Spin Precession Revisited

The cleanest physical realisation of qubit time evolution is a spin in a magnetic field. Its dynamics — Larmor precession — ties together the operator $U(t)$ , the rotation gates, and the Bloch sphere into one picture.

The Hamiltonian of a spin in a field

A magnetic moment $\vec\mu = \gamma\vec S$ in a uniform field $\vec B = B\hat z$ has energy $\hat H = -\vec\mu\cdot\vec B = -\gamma B\,\hat S_z$ . Writing $\hat S_z = \tfrac{\hbar}{2}\sigma_z$ and defining the Larmor frequency $\omega_L = -\gamma B$ (sign conventions vary; we take $\omega_L > 0$ ),

\hat H = \frac{\hbar\omega_L}{2}\,\sigma_z.

This is the same $Z$ -type Hamiltonian we have used throughout the module, now with a concrete physical origin: a real spin in a real magnetic field.

Evolution is a Z-rotation

Exponentiating,

U(t) = e^{-i\hat H t/\hbar} = e^{-i(\omega_L t/2)\sigma_z} = R_Z(\omega_L t).

Time evolution under the field is exactly the $R_Z$ rotation gate with angle $\theta = \omega_L t$ . The longer you wait, the larger the rotation angle — the spin precesses at a steady rate.

Watching the Bloch vector precess

Start with the spin pointing along $+x$ , i.e. $|+\rangle = \tfrac{1}{\sqrt2}(|0\rangle + |1\rangle)$ . Applying $U(t)$ :

|\psi(t)\rangle = \frac{1}{\sqrt2}\big(e^{-i\omega_L t/2}|0\rangle + e^{+i\omega_L t/2}|1\rangle\big).

Its Bloch vector is

\vec r(t) = \big(\cos\omega_L t,\; \sin\omega_L t,\; 0\big),

a unit arrow rotating in the equatorial plane about the field axis $\hat z$ at angular frequency $\omega_L$ . (The in-plane sense depends on the sign convention for $\omega_L$ .) The component along the field, $\langle\sigma_z\rangle = 0$ , is conserved — exactly as the previous two lessons predicted, since $\sigma_z$ commutes with $\hat H$ .

Quarter-turn example

For $\omega_L t = \pi/2$ the Bloch vector moves from $+x$ to $+y$ : the state becomes $\tfrac{1}{\sqrt2}(e^{-i\pi/4}|0\rangle + e^{+i\pi/4}|1\rangle)$ , which (up to global phase) is $\tfrac{1}{\sqrt2}(|0\rangle + i|1\rangle) = |+i\rangle$ . The probabilities of $|0\rangle$ and $|1\rangle$ stay at $1/2$ throughout — pure precession moves phase, not population.

Try it

Prepare $|+\rangle$ with a Hadamard, then precess by a quarter turn with $R_Z(\pi/2)$ . The grader checks the full statevector, which should equal $\tfrac{1}{\sqrt2}(e^{-i\pi/4}|0\rangle + e^{+i\pi/4}|1\rangle)$ .

Run your code to see the quantum state.