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Commuting operators require simultaneous eigenfunctions?

  1. Oct 26, 2014 #1

    kmm

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    Here is what I understand. The generalized uncertainty principle is: [itex] \sigma^{2}_{A} \sigma^{2}_{B} \geq ( \frac{1}{2i} \langle [ \hat{A}, \hat{B} ] \rangle )^2 [/itex]

    So if [itex] \hat{A} [/itex] and [itex] \hat{B} [/itex] commute, then the commutator [itex] [ \hat{A}, \hat{B} ] = 0 [/itex] and the operators are compatible. What I don't understand about this is why two compatible operators must have shared eigenfunctions or that in other words, incompatible operators such as the position and momentum operators of the original Heisenberg uncertainty principle cannot share eigenfunctions. I'm having a hard time completely understanding the proof of this.

    Proof: If we have [itex] \hat{A} f_{n} = \lambda_{n} f_{n} [/itex] and [itex] \hat{B}f_{n} = \mu_{n} f_{n} [/itex] so that [itex] f_{n}(x) [/itex] are eigenfunctions of both operators and the set [itex] \{f_{n}\} [/itex] are complete so that any function [itex] f(x) [/itex] can be written as a linear combination of them [itex] f = \Sigma c_{n} f{n} [/itex]. Then: [tex] [ \hat{A}, \hat{B} ] f = (\hat{A} \hat{B} - \hat{B} \hat{A}) \Sigma c_{n} f{n} = \hat{A}( \Sigma c_{n} \mu_{n} f{n} ) - \hat{B} (\Sigma c_{n} \lambda_{n} f{n}) = \Sigma c_{n} \mu_{n} \lambda_{n} f{n} - \Sigma c_{n} \mu_{n} \lambda_{n} f{n} = 0 [/tex] Since this is true for any [itex] f(x) [/itex] then [itex] [ \hat{A}, \hat{B} ] = 0 [/itex]

    I still don't quite see how this shows that incompatible operators don't share any eigenfunctions. I see how this means that if two operators share a complete set of eigenfunctions then they must commute. But I don't see how this rules out the possibility of two incompatible operators sharing some "incomplete" set of eigenfunctions. Maybe there is no such thing as an "incomplete" set of eigenfunctions? The way I understand it, the eigenfunctions of a Hermitian operator are complete, but does this mean that some subset of the eigenfunctions are also complete? If not, then it seems that this proof has not ruled out that two incompatible operators could share some set of eigenfunctions, although incomplete. I'm pretty certain that's wrong, but I'm not sure why or how I'm thinking wrong about this.
     
    Last edited: Oct 26, 2014
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  3. Oct 26, 2014 #2

    bhobba

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    You made an assumption the eigenfunctions are the same in your proof.

    You may find the following helpful:
    http://www.indiana.edu/~ssiweb/C561/PDFfiles/Simult-Exp2008.pdf

    That only works if its non degenerate.

    For the degenerate case remember physically you can relabel the outcomes anyway you like so you can easily turn a degenerate case to a non degenerate one without altering the physics.

    If two operators share the same common eigenvectors then its easy to see they must commute. So its really iff.

    I will let you think about the degenerate case which really requires using resolutions of the identity and not eigenvectors.

    Thanks
    Bill
     
    Last edited: Oct 26, 2014
  4. Oct 26, 2014 #3

    kmm

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    Yes, but what throws me off is that we assumed the set of eigenfunctions are complete. So what was proved that if two operators share a complete set of eigenfunctions, then they commute. But can operators share an "incomplete" set of eigenfunctions? So if an operator [itex] \hat{A} [/itex] has the set of eigenfunctions, [itex] f_{1}, f_{2}, f_{3} ... f_{n} [/itex] could there be another operator [itex] \hat{B} [/itex] that shares the eigenfunctions [itex] f_{1}, f_{2}, f_{3} ... f_{n' < n} [/itex]?

    That helps, but I'm definitely going to have to think about the degenerate case.

    Actually, here's a proof that seems to me to show that if two operators share any eigenfunctions, then the operators will commute. Perhaps you can tell me if you think I'm missing something here.

