Introduction

Parametric amplification pertains to oscillations, where amplification comes from modulating some parameter as opposed to injecting energy at the oscillation signal frequency. Modulating a parameter is called "pumping", and the frequency of that modulation is called the "pump frequency". Examples of parametric oscillation are numerous, for example if you have a pendulum and you modulate the pendulum length at the right frequency, you can get an increase in the system energy and thus the amplitude of the pendulum angle. Or in electronics, you can module the capacitance or inductance of an LCR circuit. Parametric amplification is used by josephson parametric amplifiers (JPAs) like in ADMX, or in the LIGO experiment to "squeeze" the uncertainty in the light's phase measurement at the expense of the uncertainty in the amplitude.

In the figure below we have a simple pendulum: a mass $m$ at the end of a massless rod with length $L_0$, but with a length $L(t)$ that can change with time. The origin is given at the equilibrium position of the pendulum as shown.

Let $x$ and $y$ label the pendlum position. Then $$\begin{align} x &= L\sin\theta\nonumber\\ y &= L_0-L\cos\theta\nonumber\\ \end{align}\label{xy}$$ with $$L=L(t)\label{loft}$$ Differentiating we get $$\begin{align} \dot x &= \dot L\sin\theta + L\dot\theta\cos\theta\nonumber\\ \dot y &= -\dot L\cos\theta + L\dot\theta\sin\theta\nonumber\\ \end{align}\label{xydot}$$ The kinetic energy is $$\begin{align} T &= \half m(\dot x^2 + \dot y^2)\nonumber\\ &= \half m([\dot L\sin\theta + L\dot\theta\cos\theta]^2 +[-\dot L\cos\theta + L\dot\theta\sin\theta]^2)\nonumber\\ &= \half m(\dot L^2 + L^2\dot\theta^2)\nonumber\\ \end{align}\label{ke}$$ and the potential energy term is $$V = mgy = mgL_0 -mgL\cos\theta\label{pot}$$ The lagrangian $\mathcal{L}=T-V$ is then $$\begin{align} \mathcal{L}&=\half m[\dot L^2 + L^2\dot\theta^2] -mgL_0 + mgL\cos\theta\nonumber\\ \end{align}\label{lagrangian}$$ Note that $L(t)$ is a variable, but not a dynamical variable, because $L(t)$ is externally pumped and not something determined by the motion. So the only variable here is $\theta$, which means we only have one Euler-Lagrange equation: $$\frac{d}{dt}\frac{\partial\mathcal{L}}{\partial\dot\theta}-\frac{\partial \mathcal{L}}{\partial\theta}=0 \nonumber$$ Since the mass $m$ is in all terms, we can form $\mathcal{L}/m$ which will also satisfy the Euler-Lagrange equation: $$ \mathcal{L}' \equiv \mathcal{L}/m = \half [\dot L^2 + L^2\dot\theta^2] - gL_0 + gL\cos\theta\label{lagrangian2}$$ which means we can use $$\frac{d}{dt}\frac{\partial\mathcal{L}'}{\partial\dot\theta}-\frac{\partial\mathcal{L}'}{\partial\theta}=0 \label{lageqn}$$ Differentiating $\ref{lagrangian2}$ with respect to $\dot\theta$ gives $$\frac{\partial\mathcal{L}'}{\partial\dot\theta} = L^2\dot\theta \label{ldot}$$ and differentiating that with respect to time gives $$\begin{align} \frac{\partial}{\partial t}\frac{\partial\mathcal{L}'}{\partial\dot\theta} &= 2L\dot L\dot\theta + L^2\ddot\theta \nonumber\\ \end{align}\label{dldtdot}$$ Differentiating $\ref{lagrangian2}$ with respect to $\theta$ gives $$\frac{\partial\mathcal{L}'}{\partial\theta} = -gL\sin\theta\label{lth}$$ Combining $\ref{dldtdot}$ and $\ref{lth}$ and cancelling common terms, and then dividing by $L^2$ gives $$\ddot\theta + 2\frac{\dot L}{L}\dot\theta + \frac{g}{L}\sin\theta = 0\nonumber$$ If we restrict ourselves to oscillations where $\sin\theta\sim \theta$ (makes the equations much more solvable, and solutions easier to see) we have the following equation: $$\ddot\theta + 2\frac{\dot L}{L}\dot\theta + \frac{g}{L}\theta = 0\label{eqnmotion}$$ To make the dynamics easier to see, let's get rid of the 1st derivative through a coordinate transformation: $$q\equiv \theta L\nonumber$$ or equivalently, $$\theta = \frac{q}{L}\label{qdef}$$ where again $L=L(t)$.

