# The de Broglie wavelength: What happens in the case of a frame change?

• A
I have a problem to understand the de Broglie wavelength. We know that also particles undergo scattering and interference at a double slit. The interference pattern is calculated by the use of the de Broglie wavelength which is defined as lambda = h / p ; p is the momentum of the particle. This seems to work for an observer who is at rest with the double slit. But as soon as the observer moves in relation to the experiment, he may calculate a very different lambda as the momentum is different in his frame. On the other hand the interference pattern will not change in the view of the observer as long as his speed is u<<c.

Isn’t this a violation of the Galilei principle that physics should be the same in any (inertial) frame?

I have a problem to understand the de Broglie wavelength. We know that also particles undergo scattering and interference at a double slit. The interference pattern is calculated by the use of the de Broglie wavelength which is defined as lambda = h / p ; p is the momentum of the particle. This seems to work for an observer who is at rest with the double slit. But as soon as the observer moves in relation to the experiment, he may calculate a very different lambda as the momentum is different in his frame. On the other hand the interference pattern will not change in the view of the observer as long as his speed is u<<c.

Isn’t this a violation of the Galilei principle that physics should be the same in any (inertial) frame?
The relevant quantity here is the particle's momentum relative to the slits, not relative to the observer. p is here defined to be the momentum of the particle in the rest frame of the double slit, no matter what frame the observer is in.

Should that be relative to the slits and the measuring screen?

Should that be relative to the slits and the measuring screen?
Yes (although I can't see that a moving screen relative to the slits would change anything fundamental, other than making the experiment into a complete mess).

"The relevant quantity here is the particle's momentum relative to the slits ..."

With this assumption there would be no problem. But it is not the solution.

1. Looking at the derivation of de Broglie in his original paper, he did not have the intention to relate this wavelength to some specific appliance like the double slit.
2. Even more essential: The use of this wavelength in the Schrödinger equation and in the Dirac function treats it as a property of the particle, not as the property of some specific interaction. If one looks on the other hand at the use of the energy relation: E = h * f for the time function, then energy is used in both equations in a correct way so that it is invariant at frame transformations. In contrast to this the use of the de Broglie relation for the wavelength and so for the spatial function cannot have correct solutions in the general case, in my understanding.

A wavelength/frequency is always frame dependant (including even light). If you want a frame independent version of de Broglie's equations, you should use four-vectors (see the section on four-vectors in https://en.m.wikipedia.org/wiki/Matter_wave).

It is right that wavelength/frequency is frame dependent. But if it is treated in the correct way then the results are Lorentz invariant (and also Galilei-invariant in the non-relativistic case). Just the wave of de Broglie does not have this property.

The exponent in the QM wave function (Schrödinger / Dirac) is -i/h(bar) *(Et – px) . E is related to the frequency f by E=h*f. When transformed into another frame, E and also f will change. But when it is subject to a Lorentz transformation, the physical results are correct again. However different for lambda=h/p. p is also different in a different frame, but the results are incorrect after Lorentz transformation.

Simple example: If this process is transformed into the frame of the electron which is going to be scattered, then E is reduced to the rest energy of the particle, which transforms correctly. p on the other hand is reduced to zero and so the wavelength is infinite. This is un-physical and the Lorentz transformation does not correct for this.

How can QM live with such behavior?

Don't t and x also transform?

It is right that wavelength/frequency is frame dependent. But if it is treated in the correct way then the results are Lorentz invariant (and also Galilei-invariant in the non-relativistic case). Just the wave of de Broglie does not have this property.
Try this one:

Robert Shuler: "Common Pedagogical Issues with De Broglie Waves: Moving Double Slits, Composite Mass, and Clock Synchronization"

Last edited by a moderator:
• Nugatory
haushofer
Ballentine's book on QM treats this in an excellent way. Basically, it amounts to the question how the wave function transforms under a Galilei boost. Well, it does not transform as a scalar, because in that case the Schrodinger eqn. is not invariant (check this out for yourself!). It transforms with an extra phase factor. This also solves your conundrum.

• vanhees71
vanhees71
Gold Member
2021 Award
Yes, and this phase factor is due to the fact that the Galileo group's Lie algebra has a non-trivial central charge, which in the representation theory of the Galileo group put in physics language, is the mass. So in quantum theory you have to represent the Galileo group as the representation of its central extension of its covering group (the latter just substitutes the SO(3) for rotational symmetry with its covering group SU(2), enabling half-integer spin, which clearly is observed in nature since all the matter around us consists of fermions of spin 1/2 (electrons, protons, and neutrons or, if you want, electrons and quarks). The additional phase factor is due to the central charge (or if seen as a representation of the classical Galileo group it's a ray representation on Hilbert space, which is of course fine, because pure quantum states are not represented by state vectors but by rays in Hilbert space).

