A Time-Dependent Harmonic Oscillator with Two Frequency Jumps: an Exact Algebraic Solution

Abstract

We consider a harmonic oscillator (HO) with a time-dependent frequency which undergoes two successive abrupt changes. By assumption, the HO starts in its fundamental state with frequency ω₀, then, at t = 0, its frequency suddenly increases to ω₁ and, after a finite time interval τ, it comes back to its original value ω₀. Contrary to what one could naively think, this problem is quite a non-trivial one. Using algebraic methods, we obtain its exact analytical solution and show that at any time t > 0 the HO is in a vacuum squeezed state. We compute explicitly the corresponding squeezing parameter (SP) relative to the initial state at an arbitrary instant and show that, surprisingly, it exhibits oscillations after the first frequency jump (from ω₀ to ω₁), remaining constant after the second jump (from ω₁ back to ω₀). We also compute the time evolution of the variance of a quadrature. Last, but not least, we calculate the vacuum (fundamental state) persistence probability amplitude of the HO, as well as its transition probability amplitude for any excited state.

Appendix: Brief Derivation of (24)

In this appendix, we shall show that for the Lie algebras defined by the commutation relations:

$$ \left[\hat{K}_{-},\hat{K}_{+}\right] = 2\epsilon\hat{K}_{3} \text{ and } \left[\hat{K}_{3},\hat{K}_{\pm}\right]=\pm\hat{K}_{\pm} , $$

where for 𝜖 = 1 we have the su(1, 1) Lie algebra and for 𝜖 = − 1 we have the su(2) Lie algebra, the following factorization is valid:

$$ e^{\lambda_{+}\hat{K}_{+}+\lambda_{-}\hat{K}_{-}+\lambda_{3}\hat{K}_{3}} = e^{{\Lambda}_{+}\hat{K}_{+}}e^{\ln({\Lambda}_{3})\hat{K}_{3}}e^{{\Lambda}_{-}\hat{K}_{-}} , $$

(53)

where

$$ {\Lambda}_{3} = \left( \cosh(\nu)-\frac{\lambda_{3}}{2\nu}\sinh(\nu)\right)^{-2} ; {\Lambda}_{\pm} = \frac{2\lambda_{\pm}\sinh(\nu)}{2\nu \cosh(\nu)-\lambda_{3}\sinh(\nu)} , $$

(54)

and $\nu ^{2} = \frac {1}{4}{\lambda _{3}^{2}}-\epsilon \lambda _{+}\lambda _{-}$. We will factorize these two Lie algebras simultaneously. In our demonstration, the following BCH-like relation:

$$ e^{\hat{A}}\hat{B} e^{-\hat{A}} = \hat{B}+\left[\hat{A},\hat{B}\right]+\frac{1}{2!}\left[\hat{A},\left[\hat{A},\hat{B}\right]\right]+\frac{1}{3!}\left[\hat{A},\left[\hat{A},\left[\hat{A},\hat{B}\right]\right]\right]+... , $$

(55)

will be used repeated times. Firstly, let us re-define the left side of (53) as the special case 𝜃 = 1 of the operator:

$$ \hat{F}_{1}(\theta)=e^{\theta(\lambda_{+}\hat{K}_{+}+\lambda_{-}\hat{K}_{-}+\lambda_{3}\hat{K}_{3})} . $$

(56)

The basic idea is to find an equivalent expression in the form:

$$ \hat{F}_{1}(\theta)=e^{{\Lambda}_{+}(\theta)\hat{K}_{+}}e^{{\Sigma}_{3}(\theta)\hat{K}_{3}}e^{{\Lambda}_{-}(\theta)\hat{K}_{-}} . $$

(57)

This means that we need to find write parameters Λ₊, Λ₋ and Σ₃ in terms of the small lambdas, so that the last two equations are equal. With this goal, we derive both expressions with respect to 𝜃 and impose the derivatives to be equal. Doing that and using the previous BCH formula, we obtain the following coupled differential equations:

$$ \begin{array}{@{}rcl@{}} {\Lambda}_{+}^{\prime}-{\Sigma}_{3}^{\prime}{\Lambda}_{+}+{\Lambda}_{-}^{\prime}e^{-{\Sigma}_{3}}\epsilon({\Lambda}_{+})^{2}&=&\lambda_{+} , \end{array} $$

