Calculate |Psi(x,t)|^2 - Merzbacher Chapter 15 Exercise 9

[IHateMayonnaise]

Homework Statement

This is Exercise 9 from chapter 15 of merzbacer. It asks to find \lvert\psi(x,t)\rvert^2 given:

<br /> <br /> \psi(x,t)=[2\pi(\Delta x)_0^2]^{-1/4}\left[1+\frac{i\hbar t}{2m(\Delta x)_0^2} \right]^{-1/2} \exp\left[\frac{-\frac{x^2}{4(\Delta x)_0^2}+ik_0x-ik_0^2\frac{\hbar t}{2m}}{1+\frac{i\hbar t}{2m(\Delta x)_0^2}}\right]<br /> <br />

Homework Equations

\lvert\psi(x,t)\rvert^2=\psi^*(x,t)\psi(x,t)

(\Delta x)^2=\langle x \rangle^2 - \langle x^2\rangle

i\hbar\frac{d}{dt}\langle A \rangle = \langle[A,H]\rangle + \left\langle\frac{\partial A}{\partial t} \right\rangle

The Attempt at a Solution

My question is quick and qualitative: is there..an easier way of doing this than the brute force way? I mean, am I not seeing something? Or is this problem as useless as it seems?

I am not allowed to use a computer in any way.

 

[zhermes]

It looks like lots of stuff will cancel out when you do it "brute force." But, you're not looking for any particular final value (e.g. the expectation value of 'x,' etc), so there isn't really anything to take a short-cut towards.

 

[javiergra24]

Remember that in order to calculate the integral

<br /> \exp[i\theta]\cdot \exp[-i\theta]=1<br />

where theta is the content of the exponential part
So calculations reduce.