Consider the following Bose-Hubbard Hamiltonian which describes a Bose-Einstein condensate confined in a two-well potential: $$ H= -T(a_L^\dagger a_R + a_L a_R^\dagger ) + \frac{U}{2}(n_L^2+n_R^2-n_L-n_R) $$ The total number of atoms in the system is $N=n_L+n_R$ and represents a conserved quantity (L stands for left, R stands for right). Suppose that the onsite interaction $U$ is attractive, i.e. $U<0$, meaning that the condensate will tend to conglomerate in a well.

If the tunnelling $T$ is zero, the ground state should be degenerate and could be whatever linear cobination of the type:

$$ |E_0\rangle = \alpha |N,0\rangle + \beta|0,N\rangle $$ because the Hamiltonian $H$ is diagonal in the Fock-state basis and both $|N,0\rangle$ and $|0,N \rangle$ are minimum-energy states.

Question 1: is this correct?

Now, let's switch on the tunnelling. The Hamiltonian is no longer diagonal in the Fock-state basis. Each of the two minimum-energy configurations modifies as follows, according to perturbation theory: $$ |N,0\rangle \rightarrow |(1^*)\rangle := c_{11} |N,0\rangle + c_{12}|N-1, 1\rangle + c_{13}|N-2, 2\rangle + \dots $$ $$ |0,N\rangle \rightarrow |(2^*)\rangle := c_{21} |0, N \rangle + c_{22}|1, N-1\rangle + c_{23}|2,N-2 \rangle + \dots $$

Of course the square modulus of $c_{12}$, $c_{13}$, ... will be bigger for bigger tunnelling amplitudes $T$ because the presence of tunnelling favours delocalization.

I suspect that, when tunnelling $T$ is present, the ground state is indeed unique and has the following expression:

$$ |E_0\rangle = \frac{1}{\sqrt{2}}\left(|(1*)\rangle + |(2*)\rangle \right) $$ This ground state can therefore be seen as an equally-weighted superposition of two states, $|(1^*)\rangle$ and $|(2^*)\rangle$ which, in turn, are a superposition of Fock states whose wideness is bigger if $T$ is bigger.

Question 2: is this reasoning correct?

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    $\begingroup$ I think both statements are correct. This concept extends to other models as well. For example, the quantum Ising model displays a similar feature: the ground state (with finite but small transverse field) is a superposition of the two possible ferromagnetic (or antiferromagnetic) states, which is also a cat state. $\endgroup$ – Harry Levine Feb 3 at 22:28
  • $\begingroup$ Harry Levine, thanks for your comment. It would be nice if you could write an answer in order to expand what you’ve mentioned in your comment. $\endgroup$ – AndreaPaco Feb 7 at 21:08

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