# Free particle – travelling wave packet

Required math: calculus

Required physics: Schrödinger equation

References: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 2.43.

We’ve looked at the stationary Gaussian wave packet for the free particle. The initial wave function in that case was

$\displaystyle \Psi(x,0)=Ae^{-ax^{2}} \ \ \ \ \ (1)$

We can turn this into a travelling Gaussian wave by adding a factor to the wave function:

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

where ${l}$ is a real constant.

Since we have added only a complex exponential, the normalization condition is the same as for the stationary case:

$\displaystyle A=\left(\frac{2a}{\pi}\right)^{1/4} \ \ \ \ \ (3)$

To find ${\Psi(x,t)}$ we follow the same procedure as in the stationary case. So we get

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

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

 $\displaystyle \phi(k)$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}\Psi(x,0)e^{-ikx}dx\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2\pi}}\left(\frac{2a}{\pi}\right)^{1/4}\int_{-\infty}^{\infty}e^{-ax^{2}-ikx+ilx}dx\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(\frac{1}{2\pi a}\right)^{1/4}e^{-(k-l)^{2}/4a} \ \ \ \ \ (7)$

So we can now find the general solution:

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

where Maple was used for the integral.

Calculating ${|\Psi(x,t)|^{2}}$ can be done using Maple, with the result:

$\displaystyle |\Psi(x,t)|^{2}=\sqrt{\frac{2}{\pi}}we^{-2w^{2}(\hbar lt/m-x)^{2}} \ \ \ \ \ (10)$

with

$\displaystyle w\equiv\left(\frac{a}{1+(2\hbar at/m)^{2}}\right)^{1/2} \ \ \ \ \ (11)$

The results above reduce to the stationary wave packet when ${l=0}$.

At ${t=0}$, ${w=\sqrt{a}}$, so ${|\Psi(x,0)|^{2}=\sqrt{2a/\pi}e^{-2ax^{2}}}$ which is correct. The wave packet at ${t=0}$ is therefore a Gaussian centred at ${x=0}$. As ${t}$ increases, ${w}$ gets smaller but in this case, the peak of the Gaussian moves according to ${x_{peak}=\hbar lt/m}$. The speed of the peak is ${x/t=\hbar l/m}$.

By direct integration we find, ${\langle x\rangle=\hbar lt/m}$. Calculating the other means requires a bit of effort but we can use Maple to do most of it. The results are:

 $\displaystyle \langle p\rangle$ $\displaystyle =$ $\displaystyle l\hbar\ \ \ \ \ (12)$ $\displaystyle \langle x^{2}\rangle$ $\displaystyle =$ $\displaystyle \frac{1+(2a\hbar t/m)^{2}+a(2\hbar lt/m)^{2}}{4a}\ \ \ \ \ (13)$ $\displaystyle \langle p^{2}\rangle$ $\displaystyle =$ $\displaystyle \hbar^{2}(a+l^{2}) \ \ \ \ \ (14)$

All these results reduce to those for the stationary wave packet from problem 2.22 when ${l=0}$.

The uncertainty principle thus becomes

 $\displaystyle \sigma_{x}\sigma_{p}$ $\displaystyle =$ $\displaystyle \sqrt{(\langle x^{2}\rangle-\langle x\rangle^{2})(\langle p^{2}\rangle-\langle p\rangle^{2})}\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\hbar}{2}\sqrt{1+(2\hbar at/m)^{2}} \ \ \ \ \ (16)$

which is the same result as in the stationary wave packet. Thus although the packet here travels with a constant speed, it spreads out at the same rate as the stationary packet.