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For $N$ spin-1/2 free fermions, the ground state is given by the Fermi sea,

$$|{\rm FS}\rangle = \prod_{|{\bf k}|<k_F} c_{{\bf k}, \uparrow}^\dagger c_{{\bf k}, \downarrow}^\dagger |0\rangle $$ while we can take Fourier transform of the fermion creation operator $$c^\dagger_{{\bf k},\sigma } = \sum_{{\bf r}_{\sigma}} {\rm e}^{-i{\bf k\cdot r}_\sigma} c^\dagger _{{\bf r}_\sigma,\sigma}$$

But how to put it in a compact form of creation operators of real space?

Machine
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  • The momentum space is much more appropriate to describe similar maths - the momentum space is a "part of the solution". It's important for the FS state that some states, those above kf, remain unoccupied. The "states above kf" is easy to be written in the energy eigenstate but they're complicated superpositions of position eigenstates. So you're effectively asking someone to write the solution without the most obvious step towards the solution. The result is guaranteed to look messy but why don't you just substitute the second formula to the first? It won't simplify. – Luboš Motl Apr 16 '13 at 13:22
  • Thanks for the comment. When it comes to metal-mott insulator transition, we need to distinguish sites with single and double occupations. Then we have to do that with real space configurations. I should have clarified this. – Machine Apr 17 '13 at 16:00

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