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

[The Head]

Homework Statement

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)

Homework Equations

For operators, in general:
(ab,c)=a(b,c)+(a,c)b
(a,bc)=(a,b)c + b(a,c)

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!

 

[DimReg]

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.

 

[The Head]

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?

 

[Dick]

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?

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.

 

[DimReg]

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.

 

[Dick]

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.

Of course not, the given formula ought to at least work for the n=2 case.

 

[andrien]

try this,xpⁿψ-pⁿxψ=-(pⁿx)ψ