Power Transfer from Signals to Atoms

The main way we will get a physical picture of how lasing (as opposed to just pumping/exciting) occurs will be to look at how the energy in one electric field, the signal, transfers power or energy to atoms to produce another electric field, and perhaps vice versa. To start, we will continue with our mechanical description of atoms as oscillators.

Table of Contents:

  1. Power Transfer to a Collection of Oscillators
  2. Time-Averaged Power Flow
  3. Poynting Derivation of Energy Transfer
  4. Reactive vs Resistive Power Flow
  5. Quality Factor
  6. Tensor Formulation of Power Flow


An electric field acting on a moving charge can be said to do work on the charge, since the net effect of the field is to displace the charge. Since work is described by $force \times distance traveled$, where the “distance traveled” here is the movement of the charge, we can write the total work as:

\begin{equation} dU = f_xdx = -e\varepsilon_xdx \end{equation}

Then the rate at which power is delivered from the field to the oscillator is given as:

\begin{equation} \frac{dU(t)}{dt} = -e\varepsilon_x(t)\frac{dx(t)}{dt} = \varepsilon_x(t)\frac{d\mu_x(t)}{dt} \end{equation}

Now then, if we sum all the power flow over all of these oscillators or atoms in a volume, then we have that the power flow per unit volume is given as:

\begin{equation} \frac{dU_a}{dt} = V^{-1}\epsilon_x(t)\sum_{i=1}^{NV}\frac{d\mu_{xi}(t)}{dt} = \varepsilon_x(t)\frac{dp_x(t)}{dt} \label{time_av_field}\end{equation}

The beauty of this equation is that it holds true for both the classical oscillator and quantum mechanical dipole moment expectation values!


Now we can consider the time-averaged flow to a collection of atoms by an applied signal:

\begin{equation} \varepsilon_x(t) = \frac{1}{2}[E(\omega)e^{j\omega{t}}+E^*(\omega)e^{-j\omega{t}}] \end{equation}

The resulting polarization would have the same time-harmonic form:

\begin{equation} p_x(t) = \frac{1}{2}[P(\omega)e^{j\omega{t}}+P^*(\omega)e^{-j\omega{t}}] \end{equation}

Then the steady-state polarization $P(\omega)$ would be related to the signal applied by:

\begin{equation} P(\omega) = \chi(\omega)\epsilon E(\omega) = [\chi{‘}(\omega)+j\chi{”}(\omega)]\epsilon{E(\omega)}\end{equation}

We can then substitute the phasor form of both the polarization and the electric field into Equation~\eqref{time_av_field} to obtain the following result:

\begin{equation} \frac{dU_a}{dt}_{av} = \frac{j\omega}{4}(E^*P-EP^*) = -\frac{1}{2}\epsilon\omega\chi{”}(\omega)|E(\omega)|^2 \end{equation}

Note here that the power absorption or emission only depends on the complex portion of the susceptibility; this is called the “resistive” or lossy part of $\chi(\omega)$ whereas $\chi{‘}(\omega)$ is the reactive part.



One thing to keep in mind with the above definition of power transfer is that transfer of power from a signal field to a medium does not necessarily mean that the medium is absorbing the energy; it could also just be storing the energy. The term that controls this is the $\chi{”}$ term: for media with $\chi{”}$ equal to zero, there will be no time-averaged energy transfer since the electric field and polarization responses would be $90^{\circ}$ out of phase.


With these physical understandings of the complex, or absorptive, part of the susceptibility, we can describe a “Q” or quality factor for the ratio of the signal energy stored in a volume to the signal power that is dissipated in the volume. This would represent a reverse “Q” factor compared to that of circuits, but it is a much more useful result for us! We can define it as:

\begin{equation} \frac{1}{Q_{circuits}} = \frac{energy~dissipated}{\omega \times energy~stored} = \frac{1}{\omega{U_{sig}}}\frac{dU_a}{dt} = -\chi{”} \end{equation}

We can thus interpret a high “Q” as being good since $\chi{”}$ represents the power absorption, so a lower value of $\chi{“}$ means that the system is storing much more energy than has to be dissipated into it. In fact, all lasers in operation have a high Q, since without it they would just absorb the incident radiation. For amplifying transitions the sign of $\chi{“}$ would flip.


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