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

[rogeralms]

Homework Statement

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

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

The Attempt at a Solution

Prove for n=1
[A,B¹]=1B^1-1 [A,B].
[A,B]=B⁰[A,B]=[A,B]

Show that it is true for n+1
[A,Bⁿ⁺¹]=[A,BⁿB]=Bⁿ[A,B]+[A,Bⁿ]B

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

Thank you.

 

[TSny]

Show that it is true for n+1
[A,Bⁿ⁺¹]=[A,BⁿB]=Bⁿ[A,B]+[A,Bⁿ]B

I am not sure how to continue.

When using induction, you want to show that the relation holds for n+1 assuming it holds for n.

 

[rogeralms]

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].

 

[TSny]

Show that it is true for n+1
[A,Bⁿ⁺¹]=[A,BⁿB]=Bⁿ[A,B]+[A,Bⁿ]B

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.

 

[rogeralms]

I overlooked the part in the problem statement that said that A and B both commute.

Then we have
Assume [A,Bⁿ]=nB^n-1[A,B] is true
Show that it is true for n+1 [A,Bⁿ⁺¹]=(n+1)Bⁿ[A,B] is true.
[A,Bⁿ⁺¹]=[A,BⁿB]=Bⁿ[A,B]+[A,Bⁿ]B=Bⁿ[A,B]+B[A,Bⁿ]
=Bⁿ[A,B]+BnB^n-1[A,B]=Bⁿ[A,B]+nBⁿ[A,B]=(n+1)Bⁿ[A,B]

 

[TSny]

I overlooked the part in the problem statement that said that A and B both commute.

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.

Then we have
Assume [A,Bⁿ]=nB^n-1[A,B] is true
Show that it is true for n+1 [A,Bⁿ⁺¹]=(n+1)Bⁿ[A,B] is true.

[A,Bⁿ⁺¹]=[A,BⁿB]=Bⁿ[A,B]+[A,Bⁿ]B=Bⁿ[A,B]+B[A,Bⁿ]
=Bⁿ[A,B]+BnB^n-1[A,B]=Bⁿ[A,B]+nBⁿ[A,B]=(n+1)Bⁿ[A,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]=nB^n-1[A,B] is true.

 

[Sunita singh]

Could anyone post the solve answer for this question

 

[Dr.AbeNikIanEdL]

Traditionally there are no full solutions shared here without showing prior work, in particular not in the homework part of the forum (though I think in this case most of the solution is above). If you are currently working on this as part of your now studies, maybe you want to share where you are stuck?

 

[WWGD]

Traditionally there are no full solutions shared here without showing prior work, in particular not in the homework part of the forum (though I think in this case most of the solution is above). If you are currently working on this as part of your now studies, maybe you want to share where you are stuck?

True, but this question was asked almost 10 years ago.

 

[Dr.AbeNikIanEdL]

So, how does that change things? I was replying to the post above mine (by someone else than the original OP, I assume someone being asked this HW question now) from this morning.

 

[Orodruin]

I mean, there is basically a full solution in this thread if you piece together the OP’s work … I don’t think we need to state it any more directly than that just so someone can copy it for their homework.

 

[berkeman]

Could anyone post the solve answer for this question

Welcome to PF. Please start a new schoolwork thread if you still need help with this question, and be sure to show your best efforts to work on the problem. Thank you.

This old thread is now closed.