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Prove commutator [A,B^n]=nB^(n-1)[A,B]

  1. Mar 12, 2015 #1
    1. The problem statement, all variables and given/known data

    Let A and B be two observables that both commute with their commutator [A,B].

    a) Show, e.g., by induction, that [A,Bn]=nBn-1 [A,B].

    3. The attempt at a solution
    Prove for n=1
    [A,B1]=1B1-1 [A,B].

    Show that it is true for n+1

    I am not sure how to continue. Could someone give me a hint what I should do next?

    Thank you.
  2. jcsd
  3. Mar 12, 2015 #2


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    When using induction, you want to show that the relation holds for n+1 assuming it holds for n.
    Last edited: Mar 13, 2015
  4. Mar 13, 2015 #3
    Thank you but I know how induction works. That is why I set up the problem with n+1.

    I need a hint on how to manipulate the commutators to show this is true, that [A,Bn+1]=(n+1)Bn [A,B].
  5. Mar 13, 2015 #4


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    You're showing the relation is true for n+1 assuming it is true for n. You have not yet used the assumption that it is true for n. Note on the right hand side you have a term [A,Bn]B. Simplify this term.
  6. Mar 15, 2015 #5
    I overlooked the part in the problem statement that said that A and B both commute.

    Then we have
    Assume [A,Bn]=nBn-1[A,B] is true
    Show that it is true for n+1 [A,Bn+1]=(n+1)Bn[A,B] is true.
  7. Mar 15, 2015 #6


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    A and B are not assumed to commute. You are given that A commutes with [A, B]. Also, B commutes with [A, B]. But A does not necessarily commute with B.

    OK. One thing that you might consider is that in getting to the end of your first line you used [A,Bn]B= B[A,Bn]. It might not be clear to the reader why this is justified. You can avoid this by rewriting [A,Bn]B in a different way using the assumption that [A,Bn]=nBn-1[A,B] is true.
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