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Homework Help: Angular momentum operators on a wave function

  1. Aug 24, 2016 #1
    1. The problem statement, all variables and given/known data
    Particle is in a state with wave function [itex]\psi (r) = A z (x+y)e^{-\lambda r}.[/itex]
    a) What is the probability that the result of the [itex]L_z[/itex] measurement is 0?
    b) What are possilble results and what are their probabilities of a [itex]L^2[/itex] measurement?
    c) What are possilble results and what are their probabilities of a [itex]L_x[/itex] measurement?

    3. The attempt at a solution
    Firstly, I tried to write wave function in spherical harmonics form:
    \psi (r) &= \alpha z (x+y); \ \ \alpha= A e^{-\lambda r}\\ &= \alpha r \cos\theta (r \sin\theta\cos\phi+r \sin\theta \sin \phi)\\ &= \alpha r^2 \cos\theta \sin \theta (\cos\phi + \sin \phi) \\ &=\alpha r^2\cos\theta \sin\theta((\frac{1}{2}+\frac{i}{2})e^{-i \phi}+(\frac{1}{2}-\frac{i}{2})e^{i \phi}) \\ &= -{\alpha \over 2} r^2 \sqrt{\frac{8 \pi}{15}} ((1+i)Y_{2,-1}+(1-i)Y_{2,1})

    Then, I normalized this function:
    [tex] |\psi,r\rangle = {\alpha \over {2c}} r^2 \sqrt{\frac{8 \pi}{15}} ((1+i) c Y_{2,-1}+(1-i) c Y_{2,1}) \\ \Longrightarrow c^2 |(1+i)|^2+c^2 |(1-i)|^2=1 \\ c={1 \over 2}[/tex]
    So finally, I get my wave function:
    [tex]\psi (r) = \alpha r^2 \sqrt{\frac{8 \pi}{15}} ({1 \over 2}(1+i) Y_{2,-1}+{1 \over 2}(1-i)Y_{2,1})[/tex]
    Can someone check this normalization, I am not sure if I did it correctly?

    a) Form definition [itex]L_z|l,m\rangle=\hbar m |l,m\rangle[/itex] I can't get 0 as a result, because there isn't any element of a wave function with m=0. So the answer is 0 probability.

    b) Form definition [itex]L^2|l,m\rangle=\hbar^2 l(l+1) |l,m\rangle[/itex] I can only get the result [itex]6 \hbar^2[/itex] with 100% probability, since [itex]l[/itex] is in both elements of a wave function 2.

    c) Form definition [itex]L_x|l,m\rangle=\frac{L_+ + L_-}{2} |l,m\rangle[/itex], I get:
    L_+ |2,1\rangle &= 2 \hbar |2,2\rangle \\
    L_+ |2,-1\rangle &= \sqrt{6} \hbar |2,0\rangle \\
    L_- |2,1\rangle &= \sqrt{6} \hbar |2,0\rangle \\
    L_- |2,-1\rangle &= 2 \hbar |2,-2\rangle\\
    \Longrightarrow L_x |2,1\rangle &= \hbar (|2,2\rangle + \frac{\sqrt{6}}{2} |2,0\rangle)\\
    \Longrightarrow L_x |2,-1\rangle &= \hbar (|2,-2\rangle + \frac{\sqrt{6}}{2} |2,0\rangle)

    So the possible results are [itex]\hbar[/itex] and [itex]\frac{\sqrt{6}}{2} \hbar[/itex]? But how can I now find their probabilities? I tried to write:
    [tex]L_x |\psi\rangle = \sqrt{\frac{8 \pi}{15}} \alpha \hbar r^2(\frac{1+i}{2}(|2,2\rangle+{\sqrt{6} \over2} |2,0\rangle)+\frac{1-i}{2}(|2,-2\rangle +{\sqrt{6} \over2} |2,0\rangle)), [/tex]
    but the squares of absolute values of coefficients are above 1.
  2. jcsd
  3. Aug 24, 2016 #2


    User Avatar

    Staff: Mentor

    Be careful with the sign in the last line.

