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Eigenvalue problem

  1. Jan 23, 2008 #1
    1. The problem statement, all variables and given/known data

    Consider the following problem:

    if [tex]\ A [/tex][tex]\psi=[/tex][tex]\lambda[/tex][tex]\psi[/tex],prove that

    [tex]\ e ^ A [/tex][tex]\psi=[/tex][tex]\ e ^\lambda[/tex][tex]\psi[/tex]

    2. Relevant equations



    3. The attempt at a solution

    This is my attempt.Please check if I am correct.

    If
    [tex]\ e ^ A [/tex][tex]\psi=[/tex][tex]\ e ^\lambda[/tex][tex]\psi[/tex]

    is correct, we should have:

    [tex]\ ln [/tex][tex]\ e^ A [/tex][tex]\psi=[/tex][tex]\ ln [/tex][tex]\ e ^\lambda[/tex][tex]\psi[/tex]

    or, [tex]\ ln \{e^A} + [/tex][tex]\ ln \psi=[/tex][tex]\ ln \{e^\lambda} + [/tex][tex]\ ln \psi[/tex]

    Now cancel [tex]\ ln \psi[/tex] from both sides and post-multiply the resulting equation by [tex]\psi[/tex]

    That is---

    [tex]\ ln \{e^A} =[/tex][tex]\ ln \{e^\lambda}[/tex]

    or, [tex]\ [ln \{e ^ A}][/tex][tex]\psi=[/tex][tex]\ [ln \{e ^\lambda}][/tex][tex]\psi[/tex]

    Alternatively,

    [tex]\ A [/tex][tex]\psi=[/tex][tex]\lambda[/tex][tex]\psi[/tex]

    So, we got the given equation from the equation to be proved.
     
    Last edited: Jan 23, 2008
  2. jcsd
  3. Jan 23, 2008 #2
    The LaTeX image is not coming good.But I think everyone will understand my procedure.
     
  4. Jan 23, 2008 #3
    Do you know the definition of the matrix [itex]e^A[/itex]?
     
  5. Jan 23, 2008 #4
    I know the definition of e^A

    Let A,B and C be three matrices.

    If I have A+C=B+C

    what is wrong if I cancel C from both sides?

    ln (psi) is a matrix if you consider psi as matrix.I am usimg psi as a complex function like psi=M+iN
     
  6. Jan 23, 2008 #5
    I did according to your hint.That works well, but still I am having problem to see why the logaritm of a matrix/operator will not work.
     
  7. Jan 23, 2008 #6

    CompuChip

    User Avatar
    Science Advisor
    Homework Helper

    Also, your way of setting up the proof is flawed, as the following direct analogy shows

    Theorem: If 1 = 1, prove that 0 = 1.

    Proof:
    If 0 = 1 is correct, then we obviously also have 1 = 0. Adding the two equations gives
    Code (Text):

    0 = 1
    1 = 0  +
    --------
    1 = 1
     
    which is the assumption given.
     
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