Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

I have recently read how the Gaussian wave packet spreads while propagating. see: http://en.wikipedia.org/wiki/Wave_packet#Gaussian_wavepackets_in_quantum_mechanics

Though I understand the mathematics I don't understand the physical explanation behind it. Can you please explain?

share|improve this question
    
If you like this question you may also enjoy reading this Phys.SE post. –  Qmechanic 3 hours ago

5 Answers 5

The Gaussian wavepacket only spreads in the free Schrodinger equation. It doesn't spread in the case where you have a harmonic oscillator, and the Gaussian width is equal to that of the ground state wavefunction. For other Gaussians in an oscillator, the width oscillates, growing and shrinking.

Spreading in the free Schrodinger equation is easiest to understand from the analyticity properties. The Schrodinger equation is analytically linked to the diffusion equation

$$ {d\over dt} \rho = {1\over 2} \nabla^2 \rho $$

The fundamental solution of the diffusion equation for initial conditions a delta function at $t=0$ and $x=0$ is the spreading Gaussian:

$$ \rho(x,t) = {1\over \sqrt{2\pi t}} e^{-{x^2\over 2t}} $$

You can see this works by substituting, but it is also obvious from a path integral. To get the spreading Schrodinger packet, substitute $it$ for t.

share|improve this answer

Ron's answer is (as always :-) definitive, and if you're going to accept an answer you should accept his. However I thought it was worth attempting a more general explanation.

Remember that the gaussian packet describes the probability distribution of the particle. When the packet spreads it doesn't mean the particle is in some sense swelling up and spreading out, it means the probability of finding the particle is spreading out.

The reason for this is that the gaussian has a spread of momentum related to the uncertainty principle $\Delta p\Delta x \ge \hbar/2$. That means there is a spread of velocities and this means the gaussian has to spread out. I'm reluctant to say that different bits of the packet are moving at different velocities because this is attempting a classical analogy that is misleading, but it hopefully gives you some physical intuition as to what is going on.

share|improve this answer

Though I understand the mathematics I don't understand the physical explanation behind it.

I'll take a stab at it.

For a free particle, momentum eigenstates are also energy eigenstates and thus have a simple time dependence, a time dependent phase with a frequency proportional to the energy of the state.

A free particle with a gaussian wave function is then a continuous superposition of momentum, and thus energy, eigenstates.

Since the phase of the different momentum eigenstates evolve at a different rate, the way the various components constructively/destructively add evolves in time.

When all the phases "line up" just so, we get the minimum uncertainty wave packet. As time evolves, the wave packet spreads since the phases evolve at different rates.

share|improve this answer

(From my book http://physics-quest.org/Book_Chapter_Klein_Gordon.pdf)


Spreading of the free field wave packet

The speed of the wave packet is given by the derivative of the Hamiltonian against the momentum.

\begin{equation} v ~~=~~ \frac{\partial H}{\partial p} ~~=~~ \frac{\partial E}{\partial p} ~~=~~ \frac{p c^2}{\sqrt{(pc)^2+(mc^2)^2}} ~~=~~ \frac{p c^2}{E~} \end{equation}

The wave-packet would not spread in the case of a single $v$ such as in the case of a massless particle which is represented by a wave-function which moves unchanged at the speed of light.

However, a localized field with a Gaussian shape has (via the Fourier transform) a Gaussian distribution of $p$ in momentum space. The relation

\begin{equation}E = \sqrt{(pc)^2+(mc^2)^2}\end{equation}

means that there will be a range of speeds instead of a single one and so, in general, the wave-packet will spread.


enter image description here figure 1.


The variation is approximately given by.

\begin{equation} \frac{\Delta v}{\Delta p}~\approx~\frac{\partial v}{\partial p} ~~\longrightarrow~\Delta v ~\approx~ \frac{\partial^2 E}{\partial\,p^2}~\Delta p \end{equation}

Given that Heisenberg's uncertainty relation $\Delta x \Delta p \geq \hbar/2$ can be derived by Fourier analysis, which in the case of a Gaussian shaped wave-function becomes $\Delta x \Delta p = \hbar/2$, the minimum value, we can write.

\begin{equation} \Delta v ~\approx~ \frac{\partial^2 E}{\partial\,p^2}~\frac{\hbar}{2\Delta x} \end{equation}

Where $\Delta x$ is the width. One can reason that the overall shape of a wave-function changes faster if $\Delta x$ is smaller for a given speed-variation $\Delta v$. We can define a dimension-less quantity shape, which has derivative in time which gives us an approximation of the relative spreading of the wave-function in time.

