# Lowest eigenstate of hopping matrix

1. Jan 30, 2016

### Fightfish

So, I was examining the ground state of a Bose-Hubbard dimer in the negligible interaction limit, which essentially amounts to constructing and diagonalizing a two-site hopping matrix that has the form
$$H_{i,i+1}^{(n)} = H_{i+1,i}^{(n)} = - \sqrt{i}\sqrt{n-i+1},$$
with all other elements zero. The superscript $n$ refers to the fixed number of particles present on the dimer, and the dimension of the matrix is given by $n+1$.

Essentially this gives rise to a hollow centrosymmetric tridiagonal matrix. Explicitly, we have:
$$H^{(2)} = \left( \begin{array}{ccc} 0 & -\sqrt{2} & 0 \\ -\sqrt{2} & 0 & -\sqrt{2} \\ 0 & -\sqrt{2} & 0 \\ \end{array} \right)$$$$H^{(3)} = \left( \begin{array}{cccc} 0 & -\sqrt{3} & 0 & 0 \\ -\sqrt{3} & 0 & -2 & 0 \\ 0 & -2 & 0 & -\sqrt{3} \\ 0 & 0 & -\sqrt{3} & 0 \\ \end{array} \right)$$$$H^{(4)} = \left( \begin{array}{ccccc} 0 & -2 & 0 & 0 & 0 \\ -2 & 0 & -\sqrt{6} & 0 & 0 \\ 0 & -\sqrt{6} & 0 & -\sqrt{6} & 0 \\ 0 & 0 & -\sqrt{6} & 0 & -2 \\ 0 & 0 & 0 & -2 & 0 \\ \end{array} \right)$$ and so on.

In examining the unnormalized eigenstate with the lowest (most negative) eigenvalue, there seems to exist a Pascal-triangle-like sequence:
$$|\psi_{g}^{(1)}\rangle= [1,1]$$$$|\psi_{g}^{(2)}\rangle= [1,\sqrt{2},1]$$$$|\psi_{g}^{(2)}\rangle= [1,\sqrt{3},\sqrt{3},1]$$$$|\psi_{g}^{(3)}\rangle= [1,\sqrt{4},\sqrt{6},\sqrt{4},1]$$$$|\psi_{g}^{(4)}\rangle= [1,\sqrt{5},\sqrt{10},\sqrt{10},\sqrt{5},1]$$
This is highly suggestive that some sort of recurrence relation or mapping to binomial expansion exists; however thus far I have not been successful in trying to extract it. Might some one be able to shed some light on this?

2. Jan 30, 2016

### Staff: Mentor

Those entries look like Pascal's triangle (with an additional square root applied).
I don't have a proof, but the numbers all match.

3. Jan 30, 2016

### Fightfish

Yes, I meant to say Pascal's triangle, but somehow ended up writing Fibonacci lol ><
Anyway, I've edited to original post to fix that error.

4. Jan 30, 2016

### Staff: Mentor

Do you have the corresponding eigenvectors? It might be easier to prove a pattern based on those.

5. Jan 30, 2016

### Fightfish

Probably I kinda presented it in a slightly confusing manner, but the entities that follow the Pascal-triangle pattern are actually the unnormalized coefficients of the eigenvector with the lowest eigenvalue for differing various of $(n)$.

6. Jan 30, 2016

### Staff: Mentor

Oh sorry, I misread your description. Then the proof should not need magic. The i'th component of $H^{(n)} |\psi_{g}^{(n)}\rangle$ is

$$-\sqrt{i-1}\sqrt{n-i+2} \sqrt{{n+1} \choose{ i-1}} + -\sqrt{i}\sqrt{n-i+1} \sqrt{{n+1} \choose {i+1}}$$
and this should be a constant multiple of
$$\sqrt{{n+1} \choose i}$$
plus minus some index errors I made.