How Does the Commutator [p-hat_x, H-hat] Reveal Quantum Mechanics Fundamentals?

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SUMMARY

The commutator [p-hat_x, H-hat] is defined as -ihbar dV/dx, where H-hat is expressed as H-hat = (p-hat^2)/(2m) + V(x). The discussion emphasizes the importance of understanding the commutator's definition, which is AB - BA, and the role of derivatives in quantum mechanics. Specifically, when applying the derivative operator to a constant, it results in zero, leaving only the term involving the potential V(x) in the commutator calculation.

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Homework Statement



by considering the action of [p-hat (subscript x), H-hat] on a general state, show that

[p-hat (subscript x), H-hat] =-ihbar dV/dx

Homework Equations



H-hat = (((p-hat)^2)/2m) +V(x)

p-hat (subscript x)= -i*h d/dx (partial derivative)


The Attempt at a Solution


tried lookin it up couldn't find anything
 
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It's right in front of you. Two hints:

1. What is the definition of the commutator of two operators? In other words, how would I write out [A, B]?

2. What is the derivative of a constant?
 
so the commutator is AB-BA

so it equals
i*h-bar d/dx(((p-hat)^2)/2m)+V(x)) -(((p-hat)^2)/2m)+V(x)) i*h-bar d/dx
=-ih-bar dV/dx ...
but why does -(((p-hat)^2)/2m)+V(x)) i*h-bar d/dx go to nothing? what do i do please, if the i*h-bar d/dx is put after the hamiltonian?
 
Last edited by a moderator:
what do i do please, if the i*h-bar d/dx is put after the hamiltonian?

In that case, you consider the derivative to be operating on 1 -- in other words d/dx = d/dx(1) = 0. So in the commutator, the only piece that doesn't vanish is the piece that features d/dx operating on a function of x -- which gives you -ih-bar dV/dx.
 

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