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?
Best Answer
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.