Two-Level Dynamics & Rabi Oscillations

This is the last lesson of Term 1, and it is where the physics becomes the machine. We take the propagator $U(t)=e^{-iHt}$ from the previous lesson and apply it to the one Hamiltonian every qubit platform implements: a driven two-level system. Out falls the Rabi oscillation — population sloshing coherently between $|0\rangle$ and $|1\rangle$ — and, with it, the punchline of the entire term: a quantum gate is just time evolution run for a chosen duration. Driving $H=\tfrac{\Omega}{2}X$ for the right time turns $|0\rangle$ into $|1\rangle$; that pulse is the $X$ gate. Everything in Term 2 (gates and circuits) and Term 4.4 (how superconducting qubits actually do this) rests here.

Learning Objectives

After this lesson you will be able to:

1. Write the two-level drive Hamiltonian and identify the on-resonance limit $H = \tfrac{\Omega}{2}X$.
2. Derive the Rabi formula $P_1(t) = \sin^2(\Omega t/2)$ for a qubit starting in $|0\rangle$, two independent ways (closed-form exponential and energy eigenbasis).
3. Generalize to nonzero detuning: $P_1(t) = \frac{\Omega^2}{\Omega_{\text{eff}}^2}\sin^2(\Omega_{\text{eff}}t/2)$ with $\Omega_{\text{eff}} = \sqrt{\Omega^2 + \Delta^2}$.
4. Interpret Rabi dynamics as Bloch-vector precession about the drive axis.
5. Show that evolving under $\tfrac{\Omega}{2}X$ for time $t$ realizes $R_x(\Omega t)$, so a $\pi$-pulse is an $X$ gate — connecting analog dynamics to the digital gate model.

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Intuition

Picture a qubit as a tiny magnet (a Bloch vector) sitting in a static field along $\hat z$ that splits the energies of $|0\rangle$ and $|1\rangle$ by $\omega_0$. Left alone, the vector just precesses about $\hat z$ and the populations never change — $|0\rangle$ and $|1\rangle$ are energy eigenstates (this was Example 1 of the previous lesson). To move population between the levels you have to push the qubit off the $z$-axis: you apply an oscillating drive (a microwave or laser tone) that, viewed in the right rotating frame, becomes a constant field along an axis in the equatorial plane. The Bloch vector then precesses about that tilted axis — and as it swings, its $\hat z$-projection runs from $|0\rangle$ down to $|1\rangle$ and back. That coherent round-trip of population is a Rabi oscillation.

The drive strength $\Omega$ (the Rabi frequency) sets how fast the vector swings. If the drive is perfectly on resonance with the qubit's energy gap, the rotation axis lies flat on the equator and the population swings all the way to $|1\rangle$. If the drive is detuned, the axis tilts up out of the plane, the precession cone is narrower, and the qubit never fully reaches $|1\rangle$. So the two knobs are: how hard you drive (amplitude $\Omega$) and how far off you are in frequency (detuning $\Delta$). Stop the drive after exactly half a swing and you have flipped the qubit — that is an $X$ gate, built out of physics.

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Theory

Symbol note. $H$ is the Hamiltonian here, never the Hadamard. $\hbar = 1$, rotation gates use the half-angle convention $R_\sigma(\theta)=e^{-i\theta\sigma/2}$ (Appendix C, E).

The driven two-level Hamiltonian

A qubit with bare energy splitting $\omega_0$ has free Hamiltonian $H_0 = \tfrac{\omega_0}{2}Z$ (energies $\pm\tfrac{\omega_0}{2}$ for $|0\rangle,|1\rangle$). Couple it to a classical field oscillating at drive frequency $\omega_d$ with amplitude (Rabi frequency) $\Omega$; the coupling is a transverse term:

$$H(t) = \frac{\omega_0}{2}Z + \Omega\cos(\omega_d t)\,X .$$

This is time-dependent, so the clean $e^{-iHt}$ machinery does not apply directly. The standard move is to pass to the frame rotating at the drive frequency $\omega_d$ — i.e. substitute $|\psi\rangle = e^{-i\frac{\omega_d}{2}Zt}|\tilde\psi\rangle$ — and drop the fast-oscillating counter-rotating term (the rotating-wave approximation, RWA). The transformed Hamiltonian becomes time-independent (derivation deferred to Term 4.4, where it is the working model of a real device):

