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…

(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. (The printed 'L₊L_y' in the question is a misprint for L₊L₋.)

(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. This is opposite to a classical oscillator, which is most likely to be found near its turning points.

(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 estimate uses E ≈ pc with p = h/2L, giving E ≈ hc/(2L) ≈ 9.9×10⁻¹¹ J ≈ 620 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.