I am trying to prove result (3.4) of the following paper:
https://doi.org/10.1103/PhysRevB.76.045302, namely, that $$-i\ \langle{u_n|\nabla_k u_n}\rangle=-\dfrac{i}{2}tr[v^\dagger(k)\nabla_k v(k) ]$$ where I am using the Einstein summation convention and $v$ is given by $$v_{nm}(k)=\langle{u_n|P\Theta|u_m\rangle}$$ where $\Theta$ is the time reversal operator and $P$ is the space inversion operator, and all the $\langle{u_n|}$ are evaluated at the same $k$ (this is a vector).

My attempt is the following:
I first write $-i\langle{u_n|\nabla_k u_n\rangle}=-\langle{u_n|r|u_n\rangle}$ because I know that both $\Theta$ takes $r\mapsto r$ and $P$ takes $r\mapsto -r$. I will also use that $v v^+=v^\dagger v=1$, $P^\dagger P=1$ and $\langle{\Theta \psi |\Theta \phi\rangle}=\langle{\phi|\psi\rangle}=\langle\psi|\phi\rangle^*$.
So, here it goes:
$$-\langle{ u_n |r|u_n \rangle}=-\langle{ P\Theta\ r\ u_n | P\Theta u_n \rangle}=\langle{ r\ P\Theta\ u_n | P\Theta u_n \rangle}=\langle{ r\ P\Theta\ u_n | u_m \rangle}\langle{u_m |P\Theta u_n \rangle}=v_{mn}\langle{ r\ P\Theta\ u_n | u_m \rangle}$$
I now use the chain rule (recalling that the $k-$space representation of $r$ is a derivative with respect to $k$) to write for example $r\langle{\psi|\phi\rangle}=\langle{r\ \psi|\phi\rangle}+\langle{\psi|r\ \phi\rangle}$. Doing so, gives:
$$\langle{ u_n |r|u_n \rangle}=v_{mn}\left(r\ \langle{P\Theta u_n|u_m\rangle}-\langle{P\Theta u_n|r\ u_m\rangle} \right)$$
I now insert the resolution of the identity in the last term: $$-\langle{ u_n |r|u_n \rangle}=v_{mn}\left( r\ v^*_{mn}-\langle{P\Theta u_n|u_s \rangle}\langle{u_s|r\ u_m\rangle} \right)=v_{mn}r\ v^*_{mn}-\underbrace{v_{mn}v^*_{sn}}_{\delta_{ms}}\langle{u_s|r\ u_m\rangle}$$
$$=tr[v\ r\ v^\dagger]-\langle{u_m|r|u_m\rangle}=tr[v\ r\ v^\dagger]-\langle{u_n|r|u_n\rangle}$$ Which is clearly wrong!

I suspect that a mistake is that $$-\langle{ P\Theta\ r\ u_n | P\Theta u_n \rangle}=\langle{ r\ P\Theta\ u_n | P\Theta u_n \rangle},$$ but even if I don't change the sign here, the last result picks up a minus sign, which gives that $$-i\ \langle{u_n|\nabla_k u_n}\rangle=-\dfrac{i}{2}tr[v(k)\nabla_k v^\dagger(k) ]=\dfrac{i}{2}tr[v^\dagger(k)\nabla_k v(k) ]$$ which differs by a sign from the result I should have gotten; so, even if that sign change when I (anti)commute $r$ past $P\Theta$ doesn't fix things.

  • $\begingroup$ First, $i\nabla_k = r$ not $-i\nabla_k = r$ (though this doesn't change any result). Second, you need to be careful about the definition of the time-reversal operator due to the fact that it sends $i\rightarrow -i$. This means that the time-reversal operator is sensitive to the choice of basis that you are working in (this probably gives you the sign change. Second, $v_{mn} r v^*_{mn} \neq tr[v\nabla_k v^\dagger]$. It should be $v_{mn} r v^*_{nm}$ (note the indices), which is the negative of what you have written due to antiunitarity of $\Theta$. $\endgroup$
    – Aaron
    Commented Oct 29, 2018 at 15:28
  • $\begingroup$ @Aaron Many thanks for the comment and sorry for the delayed reply. Note that I've indeed used $i\nabla_k=r$ without the minus sign. Also, note that $tr(AB)=A_{mn}B_{nm}$ so that $tr(AB^\dagger)=A_{mn}B^\dagger_{nm}=A_{mn}B^*_{mn}$, so I think that I am right (except if you have any objection to what I have just written). Lastly, on the comment that the TR operator is sensitive to the choice of basis, I could have done the whole thing without any $i$ and then just multiply both sides by $i$. This is indeed what I've done at first and got the same result as the one presented in the question. $\endgroup$ Commented Nov 1, 2018 at 15:32
  • $\begingroup$ @Aaron note that I have used the fact that $B^\dagger=(B^*)^T$ where the $T$ stands for transpose. $\endgroup$ Commented Nov 1, 2018 at 15:34

1 Answer 1


The Kane-Fu paper seems to have a sign error.

First, we know that $PkP^{-1} = -k$ and $\Theta k \Theta^{-1} = -k$. Hence, we have $P\Theta \nabla_k \Theta^{-1} P^{-1} |\phi\rangle = \nabla_k|\phi\rangle$ (as an operator statement, e.g. acting on a ket!). This is consistent with what you have previously wrote regarding $r$.

Let us work in reverse to prove the result. I will be using Einstein summation notation.

Let us be very careful about the derivatives. I define $\langle \nabla_k \phi | \equiv \nabla_k \langle \phi|$ and $|\nabla_k \phi\rangle \equiv \nabla_k | \phi\rangle$. \begin{align} (v^\dagger)_{mn} \nabla_k v_{nm} &= \langle P\Theta u_m|u_n\rangle \nabla_k \langle u_n|P\Theta u_m\rangle \\ &= \langle P\Theta u_m|u_n\rangle[\langle \nabla_k u_n|P\Theta u_m\rangle + \langle u_n|\nabla_k P\Theta u_m\rangle]\\ &= \langle \nabla_k u_n|P\Theta u_m\rangle\langle P\Theta u_m|u_n\rangle + \langle P\Theta u_m|u_n\rangle\langle u_n|\nabla_k P\Theta u_m\rangle \\ &= \langle\nabla_k u_n|u_n\rangle + \langle P\Theta u_m|\nabla_k P\Theta u_m\rangle \\ &= \langle\nabla_k u_n|u_n\rangle + \langle P\Theta u_m| P\Theta \nabla_k u_m\rangle \\ &= \langle\nabla_k u_n|u_n\rangle + \langle u_m| \nabla_k u_m\rangle^* \\ &= -\langle u_n| \nabla_k u_n\rangle - \langle u_m| \nabla_k u_m\rangle \text{$\hspace{3em}$ (since $\langle u_n|u_n\rangle =1$)}\\ &= -2\langle u_n|\nabla_k u_n\rangle \end{align} Hence, putting in $-i/2$, we get the desired result, though with opposite sign to the paper. I'm not exactly sure, but I believe the error of getting $0$ in your calculation comes from being a bit too careless about interchanging $r$ as an operator and $r$ as a derivative (eg not between bras and kets).


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