# Question about Deriving the de Broglie Wavelength

1. Mar 3, 2008

### peter0302

Something has always bothered me about the way the de Broglie relation is derived and I've never seen anyone address this, so I'm hoping someone here can.

As I understand it, the derivation begins with two of Einstein's equations: E=hf, and E=pc. E=hf was experimentally obtained by Einstein. E=pc is derived from special relativity, and is the special case of the more general E^2-(pc)^2=(mc^2)^2, when m=0. In other words E=pc is exactly true for a massless particle but not true for massive particles.

From E=pc and E=hf, we get pc=hc/lambda, and lambda=h/p. This is the de Broglie wavelength.

This is clearly correct for photons, but then de Broglie took it a step further and used p=ymv to get lambda=h/(ymv). However, p=ymv is true only for massive particles, yet earlier we used E=pc, which is true only for massless particles.

Isn't this a logical flaw in the derivaiton? If so, is the fact that it nonetheless leads to the right result sufficient to explain it?

2. Mar 3, 2008

### malawi_glenn

what is the question?

the de Broglie is just lambda = h/p, where p depends on the situation you have ("classical" particle or relativistic).

this was "derived" from the photon momentum and energy, in the same manner as you have demonstrated.

3. Mar 3, 2008

### peter0302

The derivation is logically flawed as the equations used are inapplicable to massive particles. So how can the final equation be right?

Why isn't it:

E=hf
ymc^2 = hf
ymc^2 = hv/lambda
lambda = hv / (ymc^2)

4. Mar 3, 2008

### malawi_glenn

your OP was not orded in a way so I could see what you meant, but now I see your point.

I have not read his PERSONAL derivation, but I believe that his derivation was under the assumption that the particles are massive. He actually called it "matter waves".

But I dont really see your general point, what is your "aim" ?

Of course p=ymv is only valid for massive particles, what is the problem?

What de Broigle did was to extend the momentum/wave lenght relation for photons to massive particles, and that step is very very very trivial. The general derivation is to use p = (E^2-m^2)^½ [I use natural units since Iam subatomic physics nerd hehe]

And it has been experimental verified and he got the Nobel prize :-)

5. Mar 3, 2008

### peter0302

I am not denying that it has been experimentally verified. But the derivation is mixing apples (massive particles) and oranges (massless particles) without regard to which equation is right for which animal.

Again, E=pc is only true for massless particles. p=ymv is only true for massive particles. Yet both equations seem to be used interchangeably.

6. Mar 3, 2008

### malawi_glenn

but as i said, he derived it UNDER THE ASSUMPTION THAT IT WAS MASSIVE PARTICLES.

De Broigle started with the momentum for massive particles, since he was only interessted in them. If you want the general debroigle wavelenght you just use p = (E^2-m^2)^½

I still dont know what you are arguing about? Can you show me a place where they are used interchangeably? Maybe you are missing something else?

de Broglie is (atleast I have never seen it) never used for photons, so thats eprhaps one thing you have missed. But as I said, he derived his relation FOR massive particles, the step towards a general de Broglie is very very very trivial.

( I have now read a bit about his own derivation)

7. Mar 3, 2008

### peter0302

E=pc *true only for massless particles*
E=hf
pc=hf
pc=hc/l
p=h/l
l=h/p
l=h/(ymv) *true only for massive particles*

See the problem?

8. Mar 3, 2008

### malawi_glenn

you are just doing unit analysis. Momentum is always momentum, hence no mixing of apples and oranges. there is no problem.

what YOU are missing is when you go from l=h/p to l=h/(ymv), since p = (E^2-m^2)^½
ALWAYS and you have to impose "i now go to massive particles only", then the derivation is ok. what you do is:

$$E=pc \rightarrow E=hf \rightarrow pc=hf \rightarrow pc=hc/l \rightarrow$$
$$l=h/p \rightarrowl h/(\gamma mv)$$

But $$E/c \nRightarrow p$$,

$$p \Rightarrow \gamma mv$$ is ok, if you state "For massive particles".
and
$$p \Rightarrow E/c$$ is ok if you state "For massles particles".

But there is NO equivalence!
$$E/c \nLeftrightarrow p \nLeftrightarrow \gamma mv$$

The thing De Broglie did was just to replace photon momentum with momentum for massive particles, he just made a guess that "this might be true", and it was. He did never said that there was equivalence between E/c and \gamma mv. He only said "what if matter has wave properties?"... and started this by inserting massive momentum istead. So in that sence, de broigle did not "derived" his forumula, he postulated it.

Last edited: Mar 3, 2008
9. Mar 3, 2008

### George Jones

Staff Emeritus
Well put malawi_glenn; this is how science is done.

Science is not just a set of logical derivations. Science is advanced by partially motivated good guesses and even wild guesses. Theory is tested by experiment; experiment motivates theory.

In this case, and after-the-fact derivation of de Broglie is given by postulating the Schrodinger equation as an equation of motion, just as F = ma is postulated, and then solving the Schrodinger equation for a free particle.

10. Mar 3, 2008

### malawi_glenn

Geroge Jones, you make me:

11. Mar 3, 2008

### peter0302

Great, so we have a guess which leads to a flawed derivation that happens to get the right result, and I'm the only one who's scatching my head.

12. Mar 3, 2008

### malawi_glenn

there is not a flawed derivation! $$E/c \nLeftrightarrow p \nLeftrightarrow \gamma mv$$
there is not even a derivation!

