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Atomic Physics - Orbital Angular Momentum Probability

  1. Apr 24, 2017 #1
    1. The problem statement, all variables and given/known data
    Consider an electron in a state described by angular wavefunction $$\psi(\theta,\phi)=\sqrt{\frac{3}{4 \pi}}\sin \theta \cos \phi$$ Here θ and φ are the polar and azimuthal angles, respectively, in the spherical coordinate system.

    i. Calculate the probability that a simultaneous measurement of the electron orbital angular momentum squared (L2) and the Z component of the electron orbital angular momentum (Lz) will give 2ħ2 and ħ respectively.

    ii. What is the probability of measuring L2 = 2ħ2 ?
    iii. What is the probability of measuring Lz = 0 ?
    2. Relevant equations
    $$\psi(\theta,\phi) = \sum_{l=0}^{l=\infty} \sum_{m=-l}^{m=+l}c_{l,m} Y_{l,m} (\theta, \phi)$$

    $$c_{l,m} = \int Y^{*}_{lm}(\theta, \phi) \psi (\theta, \phi) d\Omega$$
    3. The attempt at a solution

    In my notes I am told that $$\vert c_{l,m} \vert ^2$$ is the probability that a simultaneous measurement of L2 and Lz on a particle described by the wavefunction ψ gives l(l+1)ħ2 AND mħ.
    The probability that a measurement of L2 will give l(l+1)2 is simply the sum of the probabilities for each possible m state: $$P(l) = \sum_{m=-l}^l \vert c_{l,m} \vert ^2$$

    Looking up the wavefunction in a table, it seems to be a spherical harmonic with l=1 and m=0
    which means m can range between -1 and 1.

    Now for question i. the answer would simply be $$\vert c_{1,1} \vert ^2$$ ...but how can I calculate what c is?

    I guess my problem is similar for part ii. as the answer would be $$P(l) = \sum_{m=-l}^l \vert c_{l,m} \vert ^2$$ where m can be -1, 0 and 1...but I'm stumped on how to calculate cl,m

    Any help would be greatly appreciated !
    Thank you
     
    Last edited: Apr 24, 2017
  2. jcsd
  3. Apr 24, 2017 #2

    DrClaude

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    Staff: Mentor

    Are you sure about the parts I indicated in red?

    Why m = 0?

     
  4. Apr 24, 2017 #3
    Sorry! Edited my post, it was meant to be 2ħ2.

    I am having trouble understanding the integral though... What are Yl,m... and dΩ :(
    I have a table with values of Y0,0, Y1,0, Y1,1 etc...

    So for example $$Y_{0,0} = \sqrt{\frac{1}{4 \pi}}$$

    $$c_{0,0} = \int \sqrt{\frac{1}{4 \pi}} \sqrt{\frac{3}{4 \pi}} \sin \theta \cos \phi d\Omega$$
    where $$\int d\Omega = \int_0^{2 \pi} d\phi \int_0^{\pi} \sin \theta d\theta$$

    $$c_{0,0} = \frac{\sqrt{3}}{4\pi} \int_0^{2 \pi} \cos \phi d\phi \int_0^{\pi} \sin^2 \theta d\theta$$

    Leading to a sin φ evaluated between 2π and 0... Which is 0, so c0,0=0 ? Am I on the right track?

    If I then apply this with the other values, I could work out each value for c, and put them into the equations to get the probabilities?

    So i. would be c1,12
    ii. would be c1,-12+c1,02+c1,12

    Correct?
     
    Last edited: Apr 24, 2017
  5. Apr 24, 2017 #4
    Working through the numbers, I have got an answer of $$c_{1,0}=0$$ $$c_{1,1}=-\frac{1}{\sqrt{2}}$$ and $$c_{1,-1}=\frac{1}{\sqrt{2}}$$

    Which means the answer to i. is 1/2, ii. is 1/2+1/2 = 1
    But how do I calculate the probability of measuring Lz=0?

    The probability of m=1 AND l =1 was 1/2 (from part i.)
    and the probability of l=1 is 1 from part ii.
    This suggests that the probability of m=1 is 1/2...?
    Which I'm assuming is the same for the probability of m=-1
    As such the probability of m=0 is 0, correct?
     
    Last edited: Apr 24, 2017
  6. Apr 25, 2017 #5

    DrClaude

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    Staff: Mentor

    That's one way to do it. You can also get it directly from the fact that all coefficients ##c_{l,0}## are zero, therefore the probability of measuring ##m=0## is zero.
     
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