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Commutator Proof: Show (x,p^n)= ixp^(n-1)

  1. Feb 25, 2013 #1
    1. The problem statement, all variables and given/known data
    Using (x,p) = i (where x and p are operators and the parentheses around these operators signal a commutator), show that:

    a)(x^2,p)=2ix AND (x,p^2)=2ip
    b) (x,p^n)= ixp^(n-1), using your previous result
    c)evaluate (e^ix,p)


    2. Relevant equations
    For operators, in general:
    (ab,c)=a(b,c)+(a,c)b
    (a,bc)=(a,b)c + b(a,c)


    3. The attempt at a solution
    I have no trouble showing (a):
    (x^2,p)=x(x,p)+(x,p)x=x(i) + i(x)=2ix
    Same general strategy for (x,p^2)

    In (b), I get lost when trying to apply the result I have already found. My attempt:

    (x,p^n)=(x,p*p^(n-1)=(x,p)p^(n-1)+p(x,p^(n-1))=ip^(n-1) + p(x,p^(n-1))
    I don't know where to go from here. I have tried pulling out more factors of p from the second term, but that gets me no where (at least from what I can see). I have also just played around a lot with the commutator, even starting from the end:

    ixp^(n-1)=(x,p)p^(n-1)= xp^n - pxp^(n-1), but again, after this, whichever path I take doesn't seem to be fruitful.

    Thanks!
     
  2. jcsd
  3. Feb 25, 2013 #2
    Do you know how to do proofs by induction? It looks to me like the problem wants you to do it that way.

    An induction proof goes as follows: First you proof some base case, such as N = 1 (like part A). Then you assume it's true for some n -1, and show that implies it's also true for N. These two steps show it's true for all N.
     
  4. Feb 25, 2013 #3
    I thought about an induction proof-- I have some experience with these, but evidently not that much. After I get to where I left off:
    (x,p^n)=(x,p*p^(n-1)=(x,p)p^(n-1)+p(x,p^(n-1))=ip^(n-1) + p(x,p^(n-1))

    I have to somehow turn this into ixp(n-1). I can't figure out how to get an "x" in the final expression that is separate from the commutator. I have tried expanding the second term further:
    p(x,p^(n-1))=p(x,p*p^n-2)=p(x,p)p^(n-2) + p^2(x,p^(n-2))= p*i*p^(n-2) +p^2(x,p^(n-2))=ip^(n-1) + p^2(x,p^n-2)

    Together with the first term I have:
    ip^(n-1)+ip^(n-1) +(x,p^n-2)=2ip^(n-1) + (x,p^n-2) I feel like this only leads to an expression that equals something like nip^(n-1), but I can't find a way to make it "xip^(n-1)" Is it a mistake? Or am I just applying the strategy incorrectly?
     
  5. Feb 25, 2013 #4

    Dick

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    It looks like there is an error in the statement you are supposed to prove. [x,p^n]=inp^(n-1). There's no x in there.
     
  6. Feb 26, 2013 #5
    For induction, you are supposed to assume that the n-1 case is true. this would imply:

    (x,p^n-1) = ixp^n-2.

    On the other hand, this approach doesn't quite seem to work, so I think that Dick is right, and there isn't supposed to be an x there.
     
  7. Feb 26, 2013 #6

    Dick

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    Of course not, the given formula ought to at least work for the n=2 case.
     
  8. Feb 26, 2013 #7
    try this,xpnψ-pnxψ=-(pnx)ψ
     
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