    Proof: If operators [itex] \hat{A} [/itex] and [itex] \hat{B} [/itex] share an eigenfunction then, [itex] \hat{A} f = \lambda f [/itex] and [itex] \hat{B} f = \mu f [/itex]

    From this [itex] \hat{A} \hat{B} f = \hat{A} \mu f = \mu \hat{A} f = \mu \lambda f [/itex] and [itex] \hat{B} \hat{A} f = \hat{B} \lambda f = \lambda \hat{B} f = \lambda \mu f [/itex]. Therefore, subtracting the two equations [itex] (\hat{A} \hat{B} - \hat{B} \hat{A})f = 0 [/itex] and so [itex] [\hat{A}, \hat{B}]f = 0 [/itex] for any function [itex] f [/itex]. Therefore, [itex] [\hat{A}, \hat{B}] = 0 [/itex] if they share any eigenfunction.

    *EDIT* I see that I actually haven't avoided the case of degeneracy here.
     
    Last edited: Oct 27, 2014
  5. Oct 27, 2014 #4

    bhobba

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  6. Oct 27, 2014 #5

    Avodyne

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    This is wrong; you started out with ##f## being an eigenfunction of both ##A## and ##B##, but then later assumed ##f## was any function.

    Noncommuting operators can share eigenstates. Here is a simple example for a 3-dimensional Hilbert space:
    [tex]A=\pmatrix{ 1 & 0 & 0 \cr 0 & -1 & 0 \cr 0 & 0 & 1},
    \quad
    B=\pmatrix{ 0 & 1 & 0 \cr 1 & 0 & 0 \cr 0 & 0 & 1}.[/tex] ##A## and ##B## do not commute, but have a common eigenvector ##\pmatrix{0 \cr 0 \cr 1}##.

    An infinite-dimensional example: consider a harmonic oscillator. Then the ground state ##|0\rangle## is an eigenstate of both ##a## and ##a^\dagger a## with eigenvalue zero, but ##a## and ##a^\dagger a## do not commute. The other eigenstates of these two operators are all different.
     
  7. Oct 27, 2014 #6

    kmm

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    Thanks, I took a look at it and I think I'm going to get that.

    OK. So this seems like it would create the possibility in quantum mechanics to make a simultaneous measurement of noncommuting operators. That is, I could make a measurement and find the same eigenstate for both observables. I don't think this is actually right, I'm not sure why though. What's also confusing is that I thought it is fundamental to quantum mechanics that the generalized uncertainty principle is a consequence of the fact that two non commuting operators can't share simultaneous eigenfunctions.
     
    Last edited by a moderator: May 7, 2017
  8. Oct 27, 2014 #7

    kith

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    The crucial thing is that what you call the generalized uncertainty principle depends on the state.

    For commuting observables, the right hand side is zero for all states, so uncertainties don't play a role at all.

    For canonical conjugate observables, the commutator is equal to iħ1. Since every state is an eigenstate of the identity operator 1, the right hand side is independent of the state. So uncertainties are non-zero for all states and no state is a common eigenstate of both observables (can you prove this?).

    Cases where [A,B] is nonzero and not proportional to the identity operator are kind of intermediate to the cases above. If you find a common eigenstate of A and B, <[A,B]> is zero for this state and the uncertainties can be zero. But there are also states which aren't common eigenstates and for which <[A,B]> takes a non-zero value, which restricts the uncertainties.
     
    Last edited: Oct 27, 2014
  9. Oct 27, 2014 #8

    Avodyne

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    I agree with kith.

    And you can make simultaneous measurements of noncommuting observables if you find, as a result of the first measurement, that the system is in one of the common eigenstates. Let me modify my 3-state example to make this clearer:
    [tex]A=\pmatrix{ 1 & 0 & 0 \cr 0 & -1 & 0 \cr 0 & 0 & 2},
    \quad
    B=\pmatrix{ 0 & 1 & 0 \cr 1 & 0 & 0 \cr 0 & 0 & 3}.[/tex]
    Suppose you measure ##A## and get the result 2. Now if you measure ##B##, you are guaranteed to get the result 3.
     
  10. Oct 27, 2014 #9

    kmm

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    OK I see. I haven't been able to study entanglement much yet, but would this be an example of that?

    Thanks, this makes things a bit clearer for me. What's of interest to me now is to prove that no state is a common eigenstate for canonical conjugate observables. I'm going to have to think about this more because I'm not really sure where to begin at the moment. Qualitatively, for the position and momentum observables, I can see they have no common eigenstates since the eigenstates of position are delta functions while the eigenstates of momentum are sinusoidal waves.
     
  11. Oct 28, 2014 #10

    kith

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    Assume you have a common eigenstate and act on it with the commutator.

    I forgot to mention that the concept of eigenstates of canonical conjugate observables should be taken with a big grain of salt. These states aren't square-integrable (think of a plane wave), so they don't lie in the usual Hilbert space. They should be thought of as idealized states which can only be resembled approximately by physical states.
     
