path: root/content/know/concept/maxwell-bloch-equations/index.pdc

---
title: "Maxwell-Bloch equations"
firstLetter: "M"
publishDate: 2021-10-02
categories:
- Physics
- Quantum mechanics
- Electromagnetism

date: 2021-09-09T21:17:52+02:00
draft: false
markup: pandoc
---

# Maxwell-Bloch equations

For an electron in a two-level system with time-independent states
$\ket{g}$ (ground) and $\ket{e}$ (excited),
consider the following general solution
to the full Schrödinger equation:

$$\begin{aligned}
    \ket{\Psi}
    &= c_g \: \ket{g} \exp\!(-i E_g t / \hbar) + c_e \: \ket{e} \exp\!(-i E_e t / \hbar)
\end{aligned}$$

Perturbing this system with
an [electromagnetic wave](/know/concept/electromagnetic-wave-equation/)
introduces a time-dependent sinusoidal term $\hat{H}_1$ to the Hamiltonian.
In the [electric dipole approximation](/know/concept/electric-dipole-approximation/),
$\hat{H}_1$ is given by:

$$\begin{aligned}
    \hat{H}_1(t)
    = - \hat{\vb{p}} \cdot \vb{E}(t)
    \qquad \quad
    \vu{p}
    \equiv q \vu{x}
    \qquad \quad
    \vb{E}(t)
    = \vb{E}_0 \cos\!(\omega t)
\end{aligned}$$

Where $\vb{E}$ is an [electric field](/know/concept/electric-field/),
and $\hat{\vb{p}}$ is the dipole moment operator.
From [Rabi oscillation](/know/concept/rabi-oscillation/),
we know that the time-varying coefficients $c_g$ and $c_e$
can then be described by:

$$\begin{aligned}
    \dv{c_g}{t}
    &= i \frac{q \matrixel{g}{\vu{x}}{e} \cdot \vb{E}_0}{2 \hbar} \exp\!\big( i \omega t \!-\! i \omega_0 t \big) \: c_e
    \\
    \dv{c_e}{t}
    &= i \frac{q \matrixel{e}{\vu{x}}{g} \cdot \vb{E}_0}{2 \hbar} \exp\!\big(\!-\! i \omega t \!+\! i \omega_0 t \big) \: c_g
\end{aligned}$$

We want to rearrange these equations a bit.
Therefore, we split the electric field $\vb{E}$ like so,
where the amplitudes $\vb{E}_0^{-}$ and $\vb{E}_0^{+}$ may be slowly varying:

$$\begin{aligned}
    \vb{E}(t)
    = \vb{E}^{-}(t) + \vb{E}^{+}(t)
    = \vb{E}_0^{-} \exp\!(i \omega t) + \vb{E}_0^{+} \exp\!(-i \omega t)
\end{aligned}$$

Since $\vb{E}$ is real, $\vb{E}_0^{+} = (\vb{E}_0^{-})^*$.
Similarly, we define the transition dipole moment $\vb{p}_0^{-}$:

$$\begin{aligned}
    \vb{p}_0^{-}
    \equiv q \matrixel{e}{\vu{x}}{g}
    \qquad \quad
    \vb{p}_0^{+}
    \equiv (\vb{p}_0^{-})^*
    = q \matrixel{g}{\vu{x}}{e}
\end{aligned}$$

With these, the equations for $c_g$ and $c_e$ can be rewritten as shown below.
Note that $\vb{E}^{-}$ and $\vb{E}^{+}$ include the driving plane wave,
and the *rotating wave approximation* is still made:

$$\begin{aligned}
    \dv{c_g}{t}
    &= \frac{i}{\hbar} \vb{p}_0^{+} \cdot \vb{E}^{-} \exp\!(- i \omega_0 t) \: c_e
    \\
    \dv{c_e}{t}
    &= \frac{i}{\hbar} \vb{p}_0^{-} \cdot \vb{E}^{+} \exp\!(i \omega_0 t) \: c_g
\end{aligned}$$