It turns out that indeed the representation of the Galileo group with ##m=0## doesn't lead to anything that can be sensibly made physics sense of. This is a famous paper about this issue:

Inönü, E., Wigner, E. P.: Representations of the Galilei group, Il Nuovo Cimento 9(8), 705–718, 1952
http://dx.doi.org/10.1007/BF02782239

• haushofer
haushofer
To add to that: the centrally-extended Galilei algebra (with the mass generator) is known as the Bargmann algebra.

• vanhees71
Try this one:

Robert Shuler: "Common Pedagogical Issues with De Broglie Waves: Moving Double Slits, Composite Mass, and Clock Synchronization"

Thank you for the reference to the paper of Roberg Shuler. It was interesting to read it. But it has a logical bug.

The question treated is whether the de Broglie wave changes in case of a transformation into another frame, and as a consequence the interference pattern changes. The author does not see any change. But I cannot agree.

Let’s take a non-relativistic case, which may mean: v1 = 0.1c .

The electron has a fixed frequency f = E/h . This value is almost unchanged in the case of a transformation in the case of v1 = 0.1c .

Now, if the double slit is at rest and the electron is moving, the phase speed has the big value of vph = c2 / v1. This phase speed determines the wavelength. Lambda1 = f / vph = E/ (h*c2/ v1). This is the known de Broglie wavelength.

Now the other case, the electron at rest and the apparatus moving. So moving with speed v1 towards the electron. Now the effective wavelength is given by the frequency f = E/h and the speed v1 of the apparatus. So, lambda2 = f / v1 = E /( h*v1). - The (superluminal) phase speed assumed by de Broglie is not applicable in this case because the apparatus does not move with this speed.

Both, lambda1 and lambda2 are clearly different and deny the invariance of the de Broglie wavelength at a frame change.

Last edited by a moderator:
I don't like your phase speed. It tends to infinity as v1 tends to zero.

vanhees71
Gold Member
2021 Award
Well the plane-wave solution of the Schrödinger equation for free particles obviously is
$$u_{\vec{p}}(t,\vec{x}) = \frac{1}{(2 \pi)^3} \exp(-\mathrm{i} E t+\mathrm{i} \vec{p} \cdot \vec{x})$$
with
$$E=\frac{\vec{p}^2}{2m}.$$
So you get
$$c_{\text{ph}}=\frac{E}{p}=\frac{p}{m}.$$
What this, however, should have to do with the speed of anything in a frame, where the slits move (which you get with a Galileo transformation, where you have to take into account the non-trivial phase factor due to the central charge of the unitary transformation of the quantum version of the extended ("quantum") Galileo group).

I don't like your phase speed. It tends to infinity as v1 tends to zero.
I do not like this phase speed as well. But is is not my phase speed but the one of de Broglie.

It is interesting to look into the original paper of de Broglie where he has deduced it as the phase speed of a "ficticious wave", which in his opinion accompanies an electron. He thought to need it to solve a conflict between particle physics and relativity (and here dilation).

In my view de Broglie has misunderstood special relativity. With a correct understanding, all this is not necessary.

Well the plane-wave solution of the Schrödinger equation for free particles obviously is
$$u_{\vec{p}}(t,\vec{x}) = \frac{1}{(2 \pi)^3} \exp(-\mathrm{i} E t+\mathrm{i} \vec{p} \cdot \vec{x})$$
with
$$E=\frac{\vec{p}^2}{2m}.$$
So you get
$$c_{\text{ph}}=\frac{E}{p}=\frac{p}{m}.$$
What this, however, should have to do with the speed of anything in a frame, where the slits move (which you get with a Galileo transformation, where you have to take into account the non-trivial phase factor due to the central charge of the unitary transformation of the quantum version of the extended ("quantum") Galileo group).

E in the Schrödinger equation is the entire energy, so the sum of rest energy and kinetic energy
E in your calculation $$E=\frac{\vec{p}^2}{2m}.$$ is the kinetic energy, so different from Schrödinger.
The phase speed was deduced by de Broglie in his special way with the result $$c_{\text{ph}}=\frac{h}{p}.$$
The transformation of this phase speed into another frame is the critical thing which in my view is not correct or not feasable in de Broglie's way.

vanhees71
Gold Member
2021 Award
The "rest energy" is just an additive constant to the total energy and thus physically irrelevant (except in GR, but we are talking about non-relativistic QT here). One finds recently a lot of nonsense concerning such additive constants to the total energy (even in peer reviewed journals like the current issue of EJP), but this doