The Superfluid wavefunction in Bose Hubbard model can be written as $|\Psi_{SF}\rangle=\frac{1}{\sqrt{N!}}(b_{k=0}^{\dagger})^N|0\rangle$ in the U=0 limit, whereU is the onsite interaction, N is the total particle number, k=0 denotes the lowest energy state and $|0\rangle$is the vacuum state. By Fourier transform, $b_{k=0}^{\dagger}=\frac{1}{\sqrt{M}}\sum\limits_{i=1}^Mb_i^{\dagger}$, where M is the lattice sites number. So $|\Psi_{SF}\rangle=\frac{1}{\sqrt{N!}}(\frac{1}{\sqrt{M}}\sum\limits_{i=1}^Mb_i^{\dagger})^N|0\rangle$. How to understand in the U=0 limit, the single site many-body wave function of is equivalent to a local coherent state, or say $\vert\Psi_{SF}\rangle \propto \prod\limits_i \exp(\sqrt{\frac{N}{M}}b_i^{\dagger})|0\rangle_i $?

  • $\begingroup$ See the same question here. $\endgroup$
    – WoofDoggy
    Oct 27 '17 at 21:13

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