# Operators and commutators

1. May 25, 2013

### Cogswell

1. The problem statement, all variables and given/known data
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 $[ \hat{C}, \hat{B} ]$

2. Relevant equations

The definition of a commutator

3. 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?

2. May 25, 2013

### CompuChip

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

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