$$\boxed{\,H_{\text{RWA}} = \frac{\Delta}{2}\,Z + \frac{\Omega}{2}\,X,\qquad \Delta := \omega_0 - \omega_d\,}$$

with detuning $\Delta$ (how far the drive is off resonance) and Rabi frequency $\Omega$ (how hard we drive). This single, constant, Hermitian $2\times2$ operator generates all the dynamics below. The whole point of the rotating frame is that it converts an oscillating drive into the kind of constant-$H$ problem we solved in 1.6.1.

On resonance: deriving $P_1(t)=\sin^2(\Omega t/2)$

Set the drive exactly on resonance, $\Delta = 0$. Then

$$H = \frac{\Omega}{2}X .$$

Start in $|\psi(0)\rangle = |0\rangle$ and apply the propagator. Because $X^2 = I$, the exponential closes into trig (Appendix E, §3):

$$U(t) = e^{-i\frac{\Omega}{2}Xt} = \cos\!\frac{\Omega t}{2}\,I - i\sin\!\frac{\Omega t}{2}\,X = \begin{pmatrix}\cos\frac{\Omega t}{2} & -i\sin\frac{\Omega t}{2}\\[2pt] -i\sin\frac{\Omega t}{2} & \cos\frac{\Omega t}{2}\end{pmatrix}.$$

Apply it to $|0\rangle$:

$$|\psi(t)\rangle = U(t)|0\rangle = \cos\!\frac{\Omega t}{2}\,|0\rangle - i\sin\!\frac{\Omega t}{2}\,|1\rangle .$$

The Born rule (Appendix C) gives the excited-state population:

$$\boxed{\,P_1(t) = |\langle 1|\psi(t)\rangle|^2 = \sin^2\!\frac{\Omega t}{2} = \frac{1-\cos\Omega t}{2}\,}$$

and $P_0(t) = \cos^2(\Omega t/2)$, so $P_0 + P_1 = 1$ at all times (unitarity). The population oscillates fully between $0$ and $1$ at angular frequency $\Omega$ — these are Rabi oscillations. The period is $T = 2\pi/\Omega$; the qubit is fully inverted ($P_1=1$) at $t = \pi/\Omega$.

Second derivation (energy eigenbasis), as a cross-check. The eigenstates of $H=\tfrac{\Omega}{2}X$ are $|\pm\rangle$ with energies $E_\pm = \pm\tfrac{\Omega}{2}$. Expand $|0\rangle = \tfrac1{\sqrt2}(|+\rangle+|-\rangle)$ and evolve each by its phase (the recipe from 1.6.1):

$$|\psi(t)\rangle = \tfrac1{\sqrt2}\big(e^{-i\Omega t/2}|+\rangle + e^{+i\Omega t/2}|-\rangle\big).$$

Now $\langle 1|\pm\rangle = \pm\tfrac1{\sqrt2}$, so

$$\langle 1|\psi(t)\rangle = \tfrac1{\sqrt2}\Big(\tfrac{1}{\sqrt2}e^{-i\Omega t/2} - \tfrac{1}{\sqrt2}e^{+i\Omega t/2}\Big) = \tfrac12\big(e^{-i\Omega t/2}-e^{+i\Omega t/2}\big) = -i\sin\!\frac{\Omega t}{2},$$

giving $P_1(t) = \sin^2(\Omega t/2)$ — identical. ∎ The oscillation frequency $\Omega = E_+ - E_-$ is exactly the Bohr frequency (energy difference) of the drive Hamiltonian, as the general theory predicted.

