Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Degenerate states and wavefunction collapse

  1. Jul 19, 2012 #1
    So, in QM making a measurement collapses the state into an eigenstate of that observable. Thus, if the system is properly isolated, then the same measurement should return the same value. But the eigenvalue for that state is degenerate, then does that mean the state might actually collapse to a different state in the same eigenspace after the second measurement?

    Here is what I mean, if that's not clear:

    Suppose I make a measurement associated with the operator "O" and get a result "o". My system is now in the eigenstate |ψ0>, where O|ψ0> = o|ψ0>

    So far so good. But let's suppose the eigenvalue "o" is actually degenerate with multiplicity two. Thus, there exist linearly independent |ψ1> and |ψ2>, both having eigenvalue o under O, such that:
    0> = c11> + c22>
    Therefore, by the postulates of QM, a second measurement of O will collapse the state of the system to |ψ1> with probability |c1|2 and to |ψ2> with probability |c2|2.

    Aside from the fact that it seems strange that a repeated measurement could change the system's state (even if it yields the same value for the observable), it's even more confusing because of the arbitrariness of the basis for the eigenspace. I could just as well choose |ψ'1> and |ψ'2> such that:
    0> = c'1|ψ'1> + c'2|ψ'2>
    and now there are four different states my system could collapse to and the probabilities sum to 200% ...and you could, of course, repeat this process infinitely many times, choosing a new eigenbasis each time.

    So... clearly I'm missing something.
     
  2. jcsd
  3. Jul 19, 2012 #2
    Since both [itex]\Psi_1[/itex] and [itex]\Psi_2[/itex] have eigenvalue [itex]o[/itex], repeated application of the [itex]O[/itex] operator will simply change the normalization of the state:

    [itex]O\Psi = O(c_1\Psi_1 + c_2\Psi_2) = o(c_1\Psi_1 + c_2\Psi_2)[/itex]

    Since normalization is not a physically measurable quantity, repeated application of O doesn't change anything--it leaves it in the same state as it was coming in.
     
  4. Jul 19, 2012 #3

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    To see a difference between the two states, you have to measure a different observable - something which has different eigenvalues for the states. In this case, your choice of measurement defines the relevant basis of the system.
     
  5. Jul 19, 2012 #4
    Can you clarify this? I don't see from the eigenvalue equation you posted how the normalization is changing.
     
  6. Jul 19, 2012 #5
    Let me check to make sure I didn't misunderstand you. You say that we have [itex]\Psi_1[/itex] and [itex]\Psi_2[/itex], which are degenerate with respect to operator [itex]O[/itex], right? Therefore by definition they have the same eigenvalue, which we call [itex]o[/itex].

    If that's the case, then given a state [itex]\Psi = c_1\Psi_1 + c_2\Psi_2[/itex], operating on it with [itex]O[/itex] produces:
    [tex]O\Psi = O(c_1\Psi_1 + c_2\Psi_2) = Oc_1\Psi_1 + Oc_2\Psi_2 = c_1O\Psi_1 + c_2O\Psi_2 = c_1o\Psi_1 + c_2o\Psi_2 = o(c_1\Psi_1 + c_2\Psi_2) = o\Psi[/tex]
    So we got back out our original state, with the entire thing multiplied by a constant. Therefore, the state didn't actually change, it just got its normalization changed.

    Another way to say this is that since you are working in a degenerate basis, you are free to choose the basis vectors however you want, because any linear combination of them is still going to have eigenvalue [itex]o[/itex]. So there's no point in splitting them up into [itex]\Psi_1[/itex] and [itex]\Psi_2[/itex], you might as well just define [itex]\Psi[/itex] itself to be one of the basis vectors (with eigenvalue [itex]o[/itex], of course), and then it's obvious that operating on it with [itex]O[/itex] doesn't change the state.

    Does that answer your question, or did I misinterpret what you're asking?
     
  7. Jul 19, 2012 #6
    Actually, re-reading your question again, I think I understand what you're asking now. The answer is that "the postulates of QM" that you mentioned don't apply to degenerate eigenstates they way you're trying to force them to. Specifically, all QM says is that after observation, the system will be in an eigenstate of the operator, with definite certainty. Since any combination of degenerate eigenstates is also an eigenstate with the same value, this requirement is satisfied automatically, without having to change the value of the state at all.