\begin{equation} \frac{\partial}{\partial t}\Big\{ \mbox{shape} \Big\} ~~ \approx ~~ \frac{\Delta v}{\Delta x} ~~ \approx ~~ \frac{\partial^2 E}{\partial\, p^2}~\frac{\hbar}{2(\Delta x)^2} \end{equation}

Working out the second order derivative gives us.

\begin{equation} \frac{\partial^2 E}{\partial\, p^2} ~=~ \frac{(mc^2)^2~c^2}{~\big(~(pc)^2+(mc^2)^2~\big)^{3/2}~} ~=~ \frac{E_o^2\,c^2}{E^3} ~=~\frac{c^2}{E\gamma^2} \end{equation}

Which leads us to our final expression here.

\begin{equation} \frac{\partial}{\partial t}\Big\{ \mbox{shape} \Big\} ~~ \approx ~~ \frac{\hbar\,c^2}{2E(\gamma\Delta x)^2} \end{equation}

If we remove the gamma's then we get the expression for the rest frame.

\begin{equation} \frac{\partial}{\partial t}\Big\{ \mbox{shape} \Big\} ~~ \approx ~~ \frac{\hbar\,c^2}{2mc^2(\Delta x)^2} \end{equation}

We can summarize the results as:

  • The spreading of the wave-function is inversely proportional to the frequency (the phase change rate in time) of the particle, Higher mass particles spread slower.
  • The spreading of the wave-function is proportional to the square of the momentum spread. The smaller the initial volume in which the initial wave-function was contained the faster it spreads and keeps spreading.

enter image description here figure 2


From figure. 2 we can read the mathematical mechanism which leads to spreading. The variation $\Delta p$ of the momentum stays the same over time. It is the frequency dependency on the momentum

$E=\sqrt{(pc)^2+(mc^2)^2}$

which leads to a phase change over $p$.

The phase change is opposite at both sides of the center-momentum. These phase-changes lead to (opposite) translations of the wave-function in position space, this is the spreading. The value $\Delta x$ in our expression stays constant because $\Delta p$ stays constant, it is the initial $\Delta x$ corresponding to the pure Gaussian at $t$=$0$.

Some actual spreading rate numbers

We can work out a few numerical example to get an idea of the spreading rates. From the wide range of wavelength sizes, we can classify the Compton radius as the small size limit, although there is in principle no real barrier to go to even smaller sized wave-packets.

If we replace $\Delta x$ with twice the Compton radius $r_c=\hbar/mc$ then, assuming that our approximation is still reasonably valid in this range.

\begin{equation} \frac{\partial}{\partial t}\Big\{ \mbox{shape} \Big\} ~~ \approx ~~ \frac{c}{8\,r_c} \end{equation}

If we recall the rest-frequency of the particle: $f_o=c/(2\pi\,r_c)$ (in case of the electron $f_o=1.2355899729\,10^{20}$ Hz ), then we see that spreading speed approaches the speed of light in this range. The spread in momentum is so large that it includes velocities from close to $-c$ up to $+c$.

To confine an electron-field to a Compton radius-like volume one needs a positive charge of $~$137e, The inner-most electrons of heavy elements come close to being confined into such a small area. The Compton radius for electrons is $3.861592696\,10^{-13}$ meter.

More commonly, electrons freed from a bound state, take off with a much larger radius, comparable to the Bohr radius. ($5.291772131\,10^{-11}$ meter) This means that the spreading speed is much lower, $ v < 0.01c$, but still quite high.

The size of the wave-packet will grow fast. For instance, the famous single electron interference experiment of Akira Tomomura (see figure 3), which demonstrated the single-electron build-up of an interference pattern, shows that the electron fields in the experiment must at least be several micrometers wide. This is a factor 100,000 wider as in the confinement of the Bohr radius.


enter image description here figure 3


Hans

share|improve this answer

The explanation is really very simple to understand intuitively, and very beautiful.

Imagine that a particle an uncertainity in its velocity $v$ of $\delta v$. Suppose at $t=0$ we have $x=x_{0}$. After $t=T$, the location of the particle will be given by the range $(x_{0}+Tv-T\delta v,x_{0}+Tv+T\delta v)$, because we dont know the exact velocity the particle started with. It evident that a probability of finding a particle has changed from being localised in the beginning to being diffused after some time: this is wave packet spreading.

Note that the range (wave packet size) increases with time monotonically, this means that even if we started with a diffused particle density for general case, it will just become much more diffused by extension of the arguement.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.