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Confused again by uncertainty relation

  1. Jan 4, 2010 #1
    Say we have two non-commutative operator A, B. Now I have prepared identical systems in eigenstate of A, then I measure the observable of A, and then B immediately after. Then I must have delta A=0, so no matter what delta B is, the product is 0, seems to violate the uncertainty relation?

    Another confusion is: I know non-commutative operator A,B must have different sets of eigenfunctions, but is it possible A and B have one or two common eigenfunctions? If so, if we prepare a system in one of the common eigenstates, two observables will be well defined at the same time。
     
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  3. Jan 5, 2010 #2

    diazona

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    For two noncommuting operators A and B for which the uncertainty principle holds, the standard deviation of eigenvalues of B would be infinite.
    The uncertainty principle can be expressed as a statement about expectation values of the two operators, and expectation values must of course be computed with respect to a specific state. If the state is a common eigenstate of both operators, then the expectation value of the commutator will be zero, and thus the uncertainty principle says that the product of the uncertainties will be greater than zero.

    I had to refresh my memory from Wikipedia
    http://en.wikipedia.org/wiki/Uncertainty_principle
    so if you want more detail, that's probably a good place to look.
     
  4. Jan 5, 2010 #3
    Is there a proof for my example, I mean, not by quoting uncertainty relation itself.

    You misunderstood me, A,B do not commute,they don't share a complete set of eigenfunctions, but they can have some eigenfunctions in common(or not?)
     
    Last edited: Jan 5, 2010
  5. Jan 5, 2010 #4

    dextercioby

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    For any common eigenvector of A and B, the commutator is well defined and is zero. Thus your statement i quoted is false.
     
  6. Jan 5, 2010 #5
    But for finite dimensional matrices, obviously A,B can have some common eigenvetors but still do not commute.
    Or do you mean when we say A,B commute, we are referring to certain vectors, which may not span the whole space?
     
  7. Jan 5, 2010 #6

    dextercioby

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    One cannot formulate a precise and unique definition for the commutativity of 2 abstract linear operators in an infinite dimensional Hilbert space. This makes the case of infinite dimensions different than the one with finite dimensions (and matrices).

    So "A,B commute" is unprecise and undefined, unless A and B are self adjoint in infinite dimensions.
     
  8. Jan 5, 2010 #7
    Why can self adjoint change the situation?
     
  9. Jan 5, 2010 #8

    Demystifier

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    This is wrong. If after measuring B you measure A again, you may not obtain the same value of A as you did the first time. In fact, you cannot predict what the value of A will be. Consequently, delta A is not 0 when you measure B. The point is that measurements change the properties of the system. This is called - contextuality. For more details see e.g. Sec. 5.4 of
    http://xxx.lanl.gov/abs/quant-ph/0609163 [Found.Phys.37:1563-1611,2007]
     
  10. Jan 5, 2010 #9
    I don't understand, why can't just we keep measuring A first then B, the measurements are meaningful, aren't they?
     
  11. Jan 5, 2010 #10

    Fredrik

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    That last part is true before and right after the first measurement, but not after the second measurement. So before the second measurement the uncertainty relation is just telling you that [itex]0\cdot\Delta B\geq 0[/itex]. (What is the expectation value of [A,B] in an eigenstate of A?)

    Not sure if that's possible, but it if is, the uncertainty relation just gives you 0·0≥0 for that case.

    A measurement changes the state of the system, unless it was already in an eigenstate of the observable that we measured.
     
  12. Jan 5, 2010 #11
    Can you elaborate this a bit more?I don't quite understand what you are saying.
     
  13. Jan 5, 2010 #12

    Fredrik

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    What I'm saying is that if you actually calculate all the factors in the uncertainty relation, it all works out. This isn't too hard to do, so you should try it yourself. If you get stuck, ask about it here.

    Here's a reminder about some things that may be useful. Note in particular that every quantity that appears in the inequality depends on the state.
     
    Last edited: Jan 5, 2010
  14. Jan 5, 2010 #13

    Physics Monkey

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    Hi kof9595995,

    I think part of the confusion comes from the nature of operators like x and p satisfying [x,p] = i. A careful examination of this commutator reveals that x and p cannot be finite dimensional matrices. Furthermore, from a technical point of view, x and p don't really have eigenstates on the real line. So there is no state with [tex] \Delta x = 0 [/tex] and hence you concern does not arise.

    The Gaussian wavepacket is a state that can come as close as you like to [tex] \Delta x = 0 [/tex], but [tex] \Delta p [/tex] diverges in precisely the same way so that the uncertainty relation is always satisfied. The gaussian wavepacket is a way to make sense of the [tex]0 \cdot \infty [/tex] that appears in the uncertainty relation when considering formal position "eigenstates".

    Hope this helps.
     
