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Expectation value of Lz angular momentum

  1. May 28, 2014 #1
    1. The problem statement, all variables and given/known data

    2rdk9hw.png

    Find ##\langle L_z \rangle##. What is ##\langle L_Z \rangle## for one atom only?

    2. Relevant equations



    3. The attempt at a solution

    Using ##L_z = -i\hbar \frac{\partial }{\partial \phi}##,

    I get:

    [tex]\langle L_z\rangle = \frac{32}{3} \pi k^2 \hbar a_0^3[/tex]

    Not sure what the significance of this is. That it is non-zero? All the atoms somehow interact and produce a net angular momentum about z.

    For a single atom I think it is zero.
     
  2. jcsd
  3. May 28, 2014 #2
    You lost me at Lz = .... I can neither confirm nor deny that, since Lz is undefined in the problem (and chemists use it for magnetic quantum number, while physicists tend to use it as the z component of total angular momentum. ½(e^(ix) - e^(-ix)) = i*sin(x).
    You need to show your work...I can't follow how you got from (say) i∂(sin(Φ))/∂Φ to a number times some constants. Generally, there are lots of eigenvalues (λᵢ) for a particular operator(wave function)= λ(wave function) partial differential equation.
     
  4. May 28, 2014 #3

    strangerep

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    Unscientific,

    Like abitslow said, you need to show more of your work.

    Assuming "many atoms" means "an ensemble of atoms", then I think this is wrong. But I can't be sure since you didn't show any details of how you got the above result.

    If "for a single atom" means "a single measurement", then that sounds wrong too. What is the relationship between an operator, its eigenvalues, and a measurement of the property represented by that operator?
     
    Last edited: May 28, 2014
  5. May 29, 2014 #4
    Here's the question again:

    dlkw0o.png



    Sorry about that, I thought I shouldn't bore you guys with the maths. Here's my working:

    [tex]\frac{\partial \psi}{\partial \phi} = -ike^{-\frac{r}{2a_0}} \left( sin \theta \space e^{-i\phi}
    + sin \theta \space e^{i\phi}\right)[/tex]
    [tex]= -i sin\theta \space k e^{-\frac{r}{2a_0}} (-2i cos \phi)[/tex]
    [tex]= -2 sin\theta \space cos \phi \space k e^{-\frac{r}{2a_0}}[/tex]

    [tex]\langle \psi | L_z | \psi \rangle = k^2 (-i\hbar) \int \left( \sqrt 2 cos \theta + sin\theta \space e^{i\phi} - sin \theta \space e^{-i\phi}\right) \left(-2 sin \theta \space cos \phi \right) e^{-\frac{r}{a_0}} d^3 r[/tex]
    [tex] = 2k^2 (i\hbar) \int \left[ \sqrt 2 cos \theta \space + \space (sin \theta)(2i sin \phi) \right] \left(sin \theta \space cos \phi \right) e^{-\frac{r}{a_0}}\space r^2 sin \theta dr \space d\theta \space d\phi [/tex]
    [tex] = 2k^2 (i\hbar) \int r^2 e^{-\frac{r}{a_0}} \left(\sqrt 2 sin^2 \theta \space cos \theta cos \phi + 2i sin^3 \theta \space sin \phi cos \phi \right) dr \space d\theta \space d\phi[/tex]
    [tex] = 0[/tex]

    This gives zero, because first term has ##cos \phi##, while second term has ##sin\phi cos \phi##.
    This means that the angular momentum is spherically symmetric about the z-axis.
     

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    Last edited: May 29, 2014
  6. May 29, 2014 #5

    vanhees71

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    2016 Award

    The 0th step to the solution of a problem is its clear formulation. Sometimes it's even the most important task to ask the right question in clearly formulating the problem. You didn't even provide the question! So at least I am not able to tell you.

    The next step is to think about the concepts that may lead to the solution of the problem. So as a next step, I suggest to answer the questions by strangerep (and these are not so simple questions as it might seem on the first glance!).

    Last but not least, when you have found a solution, you should at least make simple sanity checks. Here, I'd check the units (dimensions) on both sides of your equation. What's the unit of angular momentum? What's the unit of your result on the right-hand side?
     
  7. May 29, 2014 #6
    The question is plastered as a picture, try reloading the page. I've realised that my first step is wrong so I have corrected the working - and found that ##\langle L_z \rangle = 0##!
     
    Last edited: May 29, 2014
  8. May 29, 2014 #7
    Hello unscientific et al!

