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Questions about operators and commutators

  1. Sep 28, 2015 #1
    so I have an expression here:

    [P,g(r)]= -ih dg/dr

    P is the momentum operator working on a function g(r).

    Is this true because when you expand the left hand side the expression g(r)P is zero because the del operator has nothing to work on?
     
  2. jcsd
  3. Sep 28, 2015 #2

    jfizzix

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    The equation you have here is an equation between two different operators.
    On the one hand that commutator, and on the other hand, that derivative.

    In order to test this equation, you need to act the left and right hand sides on the same wavefunction, say [itex]\psi(r)[/itex].
    If you find that [itex][P,g(r)]\psi(r) = -i\hbar \frac{dg}{dr}\psi(r)[/itex], no matter what [itex]\psi(r)[/itex] is,
    then the operators must be equal as well.
     
  4. Sep 28, 2015 #3
    Maybe I'm missing something here, because my limited knowledge of commutators says that the left hand side unpacked is

    { Pg(r)-g(r)P } Ψ(r)

    I'm not sure how to show that

    g(r)iħ dΨ/dr = 0 in order for the expression above to hold true.
     
  5. Sep 29, 2015 #4

    jfizzix

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    The expression above is not equal to zero because:
    when the momentum operator [itex]P[/itex] is before both [itex]g(r)[/itex] and [itex]\psi(r)[/itex], that gradient acts on the product [itex]g(r)\psi(r)[/itex] (and the product rule comes into affect), so you get one more term than you'd get otherwise when you compute the gradient. Indeed, that extra term is the same as what's on the right hand side of the equation:
    [itex](Pg(r) - g(r) P)\psi(r) = -i\hbar \frac{dg}{dr}\psi(r)[/itex]

    When the momentum operator [itex]P[/itex] is after [itex]g(r)[/itex], then the gradient inside [itex]P[/itex] just acts on [itex]\psi(r)[/itex].
     
  6. Sep 29, 2015 #5
    Ah I see now. So now how would I go about proving this expression:

    [g(P), r] = -iħ dg/dP

    I'm not sure how the del operator works inside a function.
     
  7. Sep 29, 2015 #6
    My thoughts are g(P) unpacks as such:

    -iħ∇(p) (the partial derivative with respect to P)

    so g(P)r would be:

    -iħdg/dP

    I mean using the logic from the previous solution this allows for the second expression to hold true. Not sure if I'm following the math rules though....
     
  8. Sep 29, 2015 #7

    jfizzix

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    There's probably many ways to deal with a situation like this, but I think the easiest would be to do the calculation in momentum space, so that [itex]g(p)[/itex] is just a function of the variable [itex]p[/itex] , while [itex]x[/itex] is then expressed as the operator [itex]i\hbar\frac{d}{dp}[/itex]. Then the problem can be solved in the same way as before.
     
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