# De Broglie Wavelength - How does this hold for slow objects?

1. Sep 7, 2009

### sam_p_r

OK so you get to the matter wave equation 'lambda = h / p' using E=cp - which describes the energy for massless particles. I can understand this holding for when cp>>mc^2 , but not for when the mc^2 is comparible. Any help?

2. Sep 7, 2009

### javierR

Well the relations between energy/momentum and frequency/wavelength hold generally:
$$E=h\nu$$, $$p=h/\lambda$$. It's the relation between E and p which is relevant in going from non-relativistic to relativistic to ultra-relativistic limits (as determined by the ratio $$pc/mc^{2}$$). You could always use $$E^{2}=(pc)^{2}+(mc^{2})^{2}$$. We ignore the second term in the ultra-relativistic limit, while in the non-relativistic limit, it simplifies to $$E=p^{2}/2m$$. Otherwise, use the full expression.

Last edited: Sep 7, 2009
3. Sep 7, 2009

### alexepascual

Could you elaborate a little more about the ultra-relativistic case?

4. Sep 7, 2009

### alexepascual

What specific problem do you see?.

5. Sep 7, 2009

### sam_p_r

That E does not equal cp, it equals the square root of (cp)^2 + (mc^2)^2. So for de broglie equation, you have to neglect mc^2, which I can see as reasonable if it is very small in comparison to cp, but there are many occasions where it wont be.

de broglie combines E=hc/lambda and E=cp,

So if you dont have E=cp you cant get the de broglie relation as it is.

6. Sep 7, 2009

### sam_p_r

So what you're saying is that it only applies (in the exact lambda=h/p) for relativistic particles? (as its the only time where you can neglect mc^2).

Because in many examples in textbooks it applies the formula to non-relativistic conditions.

7. Sep 7, 2009

### alexepascual

I think it applies to both relativistic and non-relativistic conditions. The m that you list is actually m0, the rest mass. P should be the relativistic momentum and E is the relativistic energy. These both have a correction due to speed of the particle (or refference frame).
In the limit where v=c this correction disappears.
I don't think the rest mass energy can be neglected in either the relativisti or non-relativistic case. But I may be wrong.
I think the de Broglie waves have been superceeded by the QM wave funtion, but I haven't looked at the difference. I know that the Schrodinger equation is not Lorentz-invariant and that the Dirac equation is. I would assume that there is no problem with the wave function itself. It would be nice to compare the de Broglie wave with the wave function. This must be done in many books but I can't find it in the ones I have. Maybe I can find something on the Web.

8. Sep 8, 2009

### alexepascual

It appears that there is no difference between de Broglie's waves and the wave function. I looks like de Broglie calculated things like phase velocity and wave length but he didn't come up with an expression for the wave and that Scrodinger did that. So, it would appear that de Broglie's waves and the wave function refer to the same waves.
Now, reading again about de Broglie's waves, these are faster than light. I don't remember any mention in QM about the wave funtion of a free particle being faster than light. Maybe I just didn't pay attention to that.
Maybe that's where the energy corresponding to the rest mass comes into play!

If anybody can make this clear for us I'll appreciate it.