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

A spin-$1/2$ 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.

For the physical meaning of intrinsic angular momentum, see What Spin Is and Is Not; the spinor formalism begins with Spin-1/2 Hilbert Space.

Helpful background. Two-Level Systems to work with a pair of orthonormal state amplitudes.

Two-Level Dictionary

Choose the $S_z$ eigenbasis

$$\{\lvert\uparrow\rangle,\lvert\downarrow\rangle\}.$$

The basis states satisfy

$$S_z\lvert\uparrow\rangle = \frac{\hbar}{2}\lvert\uparrow\rangle, \qquad S_z\lvert\downarrow\rangle = -\frac{\hbar}{2}\lvert\downarrow\rangle.$$

A general spin state in this two-dimensional space is

$$\lvert\psi\rangle = \alpha\lvert\uparrow\rangle +\beta\lvert\downarrow\rangle, \qquad \lvert\alpha\rvert^2+\lvert\beta\rvert^2=1.$$

This is exactly the two-level state

$$\lvert\psi\rangle = c_1\lvert1\rangle+c_2\lvert2\rangle$$

with the dictionary

$$\lvert1\rangle\leftrightarrow\lvert\uparrow\rangle, \qquad \lvert2\rangle\leftrightarrow\lvert\downarrow\rangle.$$

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.

Spin Operators And Pauli Matrices

For spin-$1/2$,

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

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

$$S_{\hat{\mathbf n}} = \hat{\mathbf n}\cdot\mathbf S = \frac{\hbar}{2} \hat{\mathbf n}\cdot\boldsymbol\sigma.$$

Its possible measurement outcomes are

$$+\frac{\hbar}{2} \qquad\text{and}\qquad -\frac{\hbar}{2}.$$

The projectors onto these outcomes are

$$P_\pm(\hat{\mathbf n}) = \frac12 \left( I\pm\hat{\mathbf n}\cdot\boldsymbol\sigma \right).$$

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

Magnetic-Field Hamiltonian

A spin magnetic moment often couples to a magnetic field through

$$H=-\boldsymbol\mu\cdot\mathbf B.$$

With

$$\boldsymbol\mu=\gamma\mathbf S,$$

the Hamiltonian becomes

$$H = -\gamma\mathbf S\cdot\mathbf B = -\frac{\gamma\hbar}{2} \mathbf B\cdot\boldsymbol\sigma.$$

This is a Pauli-matrix Hamiltonian

$$H=c_0I+\mathbf b\cdot\boldsymbol\sigma$$

with

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

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

The energy splitting is

$$\Delta E = 2\lVert\mathbf b\rVert = \hbar\lvert\gamma\rvert\,\lVert\mathbf B\rVert.$$

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.

Constant Field Along z

For a uniform field along $z$,

$$\mathbf B=B_0\hat z,$$

the Hamiltonian is

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

The two basis states $\lvert\uparrow\rangle$ and $\lvert\downarrow\rangle$ are already energy eigenstates. Their energies are

$$E_\uparrow = -\frac{\gamma\hbar B_0}{2}, \qquad E_\downarrow = \frac{\gamma\hbar B_0}{2}.$$

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

Larmor Precession As Two-Level Motion

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

$$\frac{d\mathbf r}{dt} = \frac{2}{\hbar}\, \mathbf b\times\mathbf r.$$

For the magnetic Hamiltonian,

$$\frac{d\mathbf r}{dt} = -\gamma\,\mathbf B\times\mathbf r.$$

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

$$\omega_L = \lvert\gamma\rvert\,\lVert\mathbf B\rVert.$$

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

Relation To Rabi Oscillations

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

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

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

Common Mistakes

• Treating $\lvert\uparrow\rangle$ and $\lvert\downarrow\rangle$ as tiny classical arrows instead of spin-component eigenstates.
• Forgetting the factor $\hbar/2$ between $\sigma_i$ and $S_i$.
• Confusing the effective field $\mathbf b$ in the Hamiltonian with the physical magnetic field $\mathbf B$.
• Losing the sign of the gyromagnetic ratio when deciding which state is lower in energy.
• Assuming all two-level systems are spin systems; spin-$1/2$ is one physical realization of the general algebra.
• Using the Bloch sphere as if it displayed the full $SU(2)$ spinor sign change under a $2\pi$ rotation.

Exercises

1. For $H=-(\gamma\hbar B_0/2)\sigma_z$, find the energy splitting between $\lvert\uparrow\rangle$ and $\lvert\downarrow\rangle$.

Solution

The energies are

$$E_\uparrow = -\frac{\gamma\hbar B_0}{2}, \qquad E_\downarrow = \frac{\gamma\hbar B_0}{2}.$$

The signed difference is

$$E_\downarrow-E_\uparrow = \gamma\hbar B_0.$$

The physical splitting as a positive energy is

$$\Delta E = \hbar\lvert\gamma B_0\rvert.$$

2. Show that $P_+(\hat{\mathbf n})=(I+\hat{\mathbf n}\cdot\boldsymbol\sigma)/2$ is the projector onto the $+\hbar/2$ eigenspace of $S_{\hat{\mathbf n}}$.

Solution

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

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

Therefore

$$P_+^2 = \frac14 \left( I+\hat{\mathbf n}\cdot\boldsymbol\sigma \right)^2 = \frac12 \left( I+\hat{\mathbf n}\cdot\boldsymbol\sigma \right) = P_+.$$

Also,

$$(\hat{\mathbf n}\cdot\boldsymbol\sigma)P_+=P_+.$$

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

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

Solution

For measurement along $z$,

$$p_\pm = \frac12(1\pm r_z).$$

Here $r_z=0$, so

$$p_+=\frac12, \qquad p_-=\frac12.$$

The state is sharp along $x$, not along $z$.

References

• A. Abragam, The Principles of Nuclear Magnetism, Oxford University Press, 1961.
• C. Cohen-Tannoudji, B. Diu, and F. Laloë, Quantum Mechanics, Wiley, 1977.
• D. J. Griffiths and D. F. Schroeter, Introduction to Quantum Mechanics, 3rd ed., Cambridge University Press, 2018.
• 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.