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Spin-1/2 as a Canonical System: First Encounter

A spin-1/21/2 degree of freedom is the most important physical example of a two-level system. Its spin Hilbert space is two-dimensional, its basic observables are represented by Pauli matrices, and a magnetic field produces the standard Pauli-vector Hamiltonian.

This page is only the canonical-systems first encounter. The canonical home for spin itself is What Spin Is and Is Not, and the detailed spinor formalism begins with Spin-1/2 Hilbert Space.

Choose the SzS_z eigenbasis

{∣↑⟩,∣↓⟩}.\{\lvert\uparrow\rangle,\lvert\downarrow\rangle\}.

The basis states satisfy

Sz∣↑⟩=ℏ2∣↑⟩,Sz∣↓⟩=−ℏ2∣↓⟩.S_z\lvert\uparrow\rangle = \frac{\hbar}{2}\lvert\uparrow\rangle, \qquad S_z\lvert\downarrow\rangle = -\frac{\hbar}{2}\lvert\downarrow\rangle.

A general spin state in this two-dimensional space is

∣ψ⟩=α∣↑⟩+β∣↓⟩,∣α∣2+∣β∣2=1.\lvert\psi\rangle = \alpha\lvert\uparrow\rangle +\beta\lvert\downarrow\rangle, \qquad \lvert\alpha\rvert^2+\lvert\beta\rvert^2=1.

This is exactly the two-level state

∣ψ⟩=c1∣1⟩+c2∣2⟩\lvert\psi\rangle = c_1\lvert1\rangle+c_2\lvert2\rangle

with the dictionary

∣1⟩↔∣↑⟩,∣2⟩↔∣↓⟩.\lvert1\rangle\leftrightarrow\lvert\uparrow\rangle, \qquad \lvert2\rangle\leftrightarrow\lvert\downarrow\rangle.

The special feature of the spin example is not the dimension. It is the physical meaning of the two basis states: they are eigenstates of an intrinsic angular-momentum component.

For spin-1/21/2,

Si=ℏ2σi,i=x,y,z.S_i=\frac{\hbar}{2}\sigma_i, \qquad i=x,y,z.

Thus the spin component along a unit direction n^\hat{\mathbf n} is

Sn^=n^⋅S=ℏ2n^⋅σ.S_{\hat{\mathbf n}} = \hat{\mathbf n}\cdot\mathbf S = \frac{\hbar}{2} \hat{\mathbf n}\cdot\boldsymbol\sigma.

Its possible measurement outcomes are

+ℏ2and−ℏ2.+\frac{\hbar}{2} \qquad\text{and}\qquad -\frac{\hbar}{2}.

The projectors onto these outcomes are

P±(n^)=12(I±n^⋅σ).P_\pm(\hat{\mathbf n}) = \frac12 \left( I\pm\hat{\mathbf n}\cdot\boldsymbol\sigma \right).

These projectors are the spin version of the general two-level measurement projectors used in Bloch Sphere: Wave-Mechanics Perspective.

A spin magnetic moment often couples to a magnetic field through

H=−μ⋅B.H=-\boldsymbol\mu\cdot\mathbf B.

With

μ=γS,\boldsymbol\mu=\gamma\mathbf S,

the Hamiltonian becomes

H=−γS⋅B=−γℏ2B⋅σ.H = -\gamma\mathbf S\cdot\mathbf B = -\frac{\gamma\hbar}{2} \mathbf B\cdot\boldsymbol\sigma.

This is a Pauli-matrix Hamiltonian

H=c0I+b⋅σH=c_0I+\mathbf b\cdot\boldsymbol\sigma

with

c0=0,b=−γℏ2B.c_0=0, \qquad \mathbf b = -\frac{\gamma\hbar}{2}\mathbf B.

The energy eigenstates are the spin states aligned and anti-aligned with the effective field b\mathbf b, which may be opposite to the physical magnetic field depending on the sign of γ\gamma.

The energy splitting is

ΔE=2∥b∥=ℏ∣γ∣ ∥B∥.\Delta E = 2\lVert\mathbf b\rVert = \hbar\lvert\gamma\rvert\,\lVert\mathbf B\rVert.

This is the basic scale behind Larmor precession and magnetic resonance.

The spin-volume treatment of the same Hamiltonian, with sign conventions and Zeeman language, is Spin in Magnetic Fields.

For a uniform field along zz,

B=B0z^,\mathbf B=B_0\hat z,

the Hamiltonian is

H=−γℏB02σz.H = -\frac{\gamma\hbar B_0}{2}\sigma_z.

The two basis states ∣↑⟩\lvert\uparrow\rangle and ∣↓⟩\lvert\downarrow\rangle are already energy eigenstates. Their energies are

E↑=−γℏB02,E↓=γℏB02.E_\uparrow = -\frac{\gamma\hbar B_0}{2}, \qquad E_\downarrow = \frac{\gamma\hbar B_0}{2}.

