Is the complex conjugate of the amplitude of an electron wavefunction equivalent to the amplitude of the corresponding hole?

Is the complex conjugate of the amplitude of an electron wavefunction equivalent to the amplitude of the corresponding hole? Say I consider a wavefunction of an electron that has the amplitude A. If I now take the complex conjugate of A, A*, does this quantum mechanically correspond to the amplitude of the corresponding hole? If so, is there an intuitive explanation/picture for this?

• Of which hole do you speak?
– Gert
Feb 23 '21 at 12:40
• @Gert if the electron trvales through, e.g., a metal this can be in k-space also viewed as a hole travelling in the opposite direction, I think. It is this hole I mean. Feb 23 '21 at 13:20
• @Thomas An electron travelling in one direction is not the same as a hole travellling in the opposite direction. Feb 23 '21 at 14:10

For a state of definite energy, the Schrodinger equation says that the time dependence is supposed to be $$e^{-i(E/\hbar)t}$$, where $$E$$ is an eigenvalue of the Hamiltonian. For most Hamiltonians, it is not going to be true that $$-E$$ is also an eigenvalue, so the complex conjugate $$e^{i(E/\hbar)t}$$ will not be a solution of the Schrodinger equation. Therefore it has no physical interpretation in those cases.

But in condensed matter or nuclear physics, when we have a system of fermions and there can be particle-hole excitations, one way to think about holes is that the energy of a hole is minus the energy of the corresponding particle.

The answer by Gert seems to be misunderstanding what you're asking. Taking a complex conjugate is also something we do to take a given wavefunction to its dual (ket to a bra), which is how we form a probability measure. But there we're also talking about a whole different vector space.

As far as I know, the complex conjugate $$\Psi^*$$ of a wave function $$\Psi$$ is a consequence of the Born rule which relates the probability density function $$\rho(\mathbf{r})$$ of a Normalised wave function $$\Psi$$ to the wave function by:

$$\boxed{\rho(\mathbf{r})=\Psi(\mathbf{r})^*\Psi(\mathbf{r})=|\Psi|^2}$$

If say $$A$$ is a complex number (or function), then $$A^* A$$ is a Real Number, which of course is what you expect from a probability, which has to be $$0\leq \rho\leq 1$$.

• thank you for your answer. I saw a similar term in a paper, where the psi(r) was called the amplitude of an electron and psi(r)* the amplitude of a hole. That's what motivated my question. I don't really have an intuitive picture for that in my mind. Feb 23 '21 at 13:22