The Position and Momentum Operators

In wave mechanics every physical quantity is represented by a linear operator acting on the wave function. The two most fundamental operators are position and momentum; every other observable in non-relativistic quantum mechanics is built from them.

The position operator

In one dimension the position operator $\hat{x}$ simply multiplies the wave function by the coordinate $x$ :

$\hat{x}\,\psi(x) = x\,\psi(x).$

In three dimensions each component acts the same way: $\hat{x}_i\,\psi(\mathbf{r}) = x_i\,\psi(\mathbf{r})$ . The operator is said to act by multiplication; the wave function at each point is scaled by the value of the coordinate at that point. Because multiplication by a real number is its own adjoint, $\hat{x}$ is Hermitian, which is required for any observable.

The momentum operator

The momentum operator is less obvious — it cannot be a simple multiplication in position space. The physical requirement comes from the de Broglie relation: a state of definite momentum $p$ should correspond to a plane wave $e^{ipx/\hbar}$ . For this wave to be an eigenstate of the momentum operator with eigenvalue $p$ , the operator must generate a spatial derivative. One can show (by demanding consistency with the commutation relation and Hermiticity) that the only valid choice is

$\hat{p} = -i\hbar\,\frac{\partial}{\partial x}.$

Verification in three dimensions

In three dimensions the momentum operator generalises to the gradient:

$\hat{\mathbf{p}} = -i\hbar\,\nabla.$

Acting on a plane wave $\psi(\mathbf{r}) = e^{i\mathbf{k}\cdot\mathbf{r}}$ gives

$\hat{\mathbf{p}}\,e^{i\mathbf{k}\cdot\mathbf{r}} = -i\hbar\,(i\mathbf{k})\,e^{i\mathbf{k}\cdot\mathbf{r}} = \hbar\mathbf{k}\,e^{i\mathbf{k}\cdot\mathbf{r}},$

so the eigenvalue is $\hbar\mathbf{k}$ , exactly the de Broglie momentum. This self-consistency is the derivation that fixes the form of $\hat{p}$ .

Expectation values

Given a normalised wave function $\psi(x)$ , the expectation (mean) value of position is

$\langle x \rangle = \int_{-\infty}^{\infty} \psi^*(x)\,x\,\psi(x)\,dx = \int_{-\infty}^{\infty} x\,|\psi(x)|^2\,dx,$

which is the probability-weighted average of $x$ over all space. For momentum,

$\langle p \rangle = \int_{-\infty}^{\infty} \psi^*(x)\left(-i\hbar\frac{\partial}{\partial x}\right)\psi(x)\,dx.$

Note that the derivative acts on $\psi$ , not on $\psi^*$ ; reversing the order would give the wrong (complex) answer.

Summary

| Observable | Operator in position space | |---|---| | Position $x$ | $\hat{x} = x$ (multiply by $x$ ) | | Momentum $p$ | $\hat{p} = -i\hbar\,\partial/\partial x$ |

These two operators are the building blocks of quantum mechanics: kinetic energy is $\hat{p}^2/(2m)$ , the harmonic oscillator Hamiltonian is $\hat{p}^2/(2m) + m\omega^2\hat{x}^2/2$ , and their commutator $[\hat{x},\hat{p}] = i\hbar$ encodes the Heisenberg uncertainty principle.