Energy levels in well and copper wire

Energy levels in a potential well of width L

Consider a particle of mass $m$ in a one-dimensional infinite well:

$$V(x) = 0 \ \text{ for } 0 < x < L, \qquad V(x) = \infty \ \text{ elsewhere}$$

 V = inf |                  | V = inf
         |      V = 0       |
         |                  |
         +------------------+
         0                  L      x

Outside, $\psi = 0$. Inside, the Schrödinger equation is

$$\frac{d^2\psi}{dx^2} + \frac{2mE}{\hbar^2}\psi = 0 \quad\Rightarrow\quad \frac{d^2\psi}{dx^2} + k^2\psi = 0, \quad k = \frac{\sqrt{2mE}}{\hbar}$$

General solution: $\psi(x) = A\sin kx + B\cos kx$.

Boundary conditions ($\psi$ continuous, so zero at the walls):

• $\psi(0) = 0 \Rightarrow B = 0$
• $\psi(L) = A\sin kL = 0 \Rightarrow kL = n\pi$, $n = 1, 2, 3, \ldots$

$$\begin{aligned} k_n &= \frac{n\pi}{L} \\ E_n &= \frac{\hbar^2k_n^2}{2m} = \frac{n^2\pi^2\hbar^2}{2mL^2} = \frac{n^2h^2}{8mL^2} \end{aligned}$$

Normalizing, $\psi_n(x) = \sqrt{2/L}\,\sin(n\pi x/L)$. Only discrete energies $E_1, 4E_1, 9E_1, \ldots$ are allowed.

Copper wire of length L

In the free electron model, a conduction electron in a copper wire:

• moves freely inside the wire, where the potential of the ion cores is nearly uniform (take $V = 0$);
• cannot escape at the ends, because leaving the metal needs energy equal to the work function (about 4.6 eV for Cu), which is very large compared with the energy changes involved. The ends act as nearly infinite walls.

          copper wire, length L
 surface  ======================  surface
 barrier  e-  ->    <-  e-    ->  barrier
   (high)   free electrons, V = 0   (high)
          0                      L

So the wire is exactly a one-dimensional potential well of width $L$, and the electron's wave function must vanish (be zero) at $x = 0$ and $x = L$. Applying the same Schrödinger equation and boundary conditions gives the same result:

$$E_n = \frac{n^2h^2}{8mL^2}, \qquad n = 1, 2, 3, \ldots$$

Hence the energy of an electron in a copper wire is quantized, just like a particle in a potential well.

Why it seems continuous

For a wire of $L = 1$ cm:

$$E_1 = \frac{(6.626\times10^{-34})^2}{8\times9.1\times10^{-31}\times(10^{-2})^2} = 6.03\times10^{-34}\ \text{J} = 3.8\times10^{-15}\ \text{eV}$$

Near the Fermi level of copper (about 7 eV), $n \approx 4.3\times10^{7}$ and the spacing between neighbouring levels is $\Delta E \approx (2n + 1)E_1 \approx 3\times10^{-7}$ eV, far smaller than $kT = 0.026$ eV. The levels are quantized but so closely spaced that they form a practically continuous band. Quantization becomes visible only when $L$ is a few nanometres.