2.6 Commutators

We have seen that we some operators it matters which order we apply them in. For example, if we apply $\hat{x}$ then $\hat{p}$ , we may get a different result to if we apply $\hat{p}$ then $\hat{x}$ . We can measure the degree to which two operators fail to commute using a simple expression called the commutator, denoted with square brackets.

Definition 2.3.

The commutator of two operators $\hat{A}$ and $\hat{B}$ is defined as

[\hat{A},\hat{B}]=\hat{A}\hat{B}-\hat{B}\hat{A}.

(2.120)

If $[\hat{A},\hat{B}]=0$ , then $\hat{A}$ and $\hat{B}$ commute, that is changing their order of application does not change the result.

Example 2.2.

What is the commutator of position $\hat{x}=x$ and momentum $\hat{p}=-i\hbar\frac{\partial}{\partial x}$ ?

To calculate this commutator, we need to act it on an arbitrary state $\lvert\psi\rangle$ as if it was an operator, since $\hat{p}$ contains a derivative.

$\displaystyle[\hat{x},\hat{p}]\lvert\psi\rangle$	$\displaystyle=\hat{x}\hat{p}\lvert\psi\rangle-\hat{p}\hat{x}\lvert\psi\rangle$	(2.121)
	$\displaystyle=x\left(-i\hbar\frac{\partial}{\partial x}\right)\lvert\psi% \rangle-\left(-i\hbar\frac{\partial}{\partial x}\right)(x\lvert\psi\rangle)$	(2.122)
	$\displaystyle=-i\hbar x\frac{\partial\lvert\psi\rangle}{\partial x}+i\hbar% \frac{\partial}{\partial x}(x\lvert\psi\rangle)$	(2.123)
	$\displaystyle=-i\hbar x\frac{\partial\lvert\psi\rangle}{\partial x}+i\hbar x% \frac{\partial}{\partial\lvert\psi\rangle}+i\hbar\lvert\psi\rangle$	(2.124)
	$\displaystyle=i\hbar\lvert\psi\rangle.$	(2.125)

This implies that

[\hat{x},\hat{p}]=i\hbar,

(2.126)

so $\hat{x}$ and $\hat{p}$ do not commute.

Going back to products of two operators briefly, when is the product of two Hermitian operators itself Hermitian? In section 2.5.1, we showed that

(\hat{A}\hat{B})^{\dagger}=\hat{B}^{\dagger}\hat{A}^{\dagger}.

(2.127)

Where $\hat{A}$ and $\hat{B}$ are Hermitian, we have

$\displaystyle(\hat{A}\hat{B})^{\dagger}$	$\displaystyle=\hat{B}\hat{A}$	(2.128)
	$\displaystyle=\hat{B}\hat{A}-\hat{A}\hat{B}+\hat{A}\hat{B}$	(2.129)
	$\displaystyle=\hat{A}\hat{B}-[\hat{A},\hat{B}].$	(2.130)

So the product of two Hermitian operators is Hermitian only if they commute, i.e. $[\hat{A},\hat{B}]=0$ .

Note that for two general operators that do not commute, we can switch their order in an expression by introducing the commutator:

	$\displaystyle\hat{A}\hat{B}$	$\displaystyle=\hat{A}\hat{B}-\hat{B}\hat{A}+\hat{B}\hat{A}$		(2.131)
		$\displaystyle=\hat{B}\hat{A}+[\hat{A},\hat{B}].$		(2.132)

Also, we have for any commutator that

[\hat{A},\hat{B}]=\hat{A}\hat{B}-\hat{B}\hat{A}=-(\hat{B}\hat{A}-\hat{A}\hat{B% })=-[\hat{B},\hat{A}],

(2.133)

so all commutators are antisymmetric.

You might think that two operators commuting is handy for manipulating expressions with products of operators but not much use beyond that, but actually the value of a the commutator has important implications for measurements. In particular, we can show that two observables commuting is equivalent to them sharing a common basis of eigenfunctions. This means that the two observables can be “measured simultaneously”, which means that we can know exactly which state the particle is in with respect to both observables at the same time.

Let $\hat{A}$ and $\hat{B}$ be Hermitian operators which commute.

[\hat{A},\hat{B}]=0.

(2.134)

Suppose the kets $\lvert i\rangle$ are the basis of $\hat{A}$ with eigenvalues $A_{i}$ .

