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Commutator Troubles

  1. Nov 10, 2004 #1
    I'm having difficulty trying to figure out which of the following is the correct method to properly evaluate the effect of the operators on f(x).

    Given that,
    [tex]\hat{A}f(x)=<x|\hat{A}|f>[/tex]

    If the polarity operator, [tex]\hat{U_p}[/tex], and the translation operator, [tex]\hat{U_t}(a)[/tex], act as
    [tex]\hat{U_p}f(x)=f(-x)[/tex]
    [tex]\hat{U_t}(a)f(x)=f(x-a)[/tex]

    Which of the following is the correct method of evaluating the commutator [tex][\hat{U_p},\hat{U_t}(a)]f(x)[/tex].

    [tex]\begin{array}{rl}
    \hat{U_p}\hat{U_t}(a)f(x)&=\hat{U_p}f(x-a)\\
    &=f(-x+a)[/tex]

    or

    [tex]\begin{array}{rl}
    \hat{U_p}\hat{U_t}(a)f(x)&=<x|\hat{U_p}\hat{U_t}(a)|f>\\
    &=<-x|\hat{U_t}(a)|f>\\
    &=<-x-a|f>\\
    &=f(-x-a)[/tex]

    Why do I get a different result? The order of the operators acting has obviously changed, but which is the correct order? I am tempted to believe the second case, but I can't really see the difference.
     
  2. jcsd
  3. Nov 10, 2004 #2
    They are both correct. What you have proven here is that the commutator of the two operators is NOT 0.

    marlon
     
  4. Nov 10, 2004 #3
    Hey Marlon,

    Thanks for the answer, but that's not what I was trying to get at. I realize that either way I evaluate these the commutation relation will be non-zero, I just get a different relation depending on the method. Which is what I'm wondering here.

    [tex]
    \begin{array}{rl}
    [\hat{U_p},\hat{U_t}(a)]f(x)&=<x|\hat{U_p}\hat{U_t}(a)|f>-<x|\hat{U_t}(a)\hat{U_p}|f>\\
    &=<-x|\hat{U_t}(a)|f>-<x-a|\hat{U_p}|f>\\
    &=<-x-a|f>-<-x+a|f>\\
    &=f(-x-a)-f(-x+a)
    \end{array}
    [/tex]

    and

    [tex]
    \begin{array}{rl}
    [\hat{U_p}\hat{U_t}(a)]f(x)&=\hat{U_p}\hat{U_t}(a)f(x)-\hat{U_t}(a)\hat{U_p}f(x)\\
    &=\hat{U_p}f(x-a)-\hat{U_t}(a)f(-x)\\
    &=f(-x+a)-f(-x-a)
    \end{array}
    [/tex]

    These commutation relations are clearly not equivalent. I am trying to figure out which one is the correct interpretation. I can not seem to see a flaw in either, but there must be some subtle difference between the way in which I have evaluated them. Basically, the order of the operators is reversed between the two, but I am unsure which is the correct order.
     
  5. Nov 11, 2004 #4

    James R

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    You say:

    [tex]<-x|\hat{U_t}(a)|f> = <-x-a|f>[/tex]

    But

    [tex]\hat{U_t}(a)|x> = |x - a>[/tex]?

    So, I think

    [tex]\hat{U_t}(a)|-x> = |-(x-a)> = |-x+a>[/tex]

    Is that right? If so, that's your mistake.
     
  6. Nov 11, 2004 #5
    In fact, [tex]\hat{U_t}(a)|x> = |x+a>[/tex]

    [tex]\hat{U_t}(a)[/tex] represents the translation operator, translating the current x-coordinate.

    [tex]<x|\hat{U_t}(a) = <x|\hat{U_t}^\dagger(-a) = <x-a|[/tex]

    hence, my statement that

    [tex]\hat{U_t}(a)f(x) = <x|\hat{U_t}(a)|f> = <x-a|f> = f(x-a)[/tex]

    So, if the current x coordinate is -x then the new coordinate should be -x+a, not -x-a. Following that method, I get the following though, which still doesn't match.

    [tex]
    \begin{array}{rl}
    [\hat{U_p},\hat{U_t}(a)]f(x)&=<x|\hat{U_p}\hat{U_t}(a)|f>-<x|\hat{U_t}(a)\hat{U_p}|f>\\
    &=<-x|\hat{U_t}(a)|f>-<x-a|\hat{U_t}(a)|f>\\
    &=<-(x-a)|f>-<-(x-a)|f>\\
    &=f(-x+a)-f(-x+a)
    \end{array}
    [/tex]
     
  7. Nov 11, 2004 #6

    ZapperZ

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    Have we established the validity that if A and B are each Hermitian, that AB and BA are also hermitian? If not, then even if you can write <i|A|j> and <i|B|j>, it doesn't necessarily follow that you can write <i|AB|j> etc. This is only valid for a hermitian operator, is it not? That might explain why it "doesn't match".

    Zz.

    Edit: Er.. never mind. I should have proved it to myself before I posted this. Carry on....
     
