Question about Deriving the de Broglie Wavelength

In summary, the de Broglie relation is derived by combining Einstein's equations E=hf and E=pc. However, this derivation is logically flawed as it mixes equations that are only true for either massless or massive particles. Despite this, the final equation of lambda=h/p still leads to the correct result and has been experimentally verified.
  • #1
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?
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  • #2
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
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:

ymc^2 = hf
ymc^2 = hv/lambda
lambda = hv / (ymc^2)
  • #4
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 don't 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 length 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
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

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 don't 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 that's 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
E=pc *true only for massless particles*
l=h/(ymv) *true only for massive particles*

See the problem?
  • #8
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:

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

But [tex] E/c \nRightarrow p [/tex],

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

But there is NO equivalence!
[tex] E/c \nLeftrightarrow p \nLeftrightarrow \gamma mv [/tex]

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.
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  • #9
malawi_glenn said:
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.

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
Geroge Jones, you make me: :blushing:
  • #11
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
there is not a flawed derivation! [tex] E/c \nLeftrightarrow p \nLeftrightarrow \gamma mv [/tex]
there is not even a derivation!

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

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:
[tex] E/c \Leftarrow p \Rightarrow \gamma mv [/tex]
Depending on what particle you look at.
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  • #13
it was just a wild guess.
That is my point.
  • #14
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." [Broken] --- (see 8.2, 8.3 and 8.5)Regards, Hans
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  • #15
malawi_glenn said:
Hence it was not a derivation in the formal logical sence, it was just a wild guess.

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. :bugeye:
  • #16
peter0302 said:
That is my point.

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

hiwever, it does lead to:
[tex] E/c \Leftarrow p \Rightarrow \gamma mv [/tex]
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
Ok, so the *postulate* is
lambda = h / p.

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

As long as lambda = h / p (for massive particles) is a postulate I am somewhat relieved.
  • #18
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
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 [tex] \vec{k} [/tex] and angular frequency [tex] \omega [/tex]:

[tex]\psi = exp( i (\vec{k}.\vec{r} - \omega t) ), [/tex]

where [tex]\vec{r}[/tex] = (x,y,z) is the position

Of course the wave-vector [tex]\vec{k}[/tex] and angular frequency [tex]\omega[/tex] will depend on the state of a particle, such as energy and momentum. But the value of [tex]\vec{k}.\vec{r} - \omega t[/tex] 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:

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

It is a fact of special relativity that the four-vector [tex] (k_x,k_y,k_z,\omega) [/tex] 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 [tex] E = hf = \hbar \omega [/tex] 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:

[tex] \vec{k} = (p_x,p_y,p_z,E) / \hbar [/tex]

which gives the famous wavelength-momentum relationship.
  • #20
malawi_glenn said:
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.

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:

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:


Comes easily.

Any benefits to this interpretation?
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  • #21
De Broglie firstly postulate that any matter particle has some inner periodic motion, which is in accordance with the relativistic theory.

Note of Louis de Broglie, presented by Jean Perrin.
(Translated from Comptes rendus, Vol. 177, 1923, pp. 507-510)

"Let us consider a material moving object of rest mass [itex]m_0 [/itex] moving
with respect to a fixed observer with a speed [itex]\upsilon =\beta _ c_ (\beta
<1)[/itex]. According to the principle of the inertia of energy, it should possesses
an internal energy equal to [itex]m_0 c^2[/itex]. ~ On the other hand, the quantum
principle suggests associating this internal energy with a simple periodic
phenomenon of frequency [itex]\nu _0 [/itex] such that [itex]h\nu _o =m_0 c^2[/itex], [itex]c[/itex] being,
as usual, the limiting velocity of the theory of relativity and [itex]h[/itex] Planck's
For the fixed observer, the frequency [itex]\nu =\frac{1}{h}\frac{m_0
c^2}{\sqrt {1-\beta ^2} }[/itex] corresponds to the total energy of the moving
object. But, if this fixed observer observes the internal periodic
phenomenon of the moving object, he will see it slowed down and will
attribute to it a frequency [itex]\nu _1 =\nu _0 \sqrt {1-\beta ^2} [/itex]; for him
this phenomenon varies therefore like [itex]\sin 2\pi \nu _1 t[/itex].
Now \textbf{let us suppose that at the time }[itex]t=0[/itex]\textbf{ the moving
object coincides in space with a wave of frequency }[itex]\nu [/itex] [itex]\textbf{ defined
above and propagating in the same direction as it does with the speed
} [/itex] [itex]\frac{c}{\beta }=\frac{c^2}{\upsilon }[/itex] ~ This wave, which has a
speed greater than [itex]c[/itex], cannot correspond to transport of energy; we will
only consider it as a fictitious wave associated with the motion of the
I maintain that, if at the time [itex]t=0[/itex], there is phase agreement between
the vectors of the wave and the internal phenomenon of the object, this
phase agreement will be maintained. ~ In effect, at time [itex]t[/itex] the object is
at a distance from the origin equal to [itex]\nu t=x[/itex]; its internal motion is
then represented by [itex]\sin 2\pi \nu _1 \frac{x}{\upsilon }[/itex] .
~ ~ The wave, at this point, is represented by [itex]\sin 2\pi \nu t\left(
{t-\frac{\beta x}{c}} \right)=\sin 2\pi \nu x\left( {\frac{1}{\upsilon
}-\frac{\beta }{c}} \right)[/itex] .
The two sinus are equal and the phase agreement is realized if one has
[itex]\nu _1 =\nu \left( {1-\beta ^2} \right)[/itex],
a condition that is clearly satisfied by the definitions of [itex]\nu [/itex] and [itex]\nu_1 [/itex].
The demonstration of this important result rests uniquely on the
principle of special relativity and on the correctness of the quantum
relationship as much for the fixed observer as for the moving observer.
~ Let us apply this to an atom of light. ~ I showed elsewhere [itex]^{(2)}[/itex]
that the atom of light should be considered as a moving object of a very small mass ([itex]<10^{-50}[/itex]g) that moves with a speed very nearly equal to [itex]c[/itex] (although slightly less).
We come therefore to the following conclusion:
[itex]\textit{The atom of light, which is equivalent by reason of its total energy to a radiation of frequency} [/itex] [itex]\nu [/itex] [itex]\textit{, is the seat of an internal periodic
phenomenon that, seen by the fixed observer, has at each point of space the same phase as a wave of frequency}[/itex] [itex]\nu [/itex] propagating in the same direction with a speed very nearly equal (although very slightly greater) to the constant called the speed of light". And so forth (see "Annales de la Fondation Louis de Broglie <> [Broken]

