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Commuting quantum mechanical operators

  1. Nov 30, 2014 #1
    1. The problem statement, all variables and given/known data
    Two Hermitian operators X and Y have a complete set of mutual eigenkets. Show that [X,Y]=0 and interpret this physically.

    2. Relevant equations
    [X,Y]=XY-YX
    If [X,Y]=0, XY=YX

    3. The attempt at a solution
    I have proved that [X,Y]=0, but I'm just falling a little short of understanding the physics of what is going on, so let me write down what I have.

    Suppose our system is in the state represented by the ket |Ψ> and that the complete set of mutual eigenkets is {|φn>}, with eigenvalues xn and yn for the operator X and Y respectively.

    We may write our state in terms of these eigenkets, so that
    |Ψ>=Σncnn>
    where cn represents the quantum amplitudes of the eigenkets (or eigenstates).


    Now XY=YX implies XY|Ψ>=YX|Ψ>.

    Now this means that, if we first measure the observable corresponding to the operator X, our state collapses to some eigenstate |φn> and we observe it's corresponding eigenvalue xn with probability |cn|2. We can then measure the observable corresponding to the operator Y, with the state of the system fixed in the same eigenstate, and we measure the eigenvalue yn, but now with probability 1, because we're in an eigenstate - so the overall probability of measuring these values was |cn|2. We can do this the other way around, and we still measure the same observables with the same probability. Thus the fact these observables commute means that there is no uncertainty measured in the observables corresponding to the operators X and Y.

    The bit in bold is what I don't get - I believe this link is true, but I don't see how we can associate say XY|Ψ> with first trying to measure the observable corresponding to Y (and so collapsing the state to an eigenstate) and then associate X as trying to measure the observable corresponding to X - if you mathematically do this, it doesn't collapse the state down to some eigenstage. The same applied to YX|Ψ> too. Could anybody explain this? Thankyou.
     
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  3. Nov 30, 2014 #2

    Simon Bridge

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    Mutual eigenkets would be |n.m> representing state xn and ym since n need not be the same as m.
    That would be ike wavefunction φnm. So X|nm> = xn|n,m>

    Because operators don't mean anything by themselves, XY = YX actually means that XY|ψ> = YX|ψ>
    Applying an operator to a sate vector does not "collapse" it - no. It just tells you the possible outcomes of a measurement.

    Apply that way of thinking, and compare with the case where the operators do not commute.
     
  4. Dec 1, 2014 #3
    Thanks for the reply. Let me just modify my maths to take into account what you have done and to try and understand something a bit further.

    Suppose our system is in the state represented by the ket |Ψ> and that the complete set of mutual eigenkets is {|φnm>}, with eigenvalues xn and ym for the operators X and Y respectively.

    We may write our state in terms of these eigenkets, so that
    |Ψ>=Σn,mcnmnm>
    where cn,m represents the quantum amplitudes of the eigenkets (or eigenstates).


    So I think that clears that up - I've stuck with |φnm> rather than |n,m> to stick with my convention. Now suppose I act on this with B then A, i.e compute AB|Ψ>:
    AB|Ψ>=ABΣn,mcnmnm>
    =AΣn,mcnmB|φnm>
    =AΣn,mcnmbmnm>
    n,mcnmbmA|φnm>
    n,mcnmbmannm>
    This is the same if we carry out BA|Ψ> as we have proven that A and B commute. So basically applying the operators causes the eigenvalues to enter the sum - I want to ask about this below.

    You said that 'applying an operator to a state vector does not "collapse" it - no. It just tells you the possible outcomes of a measurement.' So for AB|Ψ>=Σn,mcnmbmannm>, we are finding the observables A and B of a system. The possible states are the |φnm> and in each of these states we can measure the observables as all combinations of an,bm, the eigenvalues. Now usually if I have say for some state |Ψ> with eigenstates |φn> (which have eigenvalues φn)
    |Ψ>=Σncnn>
    and the probability of being in state |φn> and thus measuring φn is given by pn=|cn|2. So returning to the above relation, does that mean by operating on the system by AB or BA, the probabilities of being in a particular state |φnm> are now modified to pnm=|anbmcnm|2? If so, what is the meaning of this?
     
