1
$\begingroup$

I am trying to prove result (3.4) of the following paper:
http://li.mit.edu/S/2d/Paper/Fu07Kane.pdf
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.\ Any help is appreciated, thanks!

$\endgroup$
  • $\begingroup$ 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 Oct 29 '18 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. $\endgroup$ – TheQuantumMan Nov 1 '18 at 15:32
  • $\begingroup$ @Aaron note that I have used the fact that $B^\dagger=(B^*)^T$ where the $T$ stands for transpose. $\endgroup$ – TheQuantumMan Nov 1 '18 at 15:34
0
$\begingroup$

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).

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.