So, any particle is accompanied by phase wave with some frequency and wavelength. For a photon we have [itex]\lambda =\frac{h}{p}[/itex]. De Broglie shows that this relation can also be right for the particles.
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  • #22
I know I am very late to this thread and probably no one is paying any attention to it. But I would like to give a somewhat valid derivation of the de Broglie equation, as I landed on this thread by seaching for a true derivation of the same. So, it might help someone else who is also surfing the net like me to make some sense of physics.
Here goes:
In order to come up with the wavelenght of the matter wave of a particle, we need to consider the wave properties of that particle, like phase velocity, group velocity, crests, etc. Now, the phase velocity is given by V=w/k, where w is the angular frequency and k is the wave number and the group velocity is v=dw/dk, a differential. The phase velocity is associated with a single wavelength and the group velocity is associated with a wave having a number of waves, each of different frequency.
Now, first considering the virtual photon of the particle. A virtual photon is actually a quantum of energy exchanged between the electron and the proton in and atom or between electron and electron. It is emitted by all particles, like pions, neutrons, protons, etc. As it is energy and energy travels at the speed of light, we have the relation of energy in the virtual photon as E = mc^2 and also as E = hc/lambda, which gives the first result as λ = h/mc.
De Broglie hypothesized that the particle itself was not a wave, but always had with it a pilot wave, or the virtual photon or a wave that helps guide the particle through space and time. He postulated that the group velocity of the wave was equal to the actual velocity of the particle.
However, the phase velocity would be very much different. He saw that the phase velocity was equal to the angular frequency divided by the wavenumber. Since he was trying to find a velocity that fit for all particles (not just photons) he associated the phase velocity with that velocity. He equated these two equations:
V = ω/k(which the eqn for the wave) = E/p(which is the eqn for the particle)
From this new equation from the phase velocity we can derive:
V = mc^2/m v = c^2/v
Applied to Einstein’s energy equation, we have:
E = pV = mv(c^2/v)
Also, the wavelenght of the matter wave (or any wave), is given as the phase velocity divided by the frequency of the wave. Hence, λ = V/f.
Now we can get to an actual derivation of the De Broglie equation:
p = E/V
p = (hf)/V
p = h/λ
With a little algebra, we can switch this to get λ = h/mv.
I hope this helps in understanding the concept.
  • #23
welldone for a job well done,at times i also wonder about this.

1. What is the de Broglie wavelength and why is it important?

The de Broglie wavelength is a concept in quantum mechanics that describes the wave-like behavior of matter. It is important because it helps us understand the dual nature of particles, which can exhibit both wave and particle properties.

2. How do you derive the de Broglie wavelength?

The de Broglie wavelength can be derived using the formula λ = h/mv, where λ is the de Broglie wavelength, h is Planck's constant, m is the mass of the particle, and v is its velocity.

3. What is the relationship between the de Broglie wavelength and momentum?

The de Broglie wavelength is inversely proportional to the momentum of a particle. This means that as the momentum increases, the de Broglie wavelength decreases, and vice versa.

4. Can the de Broglie wavelength be applied to macroscopic objects?

No, the de Broglie wavelength is only applicable to microscopic particles, such as electrons and atoms. It is not significant enough to be observed in macroscopic objects.

5. How is the concept of the de Broglie wavelength used in real-world applications?

The de Broglie wavelength is used in a variety of real-world applications, such as electron microscopy, particle accelerators, and quantum computing. It also plays a crucial role in understanding the behavior of particles in quantum mechanics.

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