The sign of γB0\gamma B_0 determines which state is lower in energy. A global shift would not affect spin precession or transition probabilities, but the splitting does.

The Pauli-vector equation for a pure two-level Bloch vector is

drdt=2ℏ b×r.\frac{d\mathbf r}{dt} = \frac{2}{\hbar}\, \mathbf b\times\mathbf r.

For the magnetic Hamiltonian,

drdt=−γ B×r.\frac{d\mathbf r}{dt} = -\gamma\,\mathbf B\times\mathbf r.

Thus the spin expectation direction precesses around the magnetic field. The angular frequency magnitude is

ωL=∣γ∣ ∥B∥.\omega_L = \lvert\gamma\rvert\,\lVert\mathbf B\rVert.

This is not a new law beyond the two-level Hamiltonian. It is the Pauli-vector rotation formula applied to the physical spin dictionary.

A static magnetic field sets the spin splitting. A weak transverse oscillating field can drive transitions between the two spin states. In a rotating frame and under the rotating-wave approximation, the driven problem has the same effective Hamiltonian form as

Heff=ℏ2(Δσz+Ωσx).H_{\mathrm{eff}} = \frac{\hbar}{2} \left( \Delta\sigma_z+\Omega\sigma_x \right).

That is the model behind Rabi Oscillations: First Encounter. In the spin setting, Δ\Delta is the detuning from the Larmor frequency and Ω\Omega is set by the transverse drive strength and matrix element.

This page does not replace the spin volume. In particular:

The role of this page is narrower: it shows why spin-1/21/2 is a canonical two-level model and how its Hamiltonian fits the same c0I+b⋅σc_0I+\mathbf b\cdot\boldsymbol\sigma structure.

  • Treating ∣↑⟩\lvert\uparrow\rangle and ∣↓⟩\lvert\downarrow\rangle as tiny classical arrows instead of spin-component eigenstates.
  • Forgetting the factor ℏ/2\hbar/2 between σi\sigma_i and SiS_i.
  • Confusing the effective field b\mathbf b in the Hamiltonian with the physical magnetic field B\mathbf B.
  • Losing the sign of the gyromagnetic ratio when deciding which state is lower in energy.
  • Assuming all two-level systems are spin systems; spin-1/21/2 is one physical realization of the general algebra.
  • Using the Bloch sphere as if it displayed the full SU(2)SU(2) spinor sign change under a 2π2\pi rotation.
  • 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.
  • C. Cohen-Tannoudji, B. Diu, and F. Laloë, Quantum Mechanics, Wiley, 1977.
  • A. Abragam, The Principles of Nuclear Magnetism, Oxford University Press, 1961.
  1. For H=−(γℏB0/2)σzH=-(\gamma\hbar B_0/2)\sigma_z, find the energy splitting between ∣↑⟩\lvert\uparrow\rangle and ∣↓⟩\lvert\downarrow\rangle.
Solution

The energies are

E↑=−γℏB02,E↓=γℏB02.E_\uparrow = -\frac{\gamma\hbar B_0}{2}, \qquad E_\downarrow = \frac{\gamma\hbar B_0}{2}.

The signed difference is

E↓−E↑=γℏB0.E_\downarrow-E_\uparrow = \gamma\hbar B_0.

The physical splitting as a positive energy is

ΔE=ℏ∣γB0∣.\Delta E = \hbar\lvert\gamma B_0\rvert.
  1. Show that P+(n^)=(I+n^⋅σ)/2P_+(\hat{\mathbf n})=(I+\hat{\mathbf n}\cdot\boldsymbol\sigma)/2 is the projector onto the +ℏ/2+\hbar/2 eigenspace of Sn^S_{\hat{\mathbf n}}.
Solution

For a unit vector n^\hat{\mathbf n},

(n^⋅σ)2=I.(\hat{\mathbf n}\cdot\boldsymbol\sigma)^2=I.

Therefore

P+2=14(I+n^⋅σ)2=12(I+n^⋅σ)=P+.P_+^2 = \frac14 \left( I+\hat{\mathbf n}\cdot\boldsymbol\sigma \right)^2 = \frac12 \left( I+\hat{\mathbf n}\cdot\boldsymbol\sigma \right) = P_+.

Also,

(n^⋅σ)P+=P+.(\hat{\mathbf n}\cdot\boldsymbol\sigma)P_+=P_+.

Multiplying by ℏ/2\hbar/2 shows that vectors in its range have spin component +ℏ/2+\hbar/2 along n^\hat{\mathbf n}.

  1. A spin state has Bloch vector r=(1,0,0)\mathbf r=(1,0,0). What are the probabilities for measuring Sz=±ℏ/2S_z=\pm\hbar/2?
Solution

For measurement along zz,

p±=12(1±rz).p_\pm = \frac12(1\pm r_z).

Here rz=0r_z=0, so

p+=12,p−=12.p_+=\frac12, \qquad p_-=\frac12.

The state is sharp along xx, not along zz.