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Hydrogen Atom: Wavefunction collapse after measurement of Lz

  1. Oct 12, 2016 #1
    1. The problem statement, all variables and given/known data
    Suppose we have a wavefunction with n=4. If we measure the orbital angular momentum along the z-direction(no spin in this problem) and get 2*hbar then what are the possible values of the total angular momentum and what is the most general wavefunction after the measurement?

    2. Relevant equations
    l=0,1,2,..,n-1
    m=-l, -l+1,..., l-1, l
    Lz=m*hbar

    3. The attempt at a solution
    Since n=4--> l=1, 2, 3. Since we measured Lz and found it to be 2*hbar, then m=2 which could only correspond to l=2 or l=3.
    Now, since m=2 but we don't know l, the most general wavefunction after the measurement could only be C*R4l*Y[l=l m=2] where C is just a normalization constant.
    But, my professor said this is wrong and the most general wavefunction is:
    C1*R42*Y[l=2 m=2]+C2*R43*Y[l=3 m=2].
    But, I can't understand why. When we measure Lz, then doesn't the wavefunction collapse to a state of particular l(here 2 or 3 depending on the total angular momentum that the wavefunction had before measurement) and m(here 2)?

    Thanks in advance
     
  2. jcsd
  3. Oct 12, 2016 #2

    PeroK

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    How can you rule out the ##l = 3## case?
     
  4. Oct 12, 2016 #3

    PeroK

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    Or, look at it this way: Your professor's wave function would give measurements of ##n=4## and ##l=2## and it's more general than your wave function.
     
  5. Oct 12, 2016 #4
    So, if we measure the z component of the angular momentum the wavefunction collapses to a superposition of all the spherical harmonics(with their corresponding radial part) with m that corresponds the the measured value of Lz? I thought it would collapse a a particular(not superposed) wavefunction with spherical harmonic part with l being the l of the total angular momentum. Why isn't this true though? I might have a gap in my knowledge of the subject.
     
  6. Oct 12, 2016 #5

    vela

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    Suppose the answer to your question is yes. If the atom was in the state your professor said, a measurement of Lz would always corresponds to m=2. How do you propose to choose whether to keep the l=2 or l=3 piece of the wave function?
     
  7. Oct 12, 2016 #6
    Well, my answer was based on that I can't know! That's why i wrote the solution as Y[l=l, m=2]. I thought that the wavefunction would collapse to a single eigenstate of l. But, I can't understand why his answer is right and mine is wrong.
     
  8. Oct 12, 2016 #7
    I didn't rule out l=3. My answer kept l=l.
     
  9. Oct 12, 2016 #8

    vela

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    So why'd you choose l=2 instead of l=3?
     
  10. Oct 12, 2016 #9

    vela

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    You're assuming there's a definite ##l## initially. Is that justified?
     
  11. Oct 12, 2016 #10

    PeroK

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    Loosely speaking the wavefuction only collapses as much as it needs to! You mention spin: Unless you measure spin the spin component of the overall wave function will not collapse at all.
     
  12. Oct 12, 2016 #11
    Ι did not choose l=2 or l=3. I just say that we can't know.
    Isn't l the l of the total angular momentum?
     
  13. Oct 12, 2016 #12

    PeroK

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    Sorry, I didn't see that. What does ##l=l## mean?
     
  14. Oct 12, 2016 #13
    I did not understand where does spin come into play.
     
  15. Oct 12, 2016 #14
    I meant that I just wrote Y[l, 2]. I kept l general. Unknown.
     
  16. Oct 12, 2016 #15
    You mean that the initial wavefunction could be in a superposition of eigenstates of different l's and m's and upon measurement it collapsed to a superposition of those eigenstates with the measured m? That is, you are implying that my solution took as granted that the initial wavefunction was in a superposition of states but only one of those had m=0, right?
     
  17. Oct 12, 2016 #16

    PeroK

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    In this case, you would have a definite value for ##l## you just don't know what it is. That is fundamentally different from it could be either. Until you measure ##l## you can't say it has a definite value. It could still be 2 or 3, hence by definition it is a superposition of those states.

    The wave function collapse doesn't result in a definite value of all compatible observables. For example, after you measure ##n=4## you don't have a definite value of anything other than ##n##. Everything else is not unknown - in terms of definite values that you just don't know yet. They do not have a definite value into they are measured.

    Does that make sense?

    This is quite fundamental to QM.
     
  18. Oct 12, 2016 #17
    Yes, I understand what you mean. So, is my previous comment correct then(my response to vela)?
     
  19. Oct 12, 2016 #17

    PeroK

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    Possibly but I didn't understand the bit about m = 0.

    Your misunderstanding was quite subtle so it's hard to explain.

    I'm offline for a bit now.
     
  20. Oct 12, 2016 #18
    Sorry, I meant m=2.
    To rephrase, I meant that in my solution(where the wavefunction collapses to a single eigenstate of l) took as granted that the initial wavefunction(the one before the measurement) had only one eigenstate of m=0 along the superposed eigenstates. And this is wrong because the superposition(the initial wavefunction) could contain any eigenstate with m=2(in this case it could contain an eigenstate with Y[2 2] and an eigenstate with Y[3 2]) and not just one like my solution implied(because I only wrote Y[l 2] and took for granted that we also knew l).

    Did I get it right?
     
  21. Oct 12, 2016 #19

    PeroK

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    I think that's right. If you measure all three quantum numbers, then (after those measurements) you have a wavefunction of the form: ##\psi_{nlm}##, where ##n, l, m## are the results of your measurements. But, the most general state of the hydrogen atom is not ##\psi_{nlm}##, where ##n, l, m## are unknowns (*).

    (*) This would suggest that the hydrogen atom, before measurement, has definite values for the quantum numbers, which you find out by measuring them. This is not the case.

    Instead, the most general state is:

    $$\sum_{n, l, m} C_{nlm} \psi_{nlm}$$

    This implies that the quantum numbers do not have definite values until you measure them. And, if you measure only one, the others do not assume definite values until they too are measured. After a measurement of ##n=4##, the most general state is:

    $$\sum_{l, m} C_{lm} \psi_{4lm}$$

    Where ##l,m## take all the allowed values compatible with ##n = 4##.

    And, after a further measurement of ##m=2## the most general state is:

    $$\sum_{l} C_{l} \psi_{4l2}$$

    Where ##l = 2, 3##.

    That's the process by which you should have reached the answer. Hope that helps.
     
  22. Oct 13, 2016 #20
    Yes, that's how I understood this. Thanks a lot!
     
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