Dielectric, Magnetic Properties & Superconductivity
Subtopic: London Equations, Cooper Pairs & BCS Theory Basics
Question Statement:
At T = 0 K, which of the following diagram represents the occupation probability
P(E) of energy states of electrons in a BCS type superconductor?
(A)
(B)
(C)
(D)
P(E)
E
EF
1
P(E)
E
EF
1
P(E)
E
EF
1
P(E)
E
EF
1
Official Answer Key:
A
Step-by-Step Solution & Explanation
### **Official Solution & Step-by-Step Derivation (GATE PH 2022 Q.44)**
**1. BCS Superconductivity Principles**:
- In BCS theory of superconductivity, at $T = 0$ K, electrons form Cooper pairs and condense into the superconducting ground state.
- Energy spectrum of elementary quasiparticle excitations:
$E_k = \sqrt{(\epsilon_k - E_F)^2 + \Delta^2}$
- There exists an energy gap $2\Delta$ centered symmetrically at the Fermi level $E_F$. No single-particle states exist within the gap $|E - E_F| < \Delta$.
- The occupation probability $P(E)$ of electron states is:
$P(E) = v_k^2 = \frac{1}{2}\left(1 - \frac{\epsilon_k - E_F}{E_k}\right)$
- For $E \ll E_F - \Delta$, $P(E) = 1$.
- Across the energy gap around $E_F$, the occupation falls smoothly from $1$ to $0$ over the coherence energy range, matching Diagram (A).
**2. Official Verification**:
The verified official IIT Master Answer is **Option (A)**.
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