# Momentum operator in effective mass approximation

When we calculate the band structure of some solid then we often find that in the bottom of the conduction band the dispersion looks approximately quadratic with some new effective mass:

$$E(k) = \frac{\hbar^2}{2m^*} k^2 .\tag{1}$$

Now I am modelling electrons living in a crystal surface slab (i.e. bulk in 2 dimensions and confined in one dimension), where I solve for the quantized energy levels in the quantized direction. To do so I solve the free electron Hamiltonian:

$$H = -\frac{\hbar^2}{2m^*} \frac{d^2}{dx^2} \tag{2}$$

with $m^*$ being the effective mass in the crystal.

Now my question is: How can I mathematically justify the step of going from the quadratic dispersion in (1) to the operator form in (2)? I guess I should write up the Bloch wave and use that somehow. I hope my question makes sense.

You might then be worried that $m^*$ being infinite will destroy other things, but if you work out the derivation of effective mass in detail you actually see that the effective mass is actually a tensor:
$$M_{ij} = \left[ \frac{1}{\hbar^2} \frac{\partial^2 E(\vec{k})}{\partial k_i \partial k_j} \right]^{-1}$$ where $i,j = x,y$, or $z$, meaning that the mass can be (and frequently is) very different in different directions. You're lucky enough to be working in a highly symmetric geometry where your confined and deconfined directions are perpendicular and decoupled. That means you can take $m^*$ to denote the in-plane effective mass arising from the 2D band structure (which itself depends on crystal structure, etc.) and treat the out-of-plane direction as a "bare" (i.e. $m^*=m_e$) electron in whatever 1D confining quantum well structure you have.