Multiple fermions in 1D box

Suppose I have $N$ fermions in an 1D infinite potential well. At $T=0K$, the fermions will occupy the lowest $N$ states and the wave function looks like:

$$|\phi_0\rangle= \frac{1}{\sqrt{N!}}\begin{vmatrix} \psi_1^1 & \psi_2^1 & \psi_3^1 & \cdots & \psi_N^1 \\ \psi_1^2 & \psi_2^2 & \psi_3^2 & \cdots & \psi_N^2 \\ \psi_1^3 & \psi_2^3 & \psi_3^3 & \cdots & \psi_N^3 \\ \cdots & \cdots & \cdots & \cdots & \cdots \\ \psi_1^N & \psi_2^N & \psi_3^N & \cdots & \psi_N^N \\\end{vmatrix}$$

Now if I increase the temperature, there will be some occupation at $|\psi_{N+1}\rangle$ and some $|\psi_A\rangle$ will be empty. However, for particle in 1D box, we know that the wave functions are orthogonal, i.e. $$\langle\psi_A|\psi_{N+1}\rangle=0$$

Then what is it which is responsible for the transition of a fermion from $|\psi_A\rangle$ to $|\psi_{N+1}\rangle$?

• what is it which is responsible for the transition of a fermion - temperature? Or if it's not the answer you are looking for, please clarify your question. May 18, 2017 at 5:14
• If your fermions are in a thermal state, they cannot be described by a wavefunction - they are in a mixed state, normally described by a density matrix, and typically given as a thermal quantum state. May 19, 2017 at 20:06

There will be no transition between these states as long as they are eigenvectors of the Hamiltonian, i.e. as long as you treat the system as isolated. In statistical physics, the temperature of a system is imposed by the coupling with a large thermal bath. Your fermions in a box are necessarily coupled to the electromagnetic field (or to phonons for example in a solid). The electromagnetic field can play the role of a thermal bath. By increasing the temperature, you increase the average density of photons. Then, by treating perturbatively the coupling between fermions and the electromagnetic field, one can show that electrons will undergo transitions between the unperturbed states $|\ \psi_n>$ by emitting or absorbing photons.