## Optical Bloch equations

For $\ket{\Psi}$ as defined above,
the corresponding pure [density operator](/know/concept/density-operator/)
$\hat{\rho}$ is as follows:

$$\begin{aligned}
    \hat{\rho}
    = \ket{\Psi} \bra{\Psi}
    =
    \begin{bmatrix}
        c_e c_e^* & c_e c_g^* \exp\!(-i \omega_0 t) \\
        c_g c_e^* \exp\!(i \omega_0 t) & c_g c_g^*
    \end{bmatrix}
    \equiv
    \begin{bmatrix}
        \rho_{ee} & \rho_{eg} \\
        \rho_{ge} & \rho_{gg}
    \end{bmatrix}
\end{aligned}$$

Where $\omega_0 \equiv (E_e \!-\! E_g) / \hbar$ is the resonance frequency.
We take the $t$-derivative of the matrix elements,
and insert the equations for $c_g$ and $c_e$:

$$\begin{aligned}
    \dv{\rho_{gg}}{t}
    &= \dv{c_g}{t} c_g^* + c_g \dv{c_g^*}{t}
    \\
    &= \frac{i}{\hbar} \vb{p}_0^{+} \cdot \vb{E}^{-} \exp\!(- i \omega_0 t) \: c_e c_g^*
    - \frac{i}{\hbar} \vb{p}_0^{-} \cdot \vb{E}^{+} \exp\!(i \omega_0 t) \: c_g c_e^*
    \\
    \dv{\rho_{ee}}{t}
    &= \dv{c_e}{t} c_e^* + c_e \dv{c_e^*}{t}
    \\
    &= \frac{i}{\hbar} \vb{p}_0^{-} \cdot \vb{E}^{+} \exp\!(i \omega_0 t) \: c_g c_e^*
    - \frac{i}{\hbar} \vb{p}_0^{+} \cdot \vb{E}^{-} \exp\!(- i \omega_0 t) \: c_e c_g^*
    \\
    \dv{\rho_{ge}}{t}
    &= \dv{c_g}{t} c_e^* \exp\!(i \omega_0 t) + c_g \dv{c_e^*}{t} \exp\!(i \omega_0 t) + i \omega_0 c_g c_e^* \exp\!(i \omega_0 t)
    \\
    &= \frac{i}{\hbar} \vb{p}_0^{+} \cdot \vb{E}^{-} \: c_e c_e^*
    - \frac{i}{\hbar} \vb{p}_0^{+} \cdot \vb{E}^{-} \: c_g c_g^*
    + i \omega_0 c_g c_e^* \exp\!(i \omega_0 t)
    \\
    \dv{\rho_{eg}}{t}
    &= \dv{c_e}{t} c_g^* \exp\!(-i \omega_0 t) + c_e \dv{c_g^*}{t} \exp\!(-i \omega_0 t) - i \omega_0 c_e c_g^* \exp\!(- i \omega_0 t)
    \\
    &= \frac{i}{\hbar} \vb{p}_0^{-} \cdot \vb{E}^{+} \: c_g c_g^*
    - \frac{i}{\hbar} \vb{p}_0^{-} \cdot \vb{E}^{+} \: c_e c_e^*
    - i \omega_0 c_e c_g^* \: \exp\!(- i \omega_0 t)
\end{aligned}$$

Recognizing the density matrix elements allows us
to reduce these equations to:

$$\begin{aligned}
    \dv{\rho_{gg}}{t}
    &= \frac{i}{\hbar} \Big( \vb{p}_0^{+} \cdot \vb{E}^{-} \rho_{eg} - \vb{p}_0^{-} \cdot \vb{E}^{+} \rho_{ge} \Big)
    \\
    \dv{\rho_{ee}}{t}
    &= \frac{i}{\hbar} \Big( \vb{p}_0^{-} \cdot \vb{E}^{+} \rho_{ge} - \vb{p}_0^{+} \cdot \vb{E}^{-} \rho_{eg} \Big)
    \\
    \dv{\rho_{ge}}{t}
    &= i \omega_0 \rho_{ge} + \frac{i}{\hbar} \vb{p}_0^{+} \cdot \vb{E}^{-} \big( \rho_{ee} - \rho_{gg} \big)
    \\
    \dv{\rho_{eg}}{t}
    &= - i \omega_0 \rho_{eg} + \frac{i}{\hbar} \vb{p}_0^{-} \cdot \vb{E}^{+} \big( \rho_{gg} - \rho_{ee} \big)
\end{aligned}$$

These equations are correct if nothing else is affecting $\hat{\rho}$.
But in practice, these quantities decay due to various processes,
e.g. spontaneous emission (see [Einstein coefficients](/know/concept/einstein-coefficients/)).

