2 and p$_{c}$ ( t ) = m ˙ x$_{c}$ ( t ) the corresponding momentum, and S$_{c}$ ( t ) = ∫ t 0 dt ( p 2 $_{c}$/ 2 m − mω $^{2}$x2 $_{c}$/ 2 + fx$_{c}$ ) the classical ac- tion. The two variances are given by The Leipnik’s joint entropy of the Gaussian wave packet (2) is determined by the uncertainty relation as Note that the centroid ( x$_{c}$, p$_{c}$ ) does not give any contri- bution to the joint entropy. So, the question is how to find the variances for the position and momentum. There is arbitrariness in choosing a complex solution u to Eq. (3) since any linear superposition of u and u ∗ is also a solution. For instance, the Gaussian wave packet for a free particle with the minimum uncertainty at t = 0 is obtained by the solution and for a harmonic oscillator by Then, the most general solution satisfying Eq. (4) may involve two parameters r and ϑ as where r ≥ 0 and 2 π > ϑ ≥ 0. Here, the r and ϑ have the interpretation as the squeeze parameter and angle [10, 11], which are equivalent to determining two integration constants, Δ x (0) and Δ p (0). In fact, the general solution (11) leads to the Bogoliubov transformation between the Fock bases constructed from u ( t ) and u$_{0}$ ( t ): Though the free particle allows a shift of time, which corresponds to choosing a set of r and ϑ , the Hamiltonian (1) does not preserve the time-translational symmetry, so we select the wave packets at t = 0 by adjusting r and ϑ . First, we consider the joint entropy for the free parti- cle with ω = 0. As the external force only governs the centroid of the Gaussian wave packet, the joint entropy However, in the other range (2 π > ϑ > π ) the entropy has the lower bound depends not only on the squeeze parameter r but also the squeeze angle ϑ , which is drawn in Fig. 1. One interest- ing point is that the minimum entropy may be provided by the zero-squeeze angle ( ϑ = 0) even for the non-zero squeeze parameter r . The entropy (13) monotonously increases against time in the range of squeeze angle ( π ≥ ϑ ≥ 0) and has the minimum entropy S (0). This is numerically shown in Fig. 2. The initial entropy at t = 0 does not depend on the details of the external force. We thus obtain the Leipnik’s joint entropy Δ( x )( t ) = √ 〈 x $^{2}$〉 − 〈 x 〉 2 = √ ℏ u $^{∗}$( t ) u ( t ) , Δ( p )( t ) = p p 2 = m ℏ ˙ u $^{∗}$( t ) ˙ u ( t ) . (6) S ( t ) = − ∫ dx | ψ ( x, t ) | $^{2}$ln | ψ ( x, t ) | 2 − ∫ dp | ˜ ψ ( p, t ) | $^{2}$ln | ˜ ψ ( p, t ) | 2 − ln(2 π ℏ ) (7) S ( t ) = ln ( e 2 ) + ln ( 2Δ x ( t )Δ p ( t )) ℏ ) . (8) u$_{0}$ ( t ) = 1 √$_{2}$ ( 1 − i t m$_{0}$ ) , (9) u$_{0}$ ( t ) = 1 √$_{2}$$_{m}$$_{0}$$_{ω}$$_{0}$ e − iω 0 $^{t}$. (10) u ( t ) = (cosh r ) u$_{0}$ ( t ) + ( e − $^{iϑ}$sinh r ) u ∗ $_{0}$( t ) , (11) a ( t ) = (cosh r ) a$_{0}$ ( t ) − ( e $^{iϑ}$sinh r ) a † $_{0}$( t ) , a $^{†}$( t ) = (cosh r ) a † $_{0}$( t ) ( e − $^{iϑ}$sinh r ) a$_{0}$ ( t ) . (12) S (0) = ln ( e 2 ) + 1 2 ln(1 + sin 2 ϑ sinh $^{2}$2 r ) (14) S$_{min}$ = ln ( e 2 ) , (15) S ( t ) = ln ( e 2 ) + 1 2 ln ( cos 2 ϑ 2 + e 4 r sin 2 ϑ 2 ) + 1 2 ln ( (cos ϑ 2 + t m$_{0}$ sin ϑ 2 ) 2 + e − 4 $^{r}$(sin ϑ 2 − t m$_{0}$ cos ϑ 2 ) 2 ) . (13) FIG. 2: The entropy change S ( t ) − ln( e/ 2) for the free particle with ϑ = π/ 2 as the function of time T = t/m 0 and the squeeze parameter r . FIG. 1: The entropy change S (0) − ln( e/ 2) for the free particle as the function of the squeeze parameter r and the squeeze angle ϑ is drawn in the ranges 1 ≥ r ≥ 0 and 2 π > ϑ ≥ 0. 0 1 2 3 4 5 6 theta 0 0.2 0.4 0.6 0.8 1 r 0.5 1 1.5 0 0.2 0.4 0.6 0.8 1 r 0 2 4 6 8 10 T 1 2 3 4