Free-Electron Metals — Particle-in-a-Box Energies and the Fermi Energy: 17-Phys-A6 Solid State Physics, May 2015, Q4

Question 4: Free-Electron Metals — Particle-in-a-Box Energies and the Fermi Energy (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A free electron of mass $m$ confined to $0\le x\le L$ by infinite barriers; trial wave function $\psi_n=A\sin(2\pi x/\lambda_n)$ with $\psi_n(0)=\psi_n(L)=0$; Schr\ödinger equation $-\dfrac{\hbar^2}{2m}\psi_n''=\epsilon_n\psi_n$; $N$ electrons (even) fill the levels two at a time (spin up/down).

Find. (a) two reasons for electron transparency in metals; (b) $\epsilon_n$; (c) meaning of Fermi energy; (d) $\epsilon_F(N)$.

Fig. 4 — the first three particle-in-a-box levels and wave functions $\psi_1,\psi_2,\psi_3$ on the line $0\le x\le L$, matching Figure P4's shapes (one, one-and-a-half, and two full half-wavelengths respectively).

Approach. (a) recall the two textbook reasons from the free-electron/Bloch picture; (b) apply the boundary conditions to fix the allowed wavelengths, then substitute into the Schr\ödinger equation; (c)–(d) fill levels from the bottom, 2 electrons each, up to $N$.

Part (a) — why metals are transparent to free electrons. (1) Pauli exclusion: an electron can only scatter into an EMPTY final state; at low temperature almost all states well below the Fermi energy are already occupied, so the vast majority of conduction electrons have no available state to scatter into and simply cannot interact — only the thin band of electrons within $\sim k_BT$ of $\epsilon_F$ can scatter at all. (2) Wave (Bloch) nature of the electron: in a PERFECTLY periodic ion lattice, an electron propagates as a Bloch wave that passes through the crystal without attenuation, exactly as light passes through a perfect periodic medium — scattering arises only from DEVIATIONS from perfect periodicity (phonons, impurities, defects), not from the regularly-spaced ions themselves. Together these explain why conduction electrons have such remarkably long mean free paths compared to a naive classical picture of electrons colliding with every ion.
Part (b) — deriving $\epsilon_n$. The boundary condition $\psi_n(0)=A\sin(0)=0$ is automatic; $\psi_n(L)=A\sin(2\pi L/\lambda_n)=0$ requires $2\pi L/\lambda_n=n\pi$ for integer $n=1,2,3,\dots$, i.e. the wavenumber $k_n\equiv2\pi/\lambda_n=n\pi/L$ (exactly $n$ half-wavelengths fit in $L$, matching $\psi_1,\psi_2,\psi_3$ in Figure P4). Substituting $\psi_n=A\sin(k_nx)$ into the Schr\ödinger equation: $$-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}\bigl[A\sin(k_nx)\bigr]=-\frac{\hbar^2}{2m}\bigl(-k_n^2\bigr)A\sin(k_nx)=\frac{\hbar^2k_n^2}{2m}\psi_n$$ so $\epsilon_n=\hbar^2k_n^2/2m$; substituting $k_n=n\pi/L$: $$\boxed{\epsilon_n=\frac{\hbar^2}{2m}\left(\frac{n\pi}{L}\right)^2}$$ exactly the target formula — the confinement quantizes the allowed wavelengths, and the energy grows as $n^2$.
Part (c) — meaning of the Fermi energy. At absolute zero the $N$ electrons occupy the $N$ LOWEST-energy available quantum states allowed by the Pauli exclusion principle (each spatial level $n$ holding 2 electrons, spin up and down). The Fermi energy $\epsilon_F$ is the energy of the HIGHEST occupied level in this ground-state filling — the boundary between occupied and empty states at $T=0$. Physically it sets essentially every macroscopic scale in the electron gas (the electrons near $\epsilon_F$ are the only ones that can be thermally excited, scattered, or respond to a field).
Part (d) — Fermi energy for even $N$. With 2 electrons per spatial level $n$ (spin degeneracy), an even number $N$ of electrons exactly fills the lowest $N/2$ levels, $n=1,2,\dots,N/2$, with nothing left over. The highest occupied level is therefore $n_F=N/2$, and its energy (from part (b)'s formula) is the Fermi energy: $$\epsilon_F=\epsilon_{n_F}=\frac{\hbar^2}{2m}\left(\frac{n_F\pi}{L}\right)^2=\frac{\hbar^2}{2m}\left(\frac{(N/2)\pi}{L}\right)^2$$ $$\boxed{\epsilon_F=\frac{\hbar^2}{2m}\left(\frac{N\pi}{2L}\right)^2}$$ exactly the target formula — doubling $N$ quadruples $\epsilon_F$ (since $\epsilon_F\propto n_F^2\propto N^2$), a signature of Pauli-limited filling rather than everyone sharing the ground state.

Quantity	Result
(b) Level energy	$\epsilon_n=\dfrac{\hbar^2}{2m}\left(\dfrac{n\pi}{L}\right)^2$
(c) Fermi energy (definition)	energy of the highest-occupied level at $T=0$
(d) Fermi energy ($N$ even)	$\epsilon_F=\dfrac{\hbar^2}{2m}\left(\dfrac{N\pi}{2L}\right)^2$

Topic: The free-electron (Sommerfeld) model: particle-in-a-box quantization and the Fermi energy.

Key relations: $\epsilon_n=\dfrac{\hbar^2k_n^2}{2m}$, $k_n=\dfrac{n\pi}{L}$; Fermi level = highest filled level at $T=0$; $\epsilon_F\propto N^2$ in 1-D.

Why this works: confinement plus the Schr\ödinger equation forces a discrete, quantized ladder of standing-wave states; the Pauli exclusion principle then forces the $N$ electrons to stack up this ladder two at a time rather than all condensing into $n=1$, which is the origin of the Fermi energy and everything that follows from it (electronic heat capacity, Pauli paramagnetism, the long mean free path of part (a)).

Common pitfall: filling $N$ levels with 1 electron each instead of $N/2$ levels with 2 each (forgetting spin degeneracy); or mixing up $k_n=n\pi/L$ (particle in a box, standing waves) with the periodic-boundary-condition form $k_n=2\pi n/L$ used in 3-D free-electron theory.

Source: Kittel, Introduction to Solid State Physics, 8th ed., Ch. 6 (free-electron Fermi gas, Fermi energy).

Question 4 of 7: Free-Electron Metals — Particle-in-a-Box Energies and the Fermi Energy

Question 4: Free-Electron Metals — Particle-in-a-Box Energies and the Fermi Energy (20 marks)