Let $\rho_{ee}$ decays with rate $\gamma_e$.
Since the total probability $\rho_{ee} + \rho_{gg} = 1$,
we thus have:

$$\begin{aligned}
    \Big( \dv{\rho_{ee}}{t} \Big)_{e}
    = - \gamma_e \rho_{ee}
    \quad \implies \quad
    \Big( \dv{\rho_{gg}}{t} \Big)_{e}
    = \gamma_e \rho_{ee}
\end{aligned}$$

Meanwhile, for whatever reason,
let $\rho_{gg}$ decay into $\rho_{ee}$ with rate $\gamma_g$:

$$\begin{aligned}
    \Big( \dv{\rho_{gg}}{t} \Big)_{g}
    = - \gamma_g \rho_{gg}
    \quad \implies \quad
    \Big( \dv{\rho_{gg}}{t} \Big)_{g}
    = \gamma_g \rho_{gg}
\end{aligned}$$

And finally, let the diagonal (perpendicular) matrix elements
both decay with rate $\gamma_\perp$:

$$\begin{aligned}
    \Big( \dv{\rho_{eg}}{t} \Big)_{\perp}
    = - \gamma_\perp \rho_{eg}
    \qquad \quad
    \Big( \dv{\rho_{ge}}{t} \Big)_{\perp}
    = - \gamma_\perp \rho_{ge}
\end{aligned}$$

Putting everything together,
we arrive at the **optical Bloch equations** governing $\hat{\rho}$:

$$\begin{aligned}
    \boxed{
        \begin{aligned}
            \dv{\rho_{gg}}{t}
            &= \gamma_e \rho_{ee} - \gamma_g \rho_{gg}
            + \frac{i}{\hbar} \Big( \vb{p}_0^{+} \cdot \vb{E}^{-} \rho_{eg} - \vb{p}_0^{-} \cdot \vb{E}^{+} \rho_{ge} \Big)
            \\
            \dv{\rho_{ee}}{t}
            &=  \gamma_g \rho_{gg} - \gamma_e \rho_{ee}
            + \frac{i}{\hbar} \Big( \vb{p}_0^{-} \cdot \vb{E}^{+} \rho_{ge} - \vb{p}_0^{+} \cdot \vb{E}^{-} \rho_{eg} \Big)
            \\
            \dv{\rho_{ge}}{t}
            &= - \Big( \gamma_\perp - i \omega_0 \Big) \rho_{ge}
            + \frac{i}{\hbar} \vb{p}_0^{+} \cdot \vb{E}^{-} \Big( \rho_{ee} - \rho_{gg} \Big)
            \\
            \dv{\rho_{eg}}{t}
            &= - \Big( \gamma_\perp + i \omega_0 \Big) \rho_{eg}
            + \frac{i}{\hbar} \vb{p}_0^{-} \cdot \vb{E}^{+} \Big( \rho_{gg} - \rho_{ee} \Big)
        \end{aligned}
    }
\end{aligned}$$

Many authors simplify these equations a bit by choosing
$\gamma_g = 0$ and $\gamma_\perp = \gamma_e / 2$.


## Including Maxwell's equations

This two-level system has a dipole moment $\vb{p}$ as follows,
where we use [Laporte's selection rule](/know/concept/selection-rules/)
to remove diagonal terms, by assuming that
the electron's orbitals are odd or even:

$$\begin{aligned}
    \vb{p}
    &= \matrixel{\Psi}{\hat{\vb{p}}}{\Psi}
    \\
    &= q \Big( c_g c_g^* \matrixel{g}{\vu{x}}{g} + c_e c_e^* \matrixel{e}{\vu{x}}{e}
    + c_g c_e^* \matrixel{e}{\vu{x}}{g} \exp\!(i \omega_0 t) + c_e c_g^* \matrixel{g}{\vu{x}}{e} \exp\!(-i \omega_0 t) \Big)
    \\
    &= q \Big( \rho_{ge} \matrixel{e}{\vu{x}}{g} + \rho_{eg} \matrixel{g}{\vu{x}}{e} \Big)
    = \vb{p}_0^{-} \rho_{ge}(t) + \vb{p}_0^{+} \rho_{eg}(t)
    \equiv \vb{p}^{-}(t) + \vb{p}^{+}(t)
\end{aligned}$$

