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Homework Help: Problem based hermitian operator

  1. Nov 3, 2007 #1
    problem based on hermitian operator

    1. The problem statement, all variables and given/known data

    A is an hermitian operator and as we know the eigenstates a of A with eigenvalues a satisfy A psi a = a psi a.

    How do we show that lambda psi a (lambda is a non zero complex number) is an eigen state belonging to the same eigen value a. We also need to say whether the eigen value is non degenerate or not.

    2. Relevant equations

    We know the property of hermitian operator

    (-infinity to + infinity) integral phi* A psi dx = (-infinity to + infinity) psi (A phi)* dx

    3. The attempt at a solution

    (-infinity to + infinity) integral phi* A (lambda psi a) dx =

    (-infinity to + infinity) lambda psi a (A phi)* dx

    Dont know how to proceed from here.


    Last edited: Nov 3, 2007
  2. jcsd
  3. Nov 3, 2007 #2


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    Gold Member

    Your notation is confusing. Please write down the original question, EXACTLY as it was given to you, and it would help if you used either clearer notation, or the latex capability built in here. From what I can make of it, there seems to be a mistake in the question.

    Examples for latex:

    [tex]A \psi_a = a \psi_a [/tex]

    [tex]A |a \rangle = a | a \rangle [/tex]
  4. Nov 3, 2007 #3
    Here it comes.Could you please tell me where the mistake is.

    The eigenstates [tex]\psi_a[/tex] of A with eigenvalues a satisfy
    the equation [tex]A \psi_a = a \psi_a [/tex]. We need to show that then also [tex]\Lambda\psi_a[/tex] with an arbitrary complex number [tex]\Lambda[/tex] not equal to 0, is an eigenstate belonging to the same eigenvalue a.Need to check if the eigenvalue is nondegenerate or not
    Last edited: Nov 3, 2007
  5. Nov 3, 2007 #4
    I don't understand the question, obviously if [itex]\psi_a[/itex] is an eigenstate of [itex]\hat{A}[/itex] with eigenvalue a then [itex]\Lambda \psi_a[/itex] is also an eigenstate with the same eigenvalue. This just follows from the fact that A is linear.
  6. Nov 3, 2007 #5
    is there no proof for this. I mean cant we take a complex number of the form z=x+iy and prove the same.Since A is linear how can we conclude that,is there no mathematical proof for this.
  7. Nov 3, 2007 #6
    Sure. Let [itex]\hat{A}[/itex] be any linear operator with eigenvalue a and [itex]\Lambda[/itex] any complex number.


    [itex]\hat{A} (\Lambda \psi) = \Lambda \hat{A} \psi = \Lambda a \psi = a (\Lambda \psi)[/itex].
  8. Nov 3, 2007 #7
    So as I understand Hermitian operator is also a linear operator, which is why we could perform this operation.

    If B is a second commutation operator which commutes with A such that

    [A,B] = 0.

    How do we show that [tex]A \psi = a \psi [/tex] and [tex]B \psi = b \psi [/tex] ? i.e to show that these operators have simultaneous eigenstates.

    All eigen values in this problem are degenerate.

    Is this proof correct:

    if [A,B] = 0
    AB = BA
    AB-BA = 0

    [tex]AB \psi = Ab \psi = bA\psi = ba\psi = ab\psi = aB\psi = Ba\psi = BA \psi [/tex]


    [tex]A \psi = a \psi [/tex] and [tex]B \psi = b \psi [/tex]
  9. Nov 3, 2007 #8
    That doesn't seem correct to me. It looks like you've assumed your answer in your proof.

    If A,B commute then AB = BA. Suppose [itex]\psi[/itex] is an eigenstate of A with corresponding eigenvalue a. Then

    [itex]A\psi = a \psi[/itex]
    [itex]BA\psi = a B\psi[/itex]
    [itex]A(B\psi) = a (B\psi)[/itex]

    Thus [itex]B\psi[/itex] is an eigenstate of A with the same eigenvalue. Thus if the eigenvalue is non-degenerate [itex]B\psi[/itex] is proportional to [itex]\psi[/itex].
  10. Nov 3, 2007 #9
    We know that the hermitian operator S interchanges coordinate x with −x. On consider
    the Hamiltonian [tex]H = p^2/2m + V (x)[/tex] for a particle in a one dimensional
    symmetric potential V (x) = V (−x). What can be said
    about the wavefunctions in view of what we proved above i.e [tex] A \psi = a \psi and B \psi = b \psi [/tex].

    How to proceed in this case?.
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