We write a free wave packet as a linear combination of plane waves $$\Psi(x,t)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty dk\,\tilde{\varphi}(k)\,e^{-i\omega(k)t}\,e^{ikx} =\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty dk\,\tilde{\varphi}(k,t)\,e^{ikx}$$ where the time evolution of each plane wave, given by $\omega(k)=\hbar k^2/2m$, is absorbed in the momentum amplitudes $\tilde{\varphi}(k,t)$. When we chose the amplitude for the different plane waves as a Gaussian $$\tilde{\varphi}(k,t=0)=\frac{1}{\sqrt{\sqrt{2\pi}\sigma_k}}\,e^{-(k-k_0)^2/4\sigma_k^2}$$ the momentum distribution $|\tilde{\varphi}(k,t)|^2$ is a Gaussian of width $\sigma_k$ and mean $k_0$ for all $t$.

At $t=0$, where the momentum amplitudes are real, the probability density of the resulting wave packet at is a Gaussian of width $\sigma_x=1/2\sigma_k$, i.e., it is as localized as is allowed by the Heisenberg principle ($\sigma_x\,\sigma_k=1/2$). The probability distribution stays Gaussian for all $t$. As the momentum amplitudes become complex, its width $\sigma_x\sqrt{1+\omega_\sigma^2 t^2}$ increases with a characteristic time $1/\omega_\sigma=2m\sigma_x^2/\hbar$, and its center moves with the group velocity $v_g=\hbar k_0/m$.

{ "xmin":-2.5, "xmax":12.5, "ymin":0, "ymax":1.0,
"tmin":-1, "tmax":3, "dt":0.02, "t0":0,
"wp": [{"cRe": 20, "cIm": 0, "x0": 0, "k0": 4, "dk": 1}],
"prob": true
}

This animation shows the probability density for a Gaussian wave packet. It starts at $t=0$. With increasing time the packet moves to the right and broadens. What happens for negative times? Use the slider.

{ "xmin":-2.5, "xmax":12.5, "ymin":0, "ymax":1.0,
"tmin":-1, "tmax":3, "dt":0.02, "t0":0,
"wp": [{"cRe": 20, "cIm": 0, "x0": 0, "k0": 4, "dk": 0.5}],
"prob": true
}

This wave packet starts out wider than in the above animation. Can you understand why it does not broaden as quickly?

Note that the behavior at $t=0$ is the genuinely quantum mechanical feature: We cannot localize the particle more than is allowed by the uncertainty principle. Here it is a direct consequence of the wave-description following directly from the properties of the Fourier transform. The behavior for large times is what we would expect from an ensemble of classical particles with a Gaussian momentum distribution: At time $t$ a particle of momentum $\hbar k$ will be at $x(k,t)=\hbar k t/m$. Solving for $k$ and inserting into the momentum distribution, we obtain the probability density of finding a classical particle $$p(x,t)\propto e^{-(k(x(k,t))-k_0(x(k_0,t)))^2/2\sigma_k^2} = e^{-(x-x_0(t))^2/2(\hbar t\sigma_k/m)^2}$$ which is a Gaussian of width $\sigma_x\omega_\sigma t$ centered at $x_0(t)=\hbar k_0 t/m$.

So far we have plotted the probability density for the wave packets. If we want to plot the wave function we need a way to visualize complex functions. We could simply plot its real and imaginary part. It is, however, more instructive to show its modulus (the square of which gives the probability density) and phase. The modulus is easily plotted. To represent the phase we use the colors, starting with red for phase zero, going through the colors of the rainbow as the phase increases and jumping back to red for phase $2\pi$. As an illustration, below is the representation of a plane wave $e^{i(kx-\omega(k)t)}$

{ "xmin":0, "xmax":15, "ymin":0, "ymax":1.0,
"tmin":0, "tmax":3, "dt":0.02, "t0":0,
"wp": [{"cRe": 0.5, "cIm": 0, "x0": 0, "k0": 4, "dk": 0}]
}

Do you understand why the phase moves? What is its velocity?

{ "xmin":0, "xmax":15, "ymin":0, "ymax":1.0,
"tmin":0, "tmax":3, "dt":0.02, "t0":0,
"wp": [{"cRe": 0.35, "cIm": 0, "x0": 0, "k0": 3.5, "dk": 0},
{"cRe": 0.35, "cIm": 0, "x0": 0, "k0": 4.5, "dk": 0}]
}

This shows the linear combination of two plane waves. Compare the velocity of the beating pattern (the modulation of the amplitude) to that of the phase.

