# Speed of Wavepacket

For the given wavepacket

$ψ(x,t)=A*e^{-(a^2 x^2+b^2 t^2+2abxt)}$

show that it satisfies the 1D wave equation and find its speed and direction of propagation.

Wave equation is

$(∂^2 ψ)/(∂x^2 )=1/v^2 (∂^2 ψ)/(∂t^2)$

The results are:

$(∂^2 ψ)/(∂x^2 )=A(-2a^2 x-2abt)^2*e^{-(…) }-A*2a^2*e^{-(…) }$

$(∂^2 ψ)/(∂t^2 )=A(-2b^2 t-2abx)^2*e^{-(…)}-A*2b^2*e^{-(…)}$

From the above wave equation we need to have

$(2a^2 x+2abt)^2+2a^2=(1/v^2 )*[(2b^2 t+2abx)^2+2b^2]$

By comparing we get

$a=+/-(b/v)$   from last terms on the left and right

and $2ab=2b^2/v^2$  and $2a^2=2ab/v^2$ from coefficients of $x$ and $t$ in paranthesis

This basically means that only one value is acceptable: $a=b/v$

Therefore the propagation speed can be written as $v=b/a$

To find the direction of propagation one need to observe that if $a$ and $b$ are purely imaginary the initial equation transforms into

$ψ(x,t)=A*e^{-i*(a1^2*x^2+b1^2*t^2+2a1*b1*xt)}$

Comparing with the general equation of the wave propagating in the x direction

$ψ(x,t)=A*e^{-i(kx+ωt)}$

For a particular point on the wave front that maintains the same amplitude, from the expression in paranthesis to be constant, if time $t$ increases, $x$ needs to decrease.

Therefore one can tell that the original wave direction of propagation is along x to the left (towards negative x).