# De broglie wavelength and velocity

1. Nov 3, 2008

### cosmogrl

The de broglie wavelength is h/p, where p is momentum, and hence mv, so wavelenght = h/ (mv). The mass is on the denominator, so a massive object, like a baseball, will have a small wavelength, beyond detection and we can ignore it.

OK, so does the mass have to be moving to have a de broglie wavelength, because the denominator is mv. small particle can have 'detectable' wavelengths right? Do they need to be moving. What does the 'v' in the denominator mean?

Do I have a de Broglie wavelenght only when I'm moving?

I guess I'm a bit confused as to what happens with that v. What about really large particles moving really slowly?

2. Nov 4, 2008

### malawi_glenn

The wavelenght ascribed by the deBroigle wavelenght is the periodicity of the particles wavefunction, i.e probability density distribution.

If the particle is at rest, then the particle is localized.. you know where the particle is.

The momentum is not m*v either, it is p*c = (E-mc^2)^(1/2) Where m is the restmass.

Really large bodies are not covered by quantum mechanics.

3. Nov 4, 2008

### newbee

If the particle is at rest then v=0 (and p=0 for the non-relativistic situation) and it's position is spread out - a large wavelength for the wavefunction.

What makes you think QM does not apply to large bodies?

4. Nov 4, 2008

### malawi_glenn

Yes you are correct, I should not write stuff at PF before I have breakfast ;-)

Well it applies to large bodies, but the effects of QM are totally neglectable. i.e the earth moving around the sun has a deBroigle wavelenght of approx: 4E-63 m

In order to have a deBroigle wavelenght of 1m, earth needs to move at 1E-58 m/s

5. Nov 4, 2008

### cosmogrl

So, a particle at rest has a really large de broglie wavelength? Is it detectable?

6. Nov 4, 2008

### malawi_glenn

The particle or its deBroigle wavelength?

How would you measure the deBroigle wavelength of a particle (in general)?

7. Nov 4, 2008

### cosmogrl

I mean is the wavelength something detectable. I'm still not sure what happens when the particle is moving really slowly, or not moving.

8. Nov 4, 2008

### malawi_glenn

have you done quantum mechanics?

9. Nov 4, 2008

### cosmogrl

yes, I have, but its been a while, I can follow the math, but am not really sure what it all means.

10. Nov 4, 2008

### malawi_glenn

Ok then you know what a wavefunction is? The deBroigle wavelenght is related to the wavefunction of a particle, i.e the probability densitity to find the particle at a certain place.

e.g the free particle wavefunction:
http://physics.nmt.edu/~raymond/classes/ph13xbook/node94.html [Broken]

(k is equal to 2pi/lamda)

Last edited by a moderator: May 3, 2017
11. Nov 4, 2008

### cosmogrl

ok, so the de broglie wavelength isn't really a wave associated with the particle? It is just associated with the uncertainty function and the wavefunction in the sense that you can't tell where an object is. Could you measure an objects de broglie wavelenght?

My understanding of all of this was that all objects are waves and particles, and my original question was if the object had to be moving to be a wave, from de broglie wavelength equation.
Am I thinking about this the wrong way? Is the de broglie wavlength not really the wave associated with the particle?

12. Nov 4, 2008

### malawi_glenn

Yes it is the wavlength associated with the particle? The wave nature of the particle is the quantum mechanics, wavefunctions, operators etc.

Do you think the particle is moving up and down in space around an imaginary straight trajectory?

13. Nov 4, 2008

### cosmogrl

No, that's not really what I'm envisioning. I'm thinking more of a wave packet, where we really don't have a point particle, but a wave particle. But, from de broglie equation, the particle would need to be moving to be a wave packet, i.e., have a wavelength. What if it isn't moving? Is it then just a particle?

What about really large objects, like myself. I too have wave properties right? I know they are too small to be detected, but they exist nonetheless. Do I have to be moving to exhibit these wave properties?

14. Nov 4, 2008

### malawi_glenn

Wave packet, of what? what do it consists of? What is a wave particle?

What is a particle accordig to you? A small ball?

For me a particle is the wavefunction. An electron is an electron, it has properties of a classical "ball" particle and properties of a wave such as electromagnetic wave, but it is an electron. I think you must drop these classical analogies which I think you are using.

the wavelenght goes to infinity, but can you really be still? answer no. Also you are a complicated multiparticle system, you are not "one" entity.

15. Nov 5, 2008

### PhilDSP

Remember that there are a number of different interpretations within QM of exactly what particles and waves consist of or represent. Associating the particle to a location in space where the presence of the wave gives with the probability of existence at that location is but one interpretation (though a pretty common one)

De Broglie and Bohm, among others offered an interpretation that doesn't centrally involve the uncertainty principle.

The problem with measuring or even locating a De Broglie wave for a large object is that a large object consists nearly entirely of empty (or nearly empty) space. In De Broglie theory it is the movement of concentrated energy that produces the wave. If there is no substantial energy being transported you will get no substantial "matter wave".

De Broglie and Bohm, for instance, believed that a particle is accompanied by a wave, not at all that they are two aspects of one entity. But that is a minority viewpoint.

And yes, not only does a particle need to be moving to produce wave effects, it needs to be moving at a velocity faster than a certain threshold. In other words, the wavelength goes to infinity at a non-zero group velocity. But this doesn't directly take into account internal movement of energy (electron spin for instance)

Last edited: Nov 5, 2008
16. Nov 5, 2008

### cosmogrl

PhilDSP, thank you, that is exactly what I needed to know. That really clears things up for me.