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Physics 2024 Paper II 50 marks Prove

Paper II — Q2

Q2. (a) Prove that : (i) [L², Lz] = 0 (ii) [Lz, L+] = ℏL+ (iii) [L+, L-] = 2ℏLz (iv) L+ L- = L² - Lz² + ℏLz where ℏ = h/2π…

Q2. (a) Prove that :

(i)

[L², Lz] = 0

(ii)

[Lz, L+] = ℏL+

(iii)

[L+, L-] = 2ℏLz

(iv)

L+ L- = L² - Lz² + ℏLz

where ℏ = h/2π (ℏ is Planck's constant) 5+5+5+5=20 marks

(b)

The ground state wave function of a harmonic oscillator is

ψ₀(x) = (mω/ℏπ)^(1/4) exp(-mωx²/2ℏ).

(i)

At which point is the probability density maximum ?

(ii)

What is the value of the maximum probability density ? 15 marks

(c)
(i)

Assuming the potential seen by a neutron in a nucleus to be schematically represented by a one-dimensional, infinite rigid wall potential of length 10⁻¹⁵ m, estimate the minimum kinetic energy of the electron.

(ii)

Estimate the minimum kinetic energy of neutron bound within the nucleus as described above. Can an electron be confined in a nucleus ? Explain. 15 marks

हिंदी में प्रश्न पढ़ें

Q2. (a) सिद्ध कीजिए कि :

(i)

[L², Lz] = 0

(ii)

[Lz, L+] = ℏL+

(iii)

[L+, L-] = 2ℏLz

(iv)

L+ L- = L² - Lz² + ℏLz

जहाँ ℏ = h/2π (h प्लांक स्थिरांक है) 5+5+5+5=20 अंक

(b)

आध (निम्नतम) अवस्था में एक सरल आवर्ती (सनादि) दोलक का तरंग फलन

ψ₀(x) = (mω/ℏπ)^(1/4) exp(-mωx²/2ℏ) है।

(i)

इसके किस बिंदु पर प्रायिकता घनत्व अधिकतम है ?

(ii)

अधिकतम प्रायिकता घनत्व का मान क्या है ? 15 अंक

(c)
(i)

यह मानते हुए कि नाभिक में न्यूट्रॉन द्वारा अनुभव किए गए विभव को 10⁻¹⁵ मी. लंबाई के एक-आयामी, अनंत दृढ़ दीवार विभव द्वारा योजनाबद्ध रूप से दर्शाया गया है, इलेक्ट्रॉन की न्यूनतम गतिज ऊर्जा का आकलन कीजिए ।

(ii)

उपर्युक्त नाभिक में सीमित न्यूट्रॉन की न्यूनतम गतिज ऊर्जा का आकलन कीजिए । व्याख्या कीजिए कि क्या एक इलेक्ट्रॉन को नाभिक के अंदर सीमित किया जा सकता है । 15 अंक

Q2 of the 2024 UPSC Mains Physics Paper II, as printed
The question as printed in the 2024 Physics paper
Evaluate your answer to this question → See all 2024 Physics questions

Model answer

Written 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.

(a)(i) Using the angular-momentum commutation relations [Lx, Ly] = iℏLz, [Ly, Lz] = iℏLx, [Lz, Lx] = iℏLy and cyclically, and the identity [A², B] = A[A, B] + [A, B]A,

[L², Lz] = [Lx² + Ly² + Lz², Lz] = [Lx², Lz] + [Ly², Lz] + 0.

Now [Lx, Lz] = −[Lz, Lx] = −iℏLy, so [Lx², Lz] = Lx(−iℏLy) + (−iℏLy)Lx = −iℏ(LxLy + LyLx).

Also [Ly, Lz] = iℏLx, so [Ly², Lz] = Ly(iℏLx) + (iℏLx)Ly = iℏ(LyLx + LxLy).

Adding, the two terms cancel. Hence [L², Lz] = 0 .

(a)(ii) With L+ = Lx + iLy, [Lz, L+] = [Lz, Lx] + i[Lz, Ly]. Now [Lz, Lx] = iℏLy and [Lz, Ly] = −iℏLx. Therefore [Lz, L+] = iℏLy + i(−iℏLx) = ℏLx + iℏLy = ℏ(Lx + iLy) = ℏL+. [Lz, L+] = ℏL+ .

(a)(iii) [L+, L−] = [Lx + iLy, Lx − iLy] = −i[Lx, Ly] + i[Ly, Lx].

Using [Lx, Ly] = iℏLz and [Ly, Lx] = −iℏLz, [L+, L−] = −i(iℏLz) + i(−iℏLz) = ℏLz + ℏLz = 2ℏLz. [L+, L−] = 2ℏLz .

(a)(iv) L+L− = (Lx + iLy)(Lx − iLy) = Lx² + Ly² + i(LyLx − LxLy). But LyLx − LxLy = −[Lx, Ly] = −iℏLz. Hence L+L− = Lx² + Ly² + i(−iℏLz) = Lx² + Ly² + ℏLz. Since L² = Lx² + Ly² + Lz², we get L+L− = L² − Lz² + ℏLz .

