Optical Blochs Dynamics

Bloch‑vector§

let

$$\mathbf{R}=\left(u, v, w\right)$$

Where

$$\begin{cases} u &= \rho_{eg} + \rho_{ge} \\ v &= i\left(\rho_{eg} - \rho_{ge}\right) \\ w &= \rho_{ee} - \rho_{gg} \\ 1 &= \rho_{ee} + \rho_{gg} \end{cases}$$

The last line is just the trace condition

Damped Precession§

$$\dot{\mathbf{R}}=\mathbf{\Omega}\times\mathbf{R}-\Gamma_{\perp}\big(u~\hat{\mathbf{x}}+v~\hat{\mathbf{y}}\big)-\gamma\big(w-w_{\mathrm{eq}}\big)~\hat{\mathbf{z}}$$

With $\Gamma_\perp = \gamma/2 + 2\gamma_\phi$. Where in our model $w_{eq} = -1$, but other more complex model we can have different $w_{eq}$ which is the equilibrium point of the presession. And the torque vector is

$$\mathbf{\Omega}=\begin{pmatrix} \Omega_R \\ 0 \\ \Delta \end{pmatrix}$$

Undamped Special case§

We will have

$$\dot{\mathbf{R}}=\mathbf{\Omega}\times\mathbf{R}$$

Without assuming a resonant, the presession is tilted and it gives

Derivation of the w equation

Writing it in component form and using second equation to eliminate $v$

$$\begin{cases} \dot{u} = -\Delta v \\ \dot{v} = \Delta u - \Omega_R w \\ \dot{w} = \Omega_R v \end{cases} \Rightarrow \begin{cases} \ddot{u} = -\Delta \dot{v} \\ \dot{v} = \Delta u - \Omega_R w \\ \ddot{w} = \Omega_R \dot{v} \end{cases} \Rightarrow \begin{cases} \ddot{u} = \Delta \left(\Omega_R w - \Delta u \right) \\ \ddot{w} = -\Omega_R \left(\Omega_R w - \Delta u \right) \end{cases}$$

Then using the second equation, we can use it to eliminate $u$ from the system of equation, so the second equation with its double derivative becomes

$$u=\frac{1}{\Omega_R\Delta}\ddot{w}+\frac{\Omega_R}{\Delta} w \qquad \ddot{u}=\frac{1}{\Omega_R\Delta}w^{(4)}+\frac{\Omega_R}{\Delta} \ddot{w}$$

Substitute it in equation 1 gives

$$w^{(4)} + \left(\Omega_R^2 +\Delta^2\right)\ddot{w} = 0$$

Where immediately we can see the solution is

$$w = Ae^{ikt} + Be^{-ikt} + Ct + D$$

with

$$k = \sqrt{\Omega_R^2 + \Delta^2} = \Omega$$

Initial condition§

At $t=0$, $w = -1$, $u=v=0$, thus

$$\begin{cases} A + B + D &= -1 \qquad \text{for } w = -1 \\ ikA - ikB + C &= 0 \qquad \text{for } \dot{w} = \Omega_R v = 0 \\ -k^2A - k^2B &= \Omega_R^2 \qquad \text{for } \ddot{w} = -\Omega_R \left(\Omega_R w - \Delta u\right) = \Omega_R^2 \\ -ik^3A + ik^3B &= 0 \qquad \text{for } \dddot{w} = -\Omega_R \left(\Omega_R \dot{w} - \Delta \dot{u}\right) = \Omega_R\Delta \dot{u} = -\Omega_R\Delta^2 v = 0 \\ \end{cases}$$

Simplify it, we will get from the 4th equation for $k \neq 0$, we have $A = B$, substitute it in

$$\begin{cases} 2A + D = -1 \\ C = 0 \\ A = -\cfrac{\Omega_R^2}{2k^2} \\ \end{cases}$$

Lastly from the 3th equation we can substitute it into the first equation, getting us

$$A = B = -\frac{\Omega_R^2}{2\Omega^2} \qquad C = 0 \qquad D = -\frac{\Delta^2}{\Omega^2}$$

Looking at $\rho_{ee}=(w+1)/2$

$$\rho_{ee} = \frac{\Omega_R^2}{\Omega^2}\sin^2\frac{\Omega}{2} t$$

Again, we get back the Rabi oscillation, but with the power of the full equation, we can not only solve for the ideal case, we can even solve for the non-ideal case which involve damping. Or even we can solve for $\rho_{eg}$ With a different initial condition.

