2.2 Gaussian Wave Packets

We will now look at a concrete example of a valid state for a free particle constructed as a wave packet and study its properties to get a general idea of how wavefunctions behave in quantum mechanics. As discussed in section 2.1.2 above, this begins with choosing a set of wavenumbers and their amplitudes $A(k)$ . We will start with constructing the wave packet at $t=0$ , then we will discuss how it evolves with time.

2.2.1 Choosing an Amplitude $A(k)$

It would be nice to choose $A(k)$ to be a Gaussian shape, because then the Fourier transform $\psi(x,0)$ will also be a Gaussian, and Gaussians are nice to work with.

Recall that a Gaussian function has the general form

\frac{a}{\sqrt{\pi}}e^{-\frac{(x-c)^{2}}{b^{2}}},

(2.19)

which is a bell curve centered at $c$ , with height $\frac{a}{\sqrt{\pi}}$ , width $b$ , and area $a b$ . This means if we want our wavenumber probability density to be a Gaussian centered at some wavenumber $k_{0}$ with width $\Delta k$ , we should choose $a=\frac{1}{\Delta k}$ so that it will be normalised.

A(k)^{2}=\frac{1}{\Delta k\sqrt{\pi}}e^{-\frac{(k-k_{0})^{2}}{\Delta k^{2}}}.

(2.20)

The width $\Delta k$ is the distance from the centre at $e^{-1}$ of the maximum height. It is a rough measure of the range of wavenumbers of the plane waves that make up the wave packet. Note that $\Delta p=\hbar\Delta k$ , so the width $\Delta k$ is directly related to the uncertainty in momentum.

2.2.2 Calculating the Wavefunction $\psi(x,0)$

Using the Fourier transform given by equation 2.14, the wavefunction at $t=0$ is given by

	$\displaystyle\psi(x,0)$	$\displaystyle=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}A(k)e^{ikx}% \operatorname{\mathrm{d}}\!x$		(2.21)
		$\displaystyle=\frac{1}{\sqrt{\Delta k\sqrt{\pi}}\sqrt{2\pi}}\int_{-\infty}^{% \infty}e^{-\frac{(k-k_{0})^{2}}{2\Delta k^{2}}}e^{ikx}\operatorname{\mathrm{d}% }\!k,$		(2.22)

where in the second line we inserted the square root of equation 2.20 for $A(k)$ . To integrate this we use the standard result for Gaussian integrals:

\int_{-\infty}^{\infty}e^{-\alpha q^{2}}e^{-\beta q}\operatorname{\mathrm{d}}% \!q=\sqrt{\frac{\pi}{\alpha}}e^{\frac{\beta^{2}}{4\alpha}}.

(2.23)

With $q=k-k_{0}$ (so $\operatorname{\mathrm{d}}\!q=\operatorname{\mathrm{d}}\!k$ ), we get

	$\displaystyle\psi(x,0)$	$\displaystyle=\frac{1}{\sqrt{\Delta k\sqrt{\pi}}\sqrt{2\pi}}\int_{-\infty}^{% \infty}e^{-\frac{q^{2}}{2\Delta k^{2}}}e^{i(q+k_{0})x}\operatorname{\mathrm{d}% }\!q$		(2.24)
		$\displaystyle=\frac{1}{\sqrt{\Delta k\sqrt{\pi}}\sqrt{2\pi}}e^{ik_{0}x}\int_{-% \infty}^{\infty}e^{-\frac{q^{2}}{2\Delta k^{2}}}e^{iqx}\operatorname{\mathrm{d% }}\!q.$		(2.25)

We can now use the standard integral with $\alpha=\frac{1}{2\Delta k^{2}}$ and $\beta=-ix$ to get

