# Homework Help: Compute commutator of [P^m, X^n]

1. Apr 21, 2012

### Don'tKnowMuch

1. The problem statement, all variables and given/known data

Calculate [P^m, X^n]

2. Relevant equations

[P,X] = PX - XP

3. The attempt at a solution

P( P^(m-1) * X^(n-1))X - (X^n)(P^n)

=(XP +[P, X])(P^(m-1)*X^(n-1)) - (X^n)(P^n)
=(P^m)(X^n) + [P, X](P^(m-1))(X^(n-1))- (X^n)(P^n)...

I don't think the direction i am taking will be helpful. Is there some other way to do this particular commutator? I know how to do [P, X^n], but my methods fail when P is raised to some power.

2. Apr 21, 2012

### Dick

How did you do [P,X^n]? What did you get?

3. Apr 21, 2012

### Don'tKnowMuch

My first step was...

P(X^n)- (X^n)P

Then i pulled an X out of the first term and rewrote...

(PX)X^(n-1)-(X^n)P

After this, i substituted XP + [P,X] for the (PX) in the first term. Rewriting...

(XP + [P,X])*X^(n-1) - (X^n)P

Distributing the X^(n-1) and then pulling out another X to make another substitution...

X( XP + [P, X])*X^(n-2) + [P, X]*X^(n-1) - (X^n)P

After distributing, this reduces to...

(X^2)PX^(n-2) + 2[P, X]X^(n-1) - (X^n)P

This process of pulling out an X, substituting, and distributing can be repeated to obtain the pattern...

n[P, X]X^(n-1)

which can be rewritten as...

n( h-bar/ i )* X^(n-1)

4. Apr 22, 2012

### Dick

I guess I'm not really seeing what to do with the general case. In a specific case like [X^2,P^2] you can use your commutation relation to exchange X's and P's and you can get something like i*hbar*(2XP+2PX) (or as an expression in terms of PX or XP alone, but I'm not really seeing how to generalize that or what combination of operators the answer should be expressed in. I'll keep thinking about it though.

5. Apr 22, 2012

### Don'tKnowMuch

Thank you for the effort. My professor very briefly went over this, and his way went something like...

[P^m, X^n]= P(m-1)[P, X^n] + P(m-2)[P, X^n]P + P(m-3)[P, X^n]P^2...

He then jumped to the answer...

n( h-bar/i) * Ʃ(P^(m-k)X^(n-1)P^(k-1))

I am lost in the math jungle. It looks like he used some sort of expansion. Maybe a taylor, or binomial. I am principally uncertain.

6. Apr 22, 2012

### George Jones

Staff Emeritus
[AB , C] = A[B , C] + [A , C]B

Apply this repeatedly:

[P^m , X^n] = [P^(m-1)P , X^n]

= P^(m-1)[P , X^n] + [P^(m-1) , X^n]P

= P^(m-1)[P , X^n] + ([P^(m-2) P, X^n])P

= P^(m-1)[P , X^n] + (P^(m-2) [P, X^n] + [P^(m-2), X^n]P )P

Etc. Now use the result for [P , x^n].

7. Apr 22, 2012

### Dick

Your professor is using the property of the commutator that it has a kind of product rule. [AB,C] = A[B,C] + [A,C]B. C is your X^n and replace the AB with P^m=PPPP... m times. The rules extends in the obvious way if AB has more than two factors.

8. Apr 22, 2012

### Don'tKnowMuch

Gentlemen, thank you kindly. Greatly appreciated. Cool quote George; I will remember that.

cheers