    Last edited: Dec 1, 2014
  5. Dec 2, 2014 #4

    Simon Bridge

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    [Quote Now suppose I act on this with B then A, i.e compute AB|Ψ>:
    AB|Ψ>=ABΣn,mcnm|φnm>
    =AΣn,mcnmB|φnm>
    =AΣn,mcnmbm|φnm>
    =Σn,mcnmbmA|φnm>
    =Σn,mcnmbman|φnm>[\quote]
    In general the simultaneous eigenkets of X and Y will not be eigenkets of A and B... in fact, from you notation, you have implied that B must have a special relationshop with Y.
    In general you need to do a change of basis.
     
  6. Dec 2, 2014 #5
    Sorry, for some strange reason I switched from X,Y to A,B. Let me copy my above post and edit it:

    Thanks for the reply. Let me just modify my maths to take into account what you have done and to try and understand something a bit further.

    Suppose our system is in the state represented by the ket |Ψ> and that the complete set of mutual eigenkets is {|φnm>}, with eigenvalues xn and ym for the operators X and Y respectively.

    We may write our state in terms of these eigenkets, so that
    |Ψ>=Σn,mcnm|φnm>
    where cn,m represents the quantum amplitudes of the eigenkets (or eigenstates).


    So I think that clears that up - I've stuck with |φnm> rather than |n,m> to stick with my convention. Now suppose I act on this with Y then X, i.e compute XY|Ψ>:
    XY|Ψ>=XYΣn,mcnm|φnm>
    =XΣn,mcnmY|φnm>
    =XΣn,mcnmbm|φnm>
    =Σn,mcnmbmX|φnm>
    =Σn,mcnmbman|φnm>
    This is the same if we carry out YX|Ψ> as we have proven that X and Y commute. So basically applying the operators causes the eigenvalues to enter the sum - I want to ask about this below.

    You said that 'applying an operator to a state vector does not "collapse" it - no. It just tells you the possible outcomes of a measurement.' So for XY|Ψ>=Σn,mcnmbman|φnm>, we are finding the observables X and Y of a system. The possible states are the |φnm> and in each of these states we can measure the observables as all combinations of an,bm, the eigenvalues. Now usually if I have say for some state |Ψ> with eigenstates |φn> (which have eigenvalues φn)
    |Ψ>=Σncn|φn>
    and the probability of being in state |φn> and thus measuring φn is given by pn=|cn|2. So returning to the above relation, does that mean by operating on the system by XY or YX, the probabilities of being in a particular state |φnm> are now modified to pnm=|anbmcnm|2? If so, what is the meaning of this?
     
  7. Dec 3, 2014 #6

    Simon Bridge

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    I may not be making myself clear:
    The probability that a measurement will find the system described by state vector |Ψ> in a particular combined eigenstate |nm> is

    Pnm = (φnm)*φnm

    Where: φnm = <nm|ψ> : X|nm> = xn|nm> and Y|nm> = ym|nm>
    (All basis states are orthonormal.)

    So XY|nm> = YX|nm> = xnym|nm> ... you can imagine if X is the width and Y is the height, then the result of XY simultaneoulsy is to give you the area of the state. This would be like defining a new operator A: A=XY(="area") and applying that.

    This is a bit tongue-in-cheek: the exact meaning of the result depends on the exact meaning of X and Y. i.e. simultaneous measurement of H, L and S of hydrogen atoms.

    In terms of the state vector: XY|Ψ>=Σij xiyj φij|ij> ... all this is telling you is the decomposition into eigenvalues weighted by the probability amplitudes. You've seen this already with single operators - like when you looked at energy eigenvalues.

    What we are usually interested in is the result of subsequent measurements rather than simultaneous measurements. The key point is that if the operators commute, measuring one does not affect the subsequent measurement of the other - compare with measuring position then momentum. As you advance through your course, other consequences will appear. Try not to swallow too much at once.

    (I have a nagging feeling I missed something...)
     
  8. Dec 3, 2014 #7
    Thanks. I think I understand it now. I think trying to understand it in the order
    A,B have mutual eigenstates -> [A,B]=0 -> measurement of A doesn't affect B / vice versa
    threw me off. It's better to look at it as
    [A,B]=0 -> A,B have mutual eigenstates -> measurement of A doesn't affect B / vice versa (because we can write the state as |Ψ>=Σn,mcnm|φnm> and state |φnm> has probability amplitude |cnm|2 - this state corresponds to both measuring an and bm as the observables always, so whatever order we do the measurements in is irrelevant), in my opinion anyway.

    Thanks for the help :)
     
  9. Dec 3, 2014 #8

    Simon Bridge

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    Well done - if you just measured X, and get particular value, then the state vector has "collapsed" to a new state, which is a sum over the possible Y states.
     
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