# Free particle: Gaussian wave packet

Required math: calculus

Required physics: Schrödinger equation

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 2.22.

While analyzing the free particle, we saw that we could construct a normalizable combination of stationary states by writing

$\displaystyle \Psi(x,t)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}\phi(k)e^{ikx}e^{-i\hbar k^{2}t/2m}dk \ \ \ \ \ (1)$

Given the initial wave function, we can find ${\phi(k)}$ via Plancherel’s theorem:

$\displaystyle \phi(k)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}\Psi(x,0)e^{-ikx}dx \ \ \ \ \ (2)$

The integral 1 cannot usually be done in closed form, but one case where it can is if the initial wave function is a Gaussian, of the form

$\displaystyle \Psi(x,0)=Ae^{-ax^{2}} \ \ \ \ \ (3)$

To find ${A}$, we normalize:

$\displaystyle A^{2}\int_{-\infty}^{\infty}e^{-2ax^{2}}dx=1 \ \ \ \ \ (4)$

The integral comes out to ${\sqrt{\pi/2a}}$ from which we get

$\displaystyle A=\left(\frac{2a}{\pi}\right)^{1/4} \ \ \ \ \ (5)$

Finding ${\Psi(x,t)}$ requires finding ${\phi(k)}$ via equation 2. So we get (using Maple to do the integral):

 $\displaystyle \phi(k)$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2\pi}}\left(\frac{2a}{\pi}\right)^{1/4}\int_{-\infty}^{\infty}e^{-ax^{2}-ikx}dx\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(\frac{1}{2\pi a}\right)^{1/4}e^{-k^{2}/4a} \ \ \ \ \ (7)$

We can now use 1 again using Maple to do the integral:

 $\displaystyle \Psi(x,t)$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\phi(k)e^{ikx}e^{-i\hbar k^{2}t/2m}dk\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2\pi}}\left(\frac{1}{2\pi a}\right)^{1/4}\int_{-\infty}^{\infty}e^{-k^{2}/4a}e^{ikx}e^{-i\hbar k^{2}t/2m}dk\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(\frac{2a}{\pi}\right)^{1/4}\frac{e^{-ax^{2}/(1+2i\hbar at/m)}}{\sqrt{1+2i\hbar at/m}} \ \ \ \ \ (10)$

where Maple was used for the integral.

Calculating ${|\Psi(x,t)|^{2}}$ can be done using Maple, but it seems to require a bit of help. First we write out the complex conjugate:

$\displaystyle \Psi^*(x,t)=\left(\frac{2a}{\pi}\right)^{1/4}\frac{e^{-ax^{2}/(1-2i\hbar at/m)}}{\sqrt{1-2i\hbar at/m}} \ \ \ \ \ (11)$

Then we calculate ${\Psi^*\Psi}$ using the Maple command simplify(evalc(${\Psi^*\Psi}$)) assuming positive and we get

 $\displaystyle \left|\Psi(x,t)\right|^{2}$ $\displaystyle =$ $\displaystyle \sqrt{\frac{2a}{\pi}}\frac{e^{-2ax^{2}/\left[1+\left(2\hbar at/m\right)^{2}\right]}}{\sqrt{1+\left(2\hbar at/m\right)^{2}}}\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{2}{\pi}}we^{-2w^{2}x^{2}} \ \ \ \ \ (13)$

with

$\displaystyle w\equiv\left(\frac{a}{1+(2\hbar at/m)^{2}}\right)^{1/2} \ \ \ \ \ (14)$

At ${t=0}$, ${w=\sqrt{a}}$, so ${|\Psi(x,t)|^{2}=\sqrt{2a/\pi}e^{-2ax^{2}}}$ which is correct. The wave packet at ${t=0}$ is therefore a Gaussian centred at ${x=0}$. As ${t}$ increases, ${w}$ gets smaller but the centre of the Gaussian does not move from ${x=0}$ so the packet spreads out. A couple of plots show this behaviour. We’ve set ${a=1}$ in both plots. In the red plot ${t=0}$ so ${w=1}$ and in the green plot ${t}$ is larger, at a value such that ${w=0.1}$.

We can get the mean values of position and momentum by integration, although it takes a bit of work. By symmetry, ${\langle x\rangle=\langle p\rangle=0}$. To get the other two average values, we use integration with Maple.

 $\displaystyle \langle x^{2}\rangle$ $\displaystyle =$ $\displaystyle \int_{-\infty}^{\infty}x^{2}|\Psi(x,t)|^{2}dx\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{4w^{2}}\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1+(2\hbar at/m)^{2}}{4a} \ \ \ \ \ (17)$

This shows that the wave function spreads out with time. At ${t=0}$ ${\left\langle x^{2}\right\rangle =1/4a}$, but it then increases quadratically with ${t}$.

Calculating ${\langle p^{2}\rangle}$ starts with:

 $\displaystyle \langle p^{2}\rangle$ $\displaystyle =$ $\displaystyle -\hbar^{2}\int_{-\infty}^{\infty}\Psi^*\frac{\partial^{2}}{\partial x^{2}}\Psi dx \ \ \ \ \ (18)$

This can be evaluated with the Maple command simplify(evalc(int(-h^2*simplify(evalc(psixtconj*(diff(psixt, x$2)))), x = -infinity .. infinity))) assuming positive where psixtconj and psixt are the Maple expressions for ${\Psi^*}$ and ${\Psi}$ respectively. The result is: $\displaystyle \langle p^{2}\rangle=a\hbar^{2} \ \ \ \ \ (19)$ The uncertainty principle thus becomes  $\displaystyle \sigma_{x}\sigma_{p}$ $\displaystyle =$ $\displaystyle \sqrt{\langle x^{2}\rangle\langle p^{2}\rangle}\ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\hbar}{2}\sqrt{1+(2\hbar at/m)^{2}} \ \ \ \ \ (21)$ The system has the least uncertainty at ${t=0}$. Uncertainty increases with time as the wave packet spreads out. ## 5 thoughts on “Free particle: Gaussian wave packet” 1. Pero Instead of calculating using$\Psi$, I found using$\phi(k)^2\$ as the pdf for k, then got from that. This avoided any diffilcult integrals.

1. growescience

I guess it’s a bit easier, although you need to express ${x}$ in momentum space, which is

$displaystyle x=ihbarfrac{partial}{partial p} (1)$

so finding ${leftlangle x^{2}rightrangle }$ still involves taking the second derivative of ${phileft(kright)}$, although it’s true this is a bit easier than taking the second derivative of ${Psi}$.