	$\displaystyle\psi(x,0)$	$\displaystyle=\frac{\sqrt{2\pi\Delta k^{2}}}{\sqrt{\Delta k\sqrt{\pi}}\sqrt{2% \pi}}e^{ik_{0}x}e^{-\frac{\Delta k^{2}x^{2}}{2}}$		(2.26)
		$\displaystyle=\sqrt{\frac{\Delta k}{\sqrt{\pi}}}e^{ik_{0}x}e^{-\frac{x^{2}% \Delta k^{2}}{2}}.$		(2.27)

What is the probability density at $t=0$ ? We can calculate this directly by taking the magnitude square:

\lvert\psi(x,0)\rvert^{2}=\frac{\Delta k}{\sqrt{\pi}}e^{-x^{2}\Delta k^{2}}=% \frac{\Delta k}{\sqrt{\pi}}e^{-\frac{x^{2}}{\Delta x_{0}^{2}}},

(2.28)

where we have defined $\Delta x_{0}=\frac{1}{\Delta k}$ as the width at $t=0$ . So the probability density is also a Gaussian, this time centered at $x=0$ , with width $\Delta x_{0}=\frac{1}{\Delta k}$ .

Notice that $\Delta x_{0}\Delta k=1$ , which can be expressed in term of momentum via the de Broglie relation:

\Delta x_{0}\Delta p=\hbar.

(2.29)

This is a manifestation of the uncertainty principle. The factor of $1$ arises from our definition of width of a Gaussian. Notice that the width of the position probability density is inversely proportional to the width of the wavenumber probability density. This reflects the fact that due to the uncertainty principle, if the uncertainty in momentum $\Delta k$ is small then the uncertainty in position $\Delta x_{0}$ must be wide and vice versa.

2.2.3 Time-evolution of the Gaussian Wave Packet

As we found in section 2.1.2, the time evolution of our wave packet is given by

\psi(x,t)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}A(k)e^{i(kx-\omega t)}% \operatorname{\mathrm{d}}\!k.

(2.30)

To evaluate this, we need to keep in mind that $\omega$ is a function of $k$ through the Einstein and de Broglie relations. Specifically, for a free particle only we have $E=\frac{p^{2}}{2m}$ , which gives

$\displaystyle E$	$\displaystyle=\frac{p^{2}}{2m}$	(2.31)
$\displaystyle\hbar\omega$	$\displaystyle=\frac{\hbar^{2}k^{2}}{2m}$	(2.32)
$\displaystyle\implies\omega(k)$	$\displaystyle=\frac{\hbar k^{2}}{2m}.$	(2.33)

This is our dispersion relation for matter waves. Note that the $\omega$ is not proportional to $k$ , so matter waves are dispersive. This means that different wavelength will travel at different speeds. The phase velocity is given by

v_{\text{ph}}(k)=\frac{\omega}{k}=\frac{\hbar k}{2m},

(2.34)

so shorter wavelengths (larger $k$ ) travel faster. This makes physical sense as large $k$ corresponds to large momentum. Meanwhile the group velocity for a wave packet centered at $k_{0}$ is

v_{\text{gr}}(k)=\left.\frac{\operatorname{\mathrm{d}}\!\omega}{\operatorname{% \mathrm{d}}\!k}\right|_{k_{0}}=\frac{\hbar k_{0}}{2m}=2v_{\text{ph}}(k_{0}).

(2.35)

This is the speed that the envelope of the wave packet will travel at. Combining these ideas together, we expect that the wave packet will be smudged out over time, as some wavelengths in it travel faster and some slower.

Let’s see if we can show this explicitly. Substituting the square root of equation 2.20 for $A(k)$ and equation 2.33 for $\omega(k)$ into the integral, we get

\psi(x,t)=\frac{1}{\sqrt{\Delta k\sqrt{\pi}}\sqrt{2\pi}}\int_{-\infty}^{\infty% }e^{-\frac{(k-k_{0})^{2}}{2\Delta k^{2}}}e^{i\left(kx-\frac{\hbar k^{2}}{2m}t% \right)}\operatorname{\mathrm{d}}\!k.

