Solving [H,x] Operator Algebra

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The discussion focuses on solving the commutator [H,x], where H is the Hamiltonian operator defined as H = [-ih_bar(p/2m) + U(x)]. The participant initially derives the expression [H,x] = -(ih_bar/2m)[px - xp], recognizing that px - xp is equivalent to [p,x] = iħ. However, the participant is advised that the Hamiltonian used is incorrect and should be the standard nonrelativistic form H = (p²/2m) + U(x) to achieve the desired commutation relations.

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I'm trying to practice some operator algebra..

Solve [H,x] where H is the Hamiltonian operator, x is position operator, and assuming one dimensional potential energy, U(x).

I know the commutator comes out as -ih_bar(p_op)/m


Here is my work so far.

[H,x] = Hx - xH

note: H = [ -ih_bar(p/2m) + U(x) ]

so.. plug it in.. [ -ih_bar(p/2m) + U(x) ] x - x [ -ih_bar(p/2m) + U(x) ]

( x U(x) and -x U(x) cancel)

now.. -(ih_bar/2m)(p x) + (ih_bar/2m)(x p)

-(ih_bar/2m)[ p x - x p ]

x and p are both operators.. so I know they don't cancel.. I'm kind of lost at this point.
 
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Well, px - xp is just another way of writing [p,x], which by definition is i\hbar. However, this doesn't give you the answer you're looking for, because your Hamiltonian isn't right. The standard nonrelativistic Hamiltonian is \frac{\hat{p}^2}{2m} + U(\hat{x}), which should give you the commutation relations you're looking for.
 
Oh yeah.. haha thanks.
 

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