    I think that no calculation is necessary here if you just memorize the first few spherical harmonics. Then the form of the wavefunction given should look very familiar.

    The part of the question "what would you expect to be the result of" seems a little ambiguous to me. Not quite sure what they want there :confused: But I guess if you write down a list of possible outcomes and their probabilities, that should cover everything.
     
  9. May 29, 2014 #8
    You are right! It is a linear combination of ##Y_1^0##, ##Y_1^1## and ##Y_1^{-1}##. What does that say about ##\langle L_z \rangle ##?

    I know that ##L_z Y_l^{m} = mY_l^m##, so:

    [tex]\langle \psi | Y_l^m |\psi \rangle = (1 + 0 - 1) \langle \psi |\psi\rangle[/tex]

    I think that's why it is zero. If ##\langle L_z \rangle = 0 ##, it means that it is spherically symmetric about the z-axis, which is exactly what spherical harmonics are! Then I can prove it using the short working above. I think the question first wants us to show it to be zero explicitly, then use spherical harmonics to explain.
     
    Last edited: May 29, 2014
  10. May 29, 2014 #9

    strangerep

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    Well, that's not "exactly what spherical harmonics are" -- look up Wikipedia to find out what they really are.

    And what is the "it" that you say is spherically symmetric about the z-axis here? (##\Psi## is not.)

    Although you can use spherical harmonics, you really only need the azimuthal part (the ##\phi##-dependent stuff).

    If you want to delve into the problem a bit deeper, then find the eigenvalues (and associated eigenvectors) of ##L_z##, i.e., solve the equation $$ -i\hbar \, \frac{\partial \psi}{\partial \phi} ~=~ \lambda \psi ~,$$(with boundary conditions appropriate to this situation). I.e., find all possible solutions ##\psi(\phi,\lambda)## and the respective values of ##\lambda##.
     
  11. May 30, 2014 #10
    I know ##Y_l^m## is the angular part of the solution containing ##\theta,\phi##. What does the question want me to "comment on" ? That its average ##L_z## is zero? meaning its angular momentum is spherically symmetric about the z-axis.

    What about for one particle?
     
  12. May 30, 2014 #11

    strangerep

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    I would have said just that, and not much more. I.e., by making many measurements on an ensemble corresponding to this state, the expectation value should approach 0 as the number of measurements increases, even though individual measurements may give other values along the way.

    That's got me scratching my head about what the questioner intended. It depends on what the questioner meant by "expect". In a strict and literal statistical meanings, one "expects" to get the "expectation value" (if one conducts only one measurement on one atom). But a less literal meaning could also be answered by giving probabilities for the different possible measured values of ##L_z## in this state.

    The question could probably have been worded more clearly.
     
  13. May 31, 2014 #12
    That's not quite right - the expectation value is (rather confusingly) not the same as the value one "expects". Indeed, the expectation value might not even be in the spectrum of the observable! Here (if my mental math is correct at this time of the morning!?), you have a 1/3 chance of getting the expectation value, so it's no more likely than any other outcome.

    But then again, if that's the distinction that the author of the problem is looking for, you'd think that he/she would have designed things so that it's impossible to get the expectation value.

    It looks like the op cut off the beginning of this problem before posting? Perhaps seeing that might give us a context clue as to what the author is getting at?
     
  14. May 31, 2014 #13

    strangerep

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    Yeah, I know. I was just trying (unsuccessfully, it would seem) to illustrate how confusion can arise depending on how one interprets the word "expect" in advance of actually doing any measurement. :frown:
     
  15. Jun 1, 2014 #14
    Here's the full question! Not sure if the earlier parts are useful at all...
    5wkmqq.png
     
  16. Jun 2, 2014 #15
    Hmm you're right, it doesn't help much :frown:

    So my guess is that you're supposed to say that if you measure [itex]L_z[/itex] for lots and lots of atoms and then take the average, you get the expectation value (and that's the only reason there are lots of atoms in the question). And that if you measure [itex]L_z[/itex] for one atom, you have a 1/3 probability of getting each of the 3 outcomes - so you don't really "expect" any of them. Did you get those probabilities ok?

    Btw where are all these questions from that you post here? They look like British final exam problems?
     
  17. Jun 3, 2014 #16

    BvU

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    Or could it be 0.25, 0.5, 0.25 ?
     
  18. Jun 3, 2014 #17
    It looks like a sum of equally weighted spherical harmonics to me, so 1/3 - 1/3 - 1/3. But I could be wrong. :smile:
     
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