# The wave packet description

1. Nov 5, 2006

### flowerew

"The particle and the wave picture are both simplified forms of the wave packet description, a localized wave consisting of a combination of plane waves with different wavelength."

it's confusing. Can somebody explain it? particle is an object, it seems wave is also an object.How to combine the two thing together?

2. Nov 5, 2006

### masudr

flowerew: How to combine the two things together?

The quantum state vector combines wave and particle properties; one representation (in particular, the position representation) of the state vector is the wavefunction.

3. Nov 6, 2006

### FunkyDwarf

i dont think thats what hes asking. your (masudr) are refering to the wavefunction ie shrodingers equation or wavefunction for position and probility whereas i think flowerew is asking how perpendicular EM and electrical fields can be construted as a particle, which is a good question.

4. Nov 7, 2006

### XVX

I think of it as a localized wave.

Imagine the waves in a pool combining in such a way that there is a big bump in the middle of the pool that smooths out to the boundaries of the pool. A bump, that is a localized standing wave.

Pool waves never do this, because the wavenumbers are pretty much random. But for particles, there is an average wavelength, a Gaussian distribution (for a free particle) that the wavenumbers mostly hang around. Because of this, when you add together an infinite set of waves then, you get a localized standing wave "envelope" that is the particle.

Its comprised of waves, so it has wave properties, but the waves form an overall localized wave that gives it its particle properties then.

As far as EM fields and quantization, you'll need to read some Quantum Field Theory and search some "second quantization" stuff. In short, one can quantitize any normalizable field, which then becomes a QFT. The EM field is normalizable, which is why QED is so successful. But the Einstein Field Equations are not normalizable and hence the trouble.

5. Nov 7, 2006

### lightarrow

XVX,
Many people on this forum agrees on the fact that the wavefunction is just a mathematical representation and not a "real" wave.

6. Nov 7, 2006

### actionintegral

How dare you even speak such words! You obviously are not smart enough to comprehend quantum mechanics. You should be placed under house arrest and go back to studying inclined planes.

Last edited: Nov 7, 2006
7. Nov 7, 2006

### quetzalcoatl9

there are no particles, there are only waves

8. Nov 7, 2006

### lightarrow

How would you prove it?

9. Nov 7, 2006

### vanesch

Staff Emeritus
The wavefunction of 2 particles already doesn't "live" in ordinary space, but in a 6-dimensional configuration space...

10. Nov 7, 2006

### masudr

The question is asked the wrong way round. It's not that photons (or any excitations of a fundamental quantum field) are constructed out of EM fields. Instead, it's that EM fields are constructed from many zillions of photons.

11. Nov 7, 2006

### quetzalcoatl9

where in the schrodinger equation is there anything but a wavefunction?

what we call a "particle" is just the localized wavepacket in the classical limit

12. Nov 7, 2006

### Staff: Mentor

I prefer to say that there are no (pure) particles, nor (pure) waves. Quantum-mechanical objects have both particle-like and wave-like aspects.

13. Nov 7, 2006

### ueit

AFAIK you need to assume point particles in order to write down the Schrodinger's equation. So, it seems strange to deny the existence of those point particles, isn't it?

14. Nov 7, 2006

### masudr

No you don't.

You need to assume a classical system that possesses a Hamiltonian (or equivalently a Lagrangian).

15. Nov 8, 2006

### ueit

For a hydrogen atom we assume the electron and nucleus to be point charges (we know this is not true for the nucleus and there are deviations for the calculated energy because of that) so the force between them is given by the Coulomb law.

The squared amplitude of the wave function gives the probability of detecting those point charges in a certain place. So, we have particles and their motion is statistically described by a wave.

The motion of water molecules can be described by a wave but this doesn't mean that a water molecule is a wave.

16. Nov 8, 2006

### masudr

Well, technically speaking, the Hydrogen atom is treated as a system with Hamiltonian (I'll write it in Cartesians and their conjugate momenta since they are the usual canonical variables in the electrostatic case, although it'd be more concise in spherical polar co-ordinates):

$$\hat{H}( \hat{x}_1, \hat{p_x}_1, \hat{y}_1, \hat{p_y}_1, \hat{z}_1, \hat{p_z}_1, \hat{x}_2, \hat{p_x}_2, \hat{y}_2, \hat{p_y}_2, \hat{z}_2, \hat{p_z}_2) =$$

$$\frac{\hat{p_x}_1^2+ \hat{p_y}_1^2+ \hat{p_z}_1^2} {2m_1}+ \frac{\hat{p_x}_2^2+ \hat{p_y}_2^2+ \hat{p_z}_2^2} {2m_2}- \frac{1}{4\pi\epsilon_0} \frac{e^2} {\sqrt{ (\hat{x}_1 - \hat{x}_2)^2+ (\hat{y}_1 - \hat{y}_2)^2+ (\hat{z}_1 - \hat{z}_2)^2}}$$

You haven't assumed anything of the system apart from the fact that it has the above Hamiltonian. To get to that expression, you might use some classical physics of point particles and the Coulombic interaction between charged particles. However, note that the quantum mechanics of the system starts here, at the Hamiltonian.

You haven't assumed anything more, specifically, you haven't talked about particles - you've talked about systems. I'm not being pedantic: in QFT, you don't deal with particles directly, you deal with oscillating fields, where particles happen to emerge.

P.S. You usually eliminate 6 of the variables by transforming to so-called "centre-of-mass" and "relative" co-ordinates, and while this has physical content in this case, it didn't have to.

Last edited: Nov 8, 2006
17. Nov 8, 2006

### FunkyDwarf

particle physics would disagree

18. Nov 9, 2006

### quetzalcoatl9

really? i dont think so

tell me what happens to your particles when their average seperation is on the order of the de broglie thermal wavelength? are they still "particles"?

19. Nov 9, 2006

### ueit

Then please tell me what is the meaning of x1, px1 and m1? What is their physical significance?

Is there any other way?

Indeed.

I have no training in QFT so I cannot contradict you here, but my understanding was that QFT treats fields as particles and not the other way around.

20. Nov 9, 2006

### rlduncan

The SE can't be solved until you specify the potential energy function, U. For the H-atom, U(r) = -e^2/r in which you assume point charges. Is this not a contradiction to current quantum interpretations?

Last edited: Nov 9, 2006