Off resonance: detuning and the effective Rabi frequency

Restore the detuning, $H = \tfrac{\Delta}{2}Z + \tfrac{\Omega}{2}X = \tfrac12\,\vec n\cdot\vec\sigma$ with $\vec n = (\Omega,\,0,\,\Delta)$. Write $\Omega_{\text{eff}} = |\vec n| = \sqrt{\Omega^2+\Delta^2}$ and the unit axis $\hat n = \vec n/\Omega_{\text{eff}}$. Then $H = \tfrac{\Omega_{\text{eff}}}{2}\,(\hat n\cdot\vec\sigma)$, and since $(\hat n\cdot\vec\sigma)^2 = I$ the half-angle exponential applies again (Appendix E, §3):

$$U(t) = e^{-i\frac{\Omega_{\text{eff}}}{2}(\hat n\cdot\vec\sigma)\,t} = \cos\!\frac{\Omega_{\text{eff}}t}{2}\,I - i\sin\!\frac{\Omega_{\text{eff}}t}{2}\,(\hat n\cdot\vec\sigma).$$

The amplitude to land in $|1\rangle$ from $|0\rangle$ comes from the off-diagonal piece of $\hat n\cdot\vec\sigma$, which carries the factor $\Omega/\Omega_{\text{eff}}$. Squaring gives the generalized Rabi formula:

$$\boxed{\,P_1(t) = \frac{\Omega^2}{\Omega^2+\Delta^2}\,\sin^2\!\Big(\tfrac{\Omega_{\text{eff}}\,t}{2}\Big), \qquad \Omega_{\text{eff}} = \sqrt{\Omega^2+\Delta^2}\,}$$

Read off the physics: detuning makes the oscillation faster (frequency $\Omega_{\text{eff}}\ge\Omega$) but shallower — the maximum reachable population is $\Omega^2/(\Omega^2+\Delta^2) < 1$. You can no longer fully flip the qubit off resonance. Resonance ($\Delta=0$) is special precisely because it is the only setting that achieves complete population transfer. (Sweeping $\Delta$ and watching this contrast collapse is how experimentalists find a qubit's frequency — a Rabi/chevron measurement, Term 4.4.)

Bloch-sphere picture: precession about the drive axis

The Liouville/Ehrenfest machinery from 1.6.1 makes the geometry exact. For a single qubit, $H = \tfrac12\vec n\cdot\vec\sigma$ and $\rho=\tfrac12(I+\vec r\cdot\vec\sigma)$ (Appendix E, §1). Feeding $\dot\rho = -i[H,\rho]$ through the Pauli commutators $[\sigma_j,\sigma_k]=2i\varepsilon_{jkl}\sigma_l$ yields the Bloch equation:

$$\boxed{\,\frac{d\vec r}{dt} = \vec n\times\vec r\,}$$

This is exactly the equation of a vector precessing about the axis $\vec n$ at angular speed $|\vec n| = \Omega_{\text{eff}}$ — the classical-magnet picture is literally correct. On resonance $\vec n = (\Omega,0,0)$ lies on the equator along $\hat x$: starting from the north pole $\vec r(0)=(0,0,1)$ ($=|0\rangle$), the vector precesses in the $y$–$z$ plane down through the south pole ($|1\rangle$) and back. Its height is $r_z(t) = \cos\Omega t = 1 - 2P_1(t)$, reproducing $P_1=\sin^2(\Omega t/2)$. With detuning the axis tilts up to $\vec n=(\Omega,0,\Delta)$; the precession cone no longer reaches the south pole, so $r_z$ never hits $-1$ — the geometric reason the population transfer is incomplete.

The key computing connection: a gate is timed evolution

Compare the on-resonance propagator with the rotation gate from Appendix E:

$$U(t) = e^{-i\frac{\Omega}{2}Xt} = e^{-i\,(\Omega t)\,X/2} = R_x(\Omega t).$$

So letting the qubit evolve under the drive $\tfrac{\Omega}{2}X$ for a duration $t$ implements the gate $R_x(\theta)$ with rotation angle $\theta = \Omega t$. The "gate angle" is the Rabi frequency times the pulse length — you dial it with time. Special pulses:

| Pulse | Duration $t$ | Angle $\theta=\Omega t$ | Gate $R_x(\theta)$ | Effect on $|0\rangle$ | |---|---|---|---|---| | $\pi/2$-pulse | $\pi/(2\Omega)$ | $\pi/2$ | $R_x(\pi/2)$ | $\to \tfrac1{\sqrt2}(|0\rangle - i|1\rangle)$ (equator) | | $\pi$-pulse | $\pi/\Omega$ | $\pi$ | $R_x(\pi) = -iX$ | $\to -i|1\rangle \equiv |1\rangle$ | | $2\pi$-pulse | $2\pi/\Omega$ | $2\pi$ | $R_x(2\pi) = -I$ | $\to -|0\rangle \equiv |0\rangle$ |