Derivation of the u and v components

Once $w$ is known, $u$ and $v$ cost no new characteristic equation. The third equation hands $v$ directly

$$v=\frac{\dot{w}}{\Omega_R}=\frac{\Omega_R}{\Omega}\sin\Omega t$$

and the first equation hands $u$ by one integration

$$\dot{u}=-\Delta v=-\frac{\Delta\Omega_R}{\Omega}\sin\Omega t \quad\Rightarrow\quad u=\frac{\Delta\Omega_R}{\Omega^2}\cos\Omega t + C$$

Fixing $C$ with $u(0)=0$ gives $C=-\Delta\Omega_R/\Omega^2$. As a check, feeding both into $\dot{v}=\Delta u - \Omega_R w$ returns $\Omega_R\cos\Omega t$ on each side.

On resonance $\Delta=0$ kills $u$ entirely and leaves $v=\sin\Omega_R t$, $w=-\cos\Omega_R t$, a pure rotation in the $v$–$w$ plane. The detuning is exactly what tilts the precession out of that plane, by feeding a static $-\Delta\Omega_R/\Omega^2$ pedestal into $u$.

On‑resonance step response: optical nutation§

This time, let's take the full equation in resonant with decay

$$\dot{\mathbf{R}}=\mathbf{\Omega}\times\mathbf{R}-\Gamma_{\perp}\big(u~\hat{\mathbf{x}}+v~\hat{\mathbf{y}}\big)-\gamma\big(w-w_{\mathrm{eq}}\big)~\hat{\mathbf{z}}$$ $$\begin{cases} \dot{u}=-\Gamma_\perp u \\ \dot{v}=-\Omega_R w - \Gamma_\perp v \\ \dot{w} = \Omega_R v - \gamma ( w + 1 ) \end{cases}$$

Notice that the $u$ component is totally decoupled due to on resonant which is true even without decay which will make $u$ stationary. Thus, solving that for $u$ we get

$$u=u_0 e^{-\Gamma_\perp t}$$

The derivation of the w equation

We will use the third equation to express $v$ explicitly in terms of $w$. And it is

$$v=\frac{1}{\Omega_R}\dot{w}+\frac{\gamma}{\Omega_R}(w+1) \qquad \dot{v}=\frac{1}{\Omega_R}\ddot{w}+\frac{\gamma}{\Omega_R}\dot{w}$$

Substitute it in

$$\ddot{w} + \left(\gamma + \Gamma_\perp\right)\dot{w} + \left(\Omega_R^2 + \Gamma_\perp \gamma\right)w = -\Gamma_\perp \gamma$$

Again having a trivial solution

$$w = Ae^{(-\alpha+\beta)t} + Be^{(-\alpha-\beta)t} - \frac{\Gamma_\perp\gamma}{\Omega_R^2 + \Gamma_\perp\gamma} = e^{-\alpha t}\left(Ae^{\beta t} + Be^{-\beta t}\right) - \frac{\Gamma_\perp\gamma}{\Omega_R^2 + \Gamma_\perp\gamma}$$

Where $-\alpha+\beta=k$ satisfies the quadratic equation $k^2 + (\gamma+\Gamma_\perp)k + (\Omega_R^2 + \Gamma_\perp\gamma) = 0$, which then gives

$$k = \frac{-(\gamma+\Gamma_\perp) \pm \sqrt{(\gamma - \Gamma_\perp)^2 - 4\Omega_R^2}}{2}$$

Or

$$\begin{cases} \alpha = \cfrac{\gamma+\Gamma_\perp}{2} \\ \beta = \sqrt{\left(\cfrac{\gamma-\Gamma_\perp}{2}\right)^2 - \Omega_R^2} \end{cases}$$

Initial condition§

Take at $t=0$, $w=-1$ and $v=0$, Then we have

$$\begin{cases} A + B - \frac{\Gamma_\perp\gamma}{\Omega_R^2 + \Gamma_\perp\gamma} &= -1 \qquad \text{for } w=-1 \\ -\alpha\left(A + B\right) + \beta \left(A -B\right) &= 0 \qquad \text{for } \dot{w} = \Omega_R v - \gamma ( w + 1 ) = 0 \end{cases}$$