Where we have split $\vb{p}$ analogously to $\vb{E}$
by defining $\vb{p}^{+} \equiv \vb{p}_0^{+} \rho_{eg}$.
Its equation of motion can then be found from the optical Bloch equations:

$$\begin{aligned}
    \dv{\vb{p}^{+}}{t}
    = \vb{p}_0^{+} \dv{\rho_{eg}}{t}
    = - \vb{p}_0^{+} \Big( \gamma_\perp + i \omega_0 \Big) \rho_{eg}
    + \frac{i}{\hbar} \vb{p}_0^{+} \Big( \vb{p}_0^{-} \cdot \vb{E}^{+} \Big) \Big( \rho_{gg} - \rho_{ee} \Big)
\end{aligned}$$

Some authors do not bother multiplying $\rho_{ge}$ by $\vb{p}_0^{+}$.
In any case, we arrive at:

$$\begin{aligned}
    \boxed{
        \dv{\vb{p}^{+}}{t}
        = - \Big( \gamma_\perp + i \omega_0 \Big) \vb{p}^{+}
        - \frac{i}{\hbar} \Big( \vb{p}_0^{-} \cdot \vb{E}^{+} \Big) \vb{p}_0^{+} d
    }
\end{aligned}$$

Where we have defined the **population inversion** $d \in [-1, 1]$ as follows,
which quantifies the electron's excitedness:

$$\begin{aligned}
    d
    \equiv \rho_{ee} - \rho_{gg}
\end{aligned}$$

From the optical Bloch equations,
we find its equation of motion to be:

$$\begin{aligned}
    \dv{d}{t}
    &= \dv{\rho_{ee}}{t} - \dv{\rho_{gg}}{t}
    = 2 \gamma_g \rho_{gg} - 2 \gamma_e \rho_{ee}
    + \frac{i 2}{\hbar} \Big( \vb{p}^{-} \cdot \vb{E}^{+} - \vb{p}^{+} \cdot \vb{E}^{-} \Big)
\end{aligned}$$

We can rewrite the first two terms in the following intuitive form,
which describes a decay with
rate $\gamma_\parallel \equiv \gamma_g + \gamma_e$
towards an equilbrium $d_0$:

$$\begin{aligned}
    2 \gamma_g \rho_{gg} - 2 \gamma_e \rho_{ee}
    = \gamma_\parallel (d_0 - d)
    \qquad \quad
    d_0
    \equiv \frac{\gamma_g - \gamma_e}{\gamma_g + \gamma_e}
\end{aligned}$$

<div class="accordion">
<input type="checkbox" id="proof-inversion-decay"/>
<label for="proof-inversion-decay">Proof</label>
<div class="hidden">
<label for="proof-inversion-decay">Proof.</label>
We introduce some new terms, and reorganize the expression:

$$\begin{aligned}
    2 \gamma_g \rho_{gg} - 2 \gamma_e \rho_{ee}
    &= 2 \gamma_g \rho_{gg} - 2 \gamma_e \rho_{ee}
    + \gamma_g \rho_{ee} - \gamma_g \rho_{ee}
    + \gamma_e \rho_{gg} - \gamma_e \rho_{gg}
    \\
    &= \gamma_g (\rho_{gg} + \rho_{ee}) - \gamma_e (\rho_{gg} + \rho_{ee})
    + \gamma_g (\rho_{gg} - \rho_{ee}) + \gamma_e (\rho_{gg} - \rho_{ee})
\end{aligned}$$

Since the total probability $\rho_{gg} + \rho_{ee} = 1$,
and $d \equiv \rho_{ee} - \rho_{gg}$, this reduces to:

$$\begin{aligned}
    2 \gamma_g \rho_{gg} - 2 \gamma_e \rho_{ee}
    &= \gamma_g - \gamma_e - (\gamma_g + \gamma_e) d
    \\
    &= (\gamma_g + \gamma_e) \Big( \frac{\gamma_g - \gamma_e}{\gamma_g + \gamma_e} - d \Big)
    \\
    &= \gamma_\parallel ( d_0 - d )
\end{aligned}$$
</div>
</div>