A $\pi$-pulse is an $X$ gate (NOT gate): $R_x(\pi) = \cos\tfrac\pi2 I - i\sin\tfrac\pi2 X = -iX$, which equals $X$ up to the global phase $-i$ — and global phase is physically invisible (Appendix C). It sends $|0\rangle\to|1\rangle$, the very first peak of the Rabi curve $P_1(\pi/\Omega)=\sin^2(\pi/2)=1$. This is not an analogy — it is how $X$ is physically realized on real hardware, by sending a calibrated $\pi$-pulse. The "digital" gate model of Term 2.1 is an abstraction layered on top of exactly this analog control; how the pulse is engineered on a superconducting qubit (and what limits its fidelity) is the subject of Term 4.4. Choosing a different drive axis (a $Y$-quadrature drive gives $\tfrac{\Omega}{2}Y\Rightarrow R_y$; the $\Delta$ term gives $R_z$) lets a single physical mechanism realize any single-qubit rotation — which, by the Euler/ZYZ decomposition (Appendix E, §3), is all of $SU(2)$.

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Worked Examples

Example 1 — A $\pi$-pulse is an $X$ gate

Drive on resonance, $H=\tfrac{\Omega}{2}X$, for the time $t_\pi = \pi/\Omega$ (so $\theta = \Omega t_\pi = \pi$):

$$U(t_\pi) = \cos\!\tfrac\pi2\,I - i\sin\!\tfrac\pi2\,X = -iX = \begin{pmatrix}0 & -i\\ -i & 0\end{pmatrix}.$$

Acting on $|0\rangle$: $U(t_\pi)|0\rangle = -i|1\rangle$. Up to the irrelevant global phase $-i$, this is $|1\rangle$ — a complete bit flip, with $P_1(t_\pi) = \sin^2(\pi/2) = 1$. Acting on $|1\rangle$: $U(t_\pi)|1\rangle = -i|0\rangle \equiv |0\rangle$. So the $\pi$-pulse swaps $|0\rangle\leftrightarrow|1\rangle$ exactly as the $X$ (NOT) gate does. Since $X = i\,R_x(\pi)$, the standard $X$ gate and the physical $\pi$-pulse differ only by global phase — operationally the same gate.

Example 2 — Detuned drive: incomplete transfer

Drive with $\Omega = 1$ and detuning $\Delta = 1$ (so the drive is off resonance by exactly the Rabi frequency). Then $\Omega_{\text{eff}} = \sqrt{1^2+1^2} = \sqrt2$ and the maximum reachable excited population is

$$P_1^{\max} = \frac{\Omega^2}{\Omega^2+\Delta^2} = \frac{1}{1+1} = \frac12 .$$

The full curve is $P_1(t) = \tfrac12\sin^2\!\big(\tfrac{\sqrt2}{2}t\big)$. Compared to the resonant case it oscillates faster (frequency $\sqrt2$ instead of $1$) but only ever reaches halfway up — a $\pi$-pulse duration no longer produces a clean bit flip. This is why qubit drives must be carefully calibrated to resonance: a control error in $\Delta$ both speeds up and caps the rotation, corrupting the intended gate. (Detuning errors are a standard coherent-error model analyzed in Term 4.)

