Paper II — Q2
(a) Using Pauli spin matrices prove that, (i) σₓσᵧ + σᵧσₓ = 0; σᵧσᵤ + σᵤσᵧ = 0; σₓσᵤ + σᵤσₓ = 0 (ii) σ₊σ₋ = 2(1+σᵤ) (iii) σₐ + σᵦ…
Using Pauli spin matrices prove that, σₓσᵧ + σᵧσₓ = 0; σᵧσᵤ + σᵤσᵧ = 0; σₓσᵤ + σᵤσₓ = 0
σ₊σ₋ = 2(1+σᵤ)
σₐ + σᵦ = iσᵧ where α ≠ β ≠ γ 8+6+6 marks
Find the uncertainty in the momentum of a particle when its position is determined within 0·02 cm. Find also the uncertainty in the velocity of an electron and α-particle respectively when they are located within 15×10⁻⁸ cm. 15 marks
A particle is moving in a one dimensional box of width 50Å and infinite height. Calculate the probability of finding the particle within an interval of 15Å at the centres of the box when it is in its state of least energy. 15 marks
हिंदी में प्रश्न पढ़ें
पाउली प्रचक्रण आव्यूहों का उपयोग करते हुए सिद्ध कीजिए कि, σₓσᵧ + σᵧσₓ = 0; σᵧσᵤ + σᵤσᵧ = 0; σₓσᵤ + σᵤσₓ = 0
σ₊σ₋ = 2(1+σᵤ)
σₐ + σᵦ = iσᵧ जहाँ α ≠ β ≠ γ 8+6+6 अंक
एक कण के संवेग में अनिश्चितता का पता लगाइए जब उसकी स्थिति 0·02 cm के भीतर निर्धारित की जाती है। एक इलेक्ट्रॉन और अल्फा कण के वेग में अनिश्चितता का पता लगाइए जब वे 15×10⁻⁸ cm के भीतर स्थित हों। 15 अंक
एक कण 50Å चौड़ाई और अनंत ऊँचाई के एकविमीय कोष (बाक्स) में घूम रहा है। कोष (बाक्स) के केंद्र पर 15Å के अंतराल के भीतर कण को खोजने की संभावना (प्रायिकता) की गणना कीजिए जब वह अपनी न्यूनतम ऊर्जा की स्थिति में हो। 15 अंक
Model answer
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(a)(i) Take σᵤ = σ_z, the third Pauli matrix. Let σₓ = [[0,1],[1,0]], σᵧ = [[0,−i],[i,0]], σᵤ = [[1,0],[0,−1]], I = [[1,0],[0,1]].
Then σₓσᵧ = [[i,0],[0,−i]] = iσᵤ, σᵧσₓ = [[−i,0],[0,i]] = −iσᵤ, so σₓσᵧ + σᵧσₓ = 0.
Also σᵧσᵤ = [[0,i],[i,0]], σᵤσᵧ = [[0,−i],[−i,0]], so σᵧσᵤ + σᵤσᵧ = 0.
And σₓσᵤ = [[0,−1],[1,0]], σᵤσₓ = [[0,1],[−1,0]], so σₓσᵤ + σᵤσₓ = 0. Thus the Pauli matrices anticommute in pairs.
(a)(ii) Use σ₊ = σₓ + iσᵧ and σ₋ = σₓ − iσᵧ. Then σ₊σ₋ = (σₓ + iσᵧ)(σₓ − iσᵧ) = σₓ² + σᵧ² + i(σᵧσₓ − σₓσᵧ).
Now σₓ² = σᵧ² = I, and σₓσᵧ = iσᵤ, σᵧσₓ = −iσᵤ. Hence σ₊σ₋ = 2I + i(−iσᵤ − iσᵤ) = 2I + 2σᵤ = 2(1 + σᵤ). So σ₊σ₋ = 2(1 + σᵤ).
(a)(iii) As printed with a plus sign, the statement σₐ + σᵦ = iσᵧ is false. For example, for α = x, β = y, γ = u, σₓ + σᵧ = [[0, 1−i],[1+i, 0]], while iσᵤ = [[i,0],[0,−i]], so they are not equal.
The standard Pauli identity is the cyclic product relation σₐσᵦ = iσᵧ for α, β, γ cyclic. Indeed, σₓσᵧ = iσᵤ, σᵧσᵤ = iσₓ, σᵤσₓ = iσᵧ. Thus, if the intended statement is multiplication rather than addition, the proof is this cyclic product relation.
(b) By Heisenberg’s uncertainty principle, Δx Δp ≥ ħ/2, so Δp_min = ħ/(2Δx).
Given Δx = 0·02 cm = 2·0 × 10⁻⁴ m, Δp_min = (1·0545718 × 10⁻³⁴ J s)/(2 × 2·0 × 10⁻⁴ m) = 2·64 × 10⁻³¹ kg m s⁻¹.
For an electron and an α-particle located within Δx = 15 × 10⁻⁸ cm = 1·5 × 10⁻⁹ m, Δv = ħ/(2mΔx).
