I'm studying optical lattice, and I have a question with respect to the Bose-Hubbard model. The situation is as follows: two identical spinless bosons are in a double-well system.

For your convenience, my reference is the book "From Atoms Optics to Quantum Simulation: Interacting Bosons and Fermions in Three-Dimensional Optical Lattice Potentials", P 67.

One can write the Hubbard Hamiltonian as

$\hat{H}_{\rm{BH}}^{(2)} = -J \left( \hat{a}_1^{\dagger} \hat{a}_2 + \hat{a}_2 ^{\dagger} \hat{a}_1 \right) + \dfrac {U}{2} \left[ \hat{n}_1(\hat{n}_1 - 1 ) + \hat{n}_2 (\hat{n}_2 -1 ) \right]. $

Using the basis set $ \{ |2,\,0\rangle,\,|1,\,1\rangle,\,|0,\,2\rangle \} $ and diagonalizing the Hamiltonian, the ground state is

$|\psi_g\rangle \sim |2,\,0\rangle +|0,\,2\rangle + \left( \dfrac{U}{2\sqrt{2}J} +\sqrt{2+\left(\dfrac{U}{2\sqrt{2}J}\right)^2 } \right)|1,\,1\rangle. $

One can consider the following three cases: (1) $U/J \rightarrow \infty$, (2) $U/J =0$, (3) $U/J \rightarrow -\infty$.

Here is my question; when I calculated the number fluctuation in one site of the double-well $(\langle \Delta \hat{n}_1^2\rangle)$, it was the largest in the case (3), strong attractive interaction. However, in the textbook, the most delocalized state is (2), no interaction and it has the biggest number fluctuation. (Since there is no detailed calculation in the textbook I had no choice but to calculate it myself) I wonder which explanation is correct.

I'm not sure whether it is accurate but according to my calculation, the number fluctuation is larger in case (3) than (2)... Also, intuitively, I think if there is no interaction the particles should be at least slightly localized in each well. But in the strong attractive interaction limit (3), it is natural that the particles are delocalized as far as I know.

Thank you in advance for your careful reply.

  • 1
    $\begingroup$ Number fluctuation is not the same thing as delocalization. How is the latter defined in the book? $\endgroup$
    – Adam
    May 10, 2021 at 16:38
  • $\begingroup$ @Adam I solved the question. Thank you for your answer. As you mentioned, number fluctuation may not represent delocalization directly. Actually in the textbook, delocalization is not defined clearly. $\endgroup$
    – universe
    May 11, 2021 at 5:58
  • $\begingroup$ Something is not clear in the question. Is the textbook saying that case (2) has the largest number fluctuation, or only that it is the most delocalized ? $\endgroup$
    – Adam
    May 11, 2021 at 6:41
  • $\begingroup$ @Adam The textbook says the case (2) is the largest number fluctuation and the most delocalized. $\endgroup$
    – universe
    May 13, 2021 at 15:44

1 Answer 1


You have calculated delocalization in terms of the number fluctuation in one site of the well. I would calculate delocalization using number fluctuations between the two wells; i.e., I would calculate the variance of $\hat{n}=\hat{n}_1-\hat{n}_2$.

One can readily calculate that the state $\propto|2,0\rangle+|0,2\rangle$ has the largest variance of $\hat{n}$ for a 2-boson state. The same will hold true for an $N$-boson state $\propto|N,0\rangle+|0,N\rangle$, which has the largest variance for a given value of $\langle\hat{n}_1+\hat{n}_2\rangle=N$.

Edit: We can also determine which of your states has the largest variance of $\hat{n}$. Defining $\Lambda=\frac{U}{2\sqrt{2}J}$, we normalize the state to write $$|\psi\rangle=\frac{|2,0\rangle+|0,2\rangle+\left(\Lambda+\sqrt{2+\Lambda^2}\right)|1,1\rangle}{\sqrt{\left(\sqrt{\Lambda ^2+2}+\Lambda \right)^2+2}}.$$ The expectation value of $\hat{n}$ vanishes, and we can readily calculate $$\langle \hat{n}^2\rangle=\frac{4}{\left(\sqrt{\Lambda ^2+2}+\Lambda \right)^2+2}+\frac{4}{\left(\sqrt{\Lambda ^2+2}+\Lambda \right)^2+2}+0=2-\frac{2 \Lambda }{\sqrt{\Lambda ^2+2}}.$$ This is maximized by $\Lambda\to-\infty$, as you found.

Other options: We can consider other measures of delocalization. One thing to notice about the state considered maximally localized by the textbook is that it is separable in a first-quantized (ie, particle) description of the state: $$\frac{1}{2}\left(|2\rangle_L \otimes|0\rangle_R+|0\rangle_L \otimes|2\rangle_R+\sqrt{2}|1\rangle_L \otimes|1\rangle_R\right)=\frac{|L\rangle+|R\rangle}{\sqrt{2}}\otimes \frac{|L\rangle+|R\rangle}{\sqrt{2}},$$ where $L$ and $R$ refer to the left- and right-hand well, respectively. In this state, each individual particle is maximally delocalized between the two wells. Perhaps this is what the book intended?

Another standard measure of delocalization is using the inverse participation ratio $\sum_k |\psi_k|^4$. The basis in which we evaluate this is crucial, with smaller values of inverse participation ratio implying more delocalized states. If we consider the $\psi_k$ as coefficients of your state in the Fock basis as you have presented it, including normalization, we find the IPR to be $$\frac{1}{4} \left(\frac{\Lambda }{\sqrt{\Lambda ^2+2}}-\frac{3}{\Lambda ^2+2}+3\right).$$ Then maximal delocalization seems to happen at $\Lambda=-1/2$, and we indeed find that case (2) seems more delocalized than (1) or (3).

In all, the way we define delocalization has a significant impact on the resulting most delocalized state.

Further edit: Another possible measure of delocalization that the book may be using is maximizing the expectation value of $\hat{a}_1 \hat{a}_2^\dagger + \hat{a}_1^\dagger \hat{a}_2$, which is often written using the Schwinger mapping as the $x$ component of an angular momentum operator, $\hat{J}_x$. Then the state taken in the (2) limit will achieve the maximum delocalization.

In general, (2) means that each boson acts independently without interactions. For a single boson, the ground state is an equal superposition of being in each well, which could be considered the most delocalized state for that single boson. It follows that a collection of noninteracting bosons will have each boson occupy that same ground state, so this is likely what the book means is the most delocalized: each boson is independently as delocalized as possible. This is the same thing as being in the maximal eigenstate of $\hat{J}_x$.

  • 1
    $\begingroup$ I appreciate your kind reply. However, even though you indicate delocalization by the variance of n, it draws a conclusion that the case (3) is maximally delocalized as you mentioned. But the textbook describes the case (2) is maximally delocalized. So my conclusion might be that number fluctuation cannot fully indicate delocalization... $\endgroup$
    – universe
    May 11, 2021 at 6:15
  • $\begingroup$ Good point. Perhaps one of the other forms of delocalization that I've mentioned are more in line with what your textbook wanted? $\endgroup$ May 11, 2021 at 14:06
  • $\begingroup$ Or the next one I added? $\endgroup$ May 12, 2021 at 13:04
  • $\begingroup$ Your all descriptions about delocalization are convincing! Although they are too difficult for me :), I understand the rough calculation and central points you're suggesting. As you wrote, delocalization may not be defined in a specific way. $\endgroup$
    – universe
    May 13, 2021 at 15:58

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