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Confused about Shared Eigenstates

  1. Mar 17, 2010 #1
    I'm teaching myself QM and have a question about a problem in my text. (If this makes it homework, sorry to post in the wrong forum).

    Basically the question is about three observables A,B,C. They obey the following rules:

    [A,B]=0, [A,C]=0, but [B,C] not equal to 0.

    So this is like the situation with the total squared angular momentum L^2 and the axial angular momenta Lx, Ly, Lz. Now I know that if the observables commute, they "share a complete set of eigenstates." So in this situation you can have eigenstates of A that are also eigenstates of B or C, but not both at the same time [yes?]. So you can't have a single quantum state that is an eigenstate of all three [yes?]. Now the question asks about the effects of these relations on sequential measurements of the observables. If you start in an A eigenstate, and measure B or C, could the state be left unchanged since they both "contribute" eigenstates to A? If you start in B or C and measure the other, will you get an even spread across the available eigenstates since they don't commute?

    Or, am I drifting hopelessly in Hilbert Space??
     
  2. jcsd
  3. Mar 17, 2010 #2
    starting with an A's eigenstate and measuring B and C will never change the eigenstate, since the state is an eigenstate of B and C as well (as you pointed out) therefore has a definite eigenvalue (which will be measured with 100% probabillity).

    But the eigenstates of B are not eigenstates of C (again, as you pointed out)- that doesn't necesarilly means that an eigenstate of B is uniformly superpositioned from eigenstates of C. I mean, an eigenstate of B can be much more closer to one eigenstate of C and not the other. F.e.:

    [tex]|B_{1}>=\frac{\sqrt{3}}{2}|C_{1}>+\frac{1}{2}|C_{2}>[/tex]

    Perhaps the commutation property [A,B]=0,[B,C]=0 does tell that its spread uniformly, but it's not observed easily. In the general case, there is no restriction on how eigenstates are expressed in terms of other operator's eigenstates.
     
  4. Mar 17, 2010 #3
    Okay, thank you for that. I wondered if [B,C] not equaling zero made it a straight 50-50 coin flip or not.

    Now I'm definitely a beginner at QM, I'm trying to get the overall structure (I'm reading a good text, Essential Quantum Mechanics by Bowman). So in the above situation A's eigenstates (which are e.g. just eigenvectors in a matrix representation), form a complete set that spans the space (since A is an observable=Hermitian). And these are the same as B's and C', since it commutes with them. But then how can B's not be C's? I'm probably missing something basic...
     
  5. Mar 17, 2010 #4

    Matterwave

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    I believe commutation being zero only says that the two observables may share eigenstates. I have never seen a proof that they must share eigenstates or that their eigenstates must be the same. If two observables don't commute, then they literally cannot share eigenstates.

    Again, just because I've never seen a proof that commutative observables must share eigenstates, doesn't mean that no such proof exists. I may be wrong on this.
     
  6. Mar 17, 2010 #5
    Okay, my text says "If F,G share a complete set of eigenstates, then [F,G] vanishes," where "complete set" means a set of vectors that are linearly independent and span the space (i.e. a basis). I thought that since observables are Hermitian and therefore have a complete set, there was just one such complete set for each observable (i.e. its eigenvectors in the matrix representation). Now with angular momentum for example, Lz and L^2 commute, and therefore share eigenstates. But L^2 also has eigenstates with Lx and Ly. So its like L^2 has three distinct sets of eigenstates. So are these sets all complete?
     
  7. Mar 17, 2010 #6

    Matterwave

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    Well, it's certainly true that "If F,G share a complete set of eigenstates, then [F,G] vanishes"; however, does your book mention anything to the converse? "If [F,G] vanishes, then F,G share a complete set of eigenstates"?
     
  8. Mar 17, 2010 #7

    SpectraCat

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    Yes, but it is probably more correct to say that L^2 has a complete set of eigenstates that can be expressed in the basis of one of the other 3 operators, say Lz. The eigenstates of the remaining operators (Lx & y), can then be expressed as linear combinations in the Lz eigenbasis. Each of these linear combinations is of course also an eigenstate of L2, however, they are not *new* eigenstates ... they are degenerate states that are non-orthogonal to the Lz basis by definition, since they are expressed as linear combinations of these states.

    Thus, just as in the Stern-Gerlach experiment, you are free to choose one spatial axis to define eigenstates of the projection of angular momentum. If you then try to measure the projection of angular momentum on any other axis for one of these states, you will find the system in a superposition (i.e. not an eigenstate).
     
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