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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 $\partial_k$.

dsfkgjn
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1 Answers1

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It seems to me that the equation $|n,k\rangle=e^{ikx}u_{n,k}(x)$ is a bit confusing. I would agree more if it were something like $\langle x|n,k\rangle=e^{ikx}u_{n,k}(x)$. After all, 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 $$\langle n,k|x|n',k'\rangle=\int \mathrm{d}x \langle n,k| x|x\rangle \langle x|n',k'\rangle$$ and this should be equal to (after substituting $\langle x|n',k'\rangle$) $$\langle n,k|x|n',k'\rangle=\int \mathrm{d}x\,xe^{-i(k-k')x}u^*_{n,k}(x)u_{n',k'}(x).$$ Now, the variable $x$ can be written as $i\partial/\partial k$ and hence $$\langle n,k|x|n',k'\rangle =i\int \mathrm{d}x \frac{\partial}{\partial k} e^{-i(k-k')x}u^*_{n,k}(x)u_{n',k'}(x)$$ The last step involves integrating by parts (but be carefull! $u_{n,k}(x)$ also depends on $k$). So, $$\langle n,k|x|n',k'\rangle=i\frac{\partial}{\partial k}\int \mathrm{d}x e^{i(k-k')x}u^*_{n,k}(x)u_{n',k'}(x)-i\int \mathrm{d}x e^{i(k-k')x} \frac{\partial}{\partial k} u^*_{n,k}(x)u_{n',k'}(x).$$

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.

Vincent Thacker
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schris38
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