Translating Ashcroft and Mermin's "Second Proof" of Bloch's Theorem to Dirac's Notation At the end of this post I attach Ashcroft and Mermin's proof of Bloch's theorem which is not essential per se (the proof using lattice symmetries is more general), but is key in being used later as a jumping off point for the nearly free electron model.
Now I am trying to translate it to Dirac's bra-ket notation, since that always helps me think in more general, coordinate-free terms (once I pull out all of the identity operators which have been tacitly inserted). Essentially, I am trying to arrive at (8.38) by beginning at the coordinate-free eigenvalue equation for $H$.
Thus I thought to write something like
$$H|\psi\rangle=\epsilon|\psi\rangle \implies \langle \mathbf{r}|H|\psi\rangle=\langle \mathbf{r}|\int \mathrm{d}{\mathbf{k}}\left(\frac{\mathbf{p}^2}{2m}+U\right)|\mathbf{k}\rangle\langle \mathbf{k}|\psi\rangle=\int \mathrm{d}{\mathbf{k}}\left(\frac{\hbar^2\nabla^2}{2m}+U(\mathbf{r})\right)\langle \mathbf{r}|\mathbf{k}\rangle c_\mathbf{k}=\int \mathrm{d}{\mathbf{k}}\left(\frac{\hbar^2\nabla^2}{2m}+U(\mathbf{r})\right)c_\mathbf{k}e^{i \mathbf{k} \cdot\mathbf{r}}$$
where the integration is over all of momentum (k-space up to a proportionality factor). The first term in the equation above recovers (8.36), but (8.37) I cannot seem to "find". Obviously I want something of the form (8.32), so I thought to insert completeness in coordinate space ($\mathbf{r}$), but then I have some extra $\mathbf{r}$ kets in the second term which aren't there in the first.
Any help in completing the steps which I cannot would be greatly appreciated.



 A: You need to use the fact that $U$ is periodic with the same periodicity as the lattice, which means that it can be expressed as a sum
$$U(\mathbf r)=\sum_{\mathbf G\in \mathrm{RL}}u_\mathbf G e^{i\mathbf G\cdot\mathbf r}$$
where RL is the reciprocal lattice.
Also, for what it’s worth note that since Ashcroft and Mermin are working on a torus (with Born von Karmin boundary conditions) the set of possible momenta is discrete, so the integral should be a sum. You may repeat the derivation in the continuum if you wish, but it will look slightly different than it does in the text.
A: Here's how to do it by inserting an extra complete set of momentum states.
\begin{align*}
\langle \mathbf{r}|H|\psi\rangle
&=\langle \mathbf{r}|
\int \mathrm{d}{\mathbf{k}} \mathrm{d}{\mathbf{k'}}\,|\mathbf{k'}\rangle\langle \mathbf{k'}|
\left(\frac{\mathbf{p}^2}{2m}+U\right)|\mathbf{k}\rangle\langle \mathbf{k}|\psi\rangle\\
&=\int \mathrm{d}{\mathbf{k}} \mathrm{d}{\mathbf{k'}}\,
\langle \mathbf{r}|\mathbf{k'}\rangle
\left(
\langle \mathbf{k'}|\frac{\mathbf{p}^2}{2m}|\mathbf{k}\rangle
+
\langle \mathbf{k'}|U|\mathbf{k}\rangle
\right)
\langle \mathbf{k}|\psi\rangle\,.
\end{align*}
Now, because $U$ is periodic, in the lattice, it only connects momentum states if they differ by a reciprocal lattice vector $\mathbf{G}$, i.e.
$$
\langle \mathbf{k'}|U|\mathbf{k}\rangle = U_{\mathbf{G}}\delta^3(\mathbf{k}+\mathbf{G}-\mathbf{k'})\,,
$$
where $U_\mathbf{G}$ is a Fourier coefficient of $U$. Then
\begin{align*}
\langle \mathbf{r}|H|\psi\rangle
&=\int \mathrm{d}{\mathbf{k}} \mathrm{d}{\mathbf{k'}}\,
\langle \mathbf{r}|\mathbf{k'}\rangle
\left(
\frac{k^2}{2m}\delta^3(\mathbf{k}-\mathbf{k'})
+
U_{\mathbf{G}}\delta^3(\mathbf{k}+\mathbf{G}-\mathbf{k'})
\right)
\langle \mathbf{k}|\psi\rangle
\\
&=
%\frac{e^{i\mathbf{k}'\cdot\mathbf{r}}}{\sqrt{2\pi}}
\int \mathrm{d}{\mathbf{k}}\,
\langle \mathbf{r}|\mathbf{k}\rangle
\frac{k^2}{2m}
\langle \mathbf{k}|\psi\rangle
+
\int \mathrm{d}{\mathbf{k}}\,
\langle \mathbf{r}|\mathbf{k}+\mathbf{G}\rangle
U_\mathbf{G}
\langle \mathbf{k}|\psi\rangle
\\
&=
\int \mathrm{d}{\mathbf{k}}\,
\frac{e^{i\mathbf{k}\cdot\mathbf{r}}}{\sqrt{2\pi}}
\frac{k^2}{2m}
\tilde{\psi}(\mathbf{k})
+
\int \mathrm{d}{\mathbf{k}}\,
\frac{e^{i(\mathbf{k}+\mathbf{G})\cdot\mathbf{r}}}{\sqrt{2\pi}}
U_\mathbf{G}
\tilde{\psi}(\mathbf{k})
\\
&=
\int \mathrm{d}{\mathbf{k}}\,
\frac{e^{i\mathbf{k}\cdot\mathbf{r}}}{\sqrt{2\pi}}
\frac{k^2}{2m}
\tilde{\psi}(\mathbf{k})
+
\int \mathrm{d}{\mathbf{k}}\,
\frac{e^{i\mathbf{k}\cdot\mathbf{r}}}{\sqrt{2\pi}}
U_\mathbf{G}
\tilde{\psi}(\mathbf{k}-\mathbf{G})
\end{align*}
where in the last step we have used a change of variables $\mathbf{k}\to\mathbf{k}+\mathbf{G}$. We have also notated the Fourier transform of $\psi$ as $\tilde{\psi}$.
A: You can work it backwards:
$$\sum_{\mathbf{k},\mathbf{K}}U_\mathbf{K}e^{i(\mathbf{k}+\mathbf{K})\mathbf r}c_\mathbf{k}=\sum_{\mathbf{k},\mathbf{K}}U_\mathbf{K}e^{i\mathbf{K}\mathbf r}\langle \mathbf r|\mathbf{k}\rangle\langle \mathbf{k}|\psi\rangle=\sum_{\mathbf{K}}U_\mathbf{K}e^{i\mathbf{K}\mathbf r}\psi(\mathbf r)=\sum_{\mathbf{K}}U_\mathbf{K}\langle \mathbf r|\mathbf{K}\rangle\psi(\mathbf r)$$
$$=\int dr'\sum_\mathbf{K} U(\mathbf r')\langle \mathbf{K}|\mathbf r'\rangle \langle \mathbf r'|\mathbf{K}\rangle\psi(\mathbf r)=\int d^3r' U(\mathbf r')\langle \mathbf r|\mathbf r'\rangle\psi(\mathbf r) =U(\mathbf r)\psi(\mathbf r)$$