(2.36)

We would like to use the standard integral 2.23 for Gaussians again. Making the same substitution as in section 2.2.2 above ( $q=k-k_{0}$ ) gives

	$\displaystyle\psi(x,t)$	$\displaystyle=\frac{1}{\sqrt{\Delta k\sqrt{\pi}}\sqrt{2\pi}}\int_{-\infty}^{% \infty}e^{-\frac{q^{2}}{2\Delta k^{2}}}e^{i(q+k_{0})x}e^{-\frac{i\hbar(q+k_{0}% )^{2}}{2m}t}\operatorname{\mathrm{d}}\!q$		(2.37)
		$\displaystyle=\frac{1}{\sqrt{\Delta k\sqrt{\pi}}\sqrt{2\pi}}\int_{-\infty}^{% \infty}e^{-\frac{q^{2}}{2\Delta k^{2}}}e^{iqx}e^{ik_{0}x}e^{-\frac{i\hbar q^{2% }}{2m}t}e^{-\frac{i\hbar qk_{0}}{m}t}e^{-\frac{i\hbar k_{0}^{2}}{2m}t}% \operatorname{\mathrm{d}}\!q.$		(2.38)

We can simplify the notation a bit by defining $\omega_{0}=\frac{\hbar k_{0}^{2}}{2m}$ , the frequency of the state with wavenumber $k_{0}$ , and using the group velocity $v_{\text{gr}}(k_{0})=\frac{\hbar k_{0}}{m}$ . The fifth and sixth exponentials can be rewritten in terms of these quantities, and then the third and fifth exponentials can be taken out of the integral to get

\psi(x,t)=\frac{1}{\sqrt{\Delta k\sqrt{\pi}}\sqrt{2\pi}}e^{i(k_{0}x-\omega_{0}% t)}\int_{-\infty}^{\infty}e^{-\left(\frac{1}{2\Delta k^{2}}+\frac{i\hbar}{2m}t% \right)q^{2}}e^{-i(v_{\text{gr}}(k_{0})t-x)q}\operatorname{\mathrm{d}}\!q.

(2.39)

Now we can use the standard integral with

	$\displaystyle\alpha$	$\displaystyle=\frac{1}{2\Delta k^{2}}+\frac{i\hbar}{2m}t$		(2.40)
	$\displaystyle\beta$	$\displaystyle=i(v_{\text{gr}}(k_{0})t-x),$		(2.41)

to get

	$\displaystyle\psi(x,t)$	$\displaystyle=\frac{1}{\sqrt{\Delta k\sqrt{\pi}}\sqrt{2\pi}}\sqrt{\frac{\pi}{% \frac{1}{2\Delta k^{2}}+\frac{i\hbar}{2m}t}}e^{i(k_{0}x-\omega_{0}t)}e^{-\frac% {(x-v_{\text{gr}}(k_{0})t)^{2}}{4\left(\frac{1}{2\Delta k^{2}}+\frac{i\hbar}{2% m}t\right)}}$		(2.42)
		$\displaystyle=\sqrt{\frac{\Delta k}{\sqrt{\pi}\left(1+\frac{i\hbar\Delta k^{2}% t}{m}\right)}}e^{i(k_{0}x-\omega_{0}t)}e^{-\frac{\Delta k^{2}(x-v_{\text{gr}}(% k_{0})t)^{2}}{2\left(1+\frac{i\hbar\Delta k^{2}t}{m}\right)}}.$		(2.43)

Calculating the probability density, we get

\lvert\psi(x,t)\rvert^{2}=\frac{\Delta k}{\sqrt{\pi}\sqrt{1+\frac{\hbar^{2}% \Delta k^{4}t^{2}}{m^{2}}}}e^{-\frac{\Delta k^{2}(x-v_{\text{gr}}(k_{0})t)^{2}% }{1+\frac{\hbar^{2}\Delta k^{4}t^{2}}{m^{2}}}}.

