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[qm]find the angular opertor given total angular momentum wavefunction

  1. Sep 5, 2009 #1
    1. The problem statement, all variables and given/known data

    consider a system with total angular momentum, l=1 in the state

    |[tex]\psi[/tex]>=[tex]\frac{1}{\sqrt{2}}|1>-\frac{1}{2}|0>+\frac{1}{2}|-1>[/tex]
    find |[tex]^{^}L_{\psi}>[/tex]

    2. Relevant equations
    [tex]^{^}L_{z}|\psi>=\hbar m|\psi>[/tex]


    3. The attempt at a solution

    the basis in the wavefunction given are|1> , |0>, |-1> and there are orthogonal. but i'm not sure what the question is really asking. anyone care to shed some light on this question.thanks

    it shoud be L subsript y not psi. just couldnt read the handwriting
     
    Last edited: Sep 5, 2009
  2. jcsd
  3. Sep 5, 2009 #2

    kuruman

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    I cannot interpret what | Ly> might mean. The symbol for an operator is not normally placed inside a ket. Since there is handwriting involved, is it possible that you are asked to find the expectation value < Ly>? That makes more sense.
     
  4. Sep 5, 2009 #3
    i've done no work in like four months for summer but am going back soon so can somebody tell me if ive done this correctly please.

    say we wanted to find [itex]<L_y=1>[/itex], we would do:

    [itex]<L_y=1>=\int_{-\infty}^{\infty} \psi^{\star} \cdot 1 \cdot \psi dx=\int_{-1}^{1} \frac{1}{2}+\frac{1}{4}+\frac{1}{4} dx[/itex] where i have used the orthogonality of the kets.

    and so [itex]<L_y=1>=\int_{-1}^{1} dx = \frac{x^2}{2} |_{x=-1}^{x=1}=0[/itex]

    well that's definitely wrong. im pretty sure i can't just change the limits from infinity to 1 etc in this case. jeez i need to get some work done in the next couple of weeks to get back up to speed lol.
     
  5. Sep 5, 2009 #4

    kuruman

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    I do not understand what you mean by

    [tex]\left\langle L_{y} } = 1 \right\rangle [/tex]

    An expectation value is usually written as

    [tex]\left\langle \psi | L_{y} | \psi \right\rangle [/tex]

    an abbreviated form of which is

    [tex]\left\langle L_{y} \right\rangle [/tex]

    You have to "sandwich" Ly between the bra and the ket of the wavefunction |ψ> that you have then distribute it among all nine possibilities of bra-kets.
     
  6. Sep 5, 2009 #5
    bleh...it's going to be a long road back.

    ok try this:

    [itex]<\psi^{\star}|L_y|\psi>=\int_{-\infty}^{\infty} \psi^{\star} L_y \psi dx[/itex]

    when we dot product [itex]\psi^{\star}[/itex] and [itex]\psi[/itex] we get 1 though due to ket orthogonality.
    what value do I use for [itex]L_y[/itex]?
    do the limits change due to the allowed values for [itex]L_y[/itex]?

    thanks for your help.
     
  7. Sep 5, 2009 #6
    sorry for the confusion, what i mean is L[tex]_{y}[/tex]| 〉 i am going to try it out myself and post here my answer when finish.
     
  8. Sep 5, 2009 #7

    kuruman

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    Now that makes sense.
     
  9. Sep 5, 2009 #8
    If you mean <L_y>, then you compute the matrix element
    [tex]\left(\begin{array}{ccc}
    1/\sqrt{2} & -1/2 & 1/2 \end{array}\right)
    \left(\begin{array}{ccc}
    ... & ... & ...\\
    ... & ... & ...\\
    ... & ... & ...\end{array}\right)\left(\begin{array}{c}
    1/\sqrt{2}\\
    -1/2\\
    1/2
    \end{array}\right)[/tex] where the matrix is that of L_y in the 3x3 representation (spin 1). Otherwise, don't know what you meant.
     
