Waves on strings obey the wave equation. What equation do matter waves obey? Erwin Schrödinger found it in 1926. In its time-independent form it says which wave functions $\psi(x)$ a particle of energy $E$ can have in a potential energy $U(x)$. Here you will check solutions and read off their energies, using second derivatives.
Key idea
The time-independent Schrödinger equation: $$-\frac{\hbar^2}{2m}\,\frac{d^2\psi}{dx^2} + U(x)\,\psi = E\,\psi.$$ An allowed wave function must also be continuous, finite and normalizable, and it must meet the boundary conditions (for example $\psi = 0$ at a hard wall). Those conditions are what allow only certain energies. For an electron, $\frac{\hbar^2}{2m_e} = 0.0381$ eV·nm².
Why it works
Start from what we know about matter waves: $p = \hbar k$ (de Broglie) and $E = \frac{p^2}{2m} + U$ (energy). For a wave $\psi = \sin kx$, the second derivative is $\psi'' = -k^2\psi$, so $-\frac{\hbar^2}{2m}\psi'' = \frac{\hbar^2k^2}{2m}\psi = \frac{p^2}{2m}\psi$: the second-derivative term measures kinetic energy. Adding $U\psi$ and setting the total to $E\psi$ is the energy statement, written so that it works even where $k$ changes from place to place. Where $U > E$, the same equation gives $\psi'' = +\kappa^2\psi$ with $\kappa = \frac{\sqrt{2m(U - E)}}{\hbar}$, whose finite solution is a decaying exponential, $e^{-\kappa x}$: the wave leaks a little way into regions that classical physics forbids.
Worked examples
Worked example: A free electron
In a region with $U = 0$, an electron has $\psi = A\sin(kx)$ with $k = 10\ \text{nm}^{-1}$. What is its energy?
- $\psi'' = -k^2\psi$, so $-\frac{\hbar^2}{2m}\psi'' = \frac{\hbar^2k^2}{2m}\psi$. The equation holds with $E = \frac{\hbar^2k^2}{2m}$.
- $E = (0.0381\ \text{eV·nm}^2)(10\ \text{nm}^{-1})^2 = 3.81$ eV.
Answer: $E = 3.81$ eV
Worked example: Inside a barrier
An electron with $E = 3.0$ eV meets a region where $U = 5.0$ eV. Its wave function there is $\psi = Ae^{-\kappa x}$. Find $\kappa$.
- $\psi'' = \kappa^2\psi$, so the equation reads $-\frac{\hbar^2\kappa^2}{2m} + U = E$.
- $\kappa^2 = \frac{U - E}{\hbar^2/2m} = \frac{2.0\ \text{eV}}{0.0381\ \text{eV·nm}^2} = 52.5\ \text{nm}^{-2}$.
Answer: $\kappa = 7.25\ \text{nm}^{-1}$: the wave falls by a factor of $e$ every 0.14 nm
Worked example: Check a candidate
An electron in a box with walls at $x = 0$ and $x = L$ ($U = 0$ inside). Which of $\psi_1 = \sin(\pi x/L)$, $\psi_2 = \cos(\pi x/L)$ and $\psi_3 = x(L - x)$ are allowed energy states?
- $\psi_1$: $\psi_1'' = -\frac{\pi^2}{L^2}\psi_1$, so it solves the equation with $E = \frac{\hbar^2\pi^2}{2mL^2}$, and it is 0 at both walls. Allowed.
- $\psi_2$ solves the equation but is 1 at $x = 0$, not 0. Not allowed.
- $\psi_3$ is 0 at both walls, but $\psi_3'' = -2$ is not a constant times $\psi_3$, so it does not solve the equation for any single $E$.
Answer: Only $\psi_1$
Common mistake
Thinking any smooth function is an allowed wave function. It must solve the equation (with one constant $E$ everywhere) and meet the boundary conditions. $x(L - x)$ looks like a fine ground state, zero at both walls and smooth, but it fails the equation; $\cos(\pi x/L)$ solves the equation but fails at the walls. A check for any candidate: compute $-\frac{\hbar^2}{2m}\frac{\psi''}{\psi} + U$. It must come out the same number, $E$, at every $x$.