\hat{A}\lvert i\rangle=A_{i}\lvert i\rangle\quad\forall i.

(2.135)

Then because the commutator is zero, the order of application does not matter and we have

\hat{A}(\hat{B}\lvert i\rangle)=\hat{B}(\hat{A}\lvert i\rangle)=\hat{B}(A_{i}% \lvert i\rangle)=A_{i}(\hat{B}\lvert i\rangle).

(2.136)

Note that the parentheses are not strictly necessary, they are just there for clarity. The first and last parts taken together are an eigenvalue equation. If each eigenstate $\lvert i\rangle$ has a distinct eigenvalue, then this implies that $\hat{B}\lvert i\rangle$ is a scalar multiple of $\lvert i\rangle$ . This further implies that $\lvert i\rangle$ is an eigenstate of $\hat{B}$ , and so we have shown that $\hat{A}$ and $\hat{B}$ share an eigenbasis.

Conversely, suppose that we don’t know what $[\hat{A},\hat{B}]$ is but we do know that $\hat{A}$ and $\hat{B}$ share an eigenbasis. So $\hat{A}\lvert i\rangle=A_{i}\lvert i\rangle$ and $\hat{B}\lvert i\rangle=B_{i}\lvert i\rangle$ for all values of $i$ . Then we have that

$\displaystyle[\hat{A},\hat{B}]\lvert i\rangle$	$\displaystyle=\hat{A}(\hat{B}\lvert i\rangle)-\hat{B}(\hat{A}\lvert i\rangle)$	(2.137)
	$\displaystyle=B_{i}\hat{A}\lvert i\rangle-A_{i}\hat{B}\lvert i\rangle$	(2.138)
	$\displaystyle=B_{i}A_{i}\lvert i\rangle-A_{i}B_{i}\lvert i\rangle$	(2.139)
	$\displaystyle=0,$	(2.140)

for all $i$ , hence $[\hat{A},\hat{B}]$ must be zero $\hat{A}$ commutes with $\hat{B}$ .

Being able to measure two observables simultaneously makes a bit more sense now. Suppose we measure $\hat{A}$ first, then if the particle is not in an eigenstate of $\hat{A}$ already then it will collapse into one. Then, when we measure $\hat{B}$ , the particle is already in an eigenstate of $\hat{B}$ and so we will get that eigenvalue and the particle will still be in the state we measured for $\hat{A}$ . This would not be the case if $\hat{A}$ and $\hat{B}$ did not commute. Then upon measuring $\hat{B}$ , the particle would be in a superposition state with respect to $\hat{A}$ and we would no longer know its state with respect to $\hat{A}$ for certain.

As we stated in section 2.4.4, observables which do not share a set of eigenfunctions, i.e. do not commute, are called incompatible observables. The most famous set of incompatible observables is certainly position and momentum. As we saw in example 2.2, position and momentum do not commute, meaning there do not exist simultaneous eigenfunctions of $\hat{x}$ and $\hat{p}$ . There are therefore no quantum states with definite values of both position and momentum.

2.6.1 Uncertainty from Incompatibility

The fact that observables which do not commmute do not share eigenfunctions is directly responsible for the uncertainty principle. It can be shown that for two observables $\hat{A}$ and $\hat{B}$ , the product of their uncertainties obeys the relation:

\Delta A\Delta B\geq\frac{1}{2}\lvert\langle[\hat{A},\hat{B}]\rangle\rvert.

(2.141)

This is known as the general uncertainty relation. The right hand side means we find the commutator of $\hat{A}$ and $\hat{B}$ , then calculate the expectation value of the result, then take the modulus of that.

Example 2.3.

Prove the famous Heisenberg uncertainty relation 1.48:

\Delta x\Delta p\geq\frac{\hbar}{2}.

(2.142)

Putting $\hat{x}$ and $\hat{p}$ into equation 2.141 and using 2.126 gives

$\displaystyle\Delta x\Delta p$	$\displaystyle\geq\frac{1}{2}\lvert\langle[\hat{x},\hat{p}]\rangle\rvert$	(2.143)
	$\displaystyle\geq\frac{1}{2}\lvert\langle i\hbar\rangle\rvert$	(2.144)
	$\displaystyle=\geq\frac{1}{2}\lvert i\hbar\rvert$	(2.145)
	$\displaystyle=\geq\frac{\hbar}{2}.$	(2.146)