psi T^{-1}=U_{T}\Psi^{\ast}$ since the time-reversal operation is represented by an anti-unitary operator. $\endgroup$
    – FraSchelle
    Commented Nov 30, 2015 at 17:09
  • 1
    $\begingroup$ @FraSchelle Here $\Psi$ is the second-quantized operator, so you don't put $*$ on the $\Psi$'s. The equation comes from arxiv.org/abs/0912.2157, these guys probably know what they are doing:) $\endgroup$
    – Meng Cheng
    Commented Nov 30, 2015 at 18:23
  • $\begingroup$ @MengCheng Ok more certainly you're right. I still do not understand your argument in your answer, but clearly in my comment I was supposing $\Psi$ to be a wave-function. Sorry for the misunderstanding. Why the time-reversal could not change a $c$ to a $c^\dagger$ ? I'm not sure I understand this point. $\endgroup$
    – FraSchelle
    Commented Dec 1, 2015 at 21:32
  • $\begingroup$ @FraSchelle Time-reversal transformation in principle can change $\psi$ to $\psi^\dagger$, but this is a rather unusual time-reversal symmetry since it changes the total number of electrons. Certainly the physical one does not do that. $\endgroup$
    – Meng Cheng
    Commented Dec 2, 2015 at 3:28

1 Answer 1


If we define $\mathcal{T}=i\sigma_y K$ where $K$ is complex conjugation, i.e.

$\mathcal{T}\psi_\uparrow \mathcal{T}^{-1}=\psi_\downarrow, \mathcal{T}\psi_\downarrow \mathcal{T}^{-1}=-\psi_\uparrow$,

Then naively a term like $\Delta e^{i\Phi}\psi_{\uparrow}^\dagger \psi_{\downarrow}^\dagger$ is not invariant under $\mathcal{T}$. This is basically the problem you encountered, phrased a little differently. However, it does not mean that the system actually breaks $T$, as all physical observables will be invariant under $\mathcal{T}$. The resolution is in the definition of the transformation of $\psi$ under $\mathcal{T}$. Let us modify the definition to be

$\mathcal{T}\psi_\uparrow \mathcal{T}^{-1}=\psi_\downarrow e^{-i\Phi}, \mathcal{T}\psi_\downarrow \mathcal{T}^{-1}=-\psi_\uparrow e^{-i\Phi}$.

This phase is not observable (it is essentially redefining the phases of the basis states of the second-quantized Fock space, which has no consequences on physical observables), so we are free to do so. Notice that the important algebraic relation $\mathcal{T}^2=-1$ (for the classification of TI/TSC, etc.) is not affected. Then the pairing term is invariant.

Of course, this only works when $\Phi$ does not depend on positions. Otherwise (e.g. when there is a vortex) one can not get rid of the phase, since $\nabla\Phi$ is an observable, the supercurrent.


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