Timeline for When is it necessary to use more than one Slater determinant to write the wave function of a quantum system?

8 events

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May 23, 2021 at 19:02	comment	added	mike stone		@Ruslan: Yes. Exactly Also there is some deep geometry here as the Plucker relations lie behind the $\tau$ functions in the theory of integrable PDE's due to Sato, Miwa, and Jimbo.
May 23, 2021 at 18:59	history	edited	mike stone	CC BY-SA 4.0	added 22 characters in body
May 23, 2021 at 18:56	comment	added	Ruslan		Oh, I see: we can have entangled states without interactions too. This won't be representable by a single Slater determinant. I was thinking of eigenstates, that's why I thought of interactions as the reason for entanglement.
May 23, 2021 at 18:48	history	edited	mike stone	CC BY-SA 4.0	added 236 characters in body
May 23, 2021 at 18:44	comment	added	mike stone		@Ruslan. No this is not the case. The issue is simple geometry: counting the dimension of the $n$-particle Hilbert space and the number of parameters in the $n$ factors. Interactions have nothing to do with it. In particular the set of n-by-n Slaters does not form a vector space.
May 23, 2021 at 18:42	history	edited	mike stone	CC BY-SA 4.0	added 255 characters in body
May 23, 2021 at 18:39	comment	added	Ruslan		The OP has explicitly restricted the system to contain non-interacting fermions. In this case single Slater determinant is an exact representation of wavefunctions.
May 23, 2021 at 18:34	history	answered	mike stone	CC BY-SA 4.0