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Bloch's theorem can be stated as: $$ |\Psi_{n\vec{k}}\rangle=e^{i\vec{k}\cdot\vec{r}}|u_{n\vec{k}}(\vec{r})\rangle \tag{1} $$ where $|\Psi_{n\vec{k}}\rangle$ is the solution of single electron Schrodinger equation: $$ (T+V)|\psi_n(\vec{k})\rangle=H|\psi_{n\vec{k}}\rangle =E_{n\vec{k}}|\psi_{n\vec{k}}\rangle \tag{2} $$ with $V(\vec{r}+\vec{R})=V(\vec{r})$ [$\vec{R}$ is the translational vector in real space], and $u_{n\vec{k}}(\vec{r})$ in Eq.(1) is periodical in $\vec{R}$. By multiplying $e^{-i\vec{k}\cdot\vec{r}}$ on both sides of Eq.(2), we find: $$ \bar{H}|u_{n\vec{k}}(\vec{r})=E_{n\vec{k}}|u_{n\vec{k}}\rangle \tag{3} $$ where $\bar{H} \equiv e^{-i\vec{k}\cdot\vec{r}} H e^{+i\vec{k}\cdot\vec{r}}$. From this eigenequation, we can obtain the completeness relation: $$ \sum_n |u_{n\vec{k}}(\vec{r})\rangle \langle u_{n\vec{k}}(\vec{r})|=1 \tag{4} $$ and also this completeness relation: $$ \int d\vec{k} \sum_n |u_{n\vec{k}}(\vec{r})\rangle \langle u_{n\vec{k}}(\vec{r})|=1 \tag{5} $$ Now my question is which one is correct? It seems that Eq.(4) is usually used in the community of solid-state physics while Eq.(5) is generally used in the community of nonlinear optics.

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2 Answers 2

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I start from the other existing answer, but I elaborate. Also in the question, the source of confusion is in part due to the bad notation.

One can consider a wave function which depends on spin $s$ and position $x$, i.e. $\psi_{N}(x,s)$. Now, suppose you can write it as $\psi_{n\sigma}(x,s)=\phi_{n\sigma}(x)\chi_\sigma(s)$ with $N=\{n\sigma\}$ (this is possible if spin orbit coupling (SOC) is neglected in condensed matter). Then you have \begin{eqnarray} \sum_\sigma |\chi_\sigma\rangle\langle\chi_\sigma|=1 \tag{1}\\ \sum_n |\phi_{n\sigma}\rangle\langle\phi_{n\sigma}|=1 \tag{2} \end{eqnarray} and in general \begin{eqnarray} \sum_\sigma \sum_n |\psi_{n\sigma}\rangle\langle\psi_{n\sigma}|=1 \tag{3} \end{eqnarray} Notice that, for each spin channel $\sigma$ there is a complete basis set generated by the eigenstates of the spin dependent Hamiltonian $h_\sigma$

If you see the parallel with block states, you can do the same and define $\Psi_{nk}(r+R)=u_{nk}(r)e^{ik(r+R)}$. Let me define $e_k(x)=e^{ikx}$ the exponential function. You obtain \begin{eqnarray} \sum_n |u_{nk}\rangle\langle u_{nk}|=1 \tag{4}\\ \int dk |e_k\rangle\langle e_k|=1 \tag{5} \end{eqnarray} and \begin{eqnarray} \sum_n \int dk |\Psi_{nk}\rangle\langle\Psi_{nk}|=1\tag{6} \end{eqnarray}

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    $\begingroup$ +1. Welcome to our community by the way! We hope to see much more of you in the future! I added equation numbers so that people can say "Eq. 5" instead of "the second equation from the bottom" when citing this answer. $\endgroup$ Commented Jul 10, 2022 at 2:15
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"From this eigenequation, we can obtain the completeness relation"

You don't need that eigenequation to obtain the completeness relation. As long as you have a "complete" set of orthonormal vectors in a Hilbert Space (for example $n$ linearly independent vectors in an $n$-dimensional Hilbert space) the completeness relation holds. More about this has been discussed on the Physics Stack Exchange.

Some Hilbert spaces have a finite dimension (for example, the operators describing spin-1/2 particles) and the completeness relation is $\sum_n |\psi_n\rangle \langle \psi_n | $ where the number of terms in the sum is the dimension of the Hilbert space (which is finite, and equal to 2 in the case of spin-1/2 particles, equal to in the case of spin-1 particles, equal to 4 in the case of spin-3/2 particles, etc.). But some Hilbert spaces are used for continuous quantum mechanical variables such as the position $\hat{x}$, in which the completeness relation is $\int |\psi_x\rangle \langle \psi_x|\textrm{d}x$.

What you're seeing in Eq. 5 is the combination of two completeness relations. For example if you have wavefunctions which depend on spin $\hat{s}$ and position $\hat{x}$ at the same time, then we have the following two completeness relations:

\begin{eqnarray} \sum_s |\psi_{sx}\rangle \langle \psi_{sx}| &= 1 \tag{1}\\ \int |\psi_{sx}\rangle \langle \psi_{sx}| \textrm{d}x&= 1. \tag{2}\\ \end{eqnarray}

Combining the two, we get:

\begin{eqnarray} \int \sum_s |\psi_{sx}\rangle \langle \psi_{sx}|\textrm{d}x &= 1.\tag{3} \end{eqnarray}

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