Consider a quantum particle in a harmonic trap with a time-dependent frequency \[\omega = \omega(t) \,.\] In atomic units (\(\hbar = m = 1\)), the Schrödinger equation describing this system reads \[i \dot{\psi} = -\frac{1}{2} \psi'' + \frac{1}{2} \omega^2 x^2 \psi \,,\] where \(\psi = \psi(x,t)\) is the particle's wave function, and \(\; '\) and \(\dot{}\) denote \(\frac{\partial}{\partial x}\) and \(\frac{\partial}{\partial t}\), respectively.

A general treatment of this problem can be found, for instance, in Propagator for the general time-dependent harmonic oscillator with application to an ion trap

. Here we only construct an example solution to the Schrödinger equation above.

Let us look for solutions \(\psi\) in the form of a Gaussian wave packet centered at the origin: \[\psi = \left( \frac{2 \alpha}{\pi} \right)^{1/4} e^{-(\alpha + i \beta) x^2 + i \gamma} \,,\] where \(\alpha = \alpha(t)\), \(\beta = \beta(t)\), and \(\gamma = \gamma(t)\) are yet unknown, real-valued functions. The wave packet is normalized to unity. We have \[\begin{align} \psi' &= -2 (\alpha + i \beta) x \psi \,, \\ \psi'' &= \big[ 4 (\alpha + i \beta)^2 x^2 - 2 (\alpha + i \beta) \big] \psi \,, \end{align}\] and \[\dot{\psi} = \left( \frac{\dot{\alpha}}{4 \alpha} - (\dot{\alpha} + i \dot{\beta}) x^2 + i \dot{\gamma} \right) \psi \,.\] Substituting these derivatives into the Schrödinger equation, we obtain \[i \frac{\dot{\alpha}}{4 \alpha} - i (\dot{\alpha} + i \dot{\beta}) x^2 - \dot{\gamma} = -2 (\alpha + i \beta)^2 x^2 + \alpha + i \beta + \frac{1}{2} \omega^2 x^2 \,.\] Balancing terms with the same power of \(x\), we get the following system of two equations: \[\left\{ \begin{array}{rl} - i (\dot{\alpha} + i \dot{\beta}) &\!\! = \; -2 (\alpha + i \beta)^2 + \omega^2 / 2 \\[0.1cm] i \dot{\alpha} / 4 \alpha - \dot{\gamma} &\!\! = \; \alpha + i \beta \end{array} \right.\] Separating the real and imaginary parts, we obtain a system of three linearly independent, real equations: \[\left\{ \begin{array}{rl} \dot{\alpha} &\!\! = \; 4 \alpha \beta \\[0.07cm] \dot{\beta} &\!\! = \; -2 \alpha^2 + 2 \beta^2 + \omega^2 / 2 \\[0.07cm] \dot{\gamma} &\!\! = \; -\alpha \end{array} \right.\]

#### Example solution

We now use a reverse engineering approach to construct a breather type solution to the system above. Let us take \[\alpha(t) = 1 + A \sin t \,,\] where \(0 < A < 1\) is a parameter. From the first equation, it follows that \[\beta(t) = \frac{\dot{\alpha}}{4 \alpha} = \frac{A \cos t}{4 (1 + A \sin t)} \,.\] Then, the third equation yields \[\gamma(t) = -\int_0^t d\tau \, \alpha(\tau) = A (\cos t - 1) - t \,.\] Finally, the second equation determines the oscillator frequency: \[\begin{align} \omega(t) &= \sqrt{ 2 \dot{\beta} + 4 \alpha^2 - 4 \beta^2} \\[0.1cm] &= \sqrt{ 4 (1 + A \sin t)^2 - \frac{A \sin t}{2(1 + A \sin t)} - \frac{3}{4} \left( \frac{A \cos t}{1 + A \sin t} \right)^2 } \,. \end{align}\] A visualization of this solution with \(A = 1/2\) can be found here.