--- title: "Fermi energy, occupancy at 0 K, expectation value" subject: "Electrical Engineering Material" chapter: 1 key: "semester_3/electrical_engineering_material:fermi-energy-expectation-value" source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/fermi-energy-expectation-value site: BCT Engineering --- # Fermi energy, occupancy at 0 K, expectation value Electrical Engineering Material · Chapter 1: Theory of Metals - Asked in: 2072 Chaitra, 8 marks - Video: [Fermi Function Explained](https://www.youtube.com/watch?v=cBd5Kaqy7EU) — Jordan Edmunds Chetty - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/fermi-energy-expectation-value - Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-1 ## Question Define Fermi Energy. What is the probability that an electron having energy less than Fermi energy will occupy an energy level at absolute zero temperature? Determine the expectation value for any property of a particle described by a wave function Ψ. ## Short answer Fermi energy is the highest occupied electron level at 0 K. At T=0, any state with E < EF is occupied with probability 1. For an operator  representing a property, its expectation value is ⟨A⟩ = ∫ ψ*  ψ dx (normalized ψ). ## Answer #### Fermi energy The **Fermi energy** $E_F$ is the energy of the highest filled electron level in a metal at absolute zero. At 0 K all levels below $E_F$ are occupied and all above are empty. At any temperature $T > 0$, $E_F$ is the energy at which the probability of occupation is exactly $\frac{1}{2}$. For metals $E_F$ is a few eV (Cu: about 7 eV) and is given by $E_F = \dfrac{h^2}{8m}\left(\dfrac{3n}{\pi}\right)^{2/3}$. #### Occupation probability below $E_F$ at 0 K The Fermi–Dirac distribution function is $$ f(E) = \frac{1}{1 + \exp\left(\dfrac{E - E_F}{kT}\right)} $$ For $E < E_F$, $(E - E_F)$ is negative. As $T \to 0$: $$ \frac{E - E_F}{kT} \to -\infty, \qquad e^{-\infty} = 0, \qquad f(E) = \frac{1}{1 + 0} = 1 $$ So an electron state with energy less than $E_F$ is **certainly occupied** (probability = 1, i.e. 100 %) at 0 K. (For $E > E_F$, $f(E) = 1/(1 + \infty) = 0$.) ```text f(E) 1 +-----------+ T = 0 K | |\ 0.5|- - - - - -+ \ T > 0 K | | \___ 0 +-----------+------------> E E_F ``` #### Expectation value of any property $|\psi|^2dx$ is the probability of finding the particle between $x$ and $x + dx$. If a property $A$ is represented by the operator $\hat{A}$, the **expectation value** (average of many measurements on identical systems) is $$ \langle A\rangle = \frac{\int_{-\infty}^{\infty}\psi^*\,\hat{A}\,\psi\,dx}{\int_{-\infty}^{\infty}\psi^*\psi\,dx} $$ For a normalized wave function ($\int\psi^*\psi\,dx = 1$): $$ \langle A\rangle = \int_{-\infty}^{\infty}\psi^*\,\hat{A}\,\psi\,dx $$ **Derivation idea (for position):** the particle is at $x$ with probability $P(x)\,dx = |\psi|^2dx$. As with any weighted average, $$ \langle x\rangle = \frac{\sum x_iP_i}{\sum P_i} \;\rightarrow\; \langle x\rangle = \int x\,|\psi|^2dx = \int\psi^*x\,\psi\,dx $$ Replacing $x$ by the operator for any other quantity gives the general result. | Property | Expectation value | |---|---| | Position | $\langle x\rangle = \int\psi^*x\,\psi\,dx$ | | Momentum | $\langle p\rangle = \int\psi^*\left(-j\hbar\dfrac{\partial}{\partial x}\right)\psi\,dx$ | | Energy | $\langle E\rangle = \int\psi^*\left(j\hbar\dfrac{\partial}{\partial t}\right)\psi\,dx$ | | Potential energy | $\langle V\rangle = \int\psi^*V(x)\psi\,dx$ | Example: for the ground state of an infinite well of width $L$, $\langle x\rangle = L/2$ and $\langle E\rangle = h^2/8mL^2$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. At T > 0, E_F is the energy at which the occupation probability is: A. 1 B. 0 C. 1/2 D. 1/e Answer: C. f(E_F) = 1/2. (Key point: Fermi level definition) 2. The probability that a state below E_F is occupied at 0 K is: A. 0 B. 1/2 C. Depends on E D. 1 Answer: D. The exponent goes to minus infinity, so f = 1. (Key point: Probability at 0 K) 3. For a normalised psi, equals: A. Integral of psi A psi* dx without conjugation B. Integral of psi* A psi dx C. Integral of A dx D. Integral of |psi| dx Answer: B. The operator sits between psi* and psi. (Key point: Expectation formula) 4. If psi is not normalised, must be: A. Multiplied by integral of psi*psi dx B. Left unchanged C. Divided by integral of psi*psi dx D. Set to zero Answer: C. = integral psi* A psi / integral psi* psi. 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