    Last edited: Nov 11, 2004
  8. Nov 11, 2004 #7

    reilly

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    Science Advisor

    The problem lies in assuming that A f(x) and <x|A|f> are the same thing. They are not.
    In fact, <x|a|f> = integral/sum over y of <x|A|y><y|f>. So your second method is incorrect, as far as evaluating Up Ut f is concerned.
    Regards,
    Reilly Atkinson
     
  9. Nov 11, 2004 #8
    I just want to ensure that I understand you correctly. Especially since I wrote out the statements in two different orders.

    To recap how I interpret things,

    [tex]f(x)=<x|f>[/tex]

    [tex]\begin{array}{rl}
    \hat{A}f(x)&=<x|\hat{A}|f>\\
    &=\int dx' <x|\hat{A}|x'><x|f>
    \end{array}[/tex]

    Now, if I let [tex]\hat{A}=\hat{U_p}\hat{U_t}(a)[/tex] then,

    [tex]\begin{array}{rl}
    \hat{U_p}\hat{U_t}(a)f(x)&=\int dx' <x|\hat{U_p}\hat{U_t}(a)|x'><x'|f>\\
    &=\int dx' <x|\hat{U_p}|x'+a><x'|f>\\
    &=\int dx' <x|-x'-a><x'|f>\\
    &=\int dx' \delta(x+x'+a) <x'|f>\\
    &=f(-x-a)[/tex]

    since, [tex]x'=-x-a[/tex].

    Thanks for your help.
     
  10. Nov 11, 2004 #9
    This step is not correct I think:

    [tex]\begin{array}{rl}\hat{U_p}\hat{U_t}(a)f(x)&=\hat{U_p}f(x-a)\\&=f(-x+a)[/tex]​

    indeed:

    [tex]\begin{array}{rl}\hat{U_p}\hat{U_t}(a)f(x)&=\hat{U_p}f(x-a))\\&=\hat{U_p}g(x)\\&=g(-x)\\&=f(-x-a)[/tex]​

    where g(x) = f(x-a)
    and where I assume the rules of the game are as above since the definitions are valid for any functions
     
  11. Nov 11, 2004 #10
    lalbatros,

    I think you made a mistake in going from [tex]g(-x')[/tex] to [tex]f(-x-a)[/tex].

    If we define [tex]x' \equiv x-a[/tex], as you have done to go from [tex]f(x-a) \Rightarrow g(x')[/tex], then shouldn't we find that [tex]-x'=-(x-a)=-x+a[/tex]?
     
  12. Nov 12, 2004 #11
    Hi

    I did not define x' = x-a and g(x') = f(x-a)

    I defined g(x) = f(x-a) , g is another function of x

    and since the operators apply in the same way for all functions of x, the result follows.

    To summarize: check the definition of the parity operator, it applies to any function of course
     
  13. Nov 12, 2004 #12
    lalbatros,

    To manipulate it in such a manner, you assume the particle is at x, not x+a. The parity operator should effectively reflect around the axis, meaning something that is at x+a must be x+a to the left, or, at -x-a.
     
  14. Nov 12, 2004 #13
    The parity operator is simply:

    "replace x by -x"​
     
  15. Nov 12, 2004 #14
    By that definition (very different than the one I have learned) these operators do commute. Something I am quite certain they should not. Your definition leads to,

    [tex]\begin{array}{rl}
    [\hat{U_t}(a),\hat{U_p}]f(x)&=\hat{U_t}(a)\hat{U_p}f(x)-\hat{U_p}\hat{U_t}(a)f(x)\\
    &=\hat{U_t}(a)f(-x)-\hat{U_p}f(x-a)\\
    &=f(-x-a)-f(-x-a)\\
    &=0
    \end{array}
    [/tex]
     
  16. Nov 12, 2004 #15

    dextercioby

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    Why the polemics,this the correct version,in agreement with QM principles and position/momentum representations of vectors and operators...
     
  17. Nov 12, 2004 #16
    Read a definition here:


    Other starting points are possible of course. You can choose different representations.
    But in the choosen way, all this is quite simple.

    I did not make any comment on their commutator.
    You can check that they do not commute:

    UaUp f(x) = Ua f(-x) = f(-(x-a)) = f(-x+a)
    UpUa f(x) = Up f(x-a) = f(-x-a)​

    Interpretation:
    plane waves are eigenfunctions of the translation operator Ua,
    but they are obviously not eigenfunctions of the parity operator (except for a zero wavevector).​

    ps: the reason of my posts is that the formula for UpUa f(x) in the first post was not correct (see above here), not taste for polemic!
     
    Last edited: Nov 12, 2004
  18. Nov 12, 2004 #17

    reilly

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    Note, a la lalbatros, Up Ut(a) = Up Ut(a) U(-1)p Up = Ut(-a) Up, where, in spite of cumbersome notation, U(-1)p is the inverse of Up, the parity transformation. Indeed, parity and translations do not commute.
    Regards,
    Reilly Atkinson
     
  19. Nov 12, 2004 #18
    Reilly,

    Thanks for your summary:

    Up Ut(a) = Ut(-a) Up​

    This tells it crystal clear.
     
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