The variable $q$ is the arc length of the pendulum as it traces out through an angle $\theta$. For small oscillations, $q$ is very close to the $x$-coordinate of the mass.

Differentiating, we have $$\dot \theta = \frac{\dot q}{L} - \frac{q\dot L}{L^2}\label{dotth}$$ and $$\begin{align} \ddot \theta &= \frac{\ddot q}{L} - \frac{\dot q\dot L}{L^2} - \frac{\dot q\dot L}{L^2} - \frac{q\ddot L}{L^2} + 2\frac{q\dot L^2}{L^3}\nonumber\\ &= \frac{\ddot q}{L} - 2\frac{\dot q\dot L}{L^2} - \frac{q\ddot L}{L^2} + 2\frac{q\dot L^2}{L^3}\nonumber\\ \end{align}\label{ddotth}$$ Plugging $\ref{dotth}$ and $\ref{ddotth}$ into $\ref{eqnmotion}$ gives: $$\begin{align} \ddot\theta + (2\frac{\dot L}{L}\dot\theta) + [\frac{g}{L}\theta] &= \frac{\ddot q}{L} - 2\frac{\dot q\dot L}{L^2} - \frac{q\ddot L}{L^2} + 2\frac{q\dot L^2}{L^3} + (2\frac{\dot L}{L}\frac{\dot q}{L} - 2\frac{\dot L}{L}\frac{q\dot L}{L^2}) + [\frac{g}{L}\frac{q}{L}]\nonumber\\ &= \frac{\ddot q}{L} - \frac{q\ddot L}{L^2} + \frac{gq}{L^2}\nonumber \\ &= 0 \end{align}\nonumber$$ Dividing through by $L$ and rearranging gives us the final equation of motion: $$\ddot q + (\frac{g}{L} - \frac{\ddot L}{L})q = 0\label{hosc}$$ So we can see clearly that this equations describes a harmonic oscillation: $$q(t) = A\cos\omega t\nonumber$$ where the angular frequency $\omega$ is time dependent: $$\omega^2(t) = \frac{g}{L} - \frac{\ddot L}{L}\label{omega}$$ Next let's put in an explicit form for the pendulum length $L(t)$ and assume that this length is "pumped" with an angular pumping frequency $\Omega$ which changes the length by a small amount compared to the resting length $L_0$: $$L(t) = L_0(1 + \alpha\sin\Omega t)\label{loft2}$$ Then $$\dot L = L_0\alpha\Omega\cos\Omega t\nonumber$$ and $$\ddot L = -L_0\alpha\Omega^2\sin\Omega t\nonumber$$ Here the amplitude of the pumping will be given by $L_p = \alpha L_0$ with $\alpha\lt 1$, and at $t=0$, $L(t) = L_0$.

The first term $g/L$ can be expanded to first order in $\alpha$: $$\begin{align} \frac{g}{L} &= \frac{g}{L_0(1+\alpha\sin\Omega t)}\nonumber\\ &\sim \frac{g}{L_0}(1-\alpha\sin\Omega t)\nonumber\\ &= \omega_0^2(1-\alpha\sin\Omega t)\nonumber\\ \end{align}\nonumber$$ where we are using the natural of frequency of oscillation $\omega_0^2=g/L_0$. The 2nd term $\ddot L/L$ is $$\begin{align} \frac{\ddot L}{L} &= \frac{-L_0\alpha\Omega^2\sin\Omega t}{L_0(1 + \alpha\sin\Omega t)} \nonumber\\ &\sim -\alpha\Omega^2\sin\Omega t\nonumber\\ \end{align}\nonumber$$ This gives us the time dependent angular frequency $\omega(t)$: $$\begin{align} \omega^2 &= \frac{g}{L} - \frac{\ddot L}{L}\nonumber\\ &= \omega_0^2(1-\alpha\sin\Omega t) +\alpha\Omega^2\sin\Omega t \nonumber\\ &= \omega_0^2 + \alpha(\Omega^2-\omega_0^2)\sin\Omega t\nonumber\\ \end{align}\nonumber$$ which gives us the parametric equation of motion $$\ddot q + [\omega_0^2 + \alpha(\Omega^2-\omega_0^2)\sin\Omega t]q = 0\label{param}$$ This shows clearly that when you pump at the resonance frequency $\Omega=\omega_0$ there is no parametric enhancement, which makes sense. But to see what pump frequencies ($\Omega$) causes enhancement, we have to take into account the complexities of having pump frequencies and natural oscillation frequencies averaged together.