(58)

$$ \begin{array}{@{}rcl@{}} {\Sigma}_{3}^{\prime}-2\epsilon{\Lambda}_{-}^{\prime}e^{-{\Sigma}_{3}}{\Lambda}_{+}&=&{\Sigma}_{3} , \end{array} $$

(59)

$$ \begin{array}{@{}rcl@{}} {\Lambda}_{-}^{\prime}e^{-{\Sigma}_{3}}&=&\lambda_{-} , \end{array} $$

(60)

where the prime indicates derivative with respect to 𝜃. To solve the above equations, we start by decoupling them. Firstly, replacing (60) into (58) and (59) leads to:

$$ \begin{array}{@{}rcl@{}} {\Lambda}_{+}^{\prime}-{\Sigma}_{3}^{\prime}{\Lambda}_{+}+\epsilon\lambda_{-}({\Lambda}_{+})^{2}&=&\lambda_{+} , \end{array} $$

(61)

$$ \begin{array}{@{}rcl@{}} {\Sigma}_{3}^{\prime}-2\epsilon\lambda_{-}{\Lambda}_{+}&=&\lambda_{3} . \end{array} $$

(62)

Then, we obtain a differential equation for the function Λ₊ by isolating ${\Sigma }_{3}^{\prime }$ from (62) and replacing it into (61):

$$ \frac{d}{d\theta}{\Lambda}_{+}=\lambda_{+}+\lambda_{3}{\Lambda}_{+}+\in\lambda\_({\Lambda}_{+})^{2}. $$

(63)

Equation (63) is a first-order, quadratic, and non-homogeneous ordinary differential equation known as the Riccati equation. It has a unique solution and can be transformed into an ordinary, homogeneous, and second-order differential equation with the aid of the transformation:

$$ {\Lambda}_{+}=-\frac{1}{\epsilon\lambda_{-}}\frac{1}{u}\frac{du}{d\theta}, $$

(64)

leading to

$$ \frac{d^{2}u}{d\theta^{2}}+{\Gamma}\frac{du}{d\theta}+\omega^{2}_{0}u=0, $$

(65)

where we defined ${\omega _{0}^{2}} = \epsilon \lambda _{-}\lambda _{+} \text { and } {\Gamma } = -\lambda _{3}$ in order to identify it as the classical equation of a damped harmonic oscillator with natural frequency ω₀ and damped coefficient Γ. Note that by using (62) and (64), we can calculate:

$$ {\Sigma}_{3}(\theta) = \lambda_{3}\theta-2\epsilon\lambda_{-}\!(\frac{1}{\epsilon\lambda_{-}}\!)\int{\!\frac{du}{u}}+C_{1} = \lambda_{3}\theta-2\ln{u(\theta)}+C_{1}, $$

(66)

and replacing (66) into (60) we obtain:

$$ {\Lambda}_{-}(\theta) = \lambda_{-}\int{e^{{\Sigma}_{3}(\theta)}d\theta}+C_{2} , $$

(67)

where constants C₁ and C₂ are determined from the initial conditions, namely, Σ₃(𝜃 = 0) = 0 and Λ₋(𝜃 = 0) = 0. Therefore, once we find u(𝜃), we can determine the functions Λ₊, Λ₋, and Σ₃. Coming back to (65), its general solution is given by:

$$ u(\theta)=e^{-\frac{\Gamma}{2} \theta}\left( Ae^{\nu\theta}+Be^{-\nu\theta}\right) , $$

(68)

where

$$ \nu=\sqrt{\frac{1}{4}{\Gamma}^{2}-{\omega_{0}^{2}}}=\sqrt{\frac{1}{4}{\lambda_{3}^{2}}-\epsilon\lambda_{-}\lambda_{+}}, $$

(69)

and constants A and B are determined from the initial conditions. Using the above results in (64), we find:

$$ \begin{array}{@{}rcl@{}} {\Lambda}_{+}(\theta) &= -\frac{1}{\epsilon\lambda_{-}}\frac{1}{u}\left[(\nu-\frac{\Gamma}{2})u-2\nu B e^{-(\frac{\Gamma}{2}+\nu)\theta}\right]\\&=\frac{(\frac{\Gamma}{2}-\nu)}{\epsilon\lambda_{-}} + \frac{2\nu B e^{-\nu\theta}}{\epsilon\lambda_{-}\left[Ae^{\nu\theta}+Be^{-\nu\theta}\right]}, \end{array} $$

