Can [A,B^n] always equal 0 if [A,B] equals 0?

[cdot]

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.

 

[PeroK]

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?

 

[cdot]

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.

 

[Cryo]

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.

 

[PeroK]

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.

 

[cdot]

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}

 

[cdot]

Nevermind got it! Thanks for your help