Operators (or variables in quantum mechanics) do not necessarily commute. We can see our first example of that now that we have a few operators. We define the commutator to be

\bgroup\color{black}$\displaystyle [p,x]\equiv px-xp$\egroup
(using \bgroup\color{black}$p$\egroup and \bgroup\color{black}$x$\egroup as examples.)

We will now compute the commutator between \bgroup\color{black}$p$\egroup and \bgroup\color{black}$x$\egroup. Because \bgroup\color{black}$p$\egroup is represented by a differential operator, we must do this carefully. Lets think of the commutator as a (differential) operator too, as generally it will be. To make sure that we keep all the \bgroup\color{black}${\partial\over\partial x}$\egroup that we need, we will compute \bgroup\color{black}$[p,x]\psi(x)$\egroup then remove the \bgroup\color{black}$\psi(x)$\egroup at the end to see only the commutator.

={\hbar\over i}{\partial\ove...
...\partial\psi(x)\over\partial x}\right)
={\hbar\over i}\psi(x)\\

So, removing the \bgroup\color{black}$\psi(x)$\egroup we used for computational purposes, we get the commutator.
\bgroup\color{black}$\displaystyle [p,x]={\hbar\over i}$\egroup

Later we will learn to derive the uncertainty relation for two variables from their commutator. Physical variable with zero commutator have no uncertainty principle and we can know both of them at the same time.

We will also use commutators to solve several important problems.

We can compute the same commutator in momentum space.

\begin{eqnarray*}[p,x]\phi&=&[p,i\hbar{d\over dp}]\phi=i\hbar\left(p{d\over dp}\...
...=i\hbar(-\phi)={\hbar\over i}\phi\\
{[p,x]}&=&{\hbar\over i}\\

The commutator is the same in any representation.

* Example: Compute the commutator $[E,t]$.*
* Example: Compute the commutator $[E,x]$.*
* Example: Compute the commutator $[p,x^n]$.*
* Example: Compute the commutator of the angular momentum operators $[L_x,L_y]$.*

Gasiorowicz Chapter 3

Griffiths Chapter 3

Cohen-Tannoudji et al. Chapter

Jim Branson 2013-04-22