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Collapse of wave function question

  1. Jul 26, 2012 #1
    Hi all, I'm doing a practice question in which we have a hydrogen atom in the state:

    [tex]\psi = (2\psi_{100} + \psi_{210} + \sqrt{2}\psi_{211} + \sqrt{3}\psi_{21 -1})/\sqrt{10}[/tex]

    It says that, now a measurement is taken and we find the angular momentum variables to be L = 1 and L_z = 1. The question is: immediately after the measurement, what is the wave function?

    Now, I thought that we definitely observed only [itex]\psi_{211}[/itex] here, because that is the only one with these values. So, I thought that collapses the wave function to this eigenstate, and it basically stays there.

    However, it appears I'm wrong. The answer I have says that there's now a new wave function comprised of all three of the L = 1 states, and they do a bit of math to figure out their coefficients.

    Can anyone help?

    Thanks!

    PS: I never found out an answer to this question, if anyone could help me here.
     
  2. jcsd
  3. Jul 26, 2012 #2

    Simon Bridge

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    Measuring one component fixed that one in a particular state, but the others remain in superposition. iirc the picture you should have in your text/notes has moment precessing in the x-y plane.
     
  4. Jul 31, 2012 #3
    I know the drawing you're talking about, but I'm still not sure what you're saying here. I thought that if we just measured L^2 (essentially L), then after the measurement it could still be in a combination of states with that value of L^2, but it says we find L_z also. Doesn't that mean we specifically measured ψ_211? How can the others still be in superposition?
     
  5. Aug 1, 2012 #4

    Simon Bridge

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    But that would mean that on subsequent measurement of Lx, say, you'd get nothing right? But that's not what happens is it?

    Anyway ... in the vector model you've seen, which is handy for visualizing this, the angular momentum is a vector with some angle .... when you measured Lz you oriented it to the z axis (basically the measuring process defines what you mean by "the z axis".)

    When you measure Lz you are measuring only one component of the overall vector - that won't make the other components go away. It just stops that component from being uncertain ... subsequent measurements of Lz will give you the same value... if the system originally was in a superposition of Lz states then you'll have just collapsed that part of the wavefunction... since you made no measurements on the x and y components, those parts are unaffected*.

    -------------------
    * this works for commuting observables
    http://farside.ph.utexas.edu/teaching/qmech/lectures/node70.html
    http://nonlocal.com/hbar/commuting.html [Broken]
     
    Last edited by a moderator: May 6, 2017
  6. Aug 2, 2012 #5
    If in your notation with [itex]\psi_{211}[/itex] you are meaning the wave function with [itex]E=E_2 \quad L^2 = 1(1+1)= 2 \quad L_z =1[/itex], then I'd say you are right, this is the wave function after the measurement of your exercise.

    The measurement projects the state of the atom on the space spanned by [itex]\{|n,L^2=2,L_z=1\rangle, n\in \mathbb{N} \}[/itex]. The only state of the starting superposition with non zero component on this space is the one VortexLattice said.

    edit
    Exactly, it would work if it was [itex][L_i,L_j]=0[/itex], but [itex][L_i,L_j]=i \epsilon_{ijk}L_k[/itex] so they do not commute and the x and y components of angular momentum actually are affected by the measurement of [itex]L_z[/itex].

    Ilm
     
    Last edited by a moderator: May 6, 2017
  7. Aug 2, 2012 #6
    Hmmm, but from what you said it still seems like I'm right... You said "When you measure Lz you are measuring only one component of the overall vector - that won't make the other components go away. It just stops that component from being uncertain ... subsequent measurements of Lz will give you the same value...".

    But for subsequent measurements of Lz to give the same value (1, in this case), it seems like the wave function has to be in the ψ211 state! If it were in a superposition of the three different L = 1 states (for E = 2), subsequent measurements could give values of Lz = -1,0,1...right?

    I still assume the book is right and I'm missing something, but I'd like to know what.
     
    Last edited by a moderator: May 6, 2017
  8. Aug 2, 2012 #7
    Is there maybe some misunderstanding in notation or on the exercise text?

    Ilm
     
  9. Aug 2, 2012 #8
    Just to make sure, here is the problem (part e):

    Df1E6.png

    And here is the solution:

    UBm60.png

    Unless I'm missing something, they just seem to ignore the part where Lz is measured. If they only measured L = 1, this seems like it would be the right answer.
     
  10. Aug 2, 2012 #9
    As I supposed, there was a misunderstanding in the notation.

    The book says that [itex]L_z= \frac{1}{2}(L_+ + L_- )[/itex], this means that the quantum number m is the eigenvalue of [itex]L_y[/itex] ([itex]L_-[/itex] lowers m by 1).

    I hope this solves the problem^^

    Ilm
     
  11. Aug 2, 2012 #10
    Wait, what the hell? First of all, I really thought [itex]L_{\pm} = L_x \pm iL_y[/itex]. Second of all, I have never ever seen this before, I thought m was always the L_z quantum number...
     
  12. Aug 2, 2012 #11
    Well, since all [itex]L_i[/itex] commute with [itex]L^2[/itex], you can choose the angular momentum component you prefer to describe your states.

    Usually your definition of [itex]L_±[/itex] is right, because is usually used [itex]L_z[/itex] to describe the states. The useful definition of [itex]L_±[/itex] is as the operators that increase (or lowers) the value of m. So if by m you mean the eigenvalue of [itex]L_y[/itex] the operators [itex]L_±[/itex] have to change if they have to increase or lower the value of [itex]L_y[/itex].

    To be brief, since the name of the axis is arbitrary, you can simply change [itex] x→ z ,\: y→ x, \: z→ y[/itex] (to maintain the right handed orientation).

    Then [itex]L_± = L_z ±i L_x[/itex] and so [itex]L_z= \frac{1}{2}(L_+ + L_-)[/itex].

    Ilm
     
  13. Aug 2, 2012 #12
    Ok, I see that because of the symmetry, that makes sense. But it seems like nowhere in the question does it say that m is the y component of angular momentum, and it's been z everywhere else I've seen it. Is there something I'm not seeing where they make it clear that m is the y component, in the question?
     
  14. Aug 2, 2012 #13
    Absolutely not.

    I can only see it from the answer given by the book.

    I suppose it's the standard notation of the book?

    Ilm
     
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