Is the ground state a Schrödinger cat state?

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?

• 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. – Harry Levine Feb 3 at 22:28
• 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. – AndreaPaco Feb 7 at 21:08