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Change of Basis With Orbital Angular Momentum

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  1. Nov 20, 2014 #1
    1. The problem statement, all variables and given/known data
    We have the initial orbital angular momentum state in the x basis as |l,ml>x=|1,1>x. We are asked to find the column vector in the z-basis that represents the initial orbital angular momentum of the above state. It then says "hint: use an eigenvalue equation".

    2. Relevant equations
    I feel like this is a simple change of basis from the x basis to the z basis with the Lx operator. However, I am not convinced this is quite the steps I need to take.

    3. The attempt at a solution
    To start I applied the Lx operator (the 3x3 operator that I think is responsible for changing basis) it is equal to half the angular momentum ladder operators added with each other. This only gives me one column vector where there should be 3, I think, since there should be some probability with each value of the quantum number m possible for the z component.

    Another attempt was to try to find the eigenvalues of the Lx operator and to apply the 3 eigenvalues found as the probability to each vector but this didn't pan out so well. So here I am wondering how I might approach this problem in a different way. Any idea is appreciated.
     
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  3. Nov 21, 2014 #2

    stevendaryl

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    • This post contains more of the solution than advisable when posting in the homework forums. It has been left as the OP has already read it.
    You're on the right track. In terms of raising and lowering operators,
    [itex]L_x = \frac{1}{2}(L_{+} + L_{-})[/itex]

    The effect of the raising and lowering operators on eigenstates of [itex]L_z[/itex] are:
    [itex]L_{+} |\mathcal{l}, m \rangle = \sqrt{\mathcal{l}(\mathcal{l}+1) - m(m+1)} |\mathcal{l}, m+1 \rangle[/itex]
    [itex]L_{-} |\mathcal{l}, m \rangle = \sqrt{\mathcal{l}(\mathcal{l}+1) - m(m-1)} |\mathcal{l}, m-1 \rangle[/itex]

    So the effect of [itex]L_x[/itex] on eigenstates of [itex]L_z[/itex] is:
    [itex]L_x |\mathcal{l}, m \rangle [/itex]
    [itex] = \frac{1}{2} (\sqrt{\mathcal{l}(\mathcal{l}+1) - m(m+1)} |\mathcal{l}, m+1 \rangle + \sqrt{\mathcal{l}(\mathcal{l}+1) - m(m-1)} |\mathcal{l}, m-1 \rangle)[/itex]

    In the case [itex]l=1[/itex], we have 3 equations:
    1. [itex]L_{x} |1, -1\rangle = \frac{1}{\sqrt{2}}|1,0\rangle[/itex]
    2. [itex]L_{x} |1,0\rangle = \frac{1}{\sqrt{2}}(|1,1\rangle + |1,-1\rangle)[/itex]
    3. [itex]L_{x} |1,1\rangle =\frac{1}{\sqrt{2}}|1,0\rangle[/itex]

    So if you assume that [itex]|\psi\rangle[/itex] is an eigenstate of [itex]L_x[/itex] with eigenvalue [itex]+1[/itex], then write:
    [itex]|\psi\rangle = \alpha |1,-1\rangle + \beta |1,0\rangle + \gamma |1,1\rangle][/itex] and see what [itex]\alpha[/itex], [itex]\beta[/itex] and [itex]\gamma[/itex] have to be so that [itex]L_{x} |\psi\rangle = +1 |\psi\rangle[/itex]
     
  4. Nov 21, 2014 #3

    stevendaryl

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    In terms of matrices, if you represent [itex]|1, -1\rangle[/itex], [itex]|1,0\rangle[/itex], and |1,1\rangle[/itex] as
    [itex]\left(\begin{array} \\ 0 \\ 0 \\ 1 \end{array} \right)[/itex], [itex]\left(\begin{array} \\ 0 \\ 1 \\ 0 \end{array} \right)[/itex] and [itex]\left(\begin{array} \\ 1 \\ 0 \\ 0 \end{array} \right)[/itex]

    then [itex]L_x[/itex] can be represented as:

    [itex]\frac{1}{\sqrt{2}} \left(\begin{array} \\ 0 & 1 & 0 \\ 1 & 0 & 1 \\ 0 &1& 0 \end{array} \right)[/itex]

    Then the problem becomes a matrix problem:

    [itex]\frac{1}{\sqrt{2}} \left(\begin{array} \\ 0 & 1 & 0 \\ 1 & 0 & 1 \\ 0 &1& 0 \end{array} \right) \left(\begin{array} \\ \alpha \\ \beta \\ \gamma \end{array} \right)= \left(\begin{array} \\ \alpha \\ \beta \\ \gamma \end{array} \right)[/itex]
     
  5. Nov 21, 2014 #4
    Thanks that makes a lot of sense that you would act on an unknown matrix to then equal the eigenvalue times that same unknown matrix. Then we can go and expand it out into a linear combination.
     
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