35A (Principles of Quantum Mechanics)

Since

S = (ℏ/2) σ

n \cdot S = \frac{ℏ}{2} (cos θ sin θ sin θ - cos θ) .

Its normalized eigenstates are

∣ ↑_{θ} ⟩ = cos \frac{θ}{2} ∣ ↑ ⟩ + sin \frac{θ}{2} ∣ ↓ ⟩, E_{↑} = \frac{ℏ}{2},

and

∣ ↓_{θ} ⟩ = - sin \frac{θ}{2} ∣ ↑ ⟩ + cos \frac{θ}{2} ∣ ↓ ⟩, E_{↓} = - \frac{ℏ}{2} .

b

Write

c = cos (θ /2)

and

s = sin (θ /2)

. Then

∣ ↑_{θ} ↓_{θ} ⟩ - ∣ ↓_{θ} ↑_{θ} ⟩ = (c^{2} + s^{2}) (∣ ↑↓ ⟩ - ∣ ↓↑ ⟩) .

Thus

∣ ψ ⟩

is unchanged. It is the spin-one-half singlet state, so joint rotations leave it invariant. Equivalently, direct use of the Pauli matrices shows

(S_{i}^{(1)} + S_{i}^{(2)}) ∣ ψ ⟩ = 0 (i = x, y, z) .

It is therefore an eigenstate of every Cartesian component of the combined spin, with eigenvalue zero.

Using the states from part (a),

∣ χ ⟩ = \frac{1}{2} [- s ∣ ↑↑ ⟩ + c ∣ ↑↓ ⟩ - c ∣ ↓↑ ⟩ - s ∣ ↓↓ ⟩] .

The Born rule therefore gives

outcome (+ ℏ/2, + ℏ/2) (+ ℏ/2, - ℏ/2) (- ℏ/2, + ℏ/2) (- ℏ/2, - ℏ/2) probability \frac{1}{2} sin^{2} (θ /2) \frac{1}{2} cos^{2} (θ /2) \frac{1}{2} cos^{2} (θ /2) \frac{1}{2} sin^{2} (θ /2) .

After the equal-sign outcomes the state is respectively

∣ ↑↑ ⟩

∣ ↓↓ ⟩

S^{2}

2 ℏ^{2} .

The opposite-sign product states are each a nontrivial linear combination of the singlet and the

m = 0

triplet, so they are not eigenstates of

S^{2}

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