    I don't understand what you did here. Actually, compare that last line with the last line above; do you see any difference? Also, what about ##\alpha## (and the ##A## it contains)?
  4. Aug 24, 2016 #3
    Aha, I forgot that [itex]Y_l^{-m}=(-1)^m Y_l^m[/itex], so I get:
    |\psi\rangle = {\alpha \over 2} r^2 \sqrt{\frac{8 \pi}{15}} \big((1+i)Y_2^{-1}-(1-i)Y_2^{1}\big)

    Ok, here I tried something we did at lectures, but apparently I didn't understand it well. So if I go the long way:
    \langle \psi,r | \psi,r \rangle = \frac{A A^\ast e^{-2\lambda r}}{4} {8\pi \over 15}\bigg[ \big((1-i) \langle 2, -1| - (1+i)\langle 2,1|)((1+i)|2,1\rangle-(1-i)|2,1\rangle \big) \bigg] =1\\
    \frac{2 A A^\ast \pi}{15}e^{-2\lambda r} (2+2) = 1\\
    |A|=\sqrt{\frac{15}{8 \pi}}e^{-\lambda r}
    So I get my wave function:
    [tex]|\psi\rangle={r^2 \over 2} \bigg[(1+i) Y_2^{-1}-(1-i)Y_2^1\bigg].[/tex]

    EDIT: Forgot square root over fraction in [itex]|A|[/itex].
    Last edited: Aug 24, 2016
  5. Aug 25, 2016 #4
    I've got one similar exercise, but instead of angular momentum I've got spins. And I've got the same problems at [itex]S_x[/itex] measurements.
    So the exercise goes:
    We have got two particles with [itex]S_1=1[/itex] and [itex]S_2=1[/itex]. We know that [itex]S_{1z}|\psi_1\rangle=\hbar |\psi_1\rangle[/itex] and [itex] S_{2x}|\psi_2\rangle = \hbar |\psi_2\rangle. [/itex]

    a) Find wave function [itex]|\psi_1\rangle[/itex] in [itex]S_{1z}[/itex] basis and [itex]|\psi_2\rangle[/itex] in [itex]S_{2z}[/itex] basis.
    b) We measure [itex]S^2[/itex] of total spin. What are possible outcomes and what are their probabilities?
    c) Find expectation value and uncertainty of [itex]S^2[/itex].
    d) We measure x component of total spin. What are possible outcomes and what are their probabilities?

    a) [itex]|\psi_1\rangle = |11\rangle \\ |\psi_2\rangle = {1 \over 2} |1-1\rangle + {1 \over \sqrt{2}} |10\rangle+ {1 \over 2} |11\rangle.[/itex] Can someone just check this?
    |\psi_{12}\rangle&={1 \over 2}|1\rangle|-1\rangle+{1 \over \sqrt{2}} |1\rangle|0\rangle+{1 \over 2}|1\rangle|1\rangle=\\
    &={1 \over \sqrt{24}}|20\rangle+{1 \over \sqrt{12}}|00\rangle+{1 \over 2}|21\rangle+{1 \over 2}|11\rangle+{1 \over 2}|22\rangle
    For [itex]S^2|\psi_{12}\rangle=\hbar^2 s(s+1)|\psi_{12}\rangle,[/itex] we get:
    &Results \ \ \ \ &Probability\\
    &6\hbar^2 &{13\over24}\\
    &2\hbar^2 &{3 \over 8}\\
    &0 &{1 \over 12}

    c) Expectation value is [itex]\langle S^2 \rangle = \langle \psi|S^2|\psi\rangle=4\hbar^2,[/itex] but I can't find uncertainty? I am thinking in this way:
    [tex]\delta_{S^2}=\sqrt{\langle S^2\rangle- \langle S \rangle ^2} or \\
    \delta_{S^2}=\sqrt{\langle S^4\rangle- \langle S^2 \rangle ^2}?[/tex]

    d) Now, same problem as at the angular momentum. How do I find outcomes and probabilities? I tried with [itex]S_x=\frac{S_++S_-}{2}[/itex], but got some weird result, from which I can't find anything (As mentioned - like at angular momentum in previous post). Then I was thinking about Pauli matrices, so that possible outcomes would be [itex]\pm {\hbar \over 2}[/itex], but how can I apply this matrix to my wavefunction of 1x1 spins. I found something on wiki for such spins and tried but got nothing... (This two attempts are in my handwriting downthere):
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