Find psi(x,t) when psi(x,0)= Ae^(-x^2/a^2) and A, a are real constants

[Moolisa]

Homework Statement

Consider a free particle whose state at t=0 given by the gaussian wave packet. Find psi(x,t)

Relevant Equations

Gaussian wave packet at t=0, More equations in attempt at solution

EQ 1: Ψ(x,0)= Ae^-x2/a2

A. Find Ψ(x,0)

So I normalized Ψ(x,0) by squaring the function, set it equal to 1 and getting an A

I. A=(2/π)^¼ (1/√a)

B. To find Ψ(x,t)

EQ:2 Ψ(x,t)= 1/(√2π) ∫ ∅(k) e^i(kx-ωt)dk --------->when ω=(ħk²)/2m and integral from -∞ to +∞
EQ 3: ∅(k)= 1/(√2π) ∫ Ψ(x,0) e^-ikxdx -------> integral from -∞ to +∞, i is an imaginary number

Using eq3 to find ∅(k), I got

∅(k)=(2π)^1/4 (√a) e^-(ka)2/4

Using Eq2, I got
II. Ψ(x,t)=(2π)^-3/4 (√a) ∫ e^{-k2(a2/4 -iht/(2m)) +ikx} dk -------> integral from -∞ to +∞

But, in the formula to complete the square

∫ e^-(Ax2+Bx dx= (π/A) e^B2/4AI don't know how to manipulate II in order to get exp-(Ax²+Bx). That placement of the negative in II makes me think I either normalized it wrong in part a or I messed up somewhere and don't know where.

 

[Abhishek11235]

Assuming you have set up integrals correctly,we have:
$$e^{-(k^2 y^2-ikx)}$$ where $y^2= a^2/4-iht/2m$

Now compare $$(a-b)^2=a^2-2ab+b^2$$ with terms in bracket to get value of $a$ and $b$ and use it to complete square

 

[Moolisa]

Assuming you have set up integrals correctly,we have:
$$e^{-(k^2 y^2-ikx)}$$ where $y^2= a^2/4-iht/2m$

Now compare $$(a-b)^2=a^2-2ab+b^2$$ with terms in bracket to get value of $a$ and $b$ and use it to complete square

Thank you! I thought I replied when you first posted this 10 days ago, but this really helped!