Solving [H,x] Operator Algebra

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The discussion focuses on solving the commutator [H,x] where H is the Hamiltonian operator and x is the position operator in a one-dimensional potential energy context. The initial calculation leads to the expression [H,x] = -ih_bar/2m [p,x], where p is the momentum operator. A key point is that the standard nonrelativistic Hamiltonian should be correctly defined as H = (p^2)/(2m) + U(x) to yield the desired commutation relations. The confusion arises from an incorrect Hamiltonian formulation, which affects the outcome of the commutator. Correcting the Hamiltonian is essential for achieving accurate results in operator algebra.
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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.
 
Time reversal invariant Hamiltonians must satisfy ##[H,\Theta]=0## where ##\Theta## is time reversal operator. However, in some texts (for example see Many-body Quantum Theory in Condensed Matter Physics an introduction, HENRIK BRUUS and KARSTEN FLENSBERG, Corrected version: 14 January 2016, section 7.1.4) the time reversal invariant condition is introduced as ##H=H^*##. How these two conditions are identical?

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