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Fermi energy, occupancy at 0 K, expectation value

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).

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.

Fermi energy

The Fermi energy EFE_F is the energy of the highest filled electron level in a metal at absolute zero. At 0 K all levels below EFE_F are occupied and all above are empty. At any temperature T>0T > 0, EFE_F is the energy at which the probability of occupation is exactly 12\frac{1}{2}. For metals EFE_F is a few eV (Cu: about 7 eV) and is given by EF=h28m(3nπ)2/3E_F = \dfrac{h^2}{8m}\left(\dfrac{3n}{\pi}\right)^{2/3}.

Occupation probability below EFE_F at 0 K

The Fermi–Dirac distribution function is

f(E)=11+exp⁡(E−EFkT)f(E) = \frac{1}{1 + \exp\left(\dfrac{E - E_F}{kT}\right)}

For E<EFE < E_F, (E−EF)(E - E_F) is negative. As T→0T \to 0:

E−EFkT→−∞,e−∞=0,f(E)=11+0=1\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 EFE_F is certainly occupied (probability = 1, i.e. 100 %) at 0 K. (For E>EFE > E_F, f(E)=1/(1+∞)=0f(E) = 1/(1 + \infty) = 0.)

 f(E)
  1 +-----------+   T = 0 K
    |           |\
 0.5|- - - - - -+ \  T > 0 K
    |           |  \___
  0 +-----------+------------> E
               E_F

Expectation value of any property

∣ψ∣2dx|\psi|^2dx is the probability of finding the particle between xx and x+dxx + dx. If a property AA is represented by the operator A^\hat{A}, the expectation value (average of many measurements on identical systems) is

⟨A⟩=∫−∞∞ψ∗ A^ ψ dx∫−∞∞ψ∗ψ dx\langle A\rangle = \frac{\int_{-\infty}^{\infty}\psi^*\,\hat{A}\,\psi\,dx}{\int_{-\infty}^{\infty}\psi^*\psi\,dx}

For a normalized wave function (∫ψ∗ψ dx=1\int\psi^*\psi\,dx = 1):

⟨A⟩=∫−∞∞ψ∗ A^ ψ dx\langle A\rangle = \int_{-\infty}^{\infty}\psi^*\,\hat{A}\,\psi\,dx

Derivation idea (for position): the particle is at xx with probability P(x) dx=∣ψ∣2dxP(x)\,dx = |\psi|^2dx. As with any weighted average,

⟨x⟩=∑xiPi∑Pi  →  ⟨x⟩=∫x ∣ψ∣2dx=∫ψ∗x ψ dx\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 xx by the operator for any other quantity gives the general result.

PropertyExpectation value
Position⟨x⟩=∫ψ∗x ψ dx\langle x\rangle = \int\psi^*x\,\psi\,dx
Momentum⟨p⟩=∫ψ∗(−jℏ∂∂x)ψ dx\langle p\rangle = \int\psi^*\left(-j\hbar\dfrac{\partial}{\partial x}\right)\psi\,dx
Energy⟨E⟩=∫ψ∗(jℏ∂∂t)ψ dx\langle E\rangle = \int\psi^*\left(j\hbar\dfrac{\partial}{\partial t}\right)\psi\,dx
Potential energy⟨V⟩=∫ψ∗V(x)ψ dx\langle V\rangle = \int\psi^*V(x)\psi\,dx

Example: for the ground state of an infinite well of width LL, ⟨x⟩=L/2\langle x\rangle = L/2 and ⟨E⟩=h2/8mL2\langle E\rangle = h^2/8mL^2.

✓ Answer checked by Claude Opus 5.5 (high effort)

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