Potential and Kinetic energy commutator

Click For Summary
SUMMARY

The discussion centers on the commutation relation between potential and kinetic energy operators in quantum mechanics, specifically the expression [T,V]=[TV-VT]ψ. The kinetic energy operator is defined as T=(-ħ²/2μ)∂²/∂x², while the potential energy operator is V=(1/2)kx². The main issue identified is the incorrect application of the product rule instead of the chain rule during differentiation, leading to confusion in the calculation of terms involving the wave function ψ. The correct simplification yields the expression (2x∂/∂x + 1)ψ, which can be further manipulated to demonstrate its equivalence to (x∂/∂x + ∂/∂x x)ψ.

PREREQUISITES
  • Understanding of quantum mechanics principles, specifically operator algebra.
  • Familiarity with the kinetic energy operator in quantum mechanics, T=(-ħ²/2μ)∂²/∂x².
  • Knowledge of the potential energy operator, V=(1/2)kx², in harmonic oscillators.
  • Proficiency in calculus, particularly differentiation techniques including the product and chain rules.
NEXT STEPS
  • Study the derivation and implications of the commutation relations in quantum mechanics.
  • Learn about the role of operators in quantum mechanics, focusing on their physical interpretations.
  • Explore examples of harmonic oscillators and their energy operators in quantum systems.
  • Review advanced differentiation techniques, particularly in the context of quantum wave functions.
USEFUL FOR

Students of quantum mechanics, physicists working with operator theory, and anyone involved in theoretical physics or advanced calculus applications.

582153236
Messages
14
Reaction score
0

Homework Statement


[T,V]=[TV-VT]ψ

Homework Equations


T=(-ħ2/2μ)∂2/∂x2
V=(1/2)kx2[/B]

The Attempt at a Solution


[(-ħ2/2μ)∂2/∂x2((1/2)kx2ψ)]-[(1/2)kx2(-ħ2/2μ)∂2/∂x2(ψ)]

I think my problem is with executing the chain rule on the first term:

(-ħ2/2μ)[x2ψ''+2xψ'+2xψ'+2ψ-x2ψ'']
[/B]

The x2ψ'' terms cancel out but I'm left with the +4xψ' term which I makes me suspect that I've made a mistake somewhere.
 
Physics news on Phys.org
It's actually the product rule, not the chain rule. Your calculation is okay except for a few minor errors. (Where'd the k go?)

Neglecting the constant factors out front, you ended up with ##(2x\frac{\partial}{\partial x} + 1)\psi##. You can show this equals ##(x\frac{\partial}{\partial x} + \frac{\partial}{\partial x} x)\psi##, which probably looks more correct to you.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
5K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 18 ·
Replies
18
Views
2K
Replies
7
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 25 ·
Replies
25
Views
1K