What is the correct formula for [AB,C] in terms of A, B, and C?

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SUMMARY

The correct formula for the commutator [AB,C] in terms of operators A, B, and C is given by [AB,C] = A[B,C] + [A,C]B. This conclusion is derived using the properties of the momentum operator P and the position operator X, where P = -iħ(∂/∂x). The discussion emphasizes the importance of understanding the application of commutation relations in quantum mechanics to solve operator equations effectively.

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  • Understanding of quantum mechanics, specifically operator algebra
  • Familiarity with the momentum operator P and position operator X
  • Knowledge of commutation relations and their implications
  • Basic calculus, particularly differentiation with respect to wavefunctions
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Caulfield
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Homework Statement


I am solving a problem and I arrived near the end, and can't figure out what to do here:

(1/(2m)) [P^2,X]+[P^2,X]

m - mass
P - Momentum operator
X - Position operator


Homework Equations



P = -iħ(∂/∂x)
[A,B]=AB-BA
[AB,C]=A[B,C]+B[A,C] where A, B and C are any operators

The Attempt at a Solution



I used [AB,C]=A[B,C]+B[A,C] to get 1/2m (P(PX-XP) + (PX-XP)P) and blocked
 
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Caulfield said:
I used [AB,C]=A[B,C]+B[A,C] to get 1/2m (P(PX-XP) + (PX-XP)P) and blocked
For an arbitrary wavefunction ##\psi(x)##, what is ##P\psi## and ##X\psi##? If you can figure those out, then you should be able to figure out ##[X,P]\psi## ...
 
The correct formula is
[AB,C]=A[B,C]+[A,C]B.
 

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