Is the ground state a Schrödinger cat state?

Question

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?

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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?

Answer 1

Question 1: If $T = 0$, then the Hamiltonian is diagonal in the Fock state basis. If $U < 0$ is attractive, then the two Fock states with lowest energy are $|0, N\rangle$ and $|N, 0\rangle$. The ground state subspace is therefore a two-dimensional Hilbert space spanned by these two states. Any linear combination $|E_0\rangle = \alpha |0, N\rangle + \beta |N, 0\rangle$ is a ground state of the system, exactly as you wrote.

Question 2: What happens with weak tunnel coupling? Let's think about a simple example, with $N = 2$.$|0, 2\rangle$ and $|2, 0\rangle$ are degenerate with $T=0$, and for $T > 0$ they are both off-resonantly coupled to $|1, 1\rangle$. If $T$ is very small, then the population in $|1,1\rangle$ can be neglected and $|0,2 \rangle$ is effectively resonantly coupled to $|2, 0\rangle$ through a (weak) second-order process. The ground-state manifold, which previously was spanned by two degenerate states $|0, 2\rangle$ and $|2, 0\rangle$, is now perturbed by this weak resonant coupling which creates new eigenstates $(|0, 2\rangle \pm |2, 0\rangle)/\sqrt{2}$. The splitting between these two eigenstates will be given by the strength of this second order coupling.

For $N > 2$, the same principle applies, but now the resonant coupling between $|0, N\rangle$ and $|N, 0\rangle$ is determined by an $N$-th order process, so its strength will be suppressed accordingly. However, the ground state manifold will still be diagonalized as $(|0, N\rangle \pm |N, 0\rangle)/\sqrt{2}$, with increasingly small splitting for large $N$. Therefore technically there will be a well-defined unique ground state, which will be a superposition of $|0, N\rangle$ and $|N, 0\rangle$.

Side note: This same principle applies in many examples corresponding to a quantum phase transition into a symmetry-broken phase. For example, consider a quantum Ising model, $H = J \sum_i \sigma^{(i)}_x + V\sum_{i} \sigma_z^{(i)}\sigma_z^{(i+1)}$. Let's take $V < 0$, so the ferromagnetic case. If $J = 0$, then there are two degenerate ground states: $|\downarrow\rangle^{\otimes N}$ and $|\uparrow\rangle^{\otimes N}$. With small coupling $J > 0$, then the same principle applies: these two states are resonantly coupled by an $N$-th order off-resonant process, so technically they diagonalize into a symmetric and antisymmetric superposition, split by a very small energy.