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Eigenfunctions of translation operator and transposed operator property proof

  1. Oct 20, 2009 #1
    1. The problem statement, all variables and given/known data
    Find the eigenfunctions and eigenvalues of the translation operator [tex]\widehat{T_{a}}[/tex]
    Translation operator is defined as [tex]\widehat{T_{a}}\psi(x)=\psi(x+a)[/tex] (you all know that, probably you just call it differently)

    2. Relevant equations
    The eigenvalue/eigenfunction equation is given like
    [tex]\widehat{T_{a}}\psi_{n}(x)=f_{n}\psi_{n}(x)[/tex]


    3. The attempt at a solution
    I write the eigenvalues [tex]f_{n}[/tex] are in the form of [tex]c_{n}(a)[/tex] (c noting that it is a complex coefficient)
    And I don't know how to proove it correctly and clearly, but I get that only possible eigenfunctions are exponentials [tex]\psi_{n}(x)=e\ ^{x+2n\pi\i}[/tex] and the eigenvalues for [tex]\widehat{T_{a}}[/tex] are [tex]c_{n}(a)=e\ ^{a+2n\pi\i}[/tex]

    I don't have any ideas how to proove it more clearly, because this "solution" involves more thinking and assuming than solving.

    I also did similar solution for inversion operator [tex]\widehat{I_{x}}\psi(x)=\psi(-x)[/tex] by finding eigenvalues just looking at the properties of an equation which was like [tex]\psi(-x)=f\psi(x)[/tex] and using the property of odd and even functions and thus finding 2 eigenvalues of [tex]f_{1}=1[/tex] and [tex]f_{2}=-1[/tex] and thus getting infinite number of eigenfuntions. The real solution to this problem was using the property that [tex]\widehat{I_{x}}^2=\widehat{1}[/tex] and modify the equation [tex]\widehat{I_{x}}\psi(x)=f\psi(x)[/tex] by multiplying both sides from the left with [tex]\widehat{I_{x}}[/tex] and thus getting [tex]\psi(x)=f^{2}\psi(x)[/tex] and just needed to solve the quadratic equation of [tex]f^{2}=1[/tex] thus getting the same values of [tex]f_{1}=1[/tex] and [tex]f_{2}=-1[/tex] This was just an example of possible ways to solve eigenvalue equations, but in this case - the real solution shows that no other solutions are possible.


    And also I have this second problem which is more like proof of a formula which is usually given as a property.
    1. The problem statement, all variables and given/known data
    Proove that [tex](AB)^{T}=B^{T}A^{T}[/tex]


    2. Relevant equations
    The transposed operator is given in bra-ket notation as [tex]\left\langle\varphi\left|\widehat{A}^{T}\right|\psi\right\rangle=\left\langle\psi^{*}\left|\widehat{A}\right|\varphi^{*}\right\rangle[/tex] or in integral form as [tex]\int\varphi(x)\widehat{A}^{T}\psi(x)dx=\int\psi(x)\widehat{A}\varphi(x)dx[/tex]


    3. The attempt at a solution
    Well, I have completely no ideas on where to start with this one.
    Because I started with it like [tex]\left\langle\varphi\left|\widehat{AB}\right|\psi\right\rangle=\left\langle\varphi\left|\widehat{A}\right|\widehat{B}\psi\right\rangle=\left\langle\psi^{*}\left|\widehat{A}\right|\widehat{B}^{T}\varphi^{*}\right\rangle[/tex] but I really doubt that the last operation is correct and I'm allowed to do so.

    I hope I made my doubts and problems clear and thanks for help in advance!
     
  2. jcsd
  3. Oct 20, 2009 #2

    tiny-tim

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    Science Advisor
    Homework Helper

    Welcome to PF!

    Hi Muchacho! Welcome to PF! :smile:

    (have a psi: ψ :wink:)

    Where did you get n from? :confused:

    Remember, this works for any a.

    Hint: take logs of the equation. :smile:
     
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