The wave function of the propagating free Gaussian wave packet is given by $$\begin{align*} \Psi_{k_0,\sigma_k}(x,t) &=\frac{1}{\sqrt{\sqrt{2\pi}\sigma_x(1+i\omega_\sigma t)}} e^{i(k_0x-\omega(k)t)}\,e^{-(x-v_gt)^2/4\sigma_x^2(1+i\omega_\sigma t)}\\ &=\frac{e^{-(x-v_gt)^2/4\sigma_x(t)^2}}{\sqrt{\sqrt{2\pi}\sigma_x(t)}}\;e^{i(k_0x+(x-v_gt)^2\,\omega_\sigma t/4\sigma_x(t)^2+\arg(1-i\omega_\sigma t)/2-\omega(k)t)} \end{align*}$$

{ "xmin":-2.5, "xmax":12.5, "ymin":0, "ymax":1.0,
"tmin":-1, "tmax":3, "dt":0.02, "t0":0,
"wp": [{"cRe": 1, "cIm": 0, "x0": 0, "k0": 4, "dk": 0.5}]
}

For a Gaussian wave packet with narrow momentum distribution the phase is almost that of a plane wave with momentum $k_0$. Observe how the center of the probability density moves faster than the phase ($v_g=2v_{ph}$).

{ "xmin":-2.5, "xmax":12.5, "ymin":0, "ymax":1.0,
"tmin":-1, "tmax":3, "dt":0.02, "t0":0,
"wp": [{"cRe": 1, "cIm": 0, "x0": 0, "k0": 4, "dk": 1}]
}

When the momentum distribution gets wider the phase has no longer a well defined wave length. Around the maximum at $x(t)=v_gt$ the phase has still wave length $2\pi/k_0$. Observe how the wave length increases behind the maximum and is reduced in front of it. Do you see which term is responsible for this?

It is also interesting to look at linear combinations of Gaussian wave packets. Note that we are still working with single-electron wave functions, so they describe a *single* electron in a superposition state, *not* two electrons!

{ "xmin":-20, "xmax":20, "ymin":0, "ymax":1.0,
"tmin":0, "tmax":8, "dt":0.02, "t0":0,
"wp": [{"cRe": 1, "cIm": 0, "x0":-10.0, "k0": 4.0, "dk": 0.5},
{"cRe":-1, "cIm": 0, "x0": 10.0, "k0":-4.0, "dk": 0.5}]
}

This is the superposition of two Gaussian wave packets going in opposite direction: $\Psi=\mathrm{GWP}(k_0,\sigma_k,-x_0)-\mathrm{GWP}(-k_0,\sigma_k,x_0)$. What determines the wave length of the interference pattern? Why is the amplitude in the center between the wave packets always zero?

{ "xmin":-20, "xmax":20, "ymin":0, "ymax":1.0,
"tmin":0, "tmax":8, "dt":0.02, "t0":0,
"wp": [{"cRe": 30, "cIm": 0, "x0": -10.0, "k0": 4.0, "dk": 0.25},
{"cRe":-30, "cIm": 0, "x0": 10.0, "k0":-4.0, "dk": 0.25}],
"prob": true
}

This is the same wave function as above, but now showing the probability density. What happens to the wave packets after they have passed through each other? What would happen if their velocity would be smaller than the rate at which they broaden?

{ "xmin":-20, "xmax":20, "ymin":0, "ymax":1.0,
"tmin":0, "tmax":8, "dt":0.02, "t0":0,
"wp": [{"cRe": 0.8, "cIm": 0, "x0":-10.0, "k0": 2.0, "dk": 2},
{"cRe":-0.8, "cIm": 0, "x0": 10.0, "k0":-2.0, "dk": 2}]
}

This wave packet has a much wider momentum distribution and moves slower. Why do the oscillations in the wave function persist?

Looking at just one half of the above superpositions shows us how a Gaussian wave packet is reflected at a hard wall. Compare the approach to the method of image charges used in electrostatics.

{ "xmin":-10, "xmax":10, "ymin":0, "ymax":1.0,
"tmin":0, "tmax":10, "dt":0.02, "t0":0,
"wp": [{"cRe": 1, "cIm": 0, "x0":-2, "k0": 4, "dk": 0.25}], "wall":10
}

{ "xmin":-10, "xmax":10, "ymin":0, "ymax":1.0,
"tmin":0, "tmax":10, "dt":0.02, "t0":0,
"wp": [{"cRe": 30, "cIm": 0, "x0":-2, "k0": 4, "dk": 0.25}], "wall":10,
"prob": true
}

The same, now for a wave packet with much wider momentum distribution

{ "xmin":-10, "xmax":10, "ymin":0, "ymax":1.0,
"tmin":0, "tmax":10, "dt":0.02, "t0":0,
"wp": [{"cRe": 0.8, "cIm": 0, "x0":-2, "k0": 2, "dk": 2}], "wall":10
}

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