Questions: Angular Momentum Quantization

Question 1

An electron in a hydrogen atom has orbital quantum number ℓ=2. A student claims its total angular momentum magnitude equals 2ℏ. What is the correct magnitude, and why does the student's answer reflect a common misconception?

• A. 2ℏ — the magnitude equals ℏℓ
• B. 4ℏ — the magnitude equals ℏℓ²
• C. ℏ√6 — the magnitude equals ℏ√(ℓ(ℓ+1))
• D. ℏ√5 — the magnitude equals ℏ√(ℓ²−1)

Answer: (C)

The correct magnitude is ℏ√(ℓ(ℓ+1)) = ℏ√(2×3) = ℏ√6 ≈ 2.45ℏ. The student's answer of 2ℏ comes from mistaking the formula as ℏℓ. The discrepancy arises from non-commutativity: because Lx, Ly, Lz cannot simultaneously have sharp values, the total magnitude must exceed the maximum possible z-projection (which is ℓℏ). If the magnitude equaled ℓℏ, there would be no 'room' for the other components — the vector would be perfectly aligned with z, implying simultaneous knowledge of all three components, which the commutation relations forbid.

Question 2

A student argues that spin-½ is simply 'very small orbital angular momentum' and can be derived by solving the Schrödinger equation for a small rotating charge distribution. What is wrong with this picture?

• A. Nothing — spin is orbital angular momentum of the electron's self-rotation
• B. Spin-½ has no spatial wavefunction representation; it emerges from the algebra of commutation relations alone and cannot be modeled as spatial rotation
• C. Spin-½ is too large to be orbital angular momentum — the correct orbital analog would be ℓ=1
• D. The derivation is correct but only works for electrons, not other particles

Answer: (B)

Spin is fundamentally different from orbital angular momentum. Imposing single-valuedness on spatial wavefunctions only generates integer values of ℓ. Half-integer representations are forced by the algebra of commutation relations [Lᵢ, Lⱼ] = iℏεᵢⱼₖLₖ alone — no spatial picture is required or possible. An electron doesn't 'spin' in the classical sense; spin is an intrinsic degree of freedom with no classical analog. The algebraic derivation using ladder operators reveals that both integer and half-integer ℓ satisfy the commutation relations, making spin a natural extension of the same framework rather than a separate add-on.

Question 3

The quantization of angular momentum — including both integer and half-integer values — can be fully derived from the requirement that spatial wavefunctions be single-valued in spherical coordinates.

• T. True
• F. False

Answer: False

Single-valuedness of spatial wavefunctions under a 2π rotation does restrict ℓ to integer values for orbital angular momentum — you can derive ℓ = 0, 1, 2, ... this way. But it fails for half-integer values: a spin-½ state picks up a factor of −1 under 2π rotation (not +1), so it has no valid single-valued spatial wavefunction representation. The fully general derivation uses only the commutation relations and the requirement that the Lz ladder terminates at both ends, which independently forces ℓ to be integer or half-integer. Quantization from the commutation algebra is the deeper and more complete result.

Question 4

For a given angular momentum quantum number ℓ, there are 2ℓ+1 possible values of m. These states all have the same energy in a hydrogen atom and represent genuinely different physical configurations.

• T. True
• F. False

Answer: True

The 2ℓ+1 magnetic substates (m = −ℓ, −ℓ+1, ..., ℓ) are degenerate in a free hydrogen atom — they have the same energy because the choice of z-axis is arbitrary. But they are distinct physical states: each corresponds to a different projection of the angular momentum vector onto the quantization axis, or equivalently, a different spatial orientation of the orbital. This degeneracy is broken by an external magnetic field (the Zeeman effect), which distinguishes the different m values by coupling to the angular momentum. The 2ℓ+1 degeneracy directly determines the number of electrons in each subshell and underlies the structure of the periodic table.

Question 5

Why is the eigenvalue of L² equal to ℏ²ℓ(ℓ+1) rather than ℏ²ℓ², and what does this reveal about angular momentum in quantum mechanics?

Model answer: Because the three components Lx, Ly, Lz don't commute, you can never simultaneously align all angular momentum along one axis. The maximum z-projection is ℓℏ, but the remaining components have irreducible quantum uncertainty spread across x and y. This 'leftover' uncertainty adds the extra ℓ term: ℏ²ℓ(ℓ+1) = ℏ²(ℓ² + ℓ). The total magnitude always exceeds the maximum projection, which is a purely quantum mechanical result with no classical analog — a classical angular momentum vector could in principle be perfectly aligned along one axis.

This is one of the key signals that quantum angular momentum is not just 'small classical angular momentum.' In classical mechanics, you can in principle know all three components simultaneously and have |L|² = Lz² when the vector points along z. In quantum mechanics, the non-commutativity of Lx, Ly, Lz makes it impossible to simultaneously sharpen all three; there is always residual uncertainty in the components perpendicular to the measurement axis. The ℓ(ℓ+1) formula quantifies this fundamental constraint.