Lowest eigenstate of hopping matrix

In summary, the conversation revolves around examining the ground state of a Bose-Hubbard dimer in the negligible interaction limit, which involves constructing and diagonalizing a two-site hopping matrix. The matrix has a hollow centrosymmetric tridiagonal form, resembling Pascal's triangle. The unnormalized coefficients of the eigenvector with the lowest eigenvalue also follow a similar pattern. There is a discussion about proving this pattern using the eigenvectors, and a correction is made to a mistake in the original post. The conversation ends with a potential formula for the pattern.
  • #1
Fightfish
954
117
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
[tex]
H_{i,i+1}^{(n)} = H_{i+1,i}^{(n)} = - \sqrt{i}\sqrt{n-i+1},
[/tex]
with all other elements zero. The superscript [itex]n[/itex] refers to the fixed number of particles present on the dimer, and the dimension of the matrix is given by [itex]n+1[/itex].

Essentially this gives rise to a hollow centrosymmetric tridiagonal matrix. Explicitly, we have:
[tex]
H^{(2)} =
\left(
\begin{array}{ccc}
0 & -\sqrt{2} & 0 \\
-\sqrt{2} & 0 & -\sqrt{2} \\
0 & -\sqrt{2} & 0 \\
\end{array}
\right)
[/tex][tex]
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)
[/tex][tex]
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)
[/tex] and so on.

In examining the unnormalized eigenstate with the lowest (most negative) eigenvalue, there seems to exist a Pascal-triangle-like sequence:
[tex]|\psi_{g}^{(1)}\rangle= [1,1][/tex][tex]|\psi_{g}^{(2)}\rangle= [1,\sqrt{2},1][/tex][tex]|\psi_{g}^{(2)}\rangle= [1,\sqrt{3},\sqrt{3},1][/tex][tex]|\psi_{g}^{(3)}\rangle= [1,\sqrt{4},\sqrt{6},\sqrt{4},1][/tex][tex]|\psi_{g}^{(4)}\rangle= [1,\sqrt{5},\sqrt{10},\sqrt{10},\sqrt{5},1][/tex]
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?
 
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  • #2
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
mfb said:
Those entries look like Pascal's triangle.
I don't have a proof, but the numbers all match.
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
Do you have the corresponding eigenvectors? It might be easier to prove a pattern based on those.
 
  • #5
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 [itex](n)[/itex].
 
  • #6
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.
 

1. What is the "lowest eigenstate" of a hopping matrix?

The lowest eigenstate of a hopping matrix refers to the lowest energy state of the system described by the matrix. It is the state with the lowest possible energy and is also known as the ground state.

2. How is the lowest eigenstate determined for a hopping matrix?

The lowest eigenstate is determined by solving the Schrödinger equation for the hopping matrix. This involves finding the eigenvector and eigenvalue that correspond to the lowest energy state.

3. What does the lowest eigenstate of a hopping matrix represent?

The lowest eigenstate of a hopping matrix represents the most stable configuration or arrangement of particles in the system. It is the state that the system will naturally tend towards in order to minimize its energy.

4. How does the lowest eigenstate of a hopping matrix relate to other eigenstates?

The lowest eigenstate is always the state with the lowest energy, but there may be multiple eigenstates with the same energy. These are known as degenerate states and can have different spatial arrangements of particles.

5. Can the lowest eigenstate of a hopping matrix change?

Yes, the lowest eigenstate can change if the parameters of the system, such as the strength of the hopping interactions, are changed. This can lead to a different ground state and affect the overall behavior of the system.

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