Simple Commutator Identity

  • #1
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Main Question or Discussion Point

This is not a homework problem. It was stated in a textbook as trivial but I cannot prove it myself in general. If [A,B]=0 then [A,B^n] = 0 where n is a positive integer. This seems rather intuitive and I can easily see it to be true when I plug in n=2, n=3, n=4, etc. However, I cannot prove it in the general case and this really bothers me. Here's what I got so far:

[A,B^n] = [A,BB^n-1] = [A,B]B^n-1 + B[A,B^n-1] = 0 + B[A,B^n-1]

Not sure where to go from here???

Thanks so much.
 

Answers and Replies

  • #2
PeroK
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This is not a homework problem. It was stated in a textbook as trivial but I cannot prove it myself in general. If [A,B]=0 then [A,B^n] = 0 where n is a positive integer. This seems rather intuitive and I can easily see it to be true when I plug in n=2, n=3, n=4, etc. However, I cannot prove it in the general case and this really bothers me. Here's what I got so far:

[A,B^n] = [A,BB^n-1] = [A,B]B^n-1 + B[A,B^n-1] = 0 + B[A,B^n-1]

Not sure where to go from here???

Thanks so much.
Hint: ##[A, B] = 0## means that ##A## and ##B## commute.

Hint #2: do you know how to do proof by induction?
 
  • #3
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So I realize that [A,B] = 0 means AB = BA and I actually just walked myself through a case of proof by induction to show that if two linear operators A and B both commute with their commutator ( I understand what this means ) then [A,B^n]=nB^n-1[A,B]. I don't have that much experience with proof by induction (I'm just an undergrad) and even though I felt like I understood the worked out proof,I'm having a hard time applying it to this example and it's driving me crazy.
 
  • #4
Cryo
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So do you still have a problem or not?

You stated yourself that ##\left[A,\,B\right]=0##
It then follows that ##\left[A, B^2\right]=B\left[A, B\right]+\left[A, B\right]B=0##
But then it follows that ##[A, B^3]=B[A, B^2]+[A, B]B^2=0##

This can be repeated for all positive integer powers, so the general result follows.
 
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  • #5
PeroK
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So I realize that [A,B] = 0 means AB = BA and I actually just walked myself through a case of proof by induction to show that if two linear operators A and B both commute with their commutator ( I understand what this means ) then [A,B^n]=nB^n-1[A,B]. I don't have that much experience with proof by induction (I'm just an undergrad) and even though I felt like I understood the worked out proof,I'm having a hard time applying it to this example and it's driving me crazy.
If ##A## commutes with ##B## then it must commute with ##B^2## and ##B^3## etc.

Another way to look at it is that if we have ##X_1, X_2 \dots X_n## and they all commute with each other, then we can write the product of the ##X's## in any order.

In this case we have ##A, B, \dots B## that all commute with each other, so we can write the product in any order. I.e.

##AB^n = BAB^{n-1} = B^2AB^{n-2} = \dots = B^{n-1}AB = B^nA##

If you look at it this way, the result could be said to be obvious.
 
  • #6
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PeroK that makes a lot of sense. What Cryo did above also makes perfect sense. That's what I originally did, I just wasn't sure that constituted an actual proof. Also, how do you get your superscripts/subscripts to look nice on here. It doesn't seem to be working for me? Ex. x^2
 
  • #7
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Nevermind got it! Thanks for your help
 

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