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There is an exercise in the Griffiths book on QM, where a wave function:

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

is used to calculate $\psi (x,t)$, after getting $f(k)$, by Fourier analysis.

$\rho = \psi^2 (x,t)$ is a symmetric bell shaped curve, centred in zero that spreads with time.

My question is about the physical meaning of a kind of wave travelling in both directions (x- and x+) as $t$ increases.spreading wave As the probability density spreads, the curve $\rho\,$ x $t\;$ (for a given $x$ constant) has a local maximum for some $t$.

That is logical because if the probability density decreases near zero, it must increase somewhere.

The expected value of the position $\langle\psi|x|\psi\rangle$ is always zero, meaning that the particle is at rest.

But the velocity of that "wave" (except for short times) is exactly what should be expected of a particle:

$x = \sqrt{a}\frac{\hbar}{m} t$

because $\sqrt{a}\hbar$ has units of momentum.

Claudio Saspinski
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    Expected value of position being $0$ does not mean the particle is at rest. – BioPhysicist Jan 06 '20 at 01:19
  • I'm afraid I don't really understand what your question here is. Yes, a quantum particle with constant expected position does not necessarily behave like a classical particle at rest. What exactly do you want to know about that? – ACuriousMind Jan 06 '20 at 01:23
  • @ACuriousMind I know that it spreads with time. But never noted that the spreading velocity is p/m as a particle velocity. Maybe there is some physical meaning for it, or maybe it is just a coincidence? – Claudio Saspinski Jan 06 '20 at 01:32
  • If you like this question you may also enjoy reading this Phys.SE post. – Qmechanic Jan 12 '20 at 08:05

2 Answers2

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I'm not sure I fully understand the question, but I'll try to explain the physical meaning a bit.

The parameter $a$ tells us about how "localized" the wavefunction is in the initial configuration. For large $a$, the particle has a large probability of being observed near the origin (at early times). For small $a$, the particle is more likely to be observed far from the origin.

One way to think of this is that the uncertainty on the particle's position is small, so the uncertainty on its velocity must be large. For large $a$, the wavefunction rapidly spreads out, because the particle's velocity is more likely to be large (compared to the small $a$ case). This interpretation fits with the mode of $x$ you posted: $$x(t) = \frac{\sqrt{a} \hbar t}{m} . $$

For large $a$, the particle is more likely to be observed farther from the origin, than if $a$ were small.

Paul T.
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I would like to add some words.

  1. You should not mix concepts of particle and wave.
  2. In order to understand behavior of a matter wave, you'd better stick with the concept of wave, rather than the particle picture.
  3. The Gaussian wave that you gave the expression spreads out and it can be observed far away from $x=0$ after a long time with certain velocity, which is nonzero.
  4. As Aaron Stevens pointed out in the comment, expected value does not guarantee whether the wave is moving or not.
  5. In order to make your Gaussian wave packet at rest, not spreading out at the same time, you need to trap the wave in a potential with a proper depth or slope; for example, a harmonic oscillator.
  6. Without the potential, your Gaussian matter wave will spread out indefinitely, until it is collapsed to be found at certain place through observation (entanglement with environment or measurement apparatus, etc..), because it has certain probability density to have non-zero momentum always.
  7. In order to have a free natter wave at rest, you need to define indefinitely large wave packet, because the matter wave need to have only one momentum, which is identically zero.
  8. One of the inherent reason why the matter wave spread out in space is that its dispersion relation is nonlinear, so that higher momentum part of the wave spread out faster than lower momentum part (the group velocity of the matter wave packet is different from the phase velocity of the matter wave for each different momentum, in general.). Actually this is the only reason that your Gaussian wave packet spreads out in space, given that your wave is for matter, not an electromagnetic wave in vacuum.
  9. An example of a wave that does not spread out but moving together in space is an electromagnetic field, because it has a linear dispersion relation in vacuum and the group velocity of the electromagnetic field in vacuum is the same as the phase velocity of the field in vacuum.
  10. If you define chirp-free wave packet in vacuum (like the one you gave) at certain time, you will get inherent chirp due to the nonlinear dispersion relation in the case of matter over time, and you will not get any chirp in the case of electromagnetic wave.
Above Threshold
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  • Saying that the particle is at rest (as I did) seems too strong. My intention was only to point out that there is a travelling probability wave associated to a particle with a constant expected value. – Claudio Saspinski Jan 16 '20 at 01:00