I'm working with the standard Boson Hubbard model. It's Hamiltonian is defined in Fock space and commutes with total particle number N. $$[{{\hat H}_{BH}},\hat N] = 0$$

So I can simultaneously diagnonalize ${{\hat H}_{BH}}$ and ${\hat N}$. And in this process the ground state of ${{\hat H}_{BH}}$ I get will have definite total particle number.

Then for my ground state $$\left\langle {{\psi _g}} \, \middle| \,\hat b \,\middle| \, {{\psi _g}} \right\rangle = 0$$ will always holds because $$\hat b\left| {{\psi _g}} \right\rangle $$ has different total particle number thus is orthogonal to $\left| {{\psi _g}} \right\rangle $. But $\left\langle {\hat b} \right\rangle $ is the order parameter for superfluid phase. I deduce that Boson Hubbard model never has a superfluid phase, which is certainly wrong. But I don't understand what's wrong with my arguments. I'm confusing these days. Help please.

  • 2
    $\begingroup$ The fact that $[H,N]=0$ implies that there exist shared eigenstates between $H$ and $N$, not that all eigenstates of $H$ must be shared with $N$, so on that level there is no contradiction. But, on the other hand, the contention of the physical relevance of states that do not have a well-defined particle number, i.e. the possibility of coherence in superpositions of states of different $N$, was controversial for a pretty long time and its resolution (if it was achieved? I'm not sure) is definitely nontrivial. $\endgroup$ Commented Jun 13, 2018 at 10:29
  • $\begingroup$ The references in this answer might be helpful in that regard. $\endgroup$ Commented Jun 13, 2018 at 10:50
  • $\begingroup$ This question discussed the similar problem for the BCS groundstate: physics.stackexchange.com/questions/284314/… . $\endgroup$ Commented Jun 13, 2018 at 13:26
  • $\begingroup$ May I ask a little further? Yes the eigenstates can be not unique. But by such a simultaneouse diagonalization the ground state I find IS a ground state isn't it? Otherwise I can't diagonalize my Hamiltonian. So is there some limitations that I have to find the ground state in some specific way instead of by any means to get superfluid state? Why is diagonalization not free?😌 $\endgroup$
    – WSnow
    Commented Jun 14, 2018 at 16:09


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