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

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  4. 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.

    Reply
    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}.

      Reply

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