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Lowest eigenstate of hopping matrix

  1. Jan 30, 2016 #1
    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?
     
  2. jcsd
  3. Jan 30, 2016 #2

    mfb

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    2016 Award

    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.
     
  4. Jan 30, 2016 #3
    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.
     
  5. Jan 30, 2016 #4

    mfb

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    2016 Award

    Staff: Mentor

    Do you have the corresponding eigenvectors? It might be easier to prove a pattern based on those.
     
  6. Jan 30, 2016 #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].
     
  7. Jan 30, 2016 #6

    mfb

    User Avatar
    2016 Award

    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.
     
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