# Treating acceleration of charged particle in quantum mechanics

I can't find any good resources online that explain how we can derive the effect of a uniform electric field on a free electron using the quantum mechanics Hamiltonian formalism.

In our condensed matter notes, it explains that the effect of an electric field on the occupied electron states in an unfilled band is to 'tilt' it towards electron k states in the positive x-direction but I don't understand how this can be proved. Firslty it seems like the electric field would change the the band (since it would affect the energies) rather than just tilt it, also using this classical argument I would expect constant acceleration to higher and higher k states in the positive x-direction.

Therefore I wanted to go back to basics to understand exactly how electric field causing an 'current' (or acceleration of a charged particle) is treated in QM.

How does this work in QM?

• Have you considered the usual single-particle nonrelativistic model with Hamiltonian $H=p^2/2m+V(x)$ where $\nabla V=\vec E$? Jun 9, 2021 at 0:17
• This is the Hamiltonian I was thinking of yes, but I guess I would be hoping to see the perturbative approach maybe where it showed that the k-states got 'tilted' like in my question Jun 9, 2021 at 9:33

For simplicity, let's consider a one-dimensional system without any scattering process. The Hamiltonian can be written as $$\hat{{\mathcal H}} = \hat{{\mathcal H}_0} - eE\hat{x}~~(e < 0).$$ Here, the electric field is assumed to be of magnitude $$E~$$ in the $$x$$ direction.
The translation operator $$\hat{T}$$ satisfies $$\hat{T} f(x) = f(x + a),$$ for any function $$f$$ of $$x$$, where $$a$$ is the lattice constant. Using the Heisenberg equation of motion, we obtain $$\frac{\partial \hat{T}}{\partial t} = \frac{1}{i\hbar}[T, {\mathcal H}] = \frac{i}{\hbar}eEa\hat{T}.$$ On the other hand, the expected value of $$\hat{T}$$ for the state with a wavenumber $$k$$ is $$\langle \hat{T}\rangle = {\mathrm e}^{ika}$$. Therefore, $$\left\langle \frac{\partial \hat{T}}{\partial t} \right\rangle = i\frac{\partial k}{\partial t}a {\mathrm e}^{ika} = i\frac{\partial k}{\partial t}a\langle \hat{T}\rangle.$$ With these two equations, we get $$\hbar\frac{\partial k}{\partial t} = eE.$$ This is the equation of motion of the Bloch electron. From this equation, we know that electrons are accelerated by the electric field.