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Hands-on (Python)

We simulate the dynamics with scipy.linalg.expm, plot the Rabi oscillation, verify the $\pi$-pulse realizes $X$, and overlay a detuned curve to see incomplete transfer.

import numpy as np
from scipy.linalg import expm
import matplotlib.pyplot as plt

# Pauli matrices (Appendix E)
I = np.eye(2, dtype=complex)
X = np.array([[0, 1], [1, 0]], dtype=complex)
Z = np.array([[1, 0], [0, -1]], dtype=complex)

ket0 = np.array([1, 0], dtype=complex)
ket1 = np.array([0, 1], dtype=complex)

def H_drive(Omega, Delta):
    """Rotating-frame two-level Hamiltonian H = (Δ/2)Z + (Ω/2)X."""
    return 0.5 * Delta * Z + 0.5 * Omega * X

def P1(t, Omega, Delta):
    """Excited-state population starting from |0>, via U(t)=exp(-iHt)."""
    U = expm(-1j * H_drive(Omega, Delta) * t)
    psi = U @ ket0
    return np.abs(ket1 @ psi) ** 2          # Born rule |<1|ψ(t)>|²

# --- Resonant Rabi oscillations: P1(t) = sin²(Ωt/2) ---
Omega = 2.0
ts = np.linspace(0, 3 * np.pi / Omega, 500)
P1_num   = np.array([P1(t, Omega, Delta=0.0) for t in ts])
P1_exact = np.sin(Omega * ts / 2) ** 2

print("resonant: max |numeric - sin²(Ωt/2)| =", np.max(np.abs(P1_num - P1_exact)))  # ~1e-15

# --- The π-pulse maps |0> -> |1> and equals X up to global phase ---
t_pi = np.pi / Omega                         # θ = Ω t_pi = π
U_pi = expm(-1j * H_drive(Omega, 0.0) * t_pi)
psi_pi = U_pi @ ket0
print("ψ(t_π) =", np.round(psi_pi, 3))       # [0.-0.j  0.-1.j] = -i|1>
print("P1(t_π) =", round(P1(t_pi, Omega, 0.0), 12))   # 1.0  (full inversion)

# U_pi equals X up to a global phase: U_pi = -i X  ⇒  i*U_pi == X
print("U_π == X up to global phase? ", np.allclose(1j * U_pi, X))   # True

# --- Detuning: faster but shallower oscillations (incomplete transfer) ---
fig, ax = plt.subplots(figsize=(7.5, 3.4))
for Delta, style in [(0.0, "-"), (Omega, "--"), (2 * Omega, ":")]:
    Oeff = np.sqrt(Omega**2 + Delta**2)
    cap = Omega**2 / (Omega**2 + Delta**2)   # P1_max = Ω²/Ω_eff²
    curve = np.array([P1(t, Omega, Delta) for t in ts])
    ax.plot(ts, curve, style, lw=2,
            label=fr"$\Delta={Delta/Omega:.0f}\,\Omega$  ($P_1^{{\max}}={cap:.2f}$)")

ax.axhline(1.0, color="gray", ls=":", lw=0.8)
ax.set_xlabel("time  t"); ax.set_ylabel(r"$P_1(t)$")
ax.set_title(r"Rabi oscillations: resonant vs. detuned ($\Omega=2$)")
ax.legend(loc="upper right"); fig.tight_layout()
plt.show()
# On resonance the population reaches 1 (a clean X gate at t=π/Ω);
# detuned drives oscillate faster (Ω_eff) but cap below 1.

# --- The gate view: evolving (Ω/2)X for time t IS the gate R_x(Ωt) ---
def Rx(theta):                               # Appendix E half-angle convention
    return np.cos(theta/2) * I - 1j * np.sin(theta/2) * X

t = 0.37
U_t = expm(-1j * H_drive(Omega, 0.0) * t)
print("U(t) == R_x(Ω t)? ", np.allclose(U_t, Rx(Omega * t)))   # True

Braket note (optional). Because $e^{-i\frac{\Omega}{2}Xt} = R_x(\Omega t)$, the identical dynamics is a single gate in the SDK — from braket.circuits import Circuit; Circuit().rx(0, Omega * t) on a LocalSimulator — and the $\pi$-pulse is just Circuit().x(0). That is the whole bridge from this analog picture to the digital circuit model. We build Braket circuits properly in Term 2.1; the pulse-level control beneath the gate is Term 4.4.

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Exercises

E1 (easy). A qubit has Rabi frequency $\Omega = 2\pi\times 50$ MHz on resonance. How long is a $\pi$-pulse (an $X$ gate)? A $\pi/2$-pulse?

