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.

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