How Do You Compute the Commutator [P^m, X^n]?

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SUMMARY

The forum discussion centers on calculating the commutator [P^m, X^n], where P represents momentum and X represents position in quantum mechanics. Participants explore various methods, including the use of the fundamental commutation relation [P, X] = PX - XP, and discuss the application of recursive techniques to derive a general formula. The professor's approach involves using a product rule for commutators, leading to the expression n(h-bar/i) * Σ(P^(m-k)X^(n-1)P^(k-1)). This highlights the importance of understanding operator algebra in quantum mechanics.

PREREQUISITES
  • Understanding of quantum mechanics principles, specifically operator algebra.
  • Familiarity with commutation relations, particularly [P, X].
  • Knowledge of Taylor and binomial expansions in mathematical physics.
  • Experience with recursive mathematical techniques for operator manipulation.
NEXT STEPS
  • Study the derivation of the commutation relation [P, X^n] in detail.
  • Learn about the properties of commutators and their applications in quantum mechanics.
  • Explore operator algebra techniques, including the product rule for commutators.
  • Investigate the use of Taylor and binomial expansions in quantum mechanics problems.
USEFUL FOR

This discussion is beneficial for students and professionals in quantum mechanics, particularly those focusing on operator theory and commutation relations. It is also useful for educators seeking to enhance their teaching methods regarding quantum algebra.

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Homework Statement



Calculate [P^m, X^n]

Homework Equations



[P,X] = PX - XP

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.
 
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Don'tKnowMuch said:

Homework Statement



Calculate [P^m, X^n]

Homework Equations



[P,X] = PX - XP

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.

How did you do [P,X^n]? What did you get?
 
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)
 
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.
 
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.
 
[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].
 
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.
 
Gentlemen, thank you kindly. Greatly appreciated. Cool quote George; I will remember that.

cheers
 

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