To find out what frequencies $\Omega$ give any kind of parametric amplification, let's write the frequency term in the above equation as $$\begin{align} \omega^2 &= \omega_0^2 + \alpha(\Omega^2-\omega_0^2)\sin\Omega t\nonumber\\ &= \omega_0^2+\alpha f(t)\nonumber \end{align}$$ with $$f(t) = (\Omega^2-\omega_0^2)\sin\Omega t\nonumber$$ We will assume $\alpha$, which is the differential amplitude of the pendulum length oscillation, is small compared to 1. We can then rewrite the equation of motion as $$\ddot q + \omega_0^2 q + \alpha fq = 0\nonumber$$ Given how small $\alpha$ is, it's reasonable to assume that we should be able to expand $q(t)$ in powers of $\alpha$: $$q(t) = q_0(t) + \alpha q_1(t) + \alpha^2 q_2(t) +\dots\label{qexp}$$ Then the equation of motion will be given by $$\begin{align} \ddot q + \omega_0^2 q + \alpha fq =& \ddot q_0 + \alpha\ddot q_1 + \alpha^2\ddot q_2 + \dots\nonumber\\ +& \omega_0^2(q_0 + \alpha q_1 + \alpha^2 q_2 +\dots)\nonumber\\ +& \alpha f(q_0 + \alpha q_1 + \alpha^2 q_2 +\dots)\nonumber\\ &=0\nonumber\\ \end{align}\label{eqm}$$ Collecting terms per order of $\alpha$ gives: $$\begin{align} \alpha^0 &: \ddot q_0 + \omega_0^2 q_0 = 0\label{alpha0}\\ \alpha^1 &: \alpha\ddot q_1 + \alpha\omega_0^2 q_1 + \alpha fq_0 = 0\label{alpha1}\\ \alpha^2 &: \alpha^2\ddot q_2 + \alpha^2\omega_0^2 q_2 + \alpha^2 fq_1 = 0\label{alpha2} \end{align}\nonumber$$ and so on.

Equation $\ref{alpha0}$ tells us that $$q_0(t) = A\cos\omega_0 t\label{q0}$$ Equation $\ref{alpha1}$ can be rewritten $$\begin{align} \ddot q_1 + \omega_0^2 q_1 &= -fq_0\nonumber\\ &= (\Omega^2-\omega_0^2)\sin\Omega t\cdot A\cos\omega_0 t\nonumber\\ &= -\half A(\Omega^2-\omega_0^2)(\sin([\Omega+\omega_0]t)+\sin([\Omega-\omega_0]t)\nonumber\\ \end{align}$$ The solution $$q_1 = C\sin([\Omega+\omega_0]t)+D\sin([\Omega-\omega_0]t)\nonumber$$ works if $$\begin{align} C &= \frac{A}{2}\frac{\Omega^2-\omega_0^2}{(\Omega +\omega_0)^2 -\omega_0^2}\nonumber\\ &= \frac{A}{2}\frac{(\Omega^2-\omega_0^2)}{\Omega^2+2\Omega\omega_0}\nonumber\\ &= \frac{A}{2\Omega}\frac{(\Omega^2-\omega_0^2)}{\Omega + 2\omega_0}\nonumber\\ \end{align}\nonumber$$ and $$\begin{align} D &= \frac{A}{2}\frac{\Omega^2-\omega_0^2}{(\Omega -\omega_0)^2 -\omega_0^2}\nonumber\\ &= \frac{A}{2}\frac{(\Omega^2-\omega_0^2)}{\Omega^2-2\Omega\omega_0}\nonumber\\ &= \frac{A}{2\Omega}\frac{(\Omega^2-\omega_0^2)}{\Omega - 2\omega_0}\nonumber\\ \end{align}\nonumber$$ which makes gives $$q(t) = A\cos\omega_0 t - \frac{A(\Omega^2-\omega_0^2)}{2\Omega} \big[\frac{\sin([\Omega+\omega_0]t)}{\Omega +2\omega_0}+ \frac{\sin([\Omega-\omega_0]t)}{\Omega -2\omega_0}\big]\label{qt}$$ The quantity in the second demoninator $\Omega-2\omega_0$ goes to zero as $\Omega\to 2\omega_0$. This is the pumping condition for amplification: the amplitude grows as you pump more energy in at a frequency that gets close to $\Omega = 2\omega_0$.