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Commutator Evaluating Rule

  1. Apr 26, 2013 #1
    1. The problem statement, all variables and given/known data

    So we know that for two operators [itex]\hat{A}[/itex] and [itex]\hat{B} + \hat{C}[/itex] we have the following rule for the commutator of the two: [itex][\hat{A},\hat{B} + \hat{C}] = [\hat{A},\hat{B}] + [\hat{A},\hat{C}][/itex]

    However, if I'm commuting [itex][\hat{p_{x}}, \hat{H}][/itex] where [itex]\hat{H}[/itex] is the Hamiltonian and [itex]\hat{p_{x}}[/itex] is the momentum operator, can I use this operator rule to "expand" the commutator [itex][\hat{p_{x}}, \hat{H}][/itex] in the same way as [itex]\hat{A}[/itex] and [itex]\hat{B} + \hat{C}[/itex] even though [itex]V(x)[/itex] in the Hamiltonian is not an operator?

    If this can be done then we can get [itex][\hat{p_{x}}, \hat{H}][/itex] = [itex][\hat{p_{x}}, V(x)][/itex] and expand that commutator instead of messing about with the [itex]\frac{\hat{p^{2}}_{x}}{2m}[/itex] part in the Hamiltonian since [itex][\hat{p_{x}}, \hat{p^{2}}_{x}][/itex] = 0.

    Thank You
     
  2. jcsd
  3. Apr 26, 2013 #2

    Fredrik

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    Yes, you can. If you see something that looks like the sum of an operator and a number, that number should always be interpreted as "that number times the identity operator". So there's never a situation where the potential term is not an operator. Since the identity operator commutes with everything, the second commutator will be zero too. Edit: Temporary insanity here, corrected by Dick below.
     
    Last edited: Apr 26, 2013
  4. Apr 26, 2013 #3

    Dick

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    V(x) is an operator just like x is an operator. But there is nothing wrong with what you are doing. ##p_x## commutes with the ##p{_x}^2## so you only need to worry about the commutator with the potential.
     
  5. Apr 26, 2013 #4

    Dick

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    ##p_x## may commute with the identity, but it certainly doesn't commute with V(x). 'x' isn't a number. '2' is a number. There's a difference.
     
  6. Apr 26, 2013 #5
    Thank you so much! :)

    We were given a relation: [itex]\frac{d<\hat{A}>}{dt}=\frac{1}{i\hbar}<[\hat{A},\hat{H}]>[/itex] and to show that Newton's second law of motion is obtained by taking [itex]\hat{A} = \hat{p}_{x}[/itex]

    Now, if what you say is the case, how would we go about doing this? I.e, the way I did it was just the long way and the factors with the first part of the Hamiltonian in them came to zero but my lecturer did it the way shown below and I don't understand the third line, why is the commutator suddenly acting on f?

    1sl0mo.png
     
  7. Apr 26, 2013 #6
    This makes sense since I got [itex]-i\hbar \frac{\partial{V(x)}}{\partial{x}}[/itex] as the answer for the commutator with the Hamiltonian and not 0.

    Edit: Do you have any comment on my post above Dick?
     
  8. Apr 26, 2013 #7

    Dick

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    ##[p_x,V]## is an operator. So what it does is it acts on functions. Your lecturer is showing that ##[p_x,V]## acting on a function f is the same as multiplying f by -ihdV/dx. So that's what the operator is. f is a 'test function'.
     
  9. Apr 26, 2013 #8
    Is it necessary though? I.e. if you didn't have that and just expanded the commutator with (P_x * V) - (V *P_x) wouldn't you then get the same thing without the f's and then the second term would just be d/dx of 0 = 0 and the first term would just be -ihbar dV(x)/dx

    So isn't it the same thing? It seems to me like f is intentionally f = f(a != x) so it's treated as a constant or something.

    Edit: I.e. I just did this (below), is this then an incorrect way of doing it?

    http://i41.tinypic.com/2pzhcld.jpg
     
    Last edited: Apr 26, 2013
  10. Apr 26, 2013 #9

    Dick

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    Yes, it's not really correct. Just taking the d/dx part, you've written [d/dx,V(x)]=d/dx*V-V*d/dx. Any special reason to think that the V*d/dx is zero in some way? I find your lecturer's calculation much more convincing.
     
  11. Apr 26, 2013 #10

    Fredrik

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    Dick is right. Not sure what I was thinking. I stand by my first comment, that an operator plus a number doesn't really make sense and must therefore be interpreted as the operator plus "the number times the identity operator". But that isn't really relevant here. The momentum operator p is a number times the derivative operator, and because of that, it can't commute with V unless V is a constant function. So that part of my post was wrong. Sorry about that.
     
  12. Apr 26, 2013 #11
    I don't understand why it's wrong? Is that not how you evaluate a commutator? Just take A acting on B and subtract B acting on A and if they "commute" and = 0 those two terms would be equal in magnitude so you'd get 0?

    V*d/dx, operators act on everything to their right, there's nothing there except I guess a 1 so wouldn't it be V*d/dx = V*d/dx(1) which is just 0?

    If it's incorrect then I don't understand the purpose of the f because it seems sort of like you're multiplying by 2 and then dividing by 2 later. In fact, one of the questions above was to show that [itex][\hat{p}_{x},\hat{x}]f = -i\hbar f[/itex] for some general function f, and then: Hence, show that: [itex][\hat{p}_{x},\hat{x}] = -i\hbar[/itex], don't you just cancel the f's again? Or is there a subtlety I'm missing and you choose f = 1?

    We both expanded the commutator the same to show [itex][\hat{p}_{x},\hat{x}]f = -i\hbar f[/itex] but then the next line in the solutions was just "Hence, and then the same thing without the f's" so I figured they just got cancelled..
     
  13. Apr 26, 2013 #12

    Dick

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    You seem to be arguing that if I just write d/dx or ##p_x## then that will be 0 since there is nothing on the right. You often write operators with nothing on the right. You just can't evaluate them until you put something on the right. V*d/dx isn't the zero operator. If you apply it to a function f, you get V*df/dx. That's not zero. If you apply it to g then you get V*dg/dx.
     
  14. Apr 26, 2013 #13
    Okay I see, so in the third line in my lecturers solution, is f (in this case) being treated as some function that is of some variable other than x?

    Is this the approach for all of these types of questions? I can't see any other way you'd be able to pull the f out of the derivative in the first term and also have the second term = 0 unless df/dx = 0
     
  15. Apr 26, 2013 #14

    Dick

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    You expand the first term by using the product rule. Then the df/dx in the first term cancels the df/dx in the second term. df/dx doesn't have to be zero. It just cancels from the expression.
     
  16. Apr 26, 2013 #15
    Ahhh I see! That is cool! Sorry for all of this by the way but then finally I'm trying to understand what exactly happens to the f in the next line? Does it just get cancelled? Or has something else taken the place of f?
     
  17. Apr 26, 2013 #16

    Dick

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    They've shown that for any function f, [p,V]f=(-ihdV/dx)f. Since it doesn't matter what f is, why not just skip writing it and say that the operator [p,V] is equal to the operator -ihdV/dx without having to make a up a function to write after them all of the time?
     
  18. Apr 26, 2013 #17
    Ah okay, thank you once again and my apologies for being dim! We weren't actually taught much about commutators etc in our Quantum course so I'm sort of using the internet and questions to become more familiar with them!

    Thank you very much for your help!
     
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