Q1. (a) A particle limited to the x-axis has the wave function φ(x) = bx² between x = 0 and x = 2; the wave function φ(x) = 0…

(a)(i) The wave function is nonzero only between x = 0 and x = 2. Normalize it first. ∫₀² |φ(x)|² dx = |b|² ∫₀² x⁴ dx = |b|² [x⁵/5]₀² = 32|b|²/5 = 1. Hence |b|² = 5/32. Therefore the probability is P = ∫ from x=1 to x=1·5 of |φ(x)|² dx = |b|² [x⁵/5] from 1 to 1·5 = (5/32)((1·5)⁵ – 1⁵)/5 = ((3/2)⁵ – 1)/32. Since (3/2)⁵ = 243/32, P = (243/32 – 1)/32 = (211/32)/32 = 211/1024. So the probability is 211/1024, i.e. about 0·206.

(a)(ii) The expectation value of position is ⟨x⟩ = ∫₀² x |φ(x)|² dx / ∫₀² |φ(x)|² dx. Using |φ|² = |b|² x⁴, the constant |b|² cancels: ⟨x⟩ = ∫₀² x⁵ dx / ∫₀² x⁴ dx = [x⁶/6]₀² / [x⁵/5]₀² = (64/6)/(32/5). Thus ⟨x⟩ = (32/3)(5/32) = 5/3 in the same length unit as x.

(b) Let the three Cartesian components be denoted L₁ = Lₓ, L₂ = Lᵧ, L₃ = Lᵤ. The orbital angular momentum operators satisfy [L₁,L₂] = iℏL₃, [L₂,L₃] = iℏL₁, [L₃,L₁] = iℏL₂. Also L² = L₁² + L₂² + L₃². Using the operator identity [AB,C] = A[B,C] + [A,C]B, [L²,L₁] = [L₂²,L₁] + [L₃²,L₁] = L₂[L₂,L₁] + [L₂,L₁]L₂ + L₃[L₃,L₁] + [L₃,L₁]L₃. Now [L₂,L₁] = –[L₁,L₂] = –iℏL₃ and [L₃,L₁] = iℏL₂. Hence [L²,L₁] = L₂(–iℏL₃) + (–iℏL₃)L₂ + L₃(iℏL₂) + (iℏL₂)L₃ = –iℏ(L₂L₃ + L₃L₂) + iℏ(L₃L₂ + L₂L₃) = 0. By cyclic symmetry, [L²,L₂] = 0 and [L²,L₃] = 0. Hence L² commutes with every component of L.

It is possible to measure L² and any one component of L simultaneously, because L² commutes with each component. For example, simultaneous eigenfunctions of L² and L₃ exist, labelled by l and m. However, L², Lₓ, Lᵧ and Lᵤ cannot all be measured simultaneously in general, because the components do not commute with one another: [Lₓ,Lᵧ] = iℏLᵤ, and cyclic relations hold. Thus Lₓ, Lᵧ and Lᵤ cannot have definite values simultaneously except in the trivial case l = 0. Therefore only L² and one component can be specified together.

(c) For hydrogen, the energy levels are Eₙ = – μ e⁴/(8 ε₀² h² n²), where μ = mₑM/(mₑ + M) is the reduced mass of the electron–nucleus system. A photon emitted in a transition from nᵢ to n_f, with nᵢ > n_f, has energy hν = Eₙᵢ – Eₙ_f = μ e⁴/(8 ε₀² h²)(1/n_f² – 1/nᵢ²). Since ν = c/λ, 1/λ = μ e⁴/(8 ε₀² h³ c)(1/n_f² – 1/nᵢ²). This is the Rydberg formula, usually written 1/λ = R_H(1/n_f² – 1/nᵢ²). Therefore the Rydberg constant is R_H = μ e⁴/(8 ε₀² h³ c). For an infinitely heavy nucleus, R∞ = mₑe⁴/(8 ε₀² h³ c) ≈ 1·097373 × 10⁷ m⁻¹. For hydrogen, R_H is slightly smaller because of the finite proton mass. Thus the emission wavelength of any hydrogen line is inversely proportional to R_H and to the difference of inverse squares of the principal quantum numbers. For example, n_f = 1 gives the Lyman series, n_f = 2 gives the Balmer series, and n_f = 3 gives the Paschen series.

(d) Hydrogen is the most abundant element in the universe, so its spectrum is a powerful probe of cosmic structure. The 21 cm line of neutral hydrogen arises from a hyperfine spin-flip transition of the electron in the hydrogen atom. Its rest wavelength is about 21·1 cm and frequency about 1420·4 MHz. Radio waves at this wavelength pass through dust and gas, so radio interferometers can map neutral hydrogen clouds in the Milky Way and in external galaxies. Doppler shifts of the 21 cm line give line-of-sight velocities; line intensity gives column density. This has been used to map galactic rotation curves, spiral structure, gas clouds, and to infer dark matter.

Redshifted 21 cm radiation from the early universe is used for tomography of the cosmic dark ages, cosmic dawn, and the epoch of reionization. Instruments such as GMRT, FAST, VLA and SKA study these signals. The Lyman series, especially Lyman-α at about 121·6 nm, shifted into the infrared for distant objects, detects high-redshift galaxies and quasars. Absorption by intergalactic hydrogen produces the Lyman-α forest, which maps the distribution, temperature and ionization of the intergalactic medium. Balmer Hα at about 656·3 nm traces ionized hydrogen around young stars and is used to image star-forming regions and galaxy morphology. Thus hydrogen spectral lines provide position–velocity–density maps of the universe across cosmic time.

(e) The time-independent Schrödinger equation is –(ℏ²/2m) d²ψ/dx² + V(x)ψ = Eψ. Given ψ(x) = exp(– bx²), ψ′ = –2bx exp(– bx²), ψ″ = (–2b + 4b²x²) exp(– bx²). Therefore –(ℏ²/2m)ψ″ = –(ℏ²/2m)(–2b + 4b²x²)ψ = (ℏ²b/m – 2ℏ²b²x²/m)ψ. The potential term is Vψ = (2ℏ²b²x²/m)ψ. Adding, Hψ = (ℏ²b/m – 2ℏ²b²x²/m + 2ℏ²b²x²/m)ψ = (ℏ²b/m)ψ. Hence the energy eigenvalue is E = ℏ²b/m. This is the ground-state energy of the harmonic oscillator with angular frequency ω = 2ℏb/m, since E₀ = ½ℏω = ℏ²b/m. The result requires b > 0 so that ψ is normalizable.