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Do holes (as in the absence of an electron) have wavefunctions?

In my understanding, when we talk about holes, we are implicitly invoking two multiparticle wavefunctions: $$\tag{1} \Psi(x_1,...,x_N)= \left| \begin{matrix} \psi_1(x_1) & ... & \psi_N(x_1) \\ \vdots & & \vdots \\ \psi_1(x_N) & ... & \psi_N(x_N) \end{matrix} \right|$$ and $$\tag{1} \Phi(x_1,...,x_{N-1})= \left| \begin{matrix} \psi_1(x_1) & ... & \psi_{N-1}(x_1) \\ \vdots & & \vdots \\ \psi_1(x_{N-1}) & ... & \psi_{N-1}(x_{N-1}) \end{matrix} \right|$$ Then we can say that $\Phi$ is $\Psi$ with an additional hole at a single-particle orbital $N$. (I am ignoring the fact that not all multiparticle states can be written as a Slater determinant.)

I think I've heard people talk about "hole wavefunctions." How do we define a hole wavefunction? In fact, can we even define a single particle wavefunction in a multiparticle system? If it exists, does the hole wavefunction need to be antisymmetrized with electron states (i.e., for an exciton, writing a Slater determinant for the hole and the electron)?

As a side note, I could also ask similar questions about positrons.

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You have everything pretty much correct. If you have a piece of semiconductor with $10^{18}$ electrons, a full valence band would be $$\tag{3} \Psi(x_1,...,x_{10^{18}})= \left| \begin{matrix} \psi_1(x_1) & ... & \psi_{10^{18}}(x_1) \\ \vdots & & \vdots \\ \psi_1(x_{10^{18}}) & ... & \psi_{10^{18}}(x_{10^{18}}) \end{matrix} \right|$$ and then a valence band with five holes in it would be $$\tag{4} \Phi(x_1,...,x_{10^{18}-5})= \left| \begin{matrix} \psi_1(x_1) & ... & \psi_{10^{18}-5}(x_1) \\ \vdots & & \vdots \\ \psi_1(x_{10^{18}-5}) & ... & \psi_{10^{18}-5}(x_{10^{18}-5}) \end{matrix} \right|$$

Those five holes would have the wavefunctions $\psi_{10^{18}-4}(x)$, $\psi_{10^{18}-3}(x)$, $\psi_{10^{18}-2}(x)$, $\psi_{10^{18}-1}(x)$, $\psi_{10^{18}}(x)$.

So a single-particle wavefunction in a multiparticle system is just one of the entries of the Slater determinant.

Most metals and semiconductors can be described in the "single-particle approximation", i.e. there is at least one way to write the Slater determinant so that to a very good approximation you can treat each entry $\psi_i$ as a separate particle, behaving like you would expect a typical particle to behave, and only weakly interacting with the other particles (described by the other $\psi_j$). That is the situation in which people are usually talking about single-electron or single-hole wavefunctions.

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  • $\begingroup$ Ok, makes sense. So if you have an electron hole excitation (e.g. we shine light to promote an electron to the conduction band), do we need to write a Slater determinant with the electron and hole wavefunctions in it? $\endgroup$
    – ChickenGod
    Jan 30, 2014 at 6:42
  • $\begingroup$ An electron-hole excitation is another way to say "move an electron from the valence band to the conduction band". So you would replace one of the valence-band wavefunctions with a conduction-band wavefunction in the Slater determinant. $\endgroup$ Jan 30, 2014 at 13:24
  • $\begingroup$ Thanks for the reply. I guess what I meant was that, is it valid to work with a 2x2 Slater determinant with the electron and hole wavefunction rather than the giant NxN many body Slater determinant? Of course, we could bypass this whole discussion by just working with creation and annihilation operators, but lets not talk about them. $\endgroup$
    – ChickenGod
    Jan 30, 2014 at 21:04
  • $\begingroup$ Normally people bypass the whole discussion by making the single-particle approximation, i.e. not talking about Slater determinants in the first place, but instead treating each electron or hole as an independent particle. Many-particle-effects are taken into account semi-heuristically (if you need higher accuracy) by "exchange forces" and "pauli blocking" etc. etc. (You asked about Slater determinants, so I answered, but this whole topic is not the usual approach.) $\endgroup$ Jan 30, 2014 at 21:55

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