Thus we solved

$$A=-\frac{1}{2}\left(\frac{\Omega_R^2}{\Omega_R^2 + \Gamma_\perp\gamma}\right)\left(1+\frac{\alpha}{\beta}\right) \qquad B=-\frac{1}{2}\left(\frac{\Omega_R^2}{\Omega_R^2 + \Gamma_\perp\gamma}\right)\left(1-\frac{\alpha}{\beta}\right)$$ $$w = -\frac{1}{2}\left(\frac{\Omega_R^2}{\Omega_R^2 + \Gamma_\perp\gamma}\right)e^{-\alpha t}\left(\left(1+\frac{\alpha}{\beta}\right)e^{\beta t} + \left(1-\frac{\alpha}{\beta}\right)e^{-\beta t}\right) + w_{ss}$$

If $\beta$ is real, which means $\gamma - \Gamma_\perp \gt 2\Omega_R$, we have

$$w=-\left(\frac{\Omega_R^2}{\Omega_R^2 + \Gamma_\perp\gamma}\right)e^{-\alpha t}\left(\cosh\beta t + \frac{\alpha}{\beta}\sinh\beta t\right) + w_{ss} \\$$

If it is the opposite

$$w=-\left(\frac{\Omega_R^2}{\Omega_R^2 + \Gamma_\perp\gamma}\right)e^{-\alpha t}\left(\cos|\beta| t + \frac{\alpha}{|\beta|}\sin|\beta| t\right) + w_{ss} \\$$

Where

$$\begin{cases} \alpha = \cfrac{\gamma+\Gamma_\perp}{2} \\ \beta = \sqrt{\left(\cfrac{\gamma-\Gamma_\perp}{2}\right)^2 - \Omega_R^2} \end{cases}$$

Where if $\Omega_R \gg \gamma$ really, we get an underdamped oscillator,

$$w = -e^{-\alpha}\cos\Omega_Rt + w_{ss}$$

The v component§

The transverse $v$ is slaved to $w$ through the third equation, so again no new characteristic equation

$$v=\frac{1}{\Omega_R}\dot{w}+\frac{\gamma}{\Omega_R}(w+1)$$

The derivation of the v equation

Write $k_\pm = -\alpha\pm\beta$ so that $\dot{w}=Ak_+e^{k_+t}+Bk_-e^{k_-t}$. Then

$$v=\frac{1}{\Omega_R}\Big[A(k_++\gamma)e^{k_+t}+B(k_-+\gamma)e^{k_-t}\Big]+\frac{\gamma}{\Omega_R}(w_{ss}+1)$$

The constant piece uses $w_{ss}+1=\dfrac{\Omega_R^2}{\Omega_R^2+\Gamma_\perp\gamma}$, giving the steady state $v_{ss}$. For the transient, set $\delta=\gamma-\alpha=\tfrac{\gamma-\Gamma_\perp}{2}$ so that $k_\pm+\gamma=\delta\pm\beta$, insert $A,B$ from the $w$ initial conditions, and collect with the two identities

$$\alpha+\delta=\gamma \qquad \alpha\delta+\beta^2=\gamma\delta-\Omega_R^2$$

The common prefactor collapses to exactly $v_{ss}$, so that $v(0)=v_{ss}(1-1)=0$ and $v(\infty)=v_{ss}$ fall out for free.

If $\beta$ is imaginary, the same replacement $\cosh\to\cos$, $\sinh\to\sin$ applies

$$v = v_{ss}\left[1 - e^{-\alpha t}\left(\cos|\beta| t + \frac{\gamma\delta-\Omega_R^2}{\gamma|\beta|}\sin|\beta| t\right)\right]$$

And in the strong-drive limit $\Omega_R \gg \gamma, \Gamma_\perp$, $v_{ss}\to 0$, $|\beta|\to\Omega_R$, and the whole thing reduces to

$$v = e^{-\alpha t}\sin\Omega_R t$$

which is the $\sin$ partner of $w=-e^{-\alpha t}\cos\Omega_R t + w_{ss}$: the Bloch vector nutates in the $v$–$w$ plane while $u$ stays pinned at zero.

