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Quantum Mechanics: Base on which an operator is given

  1. Aug 8, 2013 #1
    The thing is that often in the problems on quantum mechanics I've found that an operator is given, but not the base on which it is represented. I'll give an especific example in a moment. So then the problem asks me to find the eigenvalues and eigenvectors for a given operator and to express it in the basis on which the operator is given, but the base is not given. I know that the information of the base is implicit in the operator it self, but I was wondering on which method I should use to find that base (I thought of using the typical matrix reduction of linear algebra, but I don't think thats the better way to do this). I'll take a problem from Cohen Tannoudji to picture what I mean.

    Page 203, Cohen Tannoudji, problem 2, complement HII (I'll do incise c):

    Consider an operator whose matrix, in an orthornormal basis ##\{ \left| 1 \right\rangle , \left| 2 \right\rangle, \left| 3 \right\rangle \} ## is:

    ##L_y=
    \begin{bmatrix}
    0 & \sqrt{2} & 0\\
    -\sqrt{2} & 0 & \sqrt{2}\\
    0 & -\sqrt{2} & 0\\
    \end{bmatrix}##

    So, the exercise asks me to find the eigenvalues and the eigenvectors, and give their normalized expansion in terms of the basis ##\{ \left| 1 \right\rangle , \left| 2 \right\rangle, \left| 3 \right\rangle \} ##.


    So, the question is, what should I do to know what the vectors of the basis ##\{ \left| 1 \right\rangle , \left| 2 \right\rangle, \left| 3 \right\rangle \} ## are? so then I can express the eigenvectors in terms of the vectors of this basis.

    From the secular equation I get the eigenvalues ##\lambda_1=0, \lambda_2=2i, \lambda_3=-2i##

    From the first eigenvalue I get the eigenvector: ##\left| \lambda_1 \right\rangle=\frac{1}{2} \begin{bmatrix}
    1\\
    0\\
    1\\
    \end{bmatrix}##

    So, what I want is to express this vector in the basis ##\{ \left| 1 \right\rangle , \left| 2 \right\rangle, \left| 3 \right\rangle \} ##, but what that kets are? how do I obtain them?

    I thought of using the fact that the matrix elements are given by ##L_{y;m,n}= \langle m | \hat L_y \left| n \right\rangle## but I've tried with a random matrix, only for one element, and get to an identity 0=0, so then I desisted, but I should try with this specific example now.
     
    Last edited: Aug 8, 2013
  2. jcsd
  3. Aug 8, 2013 #2

    TSny

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    Essentially by definition, the column matrix representation of ##|1\rangle## in the basis ##\{ \left| 1 \right\rangle , \left| 2 \right\rangle, \left| 3 \right\rangle \} ## is

    ##\begin{bmatrix}
    \langle 1 | 1 \rangle\\
    \langle 2 | 1 \rangle\\
    \langle 3 | 1\rangle\\
    \end{bmatrix}##

    etc.
     
  4. Aug 8, 2013 #3
    Thanks for your answer TSny.

    What ##| 1 \rangle## is then? the other projections will give zero, because it's an orthonormal basis, but how do I know what ##\langle 1 | 1 \rangle, \langle 2 | 2 \rangle , \langle 3 | 3 \rangle## are? they are not given, I just have the matrix representation for the given operator on that basis (which I know it must contain this information, I'm trying to know how to get it from it on the easiest way).
     
  5. Aug 8, 2013 #4

    TSny

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    Think about the meaning of orthonormal.
     
  6. Aug 8, 2013 #5
    ups! haha I'm such an idiot. I knew this was a silly question. Thanks.

    ##| 1 \rangle=\begin{bmatrix}
    1\\
    0\\
    0\\
    \end{bmatrix}, | 2 \rangle=\begin{bmatrix}
    0\\
    1\\
    0\\
    \end{bmatrix}, | 3 \rangle=\begin{bmatrix}
    0\\
    0\\
    1\\
    \end{bmatrix}##

    The basis could also be complex, right? I mean, this was trivial. Is always this way? can I choose any orthonormal basis?
     
  7. Aug 8, 2013 #6

    TSny

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    No. If you multiplied the basis vectors by complex phase factors, then it would generally change the matrix representation of your operators. So, whenever you have a matrix representation of an operator, you should assume that the representation is in terms of the real basis vectors (1, 0, 0), etc.
     
    Last edited: Aug 8, 2013
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