Radian unit mystery (damped oscillator)

Question

I would be extremely grateful for any help that anyone could offer here.

I am interested in solving the optical bloch equations for the excited state population Rabi oscillations with damping due to spontaneous decay. The problem that I have is not solving the equations mathematically, but making sense of the units.

Essentially I am solving a system of differential equations which are stated as

$$\frac{d\rho_{ee}}{dt} = i\Omega (\rho_{eg} - \rho_{ge}) - \Gamma \rho_{ee}$$ where $\rho_{ee}$ is the probability of being in the excited state.

$\Omega$ is defined to be the rabi frequency - which is an angular frequency such that the period T satisfies $\Omega T = 2\pi$ i.e. $ \Omega$ must have units of $\rm rad\ s^{-1}$. $\Gamma$ clearly has the same units. $\rho_{ee}$ is the population of the excited state.

The solution ends up with terms like $$e^{-(3\Gamma/4)t}\cos \Omega_R t$$ where $\Omega_R$ is of the same units as $\Omega$ and $\Gamma$. I am having severe difficulty understanding the units here - surely the exponent of the exponential (although dimensionless) has units of rad?

If one sets $\Omega = 0$ then one recovers $e^{-\Gamma t}$ which again has an exponent with units of rad. One would expect the excited state population to decay as $e^{-t/\tau}$ where $\tau$ is the lifetime of the state in s. I am used to this being a unitless ratio.

Does it make sense to have an exponential with a unit of radians? and how does this relate to the natural lifetime of the excited state?

Moreover the differential equation seems to suggest $\rho_{ee}$ has units of radians which is also puzzling... Are $\Gamma$ and $\rho$ actually inverse times in the differential equation and if so how do they get converted to rad s$^{-1}$ when they become the argument of the cosines.

I am confused because there is a physical difference of 2$\pi$ between frequency and angular frequency, so it matters which one $\Gamma$ is. Any light that anyone could shed would be amazing! Thanks

Answer 1

Does it make sense to have an exponential with a unit of radians?

Yes. This is the famous Euler identity:

\begin{align} e^{ix} &= 1 + ix + \frac{(ix)^2}{2!} + \frac{(ix)^3}{3!} + \cdots \\ &= \left( 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots \right) + i \cdot\left( x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots \right) \\ &= \cos x + i\sin x \end{align}

In the context of your specific question, I would prefer to say that it makes sense for the argument of the cosine to be a dimensionless number, rather than any dimensionful angle. In fact, the arguments of cosine and sine should be dimensionless, if you want to be able to use their Taylor series definitions, as above.

Surely the exponent of the exponential (although dimensionless) has units of rad? … I am used to this being a unitless ratio.

The radian is a dimensionless ratio: the ratio of an arc length to the corresponding radius.

Compare with resistor-capacitor circuits, where the capacitor has a frequency-dependent complex impedance $Z_C = (j\omega C)^{-1}$. The frequency response of the oscillating circuit has a feature at angular frequency $\omega = 1/RC$. In the DC limit, the voltage across the capacitor decays away with lifetime $\tau = 1/\Gamma = RC$. The basic relationship between the $e$-folding time and the angular frequency is the same as in your problem.

You can think about this from the basic structure of the differential equations. If you have

$$ \frac{\mathrm d^2}{\mathrm dt^2} \phi = -\omega^2 \phi, $$ then $\omega$ is an angular frequency. But with $$ \frac{\mathrm d^2}{\mathrm dt^2} \phi = +\Gamma^2 \phi, $$ the solution is exponential growth or decay with lifetime $\tau=1/\Gamma$. What you have here, with the basic shape $$ \frac{\mathrm d}{\mathrm dt}\phi = (i\omega + \Gamma)\phi $$ is, in a handwaving way, an oscillatory bit and an exponential bit. The factor of $2\pi$ comes from the way that exponentials are related to circles.

Answer 2

Does it make sense to have an exponential with a unit of radians?

Yes, because although the angle has a "label" - radians - to alert the reader as to how the angle was defined, it does not have a dimension because the radian is defined via the ratio of two quantities which have the same dimensions, eg angle subtended at the centre of a circle in radians = $\dfrac {\text {arc length in metres}}{\text{circumference of circle in metres}}$.