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I believe this is just elementary QM, but I'm getting awfully confused. The question is drawn from this paper on Wannier-Stark localization (but is self-contained): https://iopscience.iop.org/article/10.1088/0953-8984/1/8/007

Let \begin{equation} |n,k\rangle = e^{ikx}u_{n,k}(x) \end{equation} be a Bloch state, where $u_{n,k}(x+a)=u_{n,k}(x)$ is the periodic part. We wish to calculate the matrix element $\langle n,k|x|n',k'\rangle$, which the authors claim can be written as \begin{equation} \langle n,k|x|n',k'\rangle = i\delta_{n,n'}\delta_{k,k'}\frac{\partial}{\partial k} + i\delta_{k,k'}X_{n,n'}, \end{equation} where \begin{equation} X_{n,n'} = iN\int_0^ae^{i(k-k')x}u_{n,k}^*(x)\frac{\partial}{\partial k}u_{n',k'}(x) \;\mathrm{d}x. \end{equation} This latter term is physically the inter-band coupling. Note that I am not sure what $N$ is, as I can't see it defined in the paper. Also, the equation for $X_{n,n'}$ is as written in the paper, with $\partial_k$ acting on $u_{n',k'}$.

Question: show the above.

I've tried writing $x=-i\partial_k e^{ikx}$ and integrating by parts, but do not get their expression. I am also generally confused as to how a matrix element $\langle n,k|x|n',k'\rangle$ (which should be a number...?) can be equal to a derivative $\partial_k$.

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    $\begingroup$ Have you tried to find the paper they're citing? Anyway, this notation is really weird, indeed. Perhaps it might be worth to ask this question on Matter Modeling. But please make sure to state that the question is cross posted. $\endgroup$ Commented May 30, 2022 at 6:50
  • $\begingroup$ Yes, the citing paper just states the result without proof $\endgroup$
    – dsfkgjn
    Commented May 30, 2022 at 7:04
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    $\begingroup$ From the paper, the authors take a finite lattice where N is the number of unit cells. You might find it easier to reproduce the derivation if you avoid the abuse of notation by writing $|n,k⟩= \int dx e^{ikx} u_{n,k}(x) |x> and work entirely in the position basis. I'll post the complete derivation later when I have time. $\endgroup$ Commented Jun 6, 2022 at 3:11

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it seems to me that the equation $|n,k>=e^{ikx}u_{n,k}(x)$ is a bit confusing. I would agree more if it were something like $<x|n,k>=e^{ikx}u_{n,k}(x)$. Afterall, taking a quick look on the paper, they do not write down an equation like that anywhere. So, having said that (I will just offer the intermediate steps and the intermediate steps only, as I have no understanding of the subject!), one can write $$<n,k|x|n',k'>=\int dx <n,k| x|x><x|n',k'>$$ and this should be equal to (after substituting $<x|n',k'>$) $$<n,k|x|n',k'>=\int dx xe^{-i(k-k')x}u^*_{n,k}(x)u_{n',k'}(x)$$ Now, the variable $x$ can be written as $i\partial/\partial k$ and hence $$<n,k|x|n',k'>=i\int dx \frac{\partial}{\partial k} e^{-i(k-k')x}u^*_{n,k}(x)u_{n',k'}(x)$$ The last step involves integrating by parts (but be carefull! $u_{n,k}(x)$ also depends on $k$). So, $$<n,k|x|n',k'>=i\frac{\partial}{\partial k}\int dx e^{i(k-k')x}u^*_{n,k}(x)u_{n',k'}(x)-i\int dx e^{i(k-k')x} \frac{\partial}{\partial k} u^*_{n,k}(x)u_{n',k'}(x)$$ Now, you can see that the second term is of the form given by what you denote as $X_{n,n'}$ and in fact you can use the form to calculate the normalization constant $N$. Moreover, the first term can be identified with the first term provided by you in your post (there is some sort of weird orthogonality relation, but this is for you to figure out)... If there are questions, please do not hesitate to comment.

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  • $\begingroup$ This got me nearly there. Small question: can you account for the difference in signs in the last line? And the positioning of the $\partial_k$ operator? $\endgroup$
    – dsfkgjn
    Commented Jun 7, 2022 at 10:14
  • $\begingroup$ If you mean the difference of signs between the first and the second terms, the latter (difference) exists due to integration by parts. I.e. $\partial_k[e^{i(k-k')x}f(k)]=\partial_k[e^{i(k-k')x}]f(k)+e^{i(k-k')x}\partial_k[f(k)]$. I am not sure what it is you ask for the positioning of the $\partial_k$ operator... It should act on a function of $k$, so I can not move it around in an arbitrary way... Please ask freely if there are more things you want to ask $\endgroup$
    – schris38
    Commented Jun 7, 2022 at 13:06
  • $\begingroup$ Of course - I am confused why the result from the paper does not have this sign difference. $\endgroup$
    – dsfkgjn
    Commented Jun 8, 2022 at 6:53
  • $\begingroup$ Maybe this has something to do with the fact that the derivative wrt $k$ is equal to minus the derivative wrt $k'$ $\endgroup$
    – schris38
    Commented Jun 8, 2022 at 6:59
  • $\begingroup$ And I also think that I have a mistake in the calculations. I will correct it now $\endgroup$
    – schris38
    Commented Jun 8, 2022 at 7:00

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