With this, the equation for the population inversion $d$
takes the following final form:

$$\begin{aligned}
    \boxed{
        \dv{d}{t}
        = \gamma_\parallel (d_0 - d) + \frac{i 2}{\hbar} \Big( \vb{p}^{-} \cdot \vb{E}^{+} - \vb{p}^{+} \cdot \vb{E}^{-} \Big)
    }
\end{aligned}$$

Finally, we would like a relation between the polarization
and the electric field $\vb{E}$,
for which we turn to [Maxwell's equations](/know/concept/maxwells-equations/).
We start from Faraday's law,
and split $\vb{B} = \mu_0 (\vb{H} + \vb{M})$:

$$\begin{aligned}
    \nabla \cross \vb{E}
    = - \pdv{\vb{B}}{t}
    = - \mu_0 \pdv{\vb{H}}{t} - \mu_0 \pdv{\vb{M}}{t}
\end{aligned}$$

We assume that there is no magnetization $\vb{M} = 0$.
Then we we take the curl of both sides,
and replace $\nabla \cross \vb{H}$ with Ampère's circuital law:

$$\begin{aligned}
    \nabla \cross \big( \nabla \cross \vb{E} \big)
    = - \mu_0 \pdv{}{t} \big( \nabla \cross \vb{H} \big)
    = - \mu_0 \pdv{}{t} \Big( \vb{J}_\mathrm{free} + \pdv{\vb{D}}{t} \Big)
\end{aligned}$$

Inserting the definition $\vb{D} = \varepsilon_0 \vb{E} + \vb{P}$
together with Ohm's law $\vb{J}_\mathrm{free} = \sigma \vb{E}$ yields:

$$\begin{aligned}
    \boxed{
        \nabla \cross \big( \nabla \cross \vb{E} \big)
        = - \mu_0 \sigma \pdv{\vb{E}}{t} - \mu_0 \varepsilon_0 \pdv[2]{\vb{E}}{t} - \mu_0 \pdv[2]{\vb{P}}{t}
    }
\end{aligned}$$

Where $\sigma$ is the medium's conductivity, if any;
many authors assume $\sigma = 0$.
It is trivial to show that $\vb{E}$ and $\vb{P}$
can be replaced by $\vb{E}^{+}$ and $\vb{P}^{+}$.

It is also simple to convert $\vb{p}^{+}$ and $d$
into the macroscopic polarization $\vb{P}^{+}$ and total inversion $D$
by summing over the atoms:

$$\begin{aligned}
    \vb{P}^{+}(\vb{x}, t)
    &= \sum_{n} \vb{p}^{+}_n \: \delta(\vb{x} - \vb{x}_n)
    \\
    D(\vb{x}, t)
    &= \sum_{n} d_n \: \delta(\vb{x} - \vb{x}_n)
\end{aligned}$$

We thus arrive at the **Maxwell-Bloch equations**,
which are relevant for laser theory:

$$\begin{aligned}
    \boxed{
        \begin{aligned}
            \mu_0 \pdv[2]{\vb{P}^{+}}{t}
            &= - \nabla \cross \nabla \cross \vb{E}^{+} - \mu_0 \sigma \pdv{\vb{E}^{+}}{t} - \mu_0 \varepsilon_0 \pdv[2]{\vb{E}^{+}}{t}
            \\
            \pdv{\vb{P}^{+}}{t}
            &= - \Big( \gamma_\perp + i \omega_0 \Big) \vb{P}^{+}
            - \frac{i}{\hbar} \Big( \vb{p}_0^{-} \cdot \vb{E}^{+} \Big) \vb{p}_0^{+} D
            \\
            \pdv{D}{t}
            &= \gamma_\parallel (D_0 - D) + \frac{i 2}{\hbar} \Big( \vb{P}^{-} \cdot \vb{E}^{+} - \vb{P}^{+} \cdot \vb{E}^{-} \Big)
        \end{aligned}
    }
\end{aligned}$$



## References
1.  F. Kärtner,
    [Ultrafast optics: lecture notes](https://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-977-ultrafast-optics-spring-2005/lecture-notes/),
    2005, MIT.
2.  H. Haken,
    *Light: volume 2: laser light dynamics*,
    1985, North-Holland.
3.  H.J. Metcalf, P. van der Straten,
    *Laser cooling and trapping*,
    1999, Springer.