Energy Eigenvalues

When you solved the time-independent Schrödinger equation you wrote

This is an eigenvalue equation: $\psi$ is the eigenfunction and $E$ is the eigenvalue of the Hamiltonian $\hat{H}$. The allowed energies of a quantum system are precisely the eigenvalues of its Hamiltonian — nothing more, nothing less. That statement is not a definition imported from mathematics; it is a physical postulate, confirmed by the discrete spectral lines of atoms and the step-like conductance of nanowires alike.

Why only some energies are allowed

For a classical particle in a box you could give it any energy you liked. In quantum mechanics the wave function must satisfy boundary conditions — it must vanish at the walls of an infinite potential well, for example — and that requirement selects only certain solutions. Imposing the condition $\psi(0) = \psi(L) = 0$ on a particle of mass $m$ in a well of width $L$ forces the wave vector to take the discrete values $k_n = n\pi/L$ with $n = 1, 2, 3, \ldots$. The kinetic energy is $E = \hbar^2 k^2 / (2m)$, so the eigenvalues are

where $E_1 = \pi^2 \hbar^2 / (2mL^2)$ is the ground-state energy. The integer $n$ is the quantum number, and the pattern $E_n = n^2 E_1$ is the hallmark of the infinite square well.

A compact way to see where the $n^2$ comes from: the number of half-wavelengths that fit inside the box is $n$, so $\lambda_n = 2L/n$, and because the de Broglie momentum is $p = h/\lambda$, the kinetic energy $p^2/(2m)$ scales as $1/\lambda^2 \propto n^2$.

The ground state is not zero energy

A critical point: $n = 0$ is not an allowed state. Setting $n = 0$ gives $\psi = 0$ everywhere, which is not a normalizable wave function — it represents no particle at all. The lowest allowed energy, $E_1$, is strictly positive. This zero-point energy is a direct consequence of the uncertainty principle: confining the particle to a region of size $L$ forces a non-zero momentum uncertainty $\Delta p \gtrsim \hbar/(2L)$, and hence a non-zero kinetic energy.

Eigenstates and superpositions

The eigenstates $\psi_n$ form a complete orthonormal basis: any wave function inside the well can be written as $\Psi = \sum_n c_n \psi_n$. Measuring the energy in state $\Psi$ yields $E_n$ with probability $|c_n|^2$. Only if $\Psi = \psi_n$ — a pure eigenstate — is the energy outcome perfectly sharp. A superposition gives a spread of outcomes according to the Born rule; the expectation value is $\langle E \rangle = \sum_n |c_n|^2 E_n$.

This framing unifies all of quantum mechanics: every observable has an associated operator, the operator has eigenvalues (the possible measurement outcomes) and eigenstates (the states in which those outcomes are certain), and a general state is a superposition of eigenstates weighted by complex amplitudes whose squared moduli give probabilities.

Try it

This is a numerical exercise — return a number. The ground-state energy of an electron in a particular infinite square well is $E_1 = 0.38\,\text{eV}$. Using the relation $E_n = n^2 E_1$, find the energy of the $n = 3$ level in eV.

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