Rotation generators and SU(2) representations
Build angular-momentum multiplets and track the distinction between a spatial rotation and a spinor representative.
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The question for today
What you’ll understand
- Use ladder operators to identify allowed magnetic quantum numbers.
- Calculate spin-one-half rotations with a declared active convention.
- State when rotational symmetry enforces magnetic degeneracy.
Quantum angular momentum is defined by how states respond to rotations. Orbital motion is one realization; intrinsic spin is another. Their common algebra allows spectra to be organized before a detailed Hamiltonian is solved. The associated group distinction is essential: ordinary vectors return after one full turn, while a spin-one-half state vector can acquire a minus sign. The physical ray returns, and a coherent comparison can reveal the relative sign.
An algebra fixes a finite ladder
Choose active right-handed rotations with . The components of angular momentum satisfy cyclic commutation relations. They cannot all have definite values simultaneously, but their squared magnitude commutes with each component. We choose joint eigenstates of squared magnitude and the z component. The labels and are dimensionless; the measured angular momenta contain the appropriate powers of Planck's reduced constant.
The raising and lowering combinations change the magnetic label by one. Their norms must be nonnegative, and a finite-dimensional unitary irreducible representation must terminate at both ends. These conditions give and states, with a nonnegative integer or half integer. This is a quantum-algebra result, not a picture of a rigid object secretly rotating about a selected axis. An orbital wavefunction on ordinary three-dimensional space has additional single-valuedness requirements giving integer orbital labels.
Spinors represent the double cover
For spin one half, in a two-component spin basis. Squaring the Pauli matrix along any unit axis gives the identity, which separates the exponential into cosine and sine terms. A rotation by a full turn gives minus the identity; two full turns give the identity. Thus spinors give an ordinary representation of SU(2), the double cover of SO(3), and a projective representation of the spatial rotation group.
The overall minus sign is unobservable for a spinor considered alone. In an interferometer where only one coherent branch undergoes the rotation, it becomes a relative phase between branches. The Bloch vector, which consists of the three Pauli expectations, returns after one full turn because it describes the ray. Consequently a moving arrow on a sphere cannot by itself display the spinor's full phase information. A ray diagram and a state-vector calculation answer related but different questions.
What rotational invariance does and does not imply
A rotationally invariant Hamiltonian commutes with all three generators. Within a single irreducible multiplet it acts identically on every magnetic component, so these components share an energy. This guarantees at least the multiplet degeneracy when that multiplet occurs as an energy subspace. It does not force different values of total angular momentum to have the same energy. Additional radial labels and multiple copies of an irreducible representation can carry different energies, with mixing possible between equivalent copies.
A Hamiltonian commuting only with the z generator preserves axial symmetry and the magnetic label but need not preserve the full multiplet energy. A Zeeman term proportional to the z generator is the simplest counterexample. Conserved angular momentum components therefore have to be inferred from the actual symmetry group of the apparatus. The ladder algebra remains available for representing states even when an external field breaks the symmetry that would otherwise relate their energies.
Work it through
A quarter-turn of an x-polarized spin
Start from and apply . Find its spin direction and its probabilities for z and y measurements.
The two components acquire and . Factor out the common first phase to write the resulting ray as . Keeping the relative factor fixes the rotation direction.
The basis probabilities are still and . Evaluating the Pauli expectations gives , since the cross product of amplitudes is .
The state is therefore positive y spin, with positive-y probability one and negative-y probability zero. Reversing the rotation angle would reverse the y result, a useful sign-convention check.
The active positive quarter-turn sends positive x spin to positive y spin and leaves both z probabilities equal.
The living notebook
Explore the model
Interactive lab · quantum
Spin direction without the hidden common phase
For fresh paused runs Reset and set , then . Predict Bloch coordinates (1,0,0) and (0,1,0), with identical basis probabilities. Playback adds azimuth at one radian per second. This pure-state diagram does not retain an interferometric global phase and therefore cannot display the minus sign from a full spinor turn.
- Probability |0⟩
- 71.1%
- Probability |1⟩
- 28.9%
- Bloch (x, y, z)
- 0.78, 0.45, 0.42
A normalized two-level pure state, not an interacting field simulation. Playback adds one radian per second to the relative phase; basis probabilities stay fixed. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.
Look forThe Bloch vector tracks spin direction; a coherent reference is needed to expose a common spinor phase.
Pause & reason
Check your understanding
Your turn
Try it yourself
For , calculate the coefficient of and the expectation of in the resulting normalized state. Compare with the squared length of its mean angular-momentum vector.
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
Insert the initial magnetic label in the ladder factor. The normalized final state has magnetic label zero.
Compare with the solution
The coefficient is , and the normalized state is . Its squared-angular-momentum expectation is . All three mean components vanish in this state, so the squared length of the mean vector is zero. A zero mean angular momentum does not imply zero angular-momentum fluctuations or a j=0 state.
Keep the idea
Take it with you
- Finite unitary multiplets have 2j plus 1 magnetic states.
- A full turn changes a spin-one-half representative by a sign while returning its ray.
- Full rotational symmetry enforces magnetic degeneracy within a multiplet, subject to the actual Hamiltonian.
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