(70)

and from the initial condition Λ₊(𝜃 = 0) = 0 we get $A=\frac {(\nu +{\Gamma }/2)}{(\nu -{\Gamma }/2)}B$. Therefore, (70) becomes:

$$ \begin{array}{@{}rcl@{}} {\Lambda}_{+}(\theta) &=& \frac{1}{\epsilon\lambda_{-}}\left[\!\left( \frac{\Gamma}{2}-\nu\right)+\frac{2\nu\left( \nu-\frac{\Gamma}{2}\right) e^{-\nu\theta}}{\left[\left( \nu+\frac{\Gamma}{2}\right)e^{\nu\theta}+\left( \nu-\frac{\Gamma}{2}\right)e^{-\nu\theta}\right]}\right]\\ &=& \frac{1}{\epsilon\lambda_{-}}\left[\frac{2\left( \frac{{\Gamma}^{2}}{4}-\nu^{2}\right)\sinh(\nu\theta)}{2\nu \cosh(\nu\theta)+{\Gamma} \sinh(\nu\theta)}\right]. \end{array} $$

Now, using definition (69) and expressions ${\omega _{0}^{2}} = \epsilon \lambda _{-}\lambda _{+}$ and Γ = −λ₃, we obtain:

$$ {\Lambda}_{+}(\theta)=\frac{2\lambda_{+}\sinh(\nu\theta)}{2\nu \cosh(\nu\theta)-\lambda_{3} \sinh(\nu\theta)}, $$

(71)

which leads to the desired expression written in (54) if we take 𝜃 = 1.

In order to obtain Σ₃, we first replace the solution for u(𝜃), (68), into (66) to get:

$$ \begin{array}{@{}rcl@{}} {\Sigma}_{3}(\theta) &=& \lambda_{3}\theta-2\ln{\left\{e^{-\frac{\Gamma}{2} \theta}\left( \frac{(\nu+{\Gamma}/2)}{(\nu-{\Gamma}/2)}e^{\nu\theta}+e^{-\nu\theta}\right)B\right\}}+C_{1} \\ &=&(\lambda_{3}+{\Gamma})\theta-2\ln{\left( \frac{(\nu+{\Gamma}/2)}{(\nu-{\Gamma}/2)}e^{\nu\theta}+e^{-\nu\theta}\right)} + D , \end{array} $$

(72)

where all constants have been absorbed in D. Using the initial condition Σ₃(0) = 0 and that Γ = −λ₃ it can be shown that $D = 2\ln {(\frac {2\nu }{\nu -{\Gamma }/2})}$. Replacing this result into (72), we get:

$$ {\Sigma}_{3}(\theta) = \ln{\left\{\left( \cosh(\nu\theta)-\frac{\lambda_{3}}{2\nu} \sinh(\nu\theta)\right)^{-2}\right\}}, $$

(73)

which after taking 𝜃 = 1 leads to the desired result of equation in (54) since in (53) Λ₃ is defined as the argument of the logarithm.

Finally, in order to find Λ₋(𝜃), we replace (73) into (67) obtaining:

$$ \begin{array}{@{}rcl@{}} {\Lambda}_{-}(\theta) &=& \lambda_{-}\int{\frac{\text{sech}^{2}(\nu\theta)}{\left( 1-\frac{\lambda_{3}}{2\nu} \tanh(\nu\theta)\right)^{2}}d\theta}+C_{2}\\ &=& \frac{2\lambda_{-}}{\lambda_{3}}\left( \frac{2\nu \cosh(\nu\theta)}{2\nu \cosh(\nu\theta)-\lambda_{3} \sinh(\nu\theta)}\right)+C_{2}. \end{array} $$

(74)

Using the initial condition for Λ₋, it can be shown that $C_{2} = -\frac {2\lambda _{-}}{\lambda _{3}}$. Substituting this result in (74), we finally obtain:

$$ {\Lambda}_{-}(\theta) = \frac{2\lambda_{-}\sinh(\nu\theta)}{2\nu \cosh(\nu\theta)-\lambda_{3} \sinh(\nu\theta)}, $$

(75)

which leads to the desired result after taking 𝜃 = 1.