What de broigle did was NOT to say:
$$E/c \Leftrightarrow p \Leftrightarrow \gamma mv$$

he said that matter have wave properties, just as photons, and you get the forumula by replacing momentum for pure massless particles by momentum for massive particles. Hence it was not a derivation in the formal logical sence, it was just a wild guess.

What he did "prove", is that you can do this:
$$E/c \Leftarrow p \Rightarrow \gamma mv$$
Depending on what particle you look at.

Last edited: Mar 3, 2008
13. Mar 3, 2008

### peter0302

That is my point.

14. Mar 3, 2008

### Hans de Vries

The de Broglie wave length is caused by non-simultaneity and can be derived
from only E=hf and special relativity only.

The wave-front of a particle can be to the North, the West or to any direction
depending on the reference frame. The wavefront is always in the direction of
the speed, while the speed depends on the reference frame. For a particle moving
at centimeters per second this means the de Broglie wave rotates just by going
to a slightly different reference frame.

de Broglie waves from the Klein Gordon equation --- (see 8.2, 8.3 and 8.5)

Regards, Hans

Last edited: Mar 3, 2008
15. Mar 3, 2008

### Staff: Mentor

I would call it an "educated guess" based on making an analogy between massive particles and photons by way of their momentum. Schrödinger derived his famous equation similarly, by making an analogy between particle motion and optics: quantum mechanics is to classical mechanics as wave optics is to geometrical optics.

For a "wild guess" consider Kekule's inspiration for the molecular structure of benzene, via a dream about snakes swallowing their tails.

16. Mar 3, 2008

### malawi_glenn

Then everything is ok, since his guess did not lead to:
$$E/c \Leftrightarrow p \Leftrightarrow \gamma mv$$
which is totaly wrong!

hiwever, it does lead to:
$$E/c \Leftarrow p \Rightarrow \gamma mv$$
which is ok

So what de Broigle though was that lamda = h / p is the GENERAL formula, then you can use any momentum relation you want, depending on what situation you have.

17. Mar 3, 2008

### peter0302

Ok, so the *postulate* is
lambda = h / p.

The derivatio is not (as I thought previously)
E = hc / lambda,
E = pc
Therefore
pc = hc / lambda
lambda = h / p

As long as lambda = h / p (for massive particles) is a postulate I am somewhat relieved.

18. Mar 3, 2008

### malawi_glenn

well yes, you can think of it like he postulated that lambda = h / p is a general formula, and that you may use lambda = h /( \gamma * m *v) for massive particles.

19. Mar 3, 2008

### andyferris

This is my understanding of de Broglie's argument. I have not it read it myself but this is my interpretation from what I have read and may not be faithful to his approach.

He did not "guess" the result from the photon equivalent. Basically, he started with a question - if light can behave like a particle, can particles behave like a wave? What would such a wave look like?

Well, de Broglie believed in special relativity, so he knew that things you could observe, such as interference fringes, should look the same in any (non-accelerating) reference frame. We need some sort of wave that can interfere, so we introduce a wavefunction. Let's take a plane wave with wave-vector $$\vec{k}$$ and angular frequency $$\omega$$:

$$\psi = exp( i (\vec{k}.\vec{r} - \omega t) ),$$

where $$\vec{r}$$ = (x,y,z) is the position

Of course the wave-vector $$\vec{k}$$ and angular frequency $$\omega$$ will depend on the state of a particle, such as energy and momentum. But the value of $$\vec{k}.\vec{r} - \omega t$$ is the phase, measurable by interference experiments, and must be frame-independent (at least to a global factor). But note that this is essentially a product of two four-vectors:

$$\vec{k}.\vec{r} - \omega t = (k_x,k_y,k_z,\omega) . (x,y,z,t)$$

It is a fact of special relativity that the four-vector $$(k_x,k_y,k_z,\omega)$$ must therefore be a covariant four-vector.

I think at this stage it was being accepted, via the photo-electric effect and black-body radiation, that $$E = hf = \hbar \omega$$ was true for photons - or, alternatively for both these examples, one could have normal classical light interacting with matter that had quantised excitations governed by that same formula (even photon's were controversial in those days). Either way, the evidence pointed towards the fact that this formula should be true of de Broglie's matter waves.

In that case the only covariant four-vector that we could have is proportional to the four-momentum:

$$\vec{k} = (p_x,p_y,p_z,E) / \hbar$$

which gives the famous wavelength-momentum relationship.

20. Mar 4, 2008

### peter0302

I guess what just makes me uncomfortable is that that formula is so obviously correct for photons, based on a simple derivation from E=hf and E=pc, that making the sudden leap to say that the exact same formula is correct for MASSIVE particles as well (where we know E=pc is NOT correct for massive particles) is startling.

Like I said before, if I was going to start from scratch, I'd say:
E=hc/lambda
ymc^2=hc/lambda
lambda=h/(ymc)

Yet that is obviously not correct.

After thinking about it further, I've been thinking that the momentum-energy relation:

E^2=(pc)^2 + (mc^2)^2

looks an awful lot like a vector with a momentum-energy component (pc) and a rest-energy component (mc^2), whose magnitude is E. If we assume that the de Broglie wavelength depends solely on the momentum component of the "energy vector" and not at all on the rest-energy component, then the derivation:

E_momentum=hc/lambda
pc=hc/lambda
lambda=h/p

Comes easily.

Any benefits to this interpretation?

Last edited: Mar 4, 2008
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