Free induction decay (FID)§

When we starts the atom in an initial state without external field coupled to it $\Omega = 0$ ( due to $n = 0$ ), thus then

$$\begin{cases} \dot{u} = -\Gamma_\perp u -\Delta v \\ \dot{v} = \Delta u - \Gamma_\perp v \\ \dot{w} = - \gamma ( w + 1 ) \end{cases}$$

Now $w$ is decoupled out from the system and $u, v$ are coupled in a symmetric way, thus the solution is

$$\begin{cases} u(t)&=e^{-\Gamma_\perp t}\left(u_0\cos\left(\Delta t\right)-v_0 \sin(\Delta t)\right) \\ v(t)&=e^{-\Gamma_\perp t}\left(v_0 \cos(\Delta t) + u_0\sin\left(\Delta t\right)\right) \\ w(t)&=(w_0 + 1)e^{-\gamma t} - 1 \end{cases}$$

Remember that $u$ and $v$ are the dipole terms, this gives

$$\braket{d(t)}=2d_{eg}|\rho_{eg}(0)| e^{-\Gamma_\perp t}\cos\left(\omega_{eg} t + \phi\right)$$

Where the oscillation term is what cause FID light to emit.

Driven steady state: dispersion, absorption, and the phase lag§

Let's solve with all derivative set to $0$

$$\begin{cases} \dot{u}=-\Delta v -\Gamma_\perp u \\ \dot{v}= \Delta u - \Omega_R w - \Gamma_\perp v \\ \dot{w} = \Omega_R v - \gamma ( w + 1 ) \end{cases}$$

Derivation of the steady state

Setting $\dot u = \dot v = \dot w = 0$, the first and third equations hand two components for free

$$u_{ss}=-\frac{\Delta}{\Gamma_\perp}v_{ss} \qquad w_{ss}=\frac{\Omega_R}{\gamma}v_{ss}-1$$

Feeding both into the middle equation leaves only $v_{ss}$

$$0 = -\frac{\Delta^2}{\Gamma_\perp}v_{ss} - \frac{\Omega_R^2}{\gamma}v_{ss} + \Omega_R - \Gamma_\perp v_{ss} \quad\Rightarrow\quad v_{ss}=\frac{\Omega_R\Gamma_\perp}{\Delta^2 + \Gamma_\perp^2 + \frac{\Gamma_\perp}{\gamma}\Omega_R^2}$$

Writing the saturation parameter $s_0 = \Omega_R^2/\Gamma_\perp\gamma$ collapses the denominator to the power-broadened Lorentzian width $\Delta^2 + \Gamma_\perp^2(1+s_0)$.

Two things to read off immediately. First, $u_{ss}$ is odd in $\Delta$ and $v_{ss}$ is even — $u$ is the dispersive quadrature, $v$ the absorptive one, and they sit exactly $90^\circ$ apart. Second, saturation ($s_0$) only broadens the shared denominator; it cancels in the ratio

$$\frac{u_{ss}}{v_{ss}} = -\frac{\Delta}{\Gamma_\perp}$$

so the phase is a pure property of detuning-over-linewidth, independent of how hard you drive.

The phase lag§

Assembling the coherence in the linear limit $s_0 \to 0$

$$\rho_{eg}=\frac{u_{ss}-iv_{ss}}{2} = -\frac{\Omega_R}{2}\,\frac{\Delta + i\Gamma_\perp}{\Delta^2+\Gamma_\perp^2} = \frac{\Omega_R/2}{\,i\Gamma_\perp - \Delta\,}$$

which is exactly the $\dfrac{1}{i\Gamma_\perp - \Delta}$ structure of $\chi$ below — the steady-state coherence is the susceptibility, up to the field-to-$\Omega_R$ conversion. The lag is the angle of this coherence off the dispersive axis

$$\tan\phi = \frac{v_{ss}}{-u_{ss}} = \frac{\Gamma_\perp}{\Delta} \qquad\Rightarrow\qquad \phi = \arctan\frac{\Gamma_\perp}{\Delta}$$

This is the driven damped-oscillator phase curve verbatim: $\Gamma_\perp$ sets the half-width where the lag crosses $45^\circ$ and where the absorptive Lorentzian $v_{ss}$ drops to half its resonant peak. The $u_{ss}$ sign flip through resonance is the anomalous-dispersion feature. So the phase lag the coherence carries is nothing but $\arg\!\left[\dfrac{1}{i\Gamma_\perp - \Delta}\right]$, switched on the instant $\Delta \neq 0$.

