Commutators of Operators and Constants: A Quantum Mechanics Exercise

[Cogswell]

Homework Statement

Let ## \hat{A} = x ## and ## \hat{B} = \dfrac{\partial}{\partial x} ## be operators
Let ## \hat{C} ## be defined ## \hat{C} = c ## with c some complex number.

A commutator of two operators ## \hat{A} ## and ## \hat{B} ## is written ## [ \hat{A}, \hat{B} ] ## and is defined ## [ \hat{A}, \hat{B} ] = \hat{A} \hat{B} - \hat{B} \hat{A}##
A common way to evaluate commutators is to apply them to a general test function.

Evaluate ## [ \hat{A}, \hat{B} ] ##
Evaluate ## [ \hat{C}, \hat{B} ] ##

Homework Equations

The definition of a commutator

The Attempt at a Solution

I'm going to try f(x) as my tet function:

## [ \hat{A}, \hat{B} ] = x \dfrac{\partial}{\partial x} f(x) - \dfrac{\partial}{\partial x} x f(x) ##

## [ \hat{A}, \hat{B} ] = x f'(x) - x f'(x) - f(x) ##

## [ \hat{A}, \hat{B} ] = -f(x) ##

And so removing the test function:

## [ \hat{A}, \hat{B} ] = -1 ##

And for the second question:

Evaluate ## [ \hat{C}, \hat{B} ] = c \dfrac{\partial}{\partial x} f(x) - \dfrac{\partial}{\partial x} c f(x)##

Evaluate ## [ \hat{C}, \hat{B} ] = c f(x) - c f(x) = 0##

Is that right?

 

[CompuChip]

That's right.
You will see these things a lot in Quantum Mechanics, where the momentum operator
[tex]\hat p \propto \frac{\partial}{\partial x}[/tex]
(and the proportionality factor involves the imaginary unit i and the physical constant [itex]\hbar[/itex]) so that the position operator and the momentum operator can no longer be interchanged as in classical mechanics:
[tex][\hat x, \hat p] \neq 0[/tex]

Constants (complex numbers) commute with pretty much anything, which you have now shown for the special case where anything = x.