2
$\begingroup$

In his excellent book "Atomic Physics", after showing that the velocity of a group of waves equals a particle's velocity, Max Born writes that to interpret a particle of matter as a wave packet due to the superposition of a number of wave trains "comes up against insurmountable difficulties, since a wave packet of this kind is in general very soon dissipated".

Is he right?

$\endgroup$
4
  • $\begingroup$ Yes, that would be another false interpretation of quantum mechanics. Does he explain that quanta are not particles, though? If not, the book might be good as a doorstop. $\endgroup$
    – CuriousOne
    Jun 7, 2016 at 20:30
  • $\begingroup$ Well, I am yet to see some calculation involving a realistic wavepacket and not a plane wave solution. Do you know any example? The only thing that crosses my mind is a case of a coherent state when the observables distributions match the classical gaussian. Is this just a computational difficulty? Why indeed we are able to use plane waves as a substitute? $\endgroup$ Jun 7, 2016 at 20:47
  • $\begingroup$ @CuriousOne well, they are still "peddling" this idea, that say the electron would smear out in next to no time, in my less than 10 year old Q..M. Books, although I do take your point. But I don't know when Born wrote the book, or the pre QFT audience he may have been aiming it at. But if its wrong, its wrong. $\endgroup$
    – user108787
    Jun 7, 2016 at 23:24
  • $\begingroup$ @count_to_10: The electron quantum's position does "smear out" without an effective potential holding it in place, that's exactly what happens. I don't think that was a problem for Born in the early days, either (the first edition is 1935, so it is a bit long in the tooth). They were probably more at a miss to "grok" how something that effectively smears out also somehow "holds together". Classically one can't explain that, and giving up on classical mechanics completely (except as a corner case of QM) is a tough thing to do, even today. $\endgroup$
    – CuriousOne
    Jun 7, 2016 at 23:36

2 Answers 2

2
$\begingroup$

Solving the Schrödinger equation for a particle gives that wavefunction that describes it. However it's generally the case the partial differential equations that there is an infinite set of solutions, so we can't just solve the Schrödinger equation and get the equation that describes the particle - the solution gives us many equations that could describe the particle.

To find out what equation describes some specific particle we need to use the original conditions. For example we could take the form of the wavefunction $\psi(x,t)$ at time $t=0$. This what we do with a gaussian wavepacket. We take the form of the wavefunction at $t=0$ to be something like:

$$ \psi(x,0) = Ae^{-x^2/a^2} $$

Then we combine this initial condition with the solutions of the Schrödinger equation to give the full equation $\psi(x,t)$.

If you do this for a gaussian wavepacket you find that as the wavepacket evolves in time it gets broader. In effect the particle spreads out. This is an example of the uncertainty principle. The gaussian wavepacket localises the particle in space to $\Delta x \approx a$, and as a result there is an uncertainty in the momentum of around $\Delta p \approx \hbar/a$. The spread of momentum means that in effect different parts of the wavepacket are moving at different velocities, and that's why the wavepacket spreads out with time.

I'm not sure I would approve of Born's use of the word dissipated because that implies that the particle is somehow disappearing, and that isn't the case. The wavefunction gives us the probability of finding the particle, and what happens is that initially there is a high probability of finding the particle at a position $-a \lt x \lt a$. However as time goes on the region in which the probability of finding the particle is high gets bigger. In the limit of $t\rightarrow\infty$ the particle becomes completely delocalised and there is the same probability of finding the particle anywhere.

However the point of Born's statement is that all the above only makes sense in the Born interpretation i.e. that $\psi^*(x)\psi(x)dx$ gives you the probability of finding the particle in the small region between $x$ and $x+dx$. So the initial gaussian wavepacket is just an initial probability distribution for the particle. It doesn't make sense to say the wavepacket is the particle, only that the wavepacket is the description of where the particle might be found.

$\endgroup$
1
$\begingroup$

To answer this question in the spirit that Born wrote it, way back when, yes he was correct in saying that a superposition of waves with a Gaussian term included in the integral to create a wave packet would rapidly disperse.

The velocities of the individual wave packets are in general unequal when a wave's velocity depends on wavelength. Such a wave is called a dispersive wave, because a wave pulse consisting of a superposition of waves of different wavelengths will separate (disperse) into its separate wavelengths as the waves move through space at different speeds.

As far I remember, it was Schrödinger's initial impression that he could interpret the wave packet AS the electron itself, but this this did not withstand the scrutiny of others for very long, and Born introduced tbe probability wave interpretation.

As CuriousOne says, this is by no means the modern view, and if we could snip these bits of QM history out of the teaching process, and go straight to the field idea of QFT, in my view that would save a lot of time in learning material that is outdated.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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