Paper II — Q2
(a) An operator P describing the interaction of two spin 1/2 particles is P = a + bσ⃗₁·σ⃗₂, where a and b are constants, and σ⃗₁…
An operator P describing the interaction of two spin 1/2 particles is P = a + bσ⃗₁·σ⃗₂, where a and b are constants, and σ⃗₁ and σ⃗₂ are Pauli matrices of the two spins. The total spin angular momentum S⃗ = S⃗₁ + S⃗₂ = 1/2 ℏ(σ⃗₁ + σ⃗₂). Show that P, S² and Sz can be measured simultaneously. 15 marks
Consider a stream of particles of mass m each moving in the positive x-direction with kinetic energy E towards the potential barrier V(x) = 0 for x ≤ 0, V(x) = 3E/4 for x > 0. Find the fraction of particles reflected at x = 0. 15 marks
Consider the potential V(x) = { 0, 0 < x < a; ∞, elsewhere } (a) Estimate the energies of the ground state as well as those of the first and the second excited states for (i) an electron enclosed in a box of size a = 10⁻¹⁰ m. (ii) a 1 g metallic sphere which is moving in a box of size a = 10 cm.
Discuss the importance of the Quantum effects for both of these systems.
Estimate the velocities of the electron and the metallic sphere using uncertainty principle. 20 marks
हिंदी में प्रश्न पढ़ें
दो प्रचक्रण 1/2 के कणों की पारस्परिक क्रिया को एक संकारक P = a + bσ⃗₁·σ⃗₂ से दर्शाया गया है जहाँ a व b स्थिरांक एवं σ⃗₁, σ⃗₂ दोनों प्रचक्रण के पाउली आव्यूह हैं। कुल प्रचक्रण कोणीय संवेग S⃗ = S⃗₁ + S⃗₂ = 1/2 ℏ(σ⃗₁ + σ⃗₂) है। दर्शाइये कि P, S² एवं Sz का एक साथ मापन किया जा सकता है। 15 अंक
विचार कीजिए कि m द्रव्यमान के और गतिज ऊर्जा E के कणों की एक धारा धनात्मक x-दिशा में एक विभव अवरोधक की ओर गतिमय है। V(x) = 0 (x ≤ 0 के लिए), V(x) = 3E/4 (x > 0 के लिए)। x = 0 पर परावर्तित कणों का अंश ज्ञात कीजिए। 15 अंक
मान लें, विभव V(x) = { 0, 0 < x < a; ∞, बाकी सब जगह } तब (a) आद्य अवस्था तथा पहली व दूसरी उत्तेजित अवस्थाओं की ऊर्जाओं का आकलन कीजिए। जबकि (i) एक इलेक्ट्रॉन एक बाक्स के अंदर परिबद्ध है जिसका आकार a = 10⁻¹⁰ m है। (ii) 1 g की वृत्ताकार धातु जो आकार a = 10 cm के बाक्स में गतिशील है। (b) ऊपर दिये गये दोनों निकायों के लिये क्वांटम प्रभाव के महत्व की चर्चा कीजिए। (c) अनिश्चितता सिद्धांत का प्रयोग कर इलेक्ट्रॉन व वृत्ताकार धातु के वेगों का आकलन कीजिए। 20 अंक
Model answer
Drafted with AI by UPSC Answer Check against this question's marking rubric, to the expected length. UPSC does not publish answers for Mains — this is one way to score well, not an official key. How we build rubrics and answers.
(a) Simultaneous measurability of P, S² and S_z
Observables can be measured together if they commute, [A, B] = 0, because they then share a complete set of eigenstates. With S = (ħ/2)(σ₁ + σ₂) and σ₁² = σ₂² = 3, S² = (ħ²/4)(σ₁² + σ₂² + 2σ₁·σ₂) = (ħ²/4)(6 + 2σ₁·σ₂) = (ħ²/2)(3 + σ₁·σ₂), so σ₁·σ₂ = 2S²/ħ² − 3. Hence P = a + bσ₁·σ₂ = (a − 3b) + (2b/ħ²)S², a function of S² alone. Therefore [P, S²] = 0. Also [P, S_z] = (2b/ħ²)[S², S_z] = 0, because S² commutes with every component of the total spin, [S², S_z] = 0 (the Sᵢ satisfy the angular-momentum algebra). So P, S² and S_z form a mutually commuting set and can be measured simultaneously. Common eigenstates: the singlet (S = 0, S_z = 0) has P = a − 3b; the triplet (S = 1, S_z = ħ, 0, −ħ) has P = a + b.
