Berry connection and time reversal symmetry I am seeing how the Berry connection $\mathcal{A}(k)$ transforms under time reversal symmetry. I seem to have a hiccup over something simple. I may have overcomplicated things but I think it points to some misconceptions I may have.
From the definition of the Berry curvature as per usual:
\begin{align}
\mathcal{A}(k) &= i \langle \psi(k) | \frac{d}{dk} |\psi(k)\rangle \\
&= i \int  \psi^\star (k) \frac{d \psi (k)}{dk}  d\vec{r}
\end{align}
Applying time reversal $\hat{\mathcal{T}}$,
\begin{align}
\hat{\mathcal{T}} \mathcal{A}(k)  &= i \langle \hat{\mathcal{T}} \psi(k) | \frac{d}{dk} | \hat{\mathcal{T}} \psi(k)\rangle \\
&= i \int \hat{\mathcal{T}} \psi^\star (k) \frac{d  \hat{\mathcal{T}} \psi (k)}{dk}  d\vec{r} \\
&= i \int  \psi(-k) \frac{d   \psi^\star (-k)}{dk}  d\vec{r}
\end{align}
Since $\mathcal{A}$ must be real, we can conjugate it.
\begin{align}
&= -i \int  \psi^\star(-k) \frac{d   \psi(-k)}{dk}  d\vec{r} \\
&= - \mathcal{A}(-k)
\end{align}
So If I am in a time-reversal invariant system, I have
\begin{align}
\hat{\mathcal{T}} \mathcal{A} (k) &= \mathcal{A} (k) \\
\implies \mathcal{A} (k) &= - \mathcal{A}(-k)
\end{align}
Which is wrong, I am off by the minus sign.

My Questions:


*

*Is it true that the Berry connection is always real? That was my justification for conjugating it. I think they do this step in both sources listed below.

*Is this an abuse of notation if I did this (putting the derivative with the ket):
\begin{align}
\mathcal{A}(k) &= i \langle \psi(k) | \frac{d \psi(k)}{dk} \rangle \\
\hat{\mathcal{T}} \mathcal{A} (k) &= i \langle \hat{\mathcal{T}} \psi(k) |  \hat{\mathcal{T}} \frac{d \psi(k)}{dk} \rangle 
\end{align}
And if this were the case, how does $\hat{\mathcal{T}}$ act on the differential operator? Would
\begin{align}
\hat{\mathcal{T}} \frac{d }{dk} = \frac{d }{d(-k)} \hat{\mathcal{T}} 
\end{align}
be true? I feel like you can't just take the derivative out without a minus sign, or else how would the velocity operator become negative?

*Why does $\hat{\mathcal{T}} $ not act on the $i$ outside the braces in $\hat{\mathcal{T}} \mathcal{A}(k)  = i \langle \hat{\mathcal{T}} \psi(k) | \frac{d}{dk} | \hat{\mathcal{T}} \psi(k)\rangle $?

*Was there something else wrong in my derivation?



Sources:


*

*They define the Berry connection with an extra minus sign but that shouldn't matter http://www-personal.umich.edu/~sunkai/teaching/Fall_2012/chapter3_part8.pdf

*Topological States on Interfaces Protected by Symmetry, by Takahashi 2015. The derivation from that is the following:


\begin{align} \mathbf { a } ^ { \alpha } ( \mathbf { - k } ) & = - i \left\langle u ^ { \alpha } ( - \mathbf { k } ) | \nabla u ^ { \alpha } ( - \mathbf { k } ) \right\rangle \\ & = - i \left\langle \nabla \Theta u ^ { \alpha } ( - \mathbf { k } ) | \Theta u ^ { \alpha } ( - \mathbf { k } ) \right\rangle \\ & = i \left\langle \Theta u ^ { \alpha } ( - \mathbf { k } ) | \nabla \Theta u ^ { \alpha } ( - \mathbf { k } ) \right\rangle \\ & = \mathbf { a } ^ { \beta } ( \mathbf { k } ) + i \nabla \chi ( \mathbf { k } ) 
\end{align}
 A: My conclusion thus far is that time reversal does act on the derivative:
\begin{align}
\hat { \mathcal { T } } \frac { d } { d k } = \frac { d } { d ( - k ) } \hat { \mathcal { T } } =  - \frac { d } { d  k } \hat { \mathcal { T } }.
\end{align}
And that it also does act on the $i$ outside the braces. 
My negative sign mistake was that I equated 
\begin{align}
\mathcal{A}(-k)= i \langle \psi ( -k ) | \frac { d \psi ( -k ) } { d k } \rangle.
\end{align}
But it is actually:
\begin{align}
\mathcal{A}(-k)= i \langle \psi ( -k ) | \frac { d \psi ( -k ) } { d (-k) } \rangle.
\end{align}
(An easy way to check this is if $x(t)= \sin(t), v(t)= \cos(t)$, so $v(-t)= \cos(-t) = \frac{dx(-t)}{d(-t)}$ and not $\frac{dx(-t)}{dt}$).
So my derivation is now as follows:
\begin{align}
\mathcal{T} \mathcal{A} (k) &= \mathcal{T} \left( i \langle \psi ( k ) | \frac { d \psi ( k ) } { d k } \rangle \right) \\
&= \mathcal{T} \left( i \int \psi ^ { \star } ( k ) \frac { d \psi ( k ) } { d k }  d \vec { r } \right) \\
&= -i   \int \mathcal{T} \left( \psi ^ { \star } ( k ) \frac { d \psi ( k ) } { d k }\right) d \vec { r }  \\
&= -i   \int  \psi  ( -k ) \frac { d \psi^\star( -k ) } { d (-k) } d \vec { r }  \\
&= i   \int  \psi^\star  ( -k ) \frac { d \psi( -k ) } { d (-k) } d \vec { r }  \\
&= i \langle \psi ( -k ) | \frac { d \psi ( -k ) } { d (-k) } \rangle \\
&= \mathcal{A}(-k)
\end{align}
Some sources may not conjugate the $i$, but that is because they start from $\mathcal{A}(-k)$ and substitute the wave function for the time reversed wave function, which is different to applying $\mathcal{T}$ to the whole term as I have done, but both derivations are equivalent.