Electron: mₑ = 9·11 × 10⁻³¹ kg. Δvₑ = (1·0545718 × 10⁻³⁴)/(2 × 9·11 × 10⁻³¹ × 1·5 × 10⁻⁹) = 3·86 × 10⁴ m s⁻¹.
α-particle: m_α ≈ 6·64 × 10⁻²⁷ kg. Δv_α = (1·0545718 × 10⁻³⁴)/(2 × 6·64 × 10⁻²⁷ × 1·5 × 10⁻⁹) = 5·29 m s⁻¹.
These are minimum uncertainties; the actual uncertainties satisfy the corresponding ≥ inequalities.
(c) For an infinite one-dimensional box of width L = 50 Å, the normalized ground-state wavefunction is ψ₁(x) = √(2/L) sin(πx/L), 0 ≤ x ≤ L.
The interval of width 15 Å at the centre is from x = 25 − 7·5 = 17·5 Å to x = 25 + 7·5 = 32·5 Å. Hence P = ∫₁₇·₅³²·₅ (2/50) sin²(πx/50) dx.
Put y = πx/50. Then dx = (50/π)dy, and the limits become y₁ = 0·35π = 7π/20, y₂ = 0·65π = 13π/20. Thus P = (2/π)∫₇π/₂₀¹³π/₂₀ sin²y dy = (1/π)[y − (1/2) sin 2y]₇π/₂₀¹³π/₂₀ = 3/10 + (1 + √5)/(4π) ≈ 0·5575 = 55·75%.
So the probability is 3/10 + (1 + √5)/(4π) ≈ 0·5575.
How this answer will be evaluated
Framework: Principle > Setup > Derivation > Result
What earns full marks: Rigorous derivations with explicit matrices; precise unit conversions; correct integration limits.
Evaluation rubric
Each sub-part is marked on its own, against the marks and word limit printed on the paper.
- (a)(i) Prove that the Pauli matrices anticommute pairwise. 8 marks
derive— given → assumptions → stepwise derivation → result → check
Must cover
- Explicit 2x2 matrices for σx, σy, σz
- Explicit products σxσy, σyσx (and the other two pairs)
- Sum shown to be the zero matrix
Loses marks
- Result stated without matrix multiplication
- Arithmetic errors in matrix elements
Earns more
- Use σ² = I and tr σ = 0
Extra mark
- Link to fermionic anticommutation relations
- (a)(ii) Prove σ+σ- = 2(1+σz). 6 marks
derive— given → assumptions → stepwise derivation → result → check
Must cover
- Define σ± = σx ± iσy
- Expand σ+σ- using σx² = σy² = I and σxσy = iσz
- Reach 2(1+σz)
Loses marks
- Wrong definition of σ±
- Algebraic slip in simplification
Earns more
- Show the i² = -1 step
Extra mark
- Interpret as raising/lowering operators
- (a)(iii) Establish the product identity for distinct Pauli matrices (the printed sum form is a misprint). 6 marks
derive— given → assumptions → stepwise derivation → result → check
Must cover
- Note the printed sum is false; intended identity is σασβ = iσγ
- Explicit matrix check for one cyclic case
- Cyclic permutation property
Loses marks
- Missing the factor of i
- Confusing commutator with anticommutator
Earns more
- Levi-Civita form σασβ = δαβ + iεαβγσγ
Extra mark
- Connection to the cross product
- (b) Calculate the momentum and velocity uncertainties. 15 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- State ΔxΔp ≥ ħ/2
- Δp = 2.64e-31 kg m/s for Δx = 0.02 cm
- Δve ≈ 3.86e4 m/s and Δvα ≈ 5.29 m/s
Loses marks
- Unit conversion errors
- Mixing ħ and h conventions
Earns more
- Consistent SI conversion (cm to m)
- ΔxΔp ≈ h convention accepted if stated and used consistently
Extra mark
- Compare the electron and α-particle uncertainties
- (c) Find the ground-state probability in a 1-D box over the central 15 Å. 15 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- ψ1 = √(2/L) sin(πx/L) with L = 50 Å
- Limits 17.5 Å to 32.5 Å
- Evaluate to P ≈ 0.5575
Loses marks
- Using an n ≥ 2 wavefunction
- Wrong integration limits
Earns more
- Symmetric limits L/2 ± 7.5 Å
- Correct substitution y = πx/L
Extra mark
- Compare with classical value 0.30
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Establish that the statement holds for every case it claims, not for one representative case. The argument must be closed: each line follows from a definition, a hypothesis, or a named theorem you are entitled to use.
Structure that answers it
Given and to prove, restated → theorem or construction to be used, named → the argument line by line → conclusion stated as proved
Where marks are lost
Testing one example, which illustrates but proves nothing. On an if and only if claim, proving one direction and stopping forfeits that half outright, and degenerate cases — zero, the empty set, the equality case — have to be disposed of rather than assumed away.
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