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Expectation value of the angular momentum operator

  • #1

Homework Statement


Hey forum,
I copied the problem from a pdf file and uploaded the image:
http://img232.imageshack.us/img232/6345/problem4.png [Broken]

What is the probability that the measurement of L[tex]^{2}[/tex] will yield 2\hbar[tex]^{2}[/tex]

Homework Equations


[tex]\left\langle L^{2} \right\rangle[/tex] = [tex]\left\langle \Psi \left| L^{2} \right| \Psi \right\rangle[/tex]
[tex] L^{2} \Psi \right\rangle[/tex] = [tex] \hbar^{2}l(l+1) \Psi \right\rangle[/tex]

The Attempt at a Solution


So the problem I'm having is with part (b). I know how to calculate the expectation value of L[tex]^{2}[/tex]. But given a value of the expectation value, I have no idea how to calculate the probability that the expectation value will yield that given value.

Going back to the expectation value of x in a one dimensional potential, I remember that if I wanted to find the expectation value of the particle being outside, say a potential well located between the origin and x=a, my integral would be from a to infinity. but I don't see how to translate that to this case.
 
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Answers and Replies

  • #2
kuruman
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Assuming that you have done part (a), can you write the given wavefunction as a linear combination of eigenstates of L2? If so, then the absolute value squared of the coefficient of the eigenstate corresponding to L2 = 2 hbar^2 is what you want.
 
  • #3
I see. Thank you, that hint was very helpful. I also have a question for part (c).

The eigenvalue equation for L[tex]_{z}[/tex] is L[tex]_{z}[/tex][tex]\left|\right \Psi \rangle[/tex]=[tex]\hbar m \right \Psi \rangle[/tex]. The measurement of L[tex]_{z}[/tex] is only zero when m=0, but the given wavefunction is made of two eigenstates in which both have m=0. Does this mean that the measurement of L[tex]_{z}[/tex] will always give 0? If this is the case, doesn't this violate the uncertainty principle of the individual angular momentum operators Lx, Ly, and Lz? The uncertainty principle for the observables L[tex]_{x}[/tex] and L[tex]_{y}[/tex] is proportional to the expectation value of L[tex]_{z}[/tex]. If the expectation value of L[tex]_{z}[/tex] is zero, this would mean that there is no uncertainty in the measurement of Lx, Ly, or Lz. But this cannot be the case because we can't know the angular momenta in each direction simultaneously. Where exactly am I going wrong in my thought process?
 
  • #4
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The Uncertainty principle is not violated since the operators obey [[tex]L_{i}[/tex],[tex]L_{j}[/tex]] = i[tex]\hbar[/tex][tex]\epsilon_{ijk}[/tex][tex]L_{k}[/tex]
 
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  • #5
kuruman
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***redundant post, removed on edit ***
 

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