Commutators: a few theorems

Required math: calculus

Required physics: Schrödinger equation

References: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Chapter 3, Post 13.

The commutator of two operators is defined as

\displaystyle \left[A,B\right]\equiv AB-BA \ \ \ \ \ (1)

In general, a commutator is non-zero, since the order in which we apply operators can make a difference. In practice, to work out a commutator we need to apply it to a test function {f}, so that we really need to work out {\left[A,B\right]f} and then remove the test function to see the result. This is because many operators, such as the momentum, involve taking the derivative.

We’ll now have a look at a few theorems involving commutators.

Theorem 1:

\displaystyle \left[AB,C\right]=A\left[B,C\right]+\left[A,C\right]B \ \ \ \ \ (2)

Proof: The LHS is:

\displaystyle \left[AB,C\right]=ABC-CAB \ \ \ \ \ (3)

The RHS is:

\displaystyle A\left[B,C\right]+\left[A,C\right]B \displaystyle = \displaystyle ABC-ACB+ACB-CAB\ \ \ \ \ (4)
\displaystyle \displaystyle = \displaystyle ABC-CAB\ \ \ \ \ (5)
\displaystyle \displaystyle = \displaystyle \left[AB,C\right] \ \ \ \ \ (6)

QED.

Theorem 2:

\displaystyle \left[x^{n},p\right]=i\hbar nx^{n-1} \ \ \ \ \ (7)

where {p} is the momentum operator.

Proof: Using {p=\frac{\hbar}{i}\partial/\partial x} and letting the commutator operate on some arbitrary function {g}:

\displaystyle \left[x^{n},p\right]g \displaystyle = \displaystyle x^{n}\frac{\hbar}{i}\frac{\partial g}{\partial x}-\frac{\hbar}{i}\frac{\partial}{\partial x}(x^{n}g)\ \ \ \ \ (8)
\displaystyle \displaystyle = \displaystyle x^{n}\frac{\hbar}{i}\frac{\partial g}{\partial x}-\frac{\hbar}{i}nx^{n-1}g-x^{n}\frac{\hbar}{i}\frac{\partial g}{\partial x}\ \ \ \ \ (9)
\displaystyle \displaystyle = \displaystyle i\hbar nx^{n-1}g \ \ \ \ \ (10)

Removing the function {g} gives the result {\left[x^{n},p\right]=i\hbar nx^{n-1}}. QED.

Theorem 3:

\displaystyle \left[f(x),p\right]=i\hbar\frac{df}{dx} \ \ \ \ \ (11)

Again, letting the commutator operate on a function {g}:

\displaystyle \left[f(x),p\right] \displaystyle = \displaystyle f\frac{\hbar}{i}\frac{\partial g}{\partial x}-\frac{\hbar}{i}\frac{\partial}{\partial x}(fg)\ \ \ \ \ (12)
\displaystyle \displaystyle = \displaystyle f\frac{\hbar}{i}\frac{\partial g}{\partial x}-\frac{\hbar}{i}\frac{\partial f}{\partial x}g-f\frac{\hbar}{i}\frac{\partial g}{\partial x}\ \ \ \ \ (13)
\displaystyle \displaystyle = \displaystyle ih\frac{\partial f}{\partial x}g \ \ \ \ \ (14)

Removing {g} gives the result {\left[f(x),p\right]=ih\partial f/\partial x}. QED.