(b) Reflection at a potential step
Let k = √(2mE)/ħ. For x ≤ 0: ψ = A e^ikx + B e^−ikx. For x > 0 the kinetic energy is E − 3E/4 = E/4, so q = √(2m·E/4)/ħ = k/2 and ψ = C e^iqx (only a transmitted wave). Continuity of ψ and ψ′ at x = 0: A + B = C, k(A − B) = qC = (k/2)C. So A − B = (A + B)/2, which gives A = 3B and B/A = 1/3. Reflection coefficient: R = |B/A|² = ((k − q)/(k + q))² = 1/9. The transmitted fraction is T = 1 − R = 8/9, equal to (q/k)|C/A|² = (1/2)(4/3)² = 8/9. Partial reflection occurs even though E > V₀, a purely wave effect with no classical counterpart.
(c) Particle in a one-dimensional box: electron and a 1 g sphere
(a) Eₙ = n²π²ħ²/(2ma²), with ħ = 1.0546×10⁻³⁴ J s. Electron: m = 9.109×10⁻³¹ kg, a = 10⁻¹⁰ m. E₁ = π²(1.0546×10⁻³⁴)²/(2 × 9.109×10⁻³¹ × 10⁻²⁰) = 6.03×10⁻¹⁸ J = 37.6 eV. E₂ = 4E₁ = 150 eV (first excited) and E₃ = 9E₁ = 339 eV (second excited). Sphere: m = 10⁻³ kg, a = 0.10 m. E₁ = π²(1.0546×10⁻³⁴)²/(2 × 10⁻³ × 10⁻²) = 5.5×10⁻⁶³ J (3.4×10⁻⁴⁴ eV). E₂ = 4E₁ = 2.2×10⁻⁶² J and E₃ = 9E₁ = 4.9×10⁻⁶² J. (b) For the electron, the level spacing is of order 100 eV and the de Broglie wavelength is comparable to the box, so the spectrum is strongly quantised and quantum effects dominate. For the sphere, the spacing of about 10⁻⁶² J is vastly smaller than kT ≈ 4×10⁻²¹ J at 300 K, so the levels are effectively continuous and quantum effects are negligible. (c) Take Δx ≈ a, so Δp ≈ ħ/a and v ≈ ħ/(ma). Electron: v ≈ 1.05×10⁻³⁴/(9.109×10⁻³¹ × 10⁻¹⁰) ≈ 1×10⁶ m/s, non-relativistic (the ground-state rms speed √(2E₁/m) ≈ 3.6×10⁶ m/s agrees in order of magnitude). Sphere: v ≈ 1.05×10⁻³⁴/(10⁻³ × 0.10) ≈ 1×10⁻³⁰ m/s, far too small to observe.
How this answer will be evaluated
What earns full marks: Rigorous derivations with correct physical interpretation and units.
Evaluation rubric
Each sub-part is marked on its own, against the marks and word limit printed on the paper.
- (a) Show that P, S² and S_z can be measured simultaneously. 15 marks
justify— claim → 3-4 reasons → evidence → conclusion
Must cover
- Condition [A,B] = 0 for simultaneous measurement
- S² = (ħ²/2)(3 + σ₁·σ₂), so P = (a−3b) + (2b/ħ²)S²
- [P,S²] = 0
- [P,S_z] = 0 and [S²,S_z] = 0
Loses marks
- Assuming commutation without proof.
Earns more
- Marks for the commutation criterion, P as a function of S² and the commutators.
Extra mark
- Common eigenstates: singlet P = a−3b and triplet P = a+b.
- (b) Find the fraction of particles reflected at the potential step. 15 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- Wave functions for x ≤ 0 and x > 0 with q = k/2
- Continuity of ψ and ψ′ at x = 0
- B/A = 1/3
- R = 1/9
Loses marks
- Ignoring continuity of ψ′ or taking the wrong q.
Earns more
- Marks for the wave functions, matching, B/A and R.
Extra mark
- Transmission T = 8/9 and the explanation of partial reflection for E > V₀.
- (c) Find the three lowest energies and velocities for an electron and a 1 g sphere in a box, and discuss quantum effects. 20 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- Eₙ = n²π²ħ²/(2ma²)
- Electron: 37.6, 150, 339 eV
- Sphere: 5.5×10⁻⁶³, 2.2×10⁻⁶², 4.9×10⁻⁶² J
- v ≈ ħ/(ma): 10⁶ m/s and 10⁻³⁰ m/s with discussion
Loses marks
- Using the wrong mass or skipping the uncertainty principle.
Earns more
- Marks for the formula, both sets of energies, the discussion and the velocities.
Extra mark
- de Broglie wavelength and comparison of level spacing with kT.
What "Solve" is asking you to do
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Structure that answers it
Given data and what is required → method chosen, with the reason it applies → set-up (equation, circuit, free body, trial balance) → working, step by step → answer with units and any condition of validity
Where marks are lost
Doing the middle steps mentally and writing only the result. In mathematics papers, a further loss comes from giving a decimal where the exact value in surds or fractions was wanted, or from skipping the justification a part explicitly asks for.
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