# Why do wave packets spread out over time?

Why do wave functions spread out over time? Where in the math does quantum mechanics state this? As far as I've seen, the waves are not required to spread, and what does this mean if they do?

Suppose you have an infinite plane wave. To find the momentum of this wave you Fourier transform it. Because it's an infinite wave the Fourier transform is a delta function and the wave has a well defined single value for the momentum.

Now take a wave packet i.e. the same infinite plane wave but now multipled by some envelope function. When you Fourier transform this you get the original delta function but now convolved with the Fourier transform of the envelope function. The packet is made up from waves with a range of different frequencies/momenta. The longer the packet the smaller the range of momenta, but for any finite wave packet there will always be a spread of momenta.

For a massive particle the spread of momenta means there is a spread of velocities, and therefore the wave packet broadens away from its average position.

• This is incorrect; a spread of velocities does not imply that the wave packet broadens if the "back end" of the wave is locally moving faster than the "front end", as initially happens for e.g. a free particle that later evolves into a minimum-uncertainty wave packet. – tparker Sep 1 '19 at 1:37
• You can certainly construct a wave packet that initially narrows. But you can’t do that for all future time. – Bob Jacobsen Sep 1 '19 at 1:47

An easy way to make this intuitively plausible is by remarking that the Schroedinger equation in the absence of a potential is as follows

$${\partial\over\partial t}\Psi = \nabla^2\Psi$$

up to constants, which is the heat equation if we ignore the fact that the omitted constants are complex numbers rather than real and of the right sign.

If you consider your wave function to be a heat map, then it is intuitively clear that it should spread out.

• This is a very good observation, I am versed in the Schroedinger equation, but I just learned something new :D – Karl Damgaard Asmussen Sep 19 '13 at 11:53
• You can't just "omit the complex number". That factor of i makes an enormous difference, it's the reason that the Schrodinger equation is time-symmetric and the heat equation isn't, which makes this whole analogy completely break down. – knzhou Sep 1 '19 at 4:24

You can look at the beautiful physical arguments given here, or you can look at it mathematically. A wave packet consists of a combination of several solutions, in the case of your quantum mechanics problem these will be your eigenfunctions. For the sake of simplicity I will consider plane waves (these correspondend to the free particle, `particle in a box and actually every solution in quantum mechanics since wavefunctions $\psi$ should be in $L^2$ and hence we can Fourier transform them all.

So take a wave packet $\alpha(x,t)$ which is composed of plane waves:

$\alpha(x,t) = \int\limits_{-\infty}^\infty A(k)e^{i(kx-\omega(k) t)}dk,$

where A(k) is the weighing-factor which for the wavenumber and where I simply stayed in one dimension.

Wave packets are usually peaked around a certain frequency $\omega_0$ and hence around a certain wave number $k_0$, by doing this you can make for example a Gaussian out of your basefunctions which is mostly used for particles that are localised within an area.

Since the k-numbers are localised around a value $k_0$ we can do a Taylorseries of $\omega(k)$:

$\omega(k) = \omega(k_0) + \left.\frac{\partial\omega(k)}{\partial k}\right|_{k=k_0}(k-k_0)+\mathcal{O}(k^2).$

Now if we only go first order and denote the partial derivative in $k_0$ as $\omega_0'$, the wavepacket can be rewritten as:

$\alpha(x,t) = e^{it(\omega_0'k_0-\omega_0)}\int\limits_{-\infty}^\infty A(k)e^{ik(x-\omega_0't)}dk$.

Where we see that $|\alpha(x,t)|=|\alpha(x-\omega_0't,0)|$, so the wavepacket moves as a whole with a velocity $\omega_0'$, the group velocity.

Now i've only talked about the propagation of the wave-packet and not of the deformation (spreading, skewness, ...), these are effects we get if we would inspect the higher order terms of the Taylorseries of $\omega(k)$, for example the dispersion is a second order effect which will cause a spread in your wavepacket. So we see that if you have a linear dispersion relation you won't have any broadening effects, in quantum mechanics however this is not the case, consider for example the free particle or particle in a box the dispersion relations are quadratic and we will have:

$\hbar\omega = E = \frac{\hbar^2k^2}{2m} \Rightarrow \omega = \frac{\hbar k^2}{2m}$.

So you see the NONLINEAR DISPERSIONRELATIONS are the cause of the broadening of your wavepackets. The calculations for the second order term are carried out here. If you follow these calculations you will see that the time for a wave packet to double in width is given by (if it's a Gaussian):

$\tau = \frac{\sqrt{12}}{\alpha}(\Delta x)^2$,

where $\alpha = \left.\frac{\partial^2\omega(k)}{\partial k^2}\right|_{k=k_0}$, so for our free quantum particle this yields:

$\tau \sim \frac{m(\Delta x)^2}{\hbar}$, where $(\Delta x)$ is the width of your Gaussian wave packet at t=0.

For any wave propagation there is a relation between wave speed and wavelength called the dispersion relation. If wave speed does not depend on wavelength, the system is called non-dispersive (example: electromagnetic wave in free space). The Schroedinger equation is dispersive (shorter wavelengths are faster). A wave packet is necessarily a special superposition of wavelength components, and those components won't stay together if they are moving at different speeds.

First, a precision: wave functions do not necessarily spread over time. I think you mean wave packets, that is, spatially-localized wave functions. Eigenstates of the Hamiltonian obviously do not spread over time.

Then, as stated above, the fact that most wave packets spread is ultimately due to the uncertainty principle, which the time-dependent Schrödinger equation takes automatically into account.

However, it is not even true that all wave packets spread over time, or that they only spread over time. As long as the wave packet implies a superposition of only bound states of the Hamiltonian so that the motion is bounded, there will always exist a phenomena of revival, where the components of the wave packet rephase again. Finally, for special potentials (e.g. the harmonic oscillator) the wave packet does not even have to disperse.

Wave packets don't always spread out over time. For a free particle with a Gaussian wavepacket, the width actually decreases for an arbitrarily long time before eventually increasing again. The Heisenberg uncertainty product $$\Delta x \times \Delta p$$ can also decrease (although of course it can never go below $$\hbar/2$$).

• +1, this is the only current answer to this question that's correct – knzhou Sep 1 '19 at 4:24

In the Schrödinger equation it is the kinetic energy term -(d/dx)^2 that is responsible. The standard example is the following: If you start with a delta function in coordinate space at t=0 you find that it changes into a Gaussian at later times.