# Time reversal of superconducting order parameter?

This pedagogical material discusses the transformation of the superconducting order parameter $$\Delta_k$$ (matrix in spin space). Eq. (79) applies the time-reversal operator $$K=-i\sigma^yC$$ (complex conjugation operator $$C$$) to a fermion annihilation operator $$Kc_{ks}= -i\sigma^y_{ss'}c^\dagger_{-k,s'}.$$

While the $$\sigma^y$$ multiplication is fine, how is annihilation changed to creation? This looks inconsistent with the first set of equations in this answer or the similar form in this review (P44 bottom left). However, without this transform to creation, how to understand the complex conjugate in Eq. (80) $$K\Delta_{k}=\sigma^y\Delta_{-k}^*\sigma^y$$ below Eq. (79)? It looks like $$\langle c^\dagger c^\dagger \rangle=\langle cc \rangle^*$$.

This is a partial answer, but I hope it's at least helpful.

I also can't see how to reconcile the statements

(1) $$K c_{ks} = -i\sigma^y_{ss'} c^\dagger_{-k, s'}$$ and

(2) $$K \hat{c}_{\uparrow} K^{-1} = \hat{c}_{\downarrow}; K \hat{c}_{\downarrow} K^{-1} = -\hat{c}_{\uparrow}$$.

When working with antilinear objects, we first need to establish a phase convention. Suppose you have two bases for the 1D complex vector space formed by the vacuum - $$|0\rangle$$, and $$|0'\rangle = e^{i\phi} |0\rangle$$. How does $$C$$ act on these bases? If you suppose that $$C | 0 \rangle = | 0 \rangle$$, then you end up with the strange-looking

$$C | 0' \rangle = e^{-i\phi} |0\rangle = e^{2i\phi} |0'\rangle.$$

where $$C$$ is complex conjugation. This is often suppressed by picking a phase convention for the vacuum that ensures that $$C |0\rangle = |0 \rangle$$. Choosing the same convention for all basis kets, defined as $$c^\dagger_{s} |0 \rangle$$ means that $$Cc_{s} C \equiv c_s$$.

In short, the action of time reversal can change the phases in front of operators, but does not mix orthogonal states.

However, I believe that (2) is consistent with the claimed superconducting order parameter transformation.

The SC order parameter operator (as a matrix) is written such that

$$\hat{\Delta}(k) = \begin{pmatrix} \hat{c}_{-k\uparrow} & \hat{c}_{-k\downarrow} \end{pmatrix}\begin{pmatrix} \Delta_{\uparrow\uparrow}(k) & \Delta_{\uparrow\downarrow}(k) \\ \Delta_{\downarrow\uparrow}(k) & \Delta_{\downarrow\downarrow}(k) \end{pmatrix} \begin{pmatrix} \hat{c}_{k\uparrow} \\ \hat{c}_{k\downarrow} \end{pmatrix} = \Delta_{ss'}(k)\hat{c}_{-k s}\hat{c}_{k s'}$$

In second quantisation, symmetries acts on operators by the conjugation action (i.e. in the adjoint representation). Due to decades of sloppy notation, we end up with garbage like

$$\tilde{K} \hat{\Delta}(k) = K \Delta_{ss'}(k)\hat{c}_{-k s}\hat{c}_{k s'}K^{-1}$$

$$= -i \sigma^y C \Delta_{ss'}(k)\hat{c}_{-k s}\hat{c}_{k s'} i \sigma^y C \hspace{3em} \text{(*)}$$

Now there is a problem - we defined how $$C$$ acts on $$|\uparrow\rangle$$ for a single spin, but how does it act in the full Hilbert space, $$\text{Span}\{ c^\dagger_{r,s} |0\rangle ,r \in \text{Points in space} \}$$ ?

Suppose that we make the choice of phase $$C c^\dagger_{r, s}|0\rangle = c^\dagger_{r, s}|0\rangle$$. Then complex conjugation acts non-trivially on $$k$$ space eigenvectors -

$$C c^\dagger_{k,s}|0\rangle = C\sum_{r}e^{-ik\cdot r} c^\dagger _{r, s} |0\rangle$$

$$= \sum_{r}e^{ik\cdot r} C c^\dagger _{r, s} |0\rangle = \sum_{r}e^{ik\cdot r} c^\dagger _{r, s} |0\rangle = c^\dagger_{-k,s} |0\rangle$$

Using the anticommutators, this argument shows that in the standard phase convention, $$C c_{k,s} C = c_{-k, s}$$.

Using this fact gets you what you want.

$$K\hat{\Delta}(k)K = +\sigma^y \Delta_{ss'}^*(k) \hat{c}_{k, s} \hat{c}_{-k, s'} \sigma^y$$

$$= -\sigma^y \Delta_{ss'}^*(-k) \hat{c}_{-k, s'} \hat{c}_{k, s}\sigma^y$$

which is the claimed result, since $$\Delta(k) = -\Delta(-k)^T$$.

• Thank you for the helpful answer! Weirdly, I haven't received any notification from PSE and I saw it just now... This makes good sense and is consistent with what was in mind. Here, the SC order parameter is treated in the same way as a usual second-quantized Hamiltonian, then $C$ affects the c-number part as well and hence we get $\Delta^*$. Jan 14 at 3:16
• But at the end of my question, I was thinking more about another viewpoint, i.e., $\Delta$ defined from $\langle c c \rangle$ or so. Then it seems unclear how we can get $\Delta^*$ from the usual $K$ effect I linked or your (2) in the beginning because there $K$ does not mix annihilation and creation. Any thoughts? Jan 14 at 3:21
• Then you have $\langle cc \rangle^* := \langle 0 | cc 0 \rangle^* =\langle cc0 | 0 \rangle = \langle 0 | (cc)^\dagger | 0 \rangle$, where $0$ is the fermion vacuum (i.e. not the superconducting vacuum). Jan 14 at 4:06
• I mean if we have $\Delta\sim\langle cc \rangle$ and apply $K$ to the two $c$ operators, we don't get two $c^\dagger$s. Then it looks that $K$ does not change $\Delta$ to the expected $\Delta^*$. Jan 14 at 4:12