  15. Jan 5, 2010 #14
    I see, so in eigenstate of A
    [tex]\begin{array}{l}
    \left\langle {\psi }
    \mathrel{\left | {\vphantom {\psi {[A,B]\psi }}}
    \right. \kern-\nulldelimiterspace}
    {{[A,B]\psi }} \right\rangle = \left\langle {\psi }
    \mathrel{\left | {\vphantom {\psi {AB\psi }}}
    \right. \kern-\nulldelimiterspace}
    {{AB\psi }} \right\rangle - \left\langle {\psi }
    \mathrel{\left | {\vphantom {\psi {BA\psi }}}
    \right. \kern-\nulldelimiterspace}
    {{BA\psi }} \right\rangle \\
    = \left\langle {{BA\psi }}
    \mathrel{\left | {\vphantom {{BA\psi } \psi }}
    \right. \kern-\nulldelimiterspace}
    {\psi } \right\rangle - \left\langle {\psi }
    \mathrel{\left | {\vphantom {\psi {BA\psi }}}
    \right. \kern-\nulldelimiterspace}
    {{BA\psi }} \right\rangle = \lambda \left\langle {{B\psi }}
    \mathrel{\left | {\vphantom {{B\psi } \psi }}
    \right. \kern-\nulldelimiterspace}
    {\psi } \right\rangle - \lambda \left\langle {\psi }
    \mathrel{\left | {\vphantom {\psi {B\psi }}}
    \right. \kern-\nulldelimiterspace}
    {{B\psi }} \right\rangle = 0 \\
    \end{array}[/tex]
    So A, B do commute in this state.
    Then another question arises: why [x,p]!=0 for all states?
    huh, I see Physics Monkey pointed it out even before I came up with this question.
    But I don't understand this:
     
  16. Jan 5, 2010 #15

    Fredrik

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    Your calculation is correct, but this description of your result isn't. What you should be saying is that the expectation value of the commutator is 0 in this state. (Not that the commutator is 0 in this state).

    All vectors in a Hilbert space have finite norm. Now suppose that two hermitian operators have a commutator that's proportional to the identity operator: [A,B]=cI, and eigenvectors satisfying A|a>=a|a> and B|b>=b|b>.

    [tex]1=\frac c c\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-Ba)|a\rangle=0[/tex]

    (Thanks to George Jones for posting this in some other thread).

    The same thing happens if we use |b> instead of |a>. What this means is that two hermitan operators that satisfy such a commutation relation (like x and p) can't have eigenvectors. Bounded hermitian operators always do, so this is one way to see that x and p must be unbounded.

    There is a way to make sense of the notion of "eigenvectors" of operators like x and p, but it's pretty complicated, and most physicists never bother to learn it. They just pretend that all operators have eigenvectors, because it works, most of the time. See this article if you're interested in this stuff. (Thanks to strangerep for posting that).

    When we take the Hilbert space to be the usual [itex]L^2(\mathbb R^3)[/itex], the delta "function" is often said to be an eigenfunction of the position operator, but it isn't even a function. The plane wave [itex]\exp(i\vec x\cdot\vec p)[/itex] is often said to be an eigenfunction of the momentum operator, but if we try to calculate the inner product of that function with itself, using the inner product of [itex]L^2(\mathbb R^3)[/itex], the result is infinity. So it can't be a member of that Hilbert space.

    If you try to calculate the uncertainties and the expectation value of the commutator [x,p] in one of those states, you will get 0·∞ on the left, and nonsense on the right.
     
    Last edited: Jan 5, 2010
  17. Jan 5, 2010 #16
    Thanks a lot, I suppose I have a much better understanding of these stuff now.
     
  18. Jan 6, 2010 #17

    dextercioby

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    I agree only up to the last sentence. I put the statement in red. On what basis can you infer that part ?

    Here's my argument. Let x and p be operators in the Hilbert space L2 over R and the Lebsgue measure. If we postulate the commutation relation of Born & Jordan, [x,p]=i(hbar)(unit operator), then there's at least one vector in the H space for which the commutator makes sense and allows us to compute the <uncertainties>. You end up with ([tex]\delta[/tex] x times [tex]\delta[/tex]p nonzero, from which you infer that [tex]\delta[/tex] x is nonzero for that specific vector.

    Do you agree with me that [x,x]=0 as H space operators ? Try to put this also fundamental commutation relation in the so-called <uncertainty relation>. Do you get that [tex]\delta[/tex]x = 0, regardless of the vector in the domain of x (subset of vectors from H) I use to compute the averages ??

    That should prove that [tex]\delta [/tex]x can definitely be zero . And the same argument goes for [tex]\delta[/tex] p.

    So what does the <uncertainty principle> say after all ?? I mean what exactly is its relevance ?
     
    Last edited: Jan 6, 2010
  19. Jan 6, 2010 #18
    In this case uncertainty relation only tells us delax*deltax is large than or equal to 0, we don't know for sure it equals 0.
     
  20. Jan 7, 2010 #19

    dextercioby

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    yes, of course. it doesn't follow that delta x must be zero, but rather can be zero. So i stand corrected.
     
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