Let's break down and **demonstrate the relation** using physical reasoning and properties of the Fermi-Dirac distribution. --- ### **Restating the Problem** Given: - \( f(E, T) \): Fermi-Dirac distribution at temperature \( T \) - \( g(E) \): Density of states Show that: \[ \int_0^{E_F} dE\, g(E) = \int_0^{\infty} dE\, f(E, T)g(E) \] for any \( T \) and \( E_F \). --- ## **Step 1: Physical Meaning of Both Sides** ### **Left Side:** \[ N_0 = \int_0^{E_F} dE\, g(E) \] At **zero temperature (\(T=0\))**, every state with \( E < E_F \) is **fully occupied** (probability = 1), and every state with \( E > E_F \) is **empty** (probability = 0). So, this integral gives the **total number of electrons** at \( T = 0 \). --- ### **Right Side:** \[ N_T = \int_0^\infty dE\, f(E, T)g(E) \] At **finite temperature \(T\)**, the probability an energy level \( E \) is occupied is given by the **Fermi-Dirac distribution**: \[ f(E, T) = \frac{1}{e^{(E - \mu)/k_BT} + 1} \] where \( \mu \approx E_F \) for low temperatures. So, this integral gives the **total number of electrons** at temperature \( T \). --- ## **Step 2: Conservation of Particle Number** In a closed system (no particles added or removed), **the total number of electrons does not change with temperature**. The Fermi-Dirac distribution "smears" the occupation around \( E_F \) as \( T \) increases, but the area under the curve (total number) remains **constant**. Thus, \[ N_0 = N_T \] or, \[ \int_0^{E_F} dE\, g(E) = \int_0^{\infty} dE\, f(E, T)g(E) \] --- ## **Step 3: Why is This True for Any System?** - This is not dependent on the form of \( g(E) \) (density of states). - It is a consequence of **particle conservation** and the definition of Fermi-Dirac statistics. --- ## **Step 4: Mathematical Note** At \( T = 0 \): - \( f(E, 0) = 1 \) for \( E < E_F \) - \( f(E, 0) = 0 \) for \( E > E_F \) So, \[ \int_0^\infty dE\, f(E, 0)g(E) = \int_0^{E_F} dE\, g(E) \] At \( T > 0 \): - \( f(E, T) \) is a smooth function, but the **total number of particles is fixed**, so the area under the curve is the same. --- ## **Final Answer (Summary)** ### **Demonstration:** The integral \[ \int_0^{E_F} dE\, g(E) \] gives the total number of particles at \( T = 0 \). The integral \[ \int_0^\infty dE\, f(E, T)g(E) \] gives the total number of particles at finite temperature \( T \). **Since the total number of particles is conserved, these two expressions are equal for any \( T \) and \( E_F \), regardless of the material's density of states.** --- ### **Boxed Final Answer:** \[ \boxed{ \int_0^{E_F} dE\, g(E) = \int_0^\infty dE\, f(E, T)g(E) } \] for any temperature \( T \) and Fermi energy \( E_F \), because the total number of particles is conserved and the Fermi-Dirac distribution simply redistributes the occupation around \( E_F \) without changing the total number. --- **Physical Reasoning:** The Fermi-Dirac distribution at finite temperature only "smears" the sharp occupation at \( T = 0 \), but does not change the total number of particles, which is why the two integrals are always equal.