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Harmonic oscillator with a time-dependent frequency

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.