Lowest eigenstate of hopping matrix

  • Thread starter Thread starter Fightfish
  • Start date Start date
  • Tags Tags
    Eigenstate Matrix
Fightfish
Messages
953
Reaction score
118
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
<br /> H_{i,i+1}^{(n)} = H_{i+1,i}^{(n)} = - \sqrt{i}\sqrt{n-i+1},<br />
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:
<br /> H^{(2)} =<br /> \left(<br /> \begin{array}{ccc}<br /> 0 &amp; -\sqrt{2} &amp; 0 \\<br /> -\sqrt{2} &amp; 0 &amp; -\sqrt{2} \\<br /> 0 &amp; -\sqrt{2} &amp; 0 \\<br /> \end{array}<br /> \right)<br /><br /> H^{(3)} =<br /> \left(<br /> \begin{array}{cccc}<br /> 0 &amp; -\sqrt{3} &amp; 0 &amp; 0 \\<br /> -\sqrt{3} &amp; 0 &amp; -2 &amp; 0 \\<br /> 0 &amp; -2 &amp; 0 &amp; -\sqrt{3} \\<br /> 0 &amp; 0 &amp; -\sqrt{3} &amp; 0 \\<br /> \end{array}<br /> \right)<br /><br /> H^{(4)} =<br /> \left(<br /> \begin{array}{ccccc}<br /> 0 &amp; -2 &amp; 0 &amp; 0 &amp; 0 \\<br /> -2 &amp; 0 &amp; -\sqrt{6} &amp; 0 &amp; 0 \\<br /> 0 &amp; -\sqrt{6} &amp; 0 &amp; -\sqrt{6} &amp; 0 \\<br /> 0 &amp; 0 &amp; -\sqrt{6} &amp; 0 &amp; -2 \\<br /> 0 &amp; 0 &amp; 0 &amp; -2 &amp; 0 \\<br /> \end{array}<br /> \right)<br /> 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?
 
Physics news on Phys.org
Those entries look like Pascal's triangle (with an additional square root applied).
I don't have a proof, but the numbers all match.
 
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.
 
Do you have the corresponding eigenvectors? It might be easier to prove a pattern based on those.
 
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).
 
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.
 
I asked online questions about Proposition 2.1.1: The answer I got is the following: I have some questions about the answer I got. When the person answering says: ##1.## Is the map ##\mathfrak{q}\mapsto \mathfrak{q} A _\mathfrak{p}## from ##A\setminus \mathfrak{p}\to A_\mathfrak{p}##? But I don't understand what the author meant for the rest of the sentence in mathematical notation: ##2.## In the next statement where the author says: How is ##A\to...
The following are taken from the two sources, 1) from this online page and the book An Introduction to Module Theory by: Ibrahim Assem, Flavio U. Coelho. In the Abelian Categories chapter in the module theory text on page 157, right after presenting IV.2.21 Definition, the authors states "Image and coimage may or may not exist, but if they do, then they are unique up to isomorphism (because so are kernels and cokernels). Also in the reference url page above, the authors present two...
When decomposing a representation ##\rho## of a finite group ##G## into irreducible representations, we can find the number of times the representation contains a particular irrep ##\rho_0## through the character inner product $$ \langle \chi, \chi_0\rangle = \frac{1}{|G|} \sum_{g\in G} \chi(g) \chi_0(g)^*$$ where ##\chi## and ##\chi_0## are the characters of ##\rho## and ##\rho_0##, respectively. Since all group elements in the same conjugacy class have the same characters, this may be...
Back
Top