  12. Oct 28, 2014 #11

    Avodyne

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    No, this has nothing to do with entanglement.
     
  13. Oct 28, 2014 #12

    kmm

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    Well if I do that for the position and momentum operator and assume they both share an eigenfunction, then for the position operator [itex] xf = af [/itex] and for the momentum operator [itex] \frac{ \hbar}{i} \frac{d}{dx} f = bf [/itex]

    If I have the commutator act on the eigenfunction then: [itex] [x, \frac{ \hbar}{i} \frac{d}{dx}] f = x \frac{ \hbar}{i} \frac{d}{dx} f - \frac{ \hbar}{i} \frac{d}{dx} x f = xbf - \frac{ \hbar}{i} \frac{d}{dx} a f = baf - abf = 0 [/itex]

    However, if I work it out directly then: [itex] [x, \frac{ \hbar}{i} \frac{d}{dx}] f = x \frac{ \hbar}{i} \frac{d}{dx} f - \frac{ \hbar}{i} \frac{d}{dx} x f = x \frac{ \hbar}{i} \frac{d}{dx} f - \frac{ \hbar}{i} (f + x\frac{df}{dx}) = i \hbar f [/itex]

    But if [itex] i \hbar f = 0 [/itex] then the eigenfunction must be zero, but this doesn't count since then every number would be an eigenvalue. Therefore, they don't share any eigenfunctions.

    I'm not sure how to come up with something more general though. I attempted something general in post #3, but apparently that is wrong.
     
  14. Oct 28, 2014 #13

    kith

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    Why do you think this is true for any function? After all, you have used the property of f being an eigenfunction of A and B in order to derive it.

    Your result [A,B]f = 0f for a certain f is already sufficient to establish that A and B aren't canonical conjugate observables. If they were you could substitute [A,B] = iħ1 but iħf ≠ 0f.
     
  15. Oct 28, 2014 #14

    kmm

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    My thought was that f is any eigenfunction, not just any function. I shouldn't have worded it that way. A and B share the same eigenfunction f, whatever it is(this is why I thought f could be any eigenfunction), with their respective eigenvalues. Under that assumption I came to the conclusion that the commutator acting on f is zero and since f can't be zero the commutator must be. So I took this to mean that commutators that equaled zero involve operators that share any eigenfunction.

    I actually didn't understand that in general canonical conjugate observables equaled iħ1. When you refer to "a certain f" I assume you mean that we could have degeneracy. Is that correct?
     
  16. Oct 28, 2014 #15

    kith

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    That's not correct. For example if the commutator of two observables on a two-dimensional system equals [tex]\pmatrix{ 1 & 0 \cr 0 & 0}[/tex] it has one eigenvector with eigenvalue 0 and one with eigenvalue 1.

    No. With "a certain f" I meant the specific f you assumed to be a common eigenstate of the observables in your proof.

    What definition of canonical conjugate do you use?
     
  17. Oct 28, 2014 #16

    kmm

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    I see, so that shows that a commutator of two observables acting on a common eigenvector can equal something nonzero which disproves my "proof".

    I actually didn't have a specific definition. As an example, I thought that [itex] \frac{ \hbar}{i} \frac{d}{dx} [/itex] was the canonical conjugate of momentum. I thought this was the sense in which you were referring to it. I assume there is actually a more specific definition?
     
  18. Oct 28, 2014 #17

    kith

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    See canonical commutation relation.
     
  19. Oct 28, 2014 #18

    kmm

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    OK, so I see that in general the canonical commutation relation says that [itex] [x_{i}, p_{j}] = i \hbar \delta_{i,j} [/itex]. So it seems to me that what I showed in post #12 should be true. If there is such an eigenfunction that is simultaneous to x and p, then the commutator acting on that function would be zero. Since we don't get zero, they must not share simultaneous eigenfunctions.
     
  20. Oct 29, 2014 #19

    kith

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    Yes. But instead of using the specific form of P in the second line of your proof, you could have substituted iħ1 for the commutator directly. This way, the prove gets more general because it assumes only the canonical commutation relation and applies to all conjugate variables (see the section 'Generalization' in the wikipedia article above) and not only X and P.
     
  21. Oct 29, 2014 #20

    Demystifier

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    Yes, there can be common eigenstates of two non-commuting operators. But these eigenstates span a subspace on which the two operators actually commute. So yes, if the result of measurement belongs to this subspace, then you can make a simultaneous measurement of the two operators, but in that case the operators on the relevant subspace commute.
     
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