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The 1=0 paradox again!

  1. Jan 10, 2015 #1


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    I know this is raised several times in this forum but I still don't see what's the solution! So I want to discuss it again.
    Consider two self-adjoint operators A and B which obey [itex] [A,B]=c I[/itex]. Now I take the normalized state [itex] |a\rangle [/itex] such that [itex] A|a\rangle=a |a\rangle [/itex]. Now I can write:

    1=\langle a | a \rangle=\frac 1 c \langle a | cI |a \rangle=\frac 1 c \langle a | [A,B] |a \rangle=\frac 1 c [\langle a | AB |a \rangle-\langle a | BA |a \rangle]=\\ \frac 1 c [(A|a\rangle)^\dagger B|a\rangle-a\langle a |B|a\rangle]=\frac 1 c [(a|a\rangle)^\dagger B|a\rangle-a\langle a |B|a\rangle]=\frac a c [\langle a | B |a \rangle-\langle a | B |a \rangle]=0

    So there should be something wrong with the reasoning. There were several suggestions before but I don't think they work.


    1) Please DO NOT specialize to momentum and position!(Or any other pair, unless you can prove there is only a few pairs of operators that satisfy the requirements and you can say what is wrong about each pair.)

    2) I know that the commutation relation can't be realized in a finite dimensional Hilbert space. So take an infinite-dimensional Hilbert space! Can this be used here?

    3) I know that at least one of the operators should be unbounded. Can this be used here?

    4) I know domains are important. But I can still find a vector in the intersection of the domains of A and B. Can you prove their domains are disjoint? Or can you prove there is no eigenvector of A in the intersection of the domains of A and B?

    5) I know that here we should consider a rigged Hilbert space. But a rigged Hilbert space is actually a triplet like [itex] D \subset L^2(\mathbb R,dx) \subset D' [/itex]. So I can still choose vectors from D which are actually ordinary nice vectors. Can you prove there is no eigenvector of A in D? Or can you prove for such an eigenvector, B necessarily gives ill-defined results so that [itex] \langle a | B |a \rangle-\langle a | B |a \rangle [/itex] is indeterminate instead of 0?

    6) An important point here is the symmetry between A and B. So even if we can use one of the above suggestions, we then can swap the role of A and B and retain the argument. Can you break this symmetry?

    7) Hey @Greg Bernhardt , can you set a prize for solving this?:D

    I hope this time I'll get a definite solution for this and get this case closed.
    Last edited: Jan 10, 2015
  2. jcsd
  3. Jan 10, 2015 #2


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    To make this a paradox, you have to show the existence of self-adjoint operators A, B with [A,B]=cI and with real eigenvalues for A. And all the operations above have to be well-defined, of course.
    Just saying "consider" is not enough: "Consider a number x that satisfies both x=0 and x=1. Then 1=x=0".
  4. Jan 10, 2015 #3


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    Yeah, that's true but I want to know exactly which assumption is wrong or which assumptions contradict each other.
    I mean...I can't show that such a pair actually exists. But this doesn't prove it doesn't exist. I want to know how can I prove that there is no such a pair.
    Or, to put it another way, this can be seen as a proof by contradiction that such a pair does not exist. But which assumption should I drop so that I can have a pair satisfying those weakened assumptions?
    Or maybe such a pair actually exist but some part of the calculation is not correct. Which part is that?

    Also [itex] A=\frac{\hbar}{i} \frac{\partial}{\partial \varphi} [/itex] and [itex] B=\varphi [/itex] constitute such a pair because we have [itex] A e^{i\frac m \hbar \varphi}=m e^{i\frac m \hbar \varphi} [/itex]. So there exists such a pair.
    Last edited: Jan 10, 2015
  5. Jan 10, 2015 #4


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    You can basically use your "proof" that 1=0 to show that somewhere the logic failed. So at some step, there must have been some undefined quantity. As it turns out, the undefined quantity is ##\langle a|B|a\rangle## (as such, of course ##\langle a|[A,B]|a\rangle## is also undefined, the commutator is defined only on the intersection of the domains of A and B). See my previous thread on this matter: https://www.physicsforums.com/threads/commutator-expectation-value-in-an-eigenstate.747434/
  6. Jan 10, 2015 #5


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    Thank you very much man. The paper pointed to by dextercioby, was exactly the thing I was looking for all this time.
    So it turned out that
    is the the solution.
    Case Closed!
  7. Jan 10, 2015 #6

    It can't be realized as bounded operators either. They need to be unbounded.

    I think your argument proves exactly that there is no eigenvector in the intersection of the domains of A and B. As matterwave pointed out, something like ##\langle a|B|a\rangle## needs not be well-defined.

    Rigged Hilbert spaces are not important here.
  8. Jan 10, 2015 #7


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    Yeah, I saw it before but its vague to me. Because I know about unitary equivalence in terms of representation theory of groups and there, its about the similarity transformation. But I don't know how can I perform a similarity transformation on something like [itex] \frac{\hbar}{i} \frac{\partial}{\partial x} [/itex]!
    Also...does it mean that the operators [itex] \frac{\hbar}{i} \frac{\partial}{\partial \varphi} [/itex] and [itex] \varphi [/itex] defined on the space [itex] L^2([0,2\pi),d\varphi) [/itex] are unitarily equivalent to momentum and position operator? In what sense?

    I knew it. I just wanted to get some understanding of this and see why it should be the case. I mean an ordinary mathematical way of understanding this.
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