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.

2 thoughts on “Free particle – travelling wave packet

  1. Pingback: Energy-time uncertainty principle: Gaussian free particle « Physics tutorials

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