Solution

$t_\pi = \pi/\Omega = \pi/(2\pi\times 50\,\text{MHz}) = 1/(100\,\text{MHz}) = 10\ \text{ns}$. The $\pi/2$-pulse is half that, $t_{\pi/2} = \pi/(2\Omega) = 5\ \text{ns}$. (These ns-scale gate times are realistic for superconducting qubits, Term 4.4.)

E2 (easy). On resonance, at what times is the qubit in an equal superposition, $P_0 = P_1 = \tfrac12$?

Solution

$P_1 = \sin^2(\Omega t/2) = \tfrac12 \Rightarrow \sin(\Omega t/2) = \tfrac1{\sqrt2} \Rightarrow \Omega t/2 = \pi/4 + k\pi/2$, i.e.$t = \frac{\pi}{2\Omega}(2k+1)$for integer$k\ge0$. The first is the $\pi/2$-pulse $t=\pi/(2\Omega)$, which puts the Bloch vector on the equator.

E3 (medium). Derive the generalized Rabi formula's maximum, $P_1^{\max} = \Omega^2/(\Omega^2+\Delta^2)$, and explain geometrically why detuning prevents full inversion.

Solution

$P_1(t) = \frac{\Omega^2}{\Omega^2+\Delta^2}\sin^2(\Omega_{\text{eff}}t/2)$ is maximized when $\sin^2(\cdot)=1$, giving $P_1^{\max} = \Omega^2/(\Omega^2+\Delta^2)$. Geometrically the Bloch vector precesses about $\vec n=(\Omega,0,\Delta)$, which makes angle $\alpha$ with $\hat z$ where $\cos\alpha = \Delta/\Omega_{\text{eff}}$. Starting at the north pole, the vector sweeps a cone of half-angle $\alpha$ about $\vec n$; its lowest point reaches $r_z = \cos 2\alpha = 1 - 2\sin^2\alpha$, and $1-r_z^{\min} = 2P_1^{\max}$ recovers the formula. Only $\Delta=0$ ($\alpha=\pi/2$, axis on the equator) lets the cone reach the south pole. ∎

E4 (medium). Show $R_x(\pi) = -iX$ and explain in one sentence why the $\pi$-pulse counts as the $X$ gate despite the prefactor.

Solution

$R_x(\pi) = \cos\tfrac\pi2\,I - i\sin\tfrac\pi2\,X = 0\cdot I - i\cdot 1\cdot X = -iX$. The $-i$ is a global phase, which has no observable consequence (Appendix C) — $-iX$ and $X$ produce identical measurement statistics on every state — so the $\pi$-pulse is the $X$ gate physically. ∎

E5 (hard). From $H = \tfrac{\Omega}{2}X$ and $\rho=\tfrac12(I+\vec r\cdot\vec\sigma)$, derive the Bloch equation $\dot{\vec r} = (\Omega\hat x)\times\vec r$ and solve it for $\vec r(0)=(0,0,1)$.

Solution

$\dot\rho = -i[H,\rho] = -\tfrac{i\Omega}{4}[X, \vec r\cdot\vec\sigma]$. Using $[X,Y]=2iZ$, $[X,Z]=-2iY$, $[X,X]=0$: $[X,\vec r\cdot\vec\sigma] = r_y[X,Y]+r_z[X,Z] = 2i(r_y Z - r_z Y)$. So $\dot\rho = \tfrac{\Omega}{2}(r_y Z - r_z Y)\!\cdot\!\tfrac12\cdot 2 = \tfrac12(\dot{\vec r}\cdot\vec\sigma)$ forces $\dot r_x = 0,\ \dot r_y = -\Omega r_z,\ \dot r_z = +\Omega r_y$, i.e. $\dot{\vec r} = \Omega\hat x\times\vec r$. With $\vec r(0)=(0,0,1)$: $r_x(t)=0$, $r_z(t)=\cos\Omega t$, $r_y(t)=-\sin\Omega t$ — precession about $\hat x$, and $P_1 = (1-r_z)/2 = \sin^2(\Omega t/2)$ as required. ∎

E6 (hard). A $2\pi$-pulse on resonance returns every population to its start ($P_1(2\pi/\Omega)=0$), yet $U(2\pi/\Omega) = -I \neq I$. Construct a setting in which this $-1$ is physically observable, and explain why it does not contradict "global phase is unobservable."

