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. $\endgroup$ Commented Feb 3, 2019 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
    Commented Feb 7, 2019 at 21:08

1 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.

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    $\begingroup$ The true ground state of the quantum Ising model isn't just $|\downarrow\rangle^{\otimes N}+|\uparrow\rangle^{\otimes N}$. You can easily see it's not an eigenstate of $H$ $\endgroup$
    – octonion
    Commented Aug 14, 2019 at 16:47
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    $\begingroup$ I see your point and could be more explicit about admixture of other states. The point about spontaneous symmetry breaking is a good one, and would apply also to the original question about attractive bosons in a double well. But it is interesting still that for finite systems, at least in principle, a well-defined ground state exists which is a Schrödinger cat type superposition. $\endgroup$ Commented Aug 14, 2019 at 23:30
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    $\begingroup$ If two degenerate states are coupled, this is considered a 'resonant' coupling. The eigenstates are symmetric and antisymmetric superpositions of the two coupled states. If, on the other hand, the two states have a big energy gap compared to the strength of the coupling, this is 'off-resonant', and leads to only perturbative shifts in the eigenstates. $\endgroup$ Commented Aug 22, 2019 at 15:06
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    $\begingroup$ If two state $|1\rangle$ and $|2\rangle$ are degenerate but are each coupled off-resonantly to an intermediate state $|3\rangle$, then $|1\rangle$ and $|2\rangle$ can be understood to be resonantly coupled in a 'second-order' process. Here I think of second-order just as meaning it is a two-step process to understand their coupling ($|1\rangle \to |3\rangle \to |2\rangle$). This terminology is also connected with second-order perturbation theory, which can help to understand the reduced strength of the effective coupling from $|1\rangle$ to $|2\rangle$. $\endgroup$ Commented Aug 22, 2019 at 15:31
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    $\begingroup$ When studying an attractive Bose-Hubbard system (or Ising system) with many particles, an experimentalist might prepare the ground state with $T > 0$ and then ramp down $T$ to 0. If done perfectly according to the Hamiltonian, this would prepare a cat-like state. However, the phase coherence of the two components of the cat state can rarely be observed, so the usual interpretation is that the system randomly collapses into one of the two states (breaking the symmetry of the cat state). (In some cases the phase can be preserved! -- this is an ongoing effort in quantum information science.) $\endgroup$ Commented Aug 22, 2019 at 15:48

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