Expectation Values in the Box

Every measurable quantity in quantum mechanics is associated with an operator. The expectation value of an observable $\hat{A}$ in state $|\psi\rangle$ is the average outcome over many identical measurements:

\langle A \rangle = \langle \psi | \hat{A} | \psi \rangle = \int_{-\infty}^{\infty} \psi^*(x)\,\hat{A}\,\psi(x)\,dx.

Position expectation value: $\langle x \rangle$

For the $n$ -th eigenstate of the infinite square well ( $\psi_n(x) = \sqrt{2/L}\,\sin(n\pi x/L)$ , $0 \le x \le L$ ), the position expectation value is

\langle x \rangle_n = \int_0^L x\,|\psi_n(x)|^2\,dx = \frac{2}{L}\int_0^L x\,\sin^2\!\left(\frac{n\pi x}{L}\right)dx.

Replacing $\sin^2\theta = \tfrac{1}{2}(1 - \cos 2\theta)$ splits the integral into two parts:

\langle x \rangle_n = \frac{1}{L}\int_0^L x\,dx - \frac{1}{L}\int_0^L x\cos\!\left(\frac{2n\pi x}{L}\right)dx.

The first integral gives $L/2$ . The second integral, evaluated by parts with $u = x$ and $dv = \cos(2n\pi x/L)\,dx$ , produces a boundary term proportional to $\sin(2n\pi) = 0$ plus an antiderivative term proportional to $\cos(2n\pi) - \cos(0) = 1 - 1 = 0$ , so it vanishes exactly. Therefore

\langle x \rangle_n = \frac{L}{2} \quad \text{for all } n.

This result is not a coincidence: every eigenstate has the same left-right symmetry about the midpoint of the well, so the mean position is always the center regardless of the quantum number.

Energy expectation value

The energy eigenvalues are $E_n = n^2\pi^2\hbar^2/(2mL^2)$ . Because $\psi_n$ is an eigenstate of the Hamiltonian, a measurement of energy on $\psi_n$ always returns $E_n$ with certainty:

\langle E \rangle_n = E_n.

The interesting case arises when the particle is in a superposition of eigenstates. Suppose

\psi = \sum_n c_n \psi_n, \qquad \sum_n |c_n|^2 = 1.

Because the $\psi_n$ are orthonormal, the cross terms $\langle \psi_m | \hat{H} | \psi_n \rangle$ vanish for $m \ne n$ , and

\langle E \rangle = \sum_n |c_n|^2 E_n.

This is a weighted average of eigenvalues, with the probability $|c_n|^2$ serving as each weight. A measurement collapses the state to a single eigenstate $\psi_n$ , yielding energy $E_n$ with probability $|c_n|^2$ ; the expectation value is the long-run mean of those outcomes.

Equal superposition of $n = 1$ and $n = 2$

Take $\psi = (\psi_1 + \psi_2)/\sqrt{2}$ , so $|c_1|^2 = |c_2|^2 = 1/2$ . Then

\langle E \rangle = \tfrac{1}{2}E_1 + \tfrac{1}{2}E_2 = \tfrac{E_1 + E_2}{2}.

With $L = 1\,\text{nm}$ and an electron, $E_1 \approx 0.376\,\text{eV}$ and $E_2 = 4E_1 \approx 1.504\,\text{eV}$ , giving

\langle E \rangle = \frac{0.376 + 1.504}{2}\,\text{eV} \approx 0.940\,\text{eV}.

Try it

This is a numerical exercise — return a number. An electron is in the state $(\psi_1 + \psi_2)/\sqrt{2}$ inside an infinite square well of width $L = 1\,\text{nm}$ . Compute the expectation value of energy $\langle E \rangle$ in electron-volts.

Run your code to see the quantum state.

Position expectation value: ⟨x⟩\langle x \rangle⟨x⟩

Energy expectation value

Equal superposition of n=1n = 1n=1 and n=2n = 2n=2

Try it

Position expectation value: $\langle x \rangle$

Equal superposition of $n = 1$ and $n = 2$