(2.44)

This is quite a complicated expression, but there are a few key things we can pick out that tell us in simple terms how the wave packet behaves. The structure of the probability density is basically

\lvert\psi(x,t)\rvert^{2}=C(t)e^{-\frac{(x-x_{0}(t))^{2}}{\Delta x(t)^{2}}},

(2.45)

so we see that the height, width, and centre of the wave packet all change with time while retaining a Gaussian form.

Comparing the two expressions, the centre of the wave packet $x_{0}$ evolves as

x_{0}(t)=v_{\text{gr}}(k_{0})t,

(2.46)

so the position of the peak of the wavepacket moves to the right with a constant speed, the speed of the central wavenumber (or equivalently, the speed of the envelope). This makes sense, there are no forces acting on the particle so its speed stays constant with time.

The width of the wave packet is

\Delta x(t)=\frac{1}{\Delta k}\sqrt{1+\frac{\hbar^{2}\Delta k^{4}t^{2}}{m^{2}}% }=\Delta x_{0}\sqrt{1+\frac{\hbar^{2}\Delta k^{4}t^{2}}{m^{2}}},

(2.47)

where we have reintroduced $\Delta x_{0}=\frac{1}{\Delta k}$ , the width of the wave packet at $t=0$ from before. Thus the width $\Delta x$ increases over time just like we predicted.

The height has the form

C(t)=\frac{1}{\sqrt{\pi}\Delta x(t)},

(2.48)

so since the width increases with time, the height decreases. This makes sense, as to preserve the normalisation of the wave packet the amplitude must decrease as it spreads out.

Notice that in the limit $t\to 0$ , we get back our previous results for the initial wave packet $\psi(x,0)$ . If $\Delta k$ is larger at $t=0$ , i.e. if the uncertainty in momentum $\Delta p$ is larger, the the wavepacket spreads out more slowly.

Does the uncertainty in momentum $\Delta p$ increase over just like $\Delta x$ ? From equation 2.17, we have

	$\displaystyle\psi(x,t)$	$\displaystyle=\frac{1}{\sqrt{2\pi\hbar}}\int_{-\infty}^{\infty}\phi(p,t)e^{% \frac{i}{\hbar}px}\operatorname{\mathrm{d}}\!p$		(2.49)
		$\displaystyle=\frac{1}{\sqrt{2\pi\hbar}}\int_{-\infty}^{\infty}\phi(p,0)e^{% \frac{i}{\hbar}(px-Et)}\operatorname{\mathrm{d}}\!p.$		(2.50)

For the free particle, we have $E=\frac{p^{2}}{2m}$ and so

\psi(x,t)=\frac{1}{\sqrt{2\pi\hbar}}\int_{-\infty}^{\infty}\phi(p,0)e^{-\frac{% i}{\hbar}\frac{p^{2}}{2m}t}e^{\frac{i}{\hbar}px}\operatorname{\mathrm{d}}\!p.

(2.51)

Hence we have that

\lvert\phi(p,t)\rvert^{2}=\lvert\phi(p,0)e^{-\frac{i}{\hbar}\frac{p^{2}}{2m}t}% \rvert^{2}=\lvert\phi(p,0)\rvert^{2},

(2.52)

so the momentum probability density remains the same over time and $\Delta p$ does not change. This is not a general fact, it only holds for the free particle. We will find that when there are forces present, $\Delta p$ does indeed increase over time. In that sense, this is sort of a quantum analogue of Newton’s first law.

2.2 Gaussian Wave Packets

2.2.1 Choosing an Amplitude A⁢(k)

2.2.2 Calculating the Wavefunction ψ⁢(x,0)

2.2.3 Time-evolution of the Gaussian Wave Packet

2.2.1 Choosing an Amplitude $A(k)$

2.2.2 Calculating the Wavefunction $\psi(x,0)$