# What's the completeness relation of Bloch band?

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.

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 $$\begin{eqnarray} \sum_n \int dk |\Psi_{nk}\rangle\langle\Psi_{nk}|=1\tag{6} \end{eqnarray}$$

• +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. Commented Jul 10, 2022 at 2:15

"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:

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