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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?

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    $\begingroup$ Of which hole do you speak? $\endgroup$
    – Gert
    Feb 23 '21 at 12:40
  • $\begingroup$ @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. $\endgroup$
    – Thomas
    Feb 23 '21 at 13:20
  • $\begingroup$ @Thomas An electron travelling in one direction is not the same as a hole travellling in the opposite direction. $\endgroup$
    – my2cts
    Feb 23 '21 at 14:10
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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.

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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 $0\leq \rho\leq 1$.

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  • $\begingroup$ 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. $\endgroup$
    – Thomas
    Feb 23 '21 at 13:22

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