# Dipole matrix elements for Bloch wavefunctions?

Suppose we have a one-dimensional periodic system with lattice constant $$a_0$$. From Bloch's theorem, we can express the wavefunction for an electron in band $$m$$ with crystal momentum $$k$$ $$\left\langle x \middle| \psi_{m,k} \right\rangle$$ as follows:

$$\left\langle x \middle| \psi_{m,k} \right\rangle = e^{i k x} u_{m,k}(x),$$

where $$u_{m,k}(x + a_0) = u_{m,k}(x)$$. I don't understand the following expression for the matrix elements of the position operator:

$$\langle \psi_{m,k} | x | \psi_{m',k'}\rangle = i \delta_{m,m'} \delta_{k,k'} \frac{\partial}{\partial k} + i \delta_{k,k'} X_{m,m'},$$

where

$$X_{m,m'}= i N \int_0^{a_0} e^{i (k - k') x} u^*_{m,k}(x) \frac{\partial}{\partial k} u_{m',k'}(x) dx.$$

The second term is easy enough to understand. The first term, however... how can the expectation value between two eigenstates be a derivative? Am I missing something obvious here?

• Basically the same question as this. As far as I remember, the OP cross posted the question on matter modeling back then. Nov 4, 2022 at 15:25

I think that the correct expression involves $$\delta'(k-k')$$ rather than a partial derivative.

Suppose we construct a wave-packet with states taken from a single band with index $$n$$: $$\varphi({\bf r}) = \int_{Z^*} \frac{d^3k} {(2\pi)^3}a({\bf k}) \psi_{n,{\bf k}}({\bf r})= \int_{Z^*} \frac{d^3k} {(2\pi)^3}a({\bf k}) e^{i{\bf k}\cdot {\bf r}} u_{n,{\bf k}}({\bf r}).$$ Here $$Z^*$$ is the Brillouin zone.

The state is normalized if $$\int_{Z^*} \frac{d^3k} {(2\pi)^3} | a({\bf k})|^2 =1.$$ The spatially localized wave-packet is now in the domain of the unbounded operator $$\bf r$$ and
$$\langle{\psi_{n',{\bf k}}}|{r_\mu}|{\varphi}\rangle= i\left\{\frac {\partial}{\partial k_\mu} a({\bf k})\delta_{n'n} +\langle{u_{n'{\bf k}}}|{\partial_\mu u_{n,{\bf k}}}\rangle a({\bf k})\right\}.$$

• What's the difference? The derivative of the delta function is a distribution that returns the value of the function at the chosen $k$ -- which is identical to the action of a simple delta following a derivative. Nov 4, 2022 at 14:13
• @Emilio Pisanty. It's a bit of pedantry for sure, but the OP was confused by it. One expects a matrix element to be a (singular) numerical function rather than an opertor: $\langle x|\hat p| x'\rangle= -i\hbar \delta'(x-x')\ne -i\hbar \partial_x$ One needs an extra integral to convert one to the other. Nov 4, 2022 at 14:43
• I agree. It is a harmful bit of pedantry to say "I think that the correct expression is A rather than B" when the two are provably equal. The same is true between attempting to distinguish "sigular numerical function rather than an operator" when talking about $\delta'(k-k')$ versus $\delta(k-k')\partial_k$ when the two are provably equal (modulo signs). The fact that "extra integrals" are involved is completely immaterial. Nov 4, 2022 at 17:03

The second term is easy enough to understand. The first term, however... how can the expectation value between two eigenstates be a derivative? Am I missing something obvious here?

There is nothing special going on here $$-$$ this is a pretty common feature.

For contrast, consider the matrix element of the position operator between two (free-particle) momentum eigenstates: $$⟨p|\hat x|p'⟩ = i\hbar \frac{\partial}{\partial p} \delta(p-p'),$$ where you do need to take due care about where the derivative is acting on, but which is basically identical to your expression.

(Note that it is often easier to encapsulate all the complexity by "calculating" the derivative as $$\delta'(p-p')$$, as mike stone's answer does, but if you do this then it is essential to keep in mind that the derivative of the Dirac delta is a distribution and needs to be treated as such, i.e., not as a function, but as a functional.)

Something equivalent happens for the matrix element of momentum between two position eigenstates: $$⟨x|\hat p|x'⟩ = -i\hbar \frac{\partial}{\partial x} \delta(x-x').$$

In other words, the behaviour you're seeing there is generic.