Can the Schrödinger equation only give eigenvalues greater than the Fermi level in a semiconductor potential well?

Question

I'm an undergrad with no direct coursework dealing with solid state physics so please forgive misunderstandings.

Say you have a well of semiconducting material surrounded by an infinite potential barrier.

If there is no voltage applied, the valence band is full and the conduction band is empty. Solving the Schrödinger equation using $E_c = 0$ as a reference point and using the appropriate effective mass of an electron in the material produces available wavefunctions that are in the conduction band, but have essentially 0 probability of being occupied since the eigenvalues are above the Fermi level.

In some instances, I gather that you should be able to apply a voltage $V$ to the semiconductor to raise the Fermi level into the conduction band. In that case, that should also create a potential energy landscape $U = -V$, so shouldn't the solutions to the Schrodinger equation also be offset by that potential $U$, meaning that there are still no aailable wavefunctions which are thermally accessible?

Edit: let me try to rephrase because I don't think I was being clear. When you apply a gate bias voltage $V_{\text{gate}}$ to a semiconductor, you increase the potential inside the semiconductor by approximately $V_{\text{gate}}$. I've been assuming that this increases the Fermi level by $V_{\text{gate}}$ based on a paper I'm working with.

Essentially, the Schrodinger equation says that all states in a potential well have energy greater than the minimum potential in the well, but Fermi-Dirac says that those states are all unlikely to be occupied since the Fermi level is equal to the potential in the well. What am I missing?

Answer 1

Assuming what you have in mind is a field effect transistor (FET), the gate voltage causes bending of conduction and valence bands in the direction of the gate, perpendicular to the electron channel (see, for example, this figure: https://commons.wikimedia.org/wiki/File:Semiconductor_band-bending.png)

When the gate voltage is more than a threshold, there is band inversion, which means the conduction band bends below the Fermi level, allowing the conduction band to be populated by electrons upto the Fermi level. The electrons along this direction can be approximately modelled by fermions in an infinite potential well.

However, this is in the direction towards the gate. In the direction of the electron channel, which is perpendicular to the gate axis, the electrons are effectively a 2D electron gas, with parabolic dispersion. A source-drain voltage offsets the potential energy in this direction, causing the electrons to flow.

Also, there is no reason to suppose that the Fermi level is the bottom of the potential well. That is given by the conduction band energy $E_c$.