Tuesday, September 26, 2017

quantum mechanics - Do holes have wavefunctions?


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: Ψ(x1,...,xN)=|ψ1(x1)...ψN(x1)ψ1(xN)...ψN(xN)|

and Φ(x1,...,xN1)=|ψ1(x1)...ψN1(x1)ψ1(xN1)...ψN1(xN1)|
Then we can say that Φ is Ψ 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.



Answer



You have everything pretty much correct. If you have a piece of semiconductor with 1018 electrons, a full valence band would be Ψ(x1,...,x1018)=|ψ1(x1)...ψ1018(x1)ψ1(x1018)...ψ1018(x1018)|

and then a valence band with five holes in it would be Φ(x1,...,x10185)=|ψ1(x1)...ψ10185(x1)ψ1(x10185)...ψ10185(x10185)|


Those five holes would have the wavefunctions ψ10184(x), ψ10183(x), ψ10182(x), ψ10181(x), ψ1018(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 ψi 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 ψj). That is the situation in which people are usually talking about single-electron or single-hole wavefunctions.


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