Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Eigenvalue of the coherent state

  1. Feb 4, 2010 #1
    Hi all,
    the annihilation operator satisfies the equation [tex]\hat{a}[/tex]|n>=[tex]\sqrt{n}[/tex]|n-1> and [tex]\hat{a}[/tex]|0>=0
    so the matrix of [tex]\hat{a}[/tex] should be
    http://www.tuchuan.com/a/2010020418032158925.jpg [Broken]
    and zero is the only eigenvalue of this matrix.

    The coherent state is defined by [tex]\hat{a}[/tex]|[tex]\alpha[/tex]>=a|[tex]\alpha[/tex]>, yet [tex]\alpha[/tex]are not always equal to zero
    Is there anything I forgot to consider?:confused:
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Feb 4, 2010 #2

    Demystifier

    User Avatar
    Science Advisor

    I cannot see the attached matrix, but zero is not the only eigenvalue of the operator a.
     
  4. Feb 4, 2010 #3
    my picture cannot be displayed?
    try this weblink: http://www.tuchuan.com/a/2010020418032158925.jpg" [Broken]
     
    Last edited by a moderator: May 4, 2017
  5. Feb 4, 2010 #4
    Yes, the matrix is infinite dimensional.
     
  6. Feb 4, 2010 #5
    So solving the eigenvalue of matrix in infinite dimension is not the same with the process to solve the eigenvalue of matrix in finite dimension?
    I know the determinant of a-[tex]\alpha[/tex]I is [tex]\alpha[/tex][tex]^{n}[/tex] when the matrix is finite.
    How about the determinant when it is infinite? I am not very familiar with it.:blushing:
    Thanks!
     
  7. Feb 4, 2010 #6

    DrDu

    User Avatar
    Science Advisor

    You don't need the determinant! Just write down the lower right corner of your matrix eigenvalue equation: setting x1=1 you get
    [tex]\sqrt{2}x_2-\alpha x_1=0[/tex], [tex]\sqrt{3}x_3-\alpha x_2=0[/tex], etc. you find that [tex]x_i=\alpha^{(i-1)}/\sqrt{i!}[/tex] for arbitrary alpha!
     
  8. Feb 5, 2010 #7

    Demystifier

    User Avatar
    Science Advisor

    Exactly.
    By the way, for fermions the corresponding matrix is finite dimensional (only n=0 and n=1 contribute), which is why in field theory coherent states do not exist for fermions but only for bosons.
     
  9. Feb 5, 2010 #8
    That doesn't stop people using fermionic coherent states however. If you allow the eigenvalues to be anti-commuting numbers/grassmann numbers it works just fine.
    (see, eg, section 4.1.2 of Altland & Simons)
     
  10. Feb 5, 2010 #9

    Demystifier

    User Avatar
    Science Advisor

    Interesting observation, thanks!
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook