# Lorentz Transformation of the De Broglie Relation

1. May 18, 2005

### Tom Mattson

Staff Emeritus
Here is a little puzzle that I'm sure I should know the answer to, but my brain is failing me.

Consider a particle with moving with speed $u$ along the x-axis in some frame S. So its (relativistic) momentum is $p_x=\gamma_umu$. Its DeBroglie wavelength is $\lambda=h/p$.

Now consider the same particle in a frame S' which moves with speed $v$ relative to S in the x-direction. One might expect (naively perhaps) that the DeBroglie wavelength would be length-contracted according to:

$$\lambda '=\frac{\lambda}{\gamma_v}$$.

Now note that the momentum of the particle in S' is $p_x'=\gamma_v(p_x-vE)=\gamma_v\gamma_um(u-v)$. If that transformed momentum is substituted into the DeBroglie relation, we get:

$$\lambda '=\frac{h}{\gamma_v\gamma_um(u-v)}$$.

Clearly there is a conflict. I'm inclined to say that the second formula is correct because it gives an infinite result if S' is boosted to the rest frame of the particle, which is what I would expect.

So the question is, which part of the above reasoning is wrong?

Last edited: May 18, 2005
2. May 18, 2005

### marlon

I think you need to apply length contraction TWO times. In frame S, the particle has relativistic velocity u. Here you need to apply length contraction for the first time. Then, when going to frame S' with relative speed v wtr to frame S, you apply length contraction for the second time. i did not calculate this, but what do you say ?

marlon

3. May 18, 2005

### seratend

I think it may be somewhat more complicated. When you change your reference frame, you need to apply a lorentz transformation to the operator momentum.
In that case, I am not sure that the new operator commutes with the previous one (like rotations and spin), therefore having a mometum eigen state in one reference frame does not mean that you have a momentum eigen state in the new frame.

Hope this helps.

Seratend.

4. May 18, 2005

### Tom Mattson

Staff Emeritus
I don't think that's necessary at all. The particle has a speed $u$, so its momentum is $p_x=\gamma_u mu$. The observer S simply assigns to him a DeBroglie wavelength equal to $\lambda =h/p_x$.

5. May 18, 2005

### Tom Mattson

Staff Emeritus
You're definitely right about that. Momentum in one frame looks like momentum and energy in another frame. But that doesn't explain precisely why one of the two approaches above is wrong. I'm 99% certain that the first approach is wrong (the one that uses simple length contraction) but I can't see why. I mean if you plot a wave on a graph and mark off the endpoints of one cycle, and then let the plot zip past you at 0.9c, then for sure those endpoints are going to be a distance apart that is given by the length contraction formula. My question is, what gives?

6. May 18, 2005

### vanesch

Staff Emeritus
I think that the problem in the first approach is that you consider Re( <x|psi>) to be a scalar (crests and throughs of the real part of the wave function at definite positions in space, so that you can apply length contraction). It doesn't need to be.

cheers,
Patrick.

7. May 18, 2005

### seratend

I have not taken enought time to check that point, but I am almost sure that you may apply the relativistic doppler formulas (longitudinal doppler effect) to the compute the new wavelength.
To recover the relativistic doppler formula, you have to take into account the good events (space, time). (f'= gamma.(1-beta).f for the longitudinal doppler effect and f'=gamma.(1-beta.cos theta).f for other directions)

Seratend.

Last edited: May 18, 2005
8. May 18, 2005

### dextercioby

I'm sorry,Tom,but this is typically HS problem in SR...

In S:the massive particle has the speed "u",relativistic momentum $$p_{x}^{S}=m\gamma (u) u$$ and de Broglie wavelength $$\lambda^{S}=\frac{h}{p_{x}^{S}}$$

In S':the massive particle has the speed "u' ",relativistic momentum $$p_{x}^{S'}=m\gamma (u') u'$$ and the de Broglie wavelength $$\lambda^{S'}=\frac{h}{p_{x}^{S'}}$$

,where,as any HS student should know

$$u'=\frac{u-v}{1-\frac{uv}{c^{2}}}$$

Daniel.

P.S.What's this doing in QM forum...?:surprised:

9. May 18, 2005

### Tom Mattson

Staff Emeritus
You can't just answer a question without taking a dig at people, can you?

No kidding. I guess you didn't notice that that is exactly equivalent to the second of the two approaches I used. But your answer doesn't explain what is wrong with the first approach, which is what I'm asking.

Well let's see, in what course does one learn about the DeBroglie wavelength....?

10. May 18, 2005

### dextercioby

Why...?Your formula is wrong.It doesn't have even have to do with Doppler shifting (as someone wrongly suggested),because no radiation is emitted,nor absorbed.The de Broglie wavelength is an intrinsic property of every material particle (just like spin,mass and charge).Incidentally,for the photon it coincides with the classical radiation wavelength,but the photon is massless (so no Lorentz boosting from rest frame to moving ref.frame).

So,the way the de Broglie wavelength behaves under a change of inertial obsevers in SR is given by the way the momentum of the particle behaves under the same change of inertial observers and that can be seen from the very definition.

I hope u realize there's no contraction,nor dilation.U can's say the de Broglie wave contracts or dilates,right...?

Daniel.

11. May 18, 2005

### Staff: Mentor

Tom,
I think the problem is that you are treating the de Broglie wavelength measured by S as if it were an object at rest in S, where it is really only the measured length of that wavelength in S. If S measures a stick moving at speed $u$ to have a length of L, then frame S' (moving as speed $v$ with respect to S) will certainly not measure the length of that same stick as $L/\gamma_v$.

12. May 18, 2005

### Tom Mattson

Staff Emeritus
Yes, I know it's wrong. So far you haven't said anything that explains why it's wrong.

That can't be right. If the DeBroglie wavelength is intrinsic to the particle, then it would depend only on particle species, which it clearly does not. I think you are confusing the DeBroglie wavelength with the Compton wavelength, which really does only depend on particle species.

13. May 18, 2005

### marlon

That is completely correct and this is what i meant by not applying the length contraction once. I mean in a frame that moves along with the object, the deBroglie wavelength is lambda, then you will get a length contracted lambda' in frame S. Then you will also get a lambda" when looking from frame S'...these are two different situations... i mean that you need to watch what the eigenlength is and this length is defined when the object is standing still wtr the frame

marlon

Last edited: May 18, 2005
14. May 18, 2005

### marlon

.

This lambda in the RHS corresponds to an object that is NOT standing still wtr to the frame S, hence this lambda cannot be seen as an 'eigen length' (the length in frame S) and thus, the formula for length contraction does not work.

marlon

15. May 18, 2005

### Tom Mattson

Staff Emeritus
I think you're onto the answer here. I'm playing around with it and I think I'm almost there. It looks like the simple, naive length contraction really should have been a composition of Lorentz boosts. It looks like marlon was on the right track too.

Thanks,

Tom

16. May 18, 2005

### Tom Mattson

Staff Emeritus
I got that now, thanks. I didn't understand you properly the first time because I was wondering, "why the hell would S have to apply length contraction to find $\lambda$ when he knows the momentum in his own frame??"

17. May 18, 2005

### Staff: Mentor

Right. Happens to the best of us.
Yep, but I have to admit that I didn't understand marlon's point at first either. I now see that we were saying the same thing.

18. May 18, 2005

### vanesch

Staff Emeritus
Yes, but even in that case, things don't work out. The "eigenlength" of the de Broglie wavelength is infinity ! (the de Broglie wavelength of a particle in its own rest frame) So if you could Lorentz transform it as a length, it should turn out to be infinity in any frame, finite length contraction or not, once or twice applied.

The reason is simply that the de Broglie wavelength is a length property of an object that is not a scalar under Lorentz transformations (nor even under Galilean transformations). Psi doesn't transform as a scalar field under (relativistic or non-relativistic) boosts! In the rest frame, it is a constant (infinite de Broglie wavelength). So then, if it were a scalar (under Galilean, or Lorentzian transformations) it would be the same constant, no ?

cheers,
Patrick.

19. May 18, 2005

### marlon

To TOM : i reread my first answer and indeed i should have pointed out more clearly what i meant.

To Vanesch, i agree that psi does not transform as a scalar under ...

But i don't get the second part though. Are you saying that psi is a constant ? I really think i am missing something, so that is why i would like to ask if you would take the effort to elaborate a bit...

thanks

marlon

20. May 18, 2005

### Tom Mattson

Staff Emeritus
Yeah, the composition Lorentz boosts didn't work. It's clear that I have to do this quantum mechanically.

$\psi$ is a constant in the rest frame of the particle because its momentum is known with certainty, so its position is totally uncertain.