(a) Explain how the uncertainty in position is different from the uncertainty or inaccuracy of the measuring instruments. 10…

(a)

The uncertainty in position in quantum mechanics is fundamentally different from the uncertainty or inaccuracy of a measuring instrument. Instrumental uncertainty is a classical, epistemic spread in readings caused by finite resolution, noise, drift, parallax, calibration limits, or imperfect alignment. It is attached to the apparatus and the experimental procedure. A better scale, a more stable detector, repeated averaging, or calibration can reduce random uncertainty; inaccuracy, a systematic offset from the true value, can be reduced by correcting the bias. In neither case is there a universal lower bound imposed by the nature of the object being measured.

The quantum-mechanical uncertainty in position is not a defect of the instrument. It is an intrinsic spread of the state itself. For a particle described by a wavefunction, position and momentum are represented by non-commuting operators, [x,p]=iℏ, and the standard deviations of any state satisfy ΔxΔp≥ℏ/2. This follows from the Fourier relationship between ψ(x) and its momentum-space representation: a wavefunction that is sharply localised in x must contain a broad range of wavelengths, hence a broad range of momenta. If a perfect position measurement is made, the post-measurement state can be made very narrow in x, but the momentum spread necessarily becomes large. Thus even with ideal instruments, one cannot prepare or measure a state with arbitrarily small Δx and Δp simultaneously. The consequence is that quantum uncertainty is a constraint on the state, not merely on the measuring device.

(b)

For an electron in a one-dimensional infinite potential well, the wavefunction must vanish at the walls. If the full width is L, the allowed wave numbers are k_n=nπ/L, so p_n=ℏk_n and E_n=p_n²/2m=n²π²ℏ²/(2mL²)=n²h²/(8mL²). The ground state corresponds to n=1. Here L=2 Å=2×10⁻¹⁰ m. Therefore

E₁=h²/(8mL²)=(6.626×10⁻³⁴)²/[8(9.109×10⁻³¹)(2×10⁻¹⁰)²] =1.506×10⁻¹⁸ J.

Since 1 eV=1.602×10⁻¹⁹ J, E₁=9.40 eV. Since 1 aJ=10⁻¹⁸ J, this is 1.506 aJ. If the well is written from -a to +a, then a=1 Å and L=2a, so the same result is E₁=h²/[8m(2a)²]; if a is used for the full width, a=2 Å and E₁=h²/(8ma²).

(c)

The transition ¹F₃—¹D₂ is a singlet transition, so S=0, L=J, and the Landé g-factor is 1. In a magnetic field B, each level splits into 2J+1 equally spaced sublevels with energy shift ΔE=μ_B B M_J, where μ_B=eℏ/(2m_e). The upper ¹F₃ level has J=3 and M_J=-3,-2,-1,0,+1,+2,+3; the lower ¹D₂ level has J=2 and M_J=-2,-1,0,+1,+2. The electric-dipole selection rule is ΔM_J=0,±1. A labelled schematic of the allowed transitions is shown below:

All transitions with ΔM_J=0 have the same transition energy hν₀ and form the π component. All transitions with ΔM_J=+1 are shifted upward by μ_B B and form the σ⁺ component; all transitions with ΔM_J=-1 are shifted downward by μ_B B and form the σ⁻ component. In frequency form the normal Zeeman pattern is:

σ⁻ ν₀ - μ_BB/h π ν₀ σ⁺ ν₀ + μ_BB/h

The π line is linearly polarized parallel to the magnetic field. The σ components are circularly polarized when observed along the field, with opposite senses for σ⁺ and σ⁻, and are linearly polarized perpendicular to the field when observed at right angles to it.

(d)

For a rigid rotor in pure rotational states, the energy is E_J=BJ(J+1), where B is the rotational constant in energy units. The population of level J at temperature T is proportional to the degeneracy times the Boltzmann factor:

N_J∝(2J+1)exp[-BJ(J+1)/kT].

To find the level of maximum population, maximise ln N_J:

f(J)=ln(2J+1)-BJ(J+1)/kT.

Differentiating with respect to J, treating J as continuous, gives

df/dJ=2/(2J+1)-(B/kT)(2J+1)=0.

Hence (2J+1)²=2kT/B, so

J_max=½√(2kT/B)-½=√(kT/2B)-½.

Since J is quantised, the actual level is the nearest non-negative integer to this value. If the temperature is doubled, then

2J_max(2T)+1=√(2k(2T)/B)=√2√(2kT/B)=√2[2J_max(T)+1].

In the high-temperature limit, J_max is large and the -½ term is negligible, so J_max(2T)≈√2 J_max(T). Thus the most populated rotational quantum number shifts to a higher value by a factor of √2 when the temperature is doubled.

(e)

In L-S coupling the two orbital angular momenta are first coupled to give L, and the two spins are coupled to give S. Here l₁=3 and l₂=1, so

L=|l₁-l₂|, |l₁-l₂|+1, …, l₁+l₂ = 2,3,4.

The two electron spins are each ½, so

S=0 or S=1.

The total angular momentum is then J=|L-S|, |L-S|+1, …, L+S.

For S=0, J=L, giving J=2,3,4. For S=1, the allowed J values are: for L=2, J=1,2,3; for L=3, J=2,3,4; and for L=4, J=3,4,5. In term-symbol form the possible states are ¹D₂, ¹F₃, ¹G₄, ³D₁, ³D₂, ³D₃, ³F₂, ³F₃, ³F₄, ³G₃, ³G₄, and ³G₅.