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Spin Examples

Spin examples are finite-dimensional, but they are not just matrix exercises. The physical content is tied to measurement axes, rotations, tensor products, and representation conventions.

NeedLearnMain Check
First spin-1/21/2 modelSpin-One-Half First EncounterWhich axis defines the basis?
Spinor conventionsSpin-Half Hilbert SpaceNormalization and phase
Pauli algebraPauli MatricesSign and basis conventions
Rotation of a spinorSpin RotationsSU(2)SU(2) versus SO(3)SO(3) angle
Two spin-1/21/2 particlesTwo Spin-Half ParticlesProduct basis order
Clebsch-Gordan coefficientsClebsch-Gordan CoefficientsPhase convention
Compact formulaSpin-Half MatricesSi=(ℏ/2)σiS_i=(\hbar/2)\sigma_i

In the SzS_z basis,

∣+x⟩=12(11),∣+z⟩=(10).\lvert +x\rangle =\frac{1}{\sqrt2} \begin{pmatrix} 1\\ 1 \end{pmatrix}, \qquad \lvert +z\rangle = \begin{pmatrix} 1\\ 0 \end{pmatrix}.

The probability that a spin prepared in ∣+z⟩\lvert +z\rangle is found up along x^\hat x is

∣⟨+x∣+z⟩∣2=12.\lvert\langle +x\vert +z\rangle\rvert^2 =\frac12.

The number 1/21/2 is not a statement of ignorance about a preexisting SxS_x value; it is the Born-rule probability for a different measurement basis.

  • Treating a spinor’s global phase as observable.
  • Forgetting the factor of ℏ/2\hbar/2 between σi\sigma_i and SiS_i.
  • Confusing a spatial rotation angle with the spinor phase accumulated under SU(2)SU(2).
  • Changing product-basis order halfway through a two-spin calculation.
  • Comparing Clebsch-Gordan coefficients across sources without checking phase conventions.
  • J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.
  • R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.
  • D. J. Griffiths and D. F. Schroeter, Introduction to Quantum Mechanics, 3rd ed., Cambridge University Press, 2018.
  1. In the minimal worked check, what is the probability of finding spin down along x^\hat x?
Solution

The state ∣−x⟩=(∣+z⟩−∣−z⟩)/2\lvert -x\rangle=(\lvert +z\rangle-\lvert -z\rangle)/\sqrt2, so ∣⟨−x∣+z⟩∣2=1/2\lvert\langle -x\vert +z\rangle\rvert^2=1/2. The two outcomes along x^\hat x exhaust the probabilities.