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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:

  1. 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
  2. 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}

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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.

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