Solution

$U(2\pi/\Omega) = R_x(2\pi) = \cos\pi\,I - i\sin\pi\,X = -I$. On a single, isolated qubit the $-1$ is a global phase and unobservable. But make it relative: take two qubits in $\tfrac1{\sqrt2}(|0\rangle+|1\rangle)_A|0\rangle_B$ and apply the $2\pi$ drive to qubit $B$ conditioned on $A=|1\rangle$ (a controlled rotation). Then the $|1\rangle_A$ branch picks up $-1$ while the $|0\rangle_A$ branch does not, yielding $\tfrac1{\sqrt2}(|0\rangle - |1\rangle)_A|0\rangle_B = |-\rangle_A|0\rangle_B$ — a measurable phase flip on $A$, detectable by measuring $A$ in the $X$-basis. The rule "global phase is unobservable" concerns an overall phase on the whole state; once the $-1$ multiplies only part of a superposition it becomes a relative phase, which is physical. (This $4\pi$-periodicity is the spin-$\tfrac12$ signature of $SU(2)$, and the mechanism behind controlled-phase gates, Term 2.1.) ∎

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Checkpoint

1. Write the rotating-frame two-level Hamiltonian and name each parameter.
2. Derive $P_1(t)$ for $H=\tfrac{\Omega}{2}X$ starting from $|0\rangle$. What is the period?
3. Give the detuned Rabi formula and the effective frequency; what is the maximum $P_1$?
4. Geometrically, what is a Rabi oscillation on the Bloch sphere?
5. Which pulse realizes an $X$ gate, and why is that the central message of this term?

Answers

1. $H = \tfrac{\Delta}{2}Z + \tfrac{\Omega}{2}X$: $\Delta=\omega_0-\omega_d$ is the detuning, $\Omega$ is the Rabi frequency (drive amplitude).
2. $U(t)=e^{-i\frac{\Omega}{2}Xt} = \cos\frac{\Omega t}{2}I - i\sin\frac{\Omega t}{2}X$, so $|\psi(t)\rangle = \cos\frac{\Omega t}{2}|0\rangle - i\sin\frac{\Omega t}{2}|1\rangle$ and $P_1(t)=\sin^2(\Omega t/2)$; period $T=2\pi/\Omega$.
3. $P_1(t) = \frac{\Omega^2}{\Omega^2+\Delta^2}\sin^2(\Omega_{\text{eff}}t/2)$ with $\Omega_{\text{eff}}=\sqrt{\Omega^2+\Delta^2}$; maximum $\Omega^2/(\Omega^2+\Delta^2) \le 1$, equal to $1$ only on resonance.
4. Precession of the Bloch vector about the drive axis $\vec n=(\Omega,0,\Delta)$ at rate $\Omega_{\text{eff}}$ (Bloch equation $\dot{\vec r}=\vec n\times\vec r$); on resonance the axis is $\hat x$ and the vector swings from the north pole to the south pole and back.
5. The on-resonance $\pi$-pulse ($t=\pi/\Omega$): $R_x(\pi)=-iX$, an $X$ gate up to global phase. This shows a quantum gate is just time evolution run for a chosen duration — the physics (Term 1) is the gate model (Term 2).

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Further Reading

• [Sak] Sakurai & Napolitano, §5.5 — time-dependent two-level problems, the rotating-wave approximation, and the Rabi formula derived in full (with $\hbar$ explicit).
• [NC] Nielsen & Chuang, §4.2 and §7.7 — single-qubit rotations $R_x,R_y,R_z$, and how physical Hamiltonians realize gates on real hardware.
• [Gri] Griffiths & Schroeter, §11.1–11.2 — two-level systems driven by a time-dependent perturbation; Rabi flopping.
• [Pre] Preskill, Ph219, Ch. 3 — qubit dynamics, the Bloch equation, and control.

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