From the stationary Bloch vector to the radiating dipole§

The one thing to keep straight: $u,v,w$ are rotating-frame components — the carrier $e^{-i\omega t}$ was divided out to get the Bloch equations. So "stationary $(u,v,w)$" is not a dead dipole; it is the lab dipole locked to the drive, oscillating at $\omega$ with fixed amplitude and fixed phase, exactly the steady state of a forced oscillator. The stationary coherence is the phasor of that oscillation. Restore the carrier through $\rho_{eg}^{lab}=\tilde\rho_{eg}e^{-i\omega t}$ with $\tilde\rho_{eg}=(u-iv)/2$

$$\braket{d(t)}=d_{eg}\left(\rho_{eg}^{lab}+\rho_{ge}^{lab}\right)=2d_{eg}\,\mathrm{Re}\!\left[\tilde\rho_{eg}e^{-i\omega t}\right]=d_{eg}\left(u_{ss}\cos\omega t - v_{ss}\sin\omega t\right)$$

Collapsing into a single cosine

Match $u_{ss}\cos\omega t - v_{ss}\sin\omega t = R\cos(\omega t + \phi)$ by expanding the right side

$$R\cos(\omega t+\phi)=R\cos\phi\cos\omega t - R\sin\phi\sin\omega t \quad\Rightarrow\quad R\cos\phi = u_{ss},\quad R\sin\phi = v_{ss}$$

so $R=\sqrt{u_{ss}^2+v_{ss}^2}$ and $\tan\phi = v_{ss}/u_{ss}$. Feeding the steady state $u_{ss}=-\Omega_R\Delta/D$, $v_{ss}=\Omega_R\Gamma_\perp/D$ with $D=\Delta^2+\Gamma_\perp^2(1+s_0)$, the common $D$ cancels in the ratio, so the phase never sees the saturation

$$\tan\phi = \frac{v_{ss}}{u_{ss}} = -\frac{\Gamma_\perp}{\Delta} \qquad R = \frac{\Omega_R\sqrt{\Delta^2+\Gamma_\perp^2}}{\Delta^2+\Gamma_\perp^2(1+s_0)}$$

This is the $\cos(\omega t + \phi)$ we were after: the dipole radiates at the drive frequency, indefinitely, and $\phi$ is its phase against the drive. It is the same structure as the FID line $2d_{eg}|\rho_{eg}|\cos(\omega_{eg}t + \phi)$ — but there the phasor decays and rides the atomic frequency, whereas here it is pinned and rides the drive. The amplitude is the driven-oscillator response curve, and the phase sweeps

$$\phi=\begin{cases} \to 0 & |\Delta|\gg\Gamma_\perp,\ \Delta<0 \quad \text{(in phase — dipole follows the drive)} \\ = \dfrac{\pi}{2} & \Delta = 0 \qquad\qquad\;\; \text{(quadrature — maximal absorption)} \\ \to \pi & |\Delta|\gg\Gamma_\perp,\ \Delta>0 \quad \text{(anti-phase)} \end{cases}$$

exactly a forced damped oscillator tuned through resonance. Note the phase is saturation-independent ($s_0$ cancels in $u_{ss}/v_{ss}$): driving harder grows the amplitude but never moves the lag. And since $\phi=\arg(i\Gamma_\perp-\Delta)=-\arg\chi$, this lag is the argument of the susceptibility below — the dipole is never doing nothing, it is a steadily driven antenna whose fixed amplitude and phase are precisely what make the single-frequency ratio $P(\omega)/E(\omega)$ well defined.

Suspectibily§

In some sense, we already have the full intuition about the suspectibility since it is just properties of the dipole under different cases, but if we solve it explicitly

$$\chi(\omega)=\frac{d_{eg}^2}{\epsilon_0\hbar}\frac{1}{i\Gamma_\perp-\Delta}$$

Where the phase lag comes from the phase of the coherent term and the absorbtion of light comes from the resonant term.

$$\mathrm{Im}[\chi]=\frac{d_{eg}^{2}}{\epsilon_{0}\hbar}\frac{\Gamma_{\perp}}{\Gamma_{\perp}^{2}+\Delta^{2}},\qquad\mathrm{Re}[\chi]=\frac{d_{eg}^{2}}{\epsilon_{0}\hbar}\frac{-\Delta}{\Gamma_{\perp}^{2}+\Delta^{2}}$$