  10. Sep 6, 2009 #9
    alright here is my solution, please comment.

    [tex]L_{\pm}=L_{x}+iL_{y}[/tex]
    [tex]L_{\pm}|l,m>=\hbar\sqrt{(l_{\mp}m)(l_{\pm}m+1)}|l,m_{\pm}1>[/tex]

    [tex]L_{y}=(L_{+}-L{-})/2i[/tex]

    [tex]L_{y}=<\psi|L_{y}|\psi>=\frac{1}{2i}[<\psi|L_{+}|\psi>-<\psi|L_{-}|\psi>][/tex]
    [tex]=\frac{1}{2i}[\hbar\sqrt{(l_{\mp}m)(l_{\pm}m+1)}(\frac{1}{\sqrt{2}}[<1|L_{+}|1>-\frac{1}{2}<0|L_{+}|0>+\frac{1}{2}<-1|L_{+}|-1>
    -\hbar\sqrt{(l_{\mp}m)(l_{\pm}m+1)}(\frac{1}{\sqrt{2}]<1|L_{-}|1>-
    \frac{1}{2}<0|L_{-}|0>+\frac{1}{2}<-1|L_{-}|-1>]] [/tex]
    [tex]=\frac{1}{2i}[\frac{1}{\sqrt{2}}<1|2>-\frac{1}{2}<0|1>+\frac{1}{2}<-1|0>][/tex]
    [tex]-\frac{1}{\sqrt{2}}[<1|0>+\frac{1}{2}<0|-1>-\frac{1}{2}<-1|-2>]=0[/tex]
     
    Last edited: Sep 6, 2009
  11. Sep 6, 2009 #10
    interesting this is another solution, beside my way of doing it
     
  12. Sep 6, 2009 #11

    gabbagabbahey

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    Is there a reason why you've written the [itex]\pm[/itex] as subscripts on the RHS?....Surely you mean

    [tex]L_{\pm}|l,m\rangle=\hbar\sqrt{(l\mp m)(l \pm m+1)}|l,m \pm 1\rangle[/tex]

    Right?

    First, [itex]L_y[/itex] is just an operator, [itex]\langle L_y\rangle=\langle\psi\vert\L_y\vert\psi\rangle[/itex] is its expectation value.

    Secondly, why are you calculating the expectation value? I thought you said the problem asked you to calculate [itex]L_y\vert\psi\rangle[/tex]....

    You need to be careful here, the states [itex]|0\rangle[/itex], [itex]|1\rangle[/itex] and [itex]|-1\rangle[/itex] correspond to the different values of [itex]m[/itex], so the coefficient [itex]\sqrt{(l\mp m)(l \pm m+1)}[/itex] will have different values when [itex]L_y[/itex] operates on each state. For example,

    [tex]L_+\vert0\rangle=\hbar\sqrt{(1-0)(1+ 0+1)}|1,0+1\rangle=\sqrt{2}\hbar|1,1\rangle[/tex]

    while

    [tex]L_+\vert1\rangle=\hbar\sqrt{(1-1)(1+ 1+1)}|1,1+1\rangle=0[/tex].

    You also need to operate on the state [itex]|\psi\rangle[/itex] with [itex]L_y[/itex], before you multiply by [itex]\langle\psi|=\frac{1}{\sqrt{2}}\langle1|-\frac{1}{2}\langle0|+\frac{1}{2}\langle-1|[/itex] and distribute the different inner products. For example,

    [tex]\left(\langle0|+\langle1|\right)L_{-}\left(|0\rangle+|1\rangle\right)=\langle0|L_{-}|0\rangle+\langle0|L_{-}|1\rangle+\langle1|L_{-}|0\rangle+\langle1|L_{-}|1\rangle\neq\langle0|L_{-}|0\rangle+\langle1|L_{-}|1\rangle[/tex]
     
    Last edited: Sep 6, 2009
  13. Sep 6, 2009 #12

    gabbagabbahey

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    First, the integral on the RHS is what you get when you expand [itex]\psi[/itex], [itex]\psi^*[/itex] and [itex]L_y[/itex] in the x-basis....if you don't know what those expanded versions are, there is not much point in using this method.

    Second, [itex]L_y[/itex] operates on [itex]\psi(x)[/itex] before you take the product with [itex]\psi^*[/itex], so unless the effect of the operator is to simply multiply by a scalar (say,[itex]\alpha[/itex] ), you can't
    say that [itex]\oint\psi^*(x)L_y\psi(x)dx=\alpha\oint \psi^*(x)\psi(x)dx[/itex].
     
    Last edited: Sep 6, 2009
  14. Sep 6, 2009 #13
    thank for the thorough explanation,
    to answer part 1, the +/- at RHS is not a subscript, the latex formatted it that, it doesn't meant to be. after reading your code, i understand how to format it already.

    part II, i am calculating the expectation value, not the
    [itex]L_y\vert\psi\rangle[/itex]. as i read the question wrongly (it was a handwritten one)

    part III, the operation is distributive.Noted. thanks for the good effort ;)
     
  15. Sep 6, 2009 #14
    thanks a lot. had a look over some notes from last year as well as your reply - helped a lot!
     
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