(b)(i) The probability density is P(x) = |ψ₀(x)|² = √(mω/(πℏ)) exp(−mωx²/ℏ). The exponential is maximum when x = 0. Therefore the probability density is maximum at x = 0 .

(b)(ii) At x = 0, Pmax = P(0) = √(mω/(πℏ)). Using ℏ = h/2π, this is also Pmax = √(2mω/h). Pmax = √(mω/(πℏ)) , with unit m⁻¹.

(c)(i) For a particle in an infinite square well of width L, Eₙ = n²π²ℏ²/(2mL²) = n²h²/(8mL²), n = 1, 2, 3, ... The minimum kinetic energy is for n = 1. Taking L = 10⁻¹⁵ m and electron mass mₑ = 9.109 × 10⁻³¹ kg,

Eₑ = h²/(8mₑL²) = (6.626 × 10⁻³⁴)²/[8(9.109 × 10⁻³¹)(10⁻¹⁵)²] J ≈ 6.02 × 10⁻⁸ J ≈ 3.76 × 10¹¹ eV = 3.76 × 10⁵ MeV. Eₑ(min) ≈ 376 GeV .

This is the non-relativistic infinite-square-well result. Since it exceeds mₑc² ≈ 0.511 MeV by many orders, the non-relativistic formula is not quantitatively valid for the electron; a relativistic order-of-magnitude estimate would be lower, of order hundreds of MeV, but still far above nuclear binding energies.

(c)(ii) For a neutron, mₙ = 1.675 × 10⁻²⁷ kg. Thus Eₙ = h²/(8mₙL²) = (6.626 × 10⁻³⁴)²/[8(1.675 × 10⁻²⁷)(10⁻¹⁵)²] J ≈ 3.28 × 10⁻¹¹ J ≈ 2.05 × 10⁸ eV = 205 MeV. Eₙ(min) ≈ 205 MeV .

Here Eₙ < mₙc² ≈ 940 MeV, so the non-relativistic estimate is reasonable as an order of magnitude.

An electron cannot be confined inside a nucleus. Its zero-point kinetic energy in a nuclear-sized well would be enormously larger than typical nuclear binding energies, which are only of the order of MeV. Moreover, the electron does not feel the strong nuclear force, and electrostatic attraction by protons cannot supply such enormous confinement energy. In beta decay, electrons are created and emitted; they are not bound inside the nucleus.

What "Prove" is asking you to do

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.

All UPSC directive words, compared →

How this answer will be evaluated

Approach

Framework: Quantum Mechanics: Angular Momentum Algebra & Particle in a Box. (a) derive: given > assumptions > stepwise derivation > result > check | (b) derive: given > assumptions > stepwise derivation > result > check | (c) explain: definition/context > points in order > small example > short close Full marks: Rigorous derivation in (a), precise calculus in (b), and physically sound estimation in (c).

Key points expected

  • Angular momentum commutator algebra
  • Gaussian probability density maximum
  • Particle in a box energy estimation
  • Nuclear scale energy comparison

Evaluation rubric

Each sub-part is marked on its own, against the marks and word limit printed on the paper.

  1. (a) Prove the four commutation relations and operator identity using fundamental angular momentum commutators. 20 marks

    derive— given → assumptions → stepwise derivation → result → check

    Must cover

    • State fundamental commutators [Li, Lj] = iℏεijkLk
    • Define L± = Lx ± iLy explicitly
    • Show [L², Lz] = 0 via linearity
    • Derive L+L- identity by expansion

    Loses marks

    • Quoting results without derivation
    • Sign errors in commutator expansion

    Earns more

    • Correct use of Jacobi identity
    • Clear step-by-step expansion of products
    • Consistent use of ℏ notation

    Extra mark

    • Mention of SU(2) algebra structure
  2. (b) Determine the location and value of the maximum probability density for the ground state. 15 marks

    derive— given → assumptions → stepwise derivation → result → check

    Must cover

    • Write probability density |ψ₀(x)|²
    • Differentiate w.r.t x and set to zero
    • Identify maximum at x = 0
    • Calculate value (mω/ℏπ)^(1/2)

    Loses marks

    • Confusing wave function with probability density
    • Missing the square root in normalization

    Earns more

    • Explicit calculation of the derivative
    • Verification of second derivative for maximum

    Extra mark

    • Comparison with classical probability distribution
  3. (c) Estimate minimum kinetic energy for neutron and electron in a 10⁻¹⁵ m box; explain confinement. 15 marks

    explain— definition/context → points in order → small example → short close

    Must cover

    • Use E = p²/2m with p = h/2L
    • Calculate neutron energy (~20 MeV)
    • Calculate electron energy (~100 GeV)
    • Compare with nuclear binding energy

    Loses marks

    • Using wrong mass for electron/neutron
    • Ignoring the 'infinite wall' boundary condition

    Earns more

    • Correct unit conversion (MeV/GeV)
    • Logical argument for electron exclusion

    Extra mark

    • Mention of relativistic effects for electron

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