# Why is the de Broglie wavelength sometimes larger than the particle it describes?

1. Jun 28, 2011

### jaketodd

Why is the de Broglie wavelength sometimes larger than the particle it describes?

Thanks,

Jake

2. Jun 28, 2011

### SpectraCat

Because that's what QM (the most successful physical theory to date) predicts, and those predictions agree with experiments.

Also for some particles (i.e. electrons & neutrinos), there is no known "size" that can be associated with the particles themselves, so if the de Broglie wavelength is measurable, then it is automatically "larger" than the particles.

3. Jun 28, 2011

### Lapidus

If you have cold (i.e. little momentum) big molecules/ atoms, than for example Broglie wave length will exceed the size of the particle.

4. Jun 28, 2011

### jaketodd

Should the particle size be the de Broglie wavelength, or does only part of the wavelength exist at a time (what can fit inside the particle's diameter)?

5. Jun 28, 2011

### Bill_K

Jaketodd, There were two points made in the responses above that bear repeating. One, that the de Broglie wavelength h/p is not related in any way to the size of the particle. Only to its momentum, and is a measure of how localized its wavefunction is. Two, that all elementary particles including electrons, quarks, etc are, to our best knowledge, point particles.

6. Jun 28, 2011

### jaketodd

So is a particle's wavefunction size/localization the size of the de Broglie wavelength?

Thanks,

Jake

7. Jul 1, 2011

### Jacques_L

Please let us know what a "size of a particle" could be.
Radiocrystallographists use the width of X diffraction rays to evaluate the size of the diffracting crystallites. So clays give broad diffration peaks on a Debye-Scherrer diffractograms, though silts give fine peaks. I could confuse in justice an international crook by such facts. To give precise spots or peaks, the Bragg law demands large crystallites, and monochromatic waves, so with long and broad spindle of each incident quanton, photon or neutron or electron.

For instance, we obtained broad spots by diffracting electrons on a carbide inclusion in a Laue diffractogram, in a Siemens electronic microscope. The monochromaticicty of the electrons was not so perfect, and maybe the carbide inclusion was not a so perfect crystal.

When you write or say "size of a particle", you surrepticiously mean that it is or can be a corpuscle. But where are the experiments that could support such a postulate ?

A little more has to be known : the broglian period, frequency and when moving the broglian wavelength, suffice when an electron interferes with itself - in an Aharanov-Bohm experiment, for instance.
But when an electron interferes whith electromagnetic fields, a photon for instance in a Compton scattering, then the intervening period is the Dirac-Schrödinger, $\frac{h}{2mc^2}$, half of the broglian one. So is the wavelength, too.

Last edited: Jul 1, 2011
8. Jul 1, 2011

### SpectraCat

This is the second time you have mentioned "Compton diffusion" ... the context seems to suggest you mean "Compton scattering" .. is that correct? I think in French, the word for scattering might be "diffusion", however in English, diffusion has a different meaning in the context of physics.

9. Jul 1, 2011

### Jacques_L

Thank you for correcting my english. You are right.

10. Jul 1, 2011

### SpectraCat

No problem .. I didn't think there was any such thing as "Compton diffusion", but I wasn't completely sure, so I figured I check.

Also, I should have done this before: Welcome to PF!

11. Jul 1, 2011

### Jacques_L

Thanks for the welcome.
It is a pleasure to exchange in a so cooperative spirit.

I came here because of a cross reference to the thread initiated by Ruth Kastner.
I am among those who rediscovered independently the TIQM, but I came far later than John G. Cramer. Anyway I was the first to give a first approximation of the diameter of Fermat's spindle between emitter and absorber. An approximation which could be much more rigourous.

I think in french, then translate, and the results may be so-so... My english has rusted, in the meantime.