    So, if the state starts out as [itex]c_1\Psi_1 + c_2\Psi_2[/itex], then that's just [itex]\Psi[/itex] in a different basis, so the wavefunction is already "collapsed" into an eigenstate of [itex]O[/itex], and it won't change at all. So, no matter how many times you apply [itex]O[/itex], it'll stay in exactly the same state as it started out in. It won't collapse into [itex]\Psi_1[/itex] or [itex]\Psi_2[/itex], because, as you mentioned, those are arbitrary anyway, and it wouldn't have any idea which one to go into.
     
  8. Jul 19, 2012 #7
    I think you misunderstood me a bit... I'll try to explain what I mean a bit better.

    It's my understanding that one of the fundamental postulates of quantum mechanics is as follows: "Let the state of a system, [itex]|\phi>[/itex], be expressed as a linear combination of eigenkets of some operator [itex]O[/itex] as follows: [itex]|\phi> = c_1|\psi_1> + c_2|\psi_2> + ... + c_n|\psi_n>[/itex], where [itex]\psi_i[/itex] is an eigenket with eigenvalue [itex]\lambda_i[/itex]. Then the measurement associated with [itex]O[/itex] will yield one of the eigenvalues and will collapse the state of the system to [itex]|\psi_i>[/itex] with probability [itex]|c_i|^2[/itex]."

    But supposed the system is twice-degenerate with [itex]\lambda_1 = \lambda_2[/itex] and the measurement associated with [itex]O[/itex] yielded this eigenvalue. Additionally, suppose we have a commuting operator for which [itex]|\psi_1>[/itex] and [itex]|\psi_2>[/itex] correspond to different eigenvalues. Thus, we do that commuting measurement and determine that the system has in fact collapsed to [itex]|\psi_1>[/itex]. So far, so good. But I can now choose a new basis for the eigenspace associated with [itex]\lambda_1[/itex] and write:
    [itex]|\psi_1> = c'_1|\psi'_1> + c_2'|\psi'_2>[/itex]

    Now suppose I do the measurement associated with [itex]O[/itex] again. Instead of going straight to the eigenvalue equation like you did, just apply the postulate I stated with to the above equation. We have:
    [itex]|\psi_1> = c'_1|\psi'_1> + c_2'|\psi'_2> + 0|\psi_3> + ... + 0|\psi_n>[/itex]
    Hence, according to the postulate, after I make measurement (which, yes, will yield [itex]\lambda_1[/itex] with probability 1, but I'm concerned about state) the state will collapse to [itex]|\psi'_1>[/itex] with probability [itex]|c'_1|^2[/itex] and [itex]|\psi'_2>[/itex] with probability [itex]|c'_2|^2[/itex]. Thus, even though the system is already in an eigenstate of [itex]O[/itex], doing the associated measurement changes the state to a new eigenstate (albeit, one with the same eigenvalue).

    Aside from the fact that this flies in the face of my understanding of QM, it completely breaks down if instead I choose another basis (say, double primed) for the eigenspace and get two more states. These may be linearly dependent on the primed states, but they're still distinct. This is obviously ridiculous since the probability of ending up in one of the primed states was already 1.

    I hope that clarifies my question, and, I believe also shows why I don't think mfb's explanation solves the problem.
     
  9. Jul 19, 2012 #8
    Ah, you snuck in your reply before I could make mine. OK, I think I see. So in my above reply, what I stated as a postulate of QM is not correct?
     
  10. Jul 19, 2012 #9
    Correct. You're trying to mechanically apply a principle in a situation that it isn't meant for. All the rules say is "if a state is in a linear combination of two distinct eigenstates, observation will collapse the state into one or the other, with a probability related to the relative coefficients of the two states". The operative word here is "distinct"--the point of this rule is to ensure that the resulting state has a definite eigenvalue, so it only applies when there are states with different values mixed together.

    In this case, your state [itex]c_1\Psi_1 + c_2\Psi_2[/itex] could just as well be written as [itex]\Psi[/itex] by a change of basis, so there's no collapsing to be done.
     
  11. Jul 19, 2012 #10
    OK, I get it. Thank you!

    If I could trouble you (or anyone else reading this thread) further, I recently posted another question about degenerate eigenspaces (as they show up in representation theory) here: https://www.physicsforums.com/showthread.php?t=621178 (it's the question in my second post that's causing me problems, not my first). Unfortunately, I haven't gotten any help from the people in the math forum. As my progress there has ground to halt due to this problem, would you mind taking a look and seeing if you know the answer? If I can get that last bit down, I think that will take care of all the points of degenerate eigenvalues that I've found confusing.

    Thanks again for your help!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Degenerate states and wavefunction collapse
  1. Collapse Wavefunction (Replies: 2)

  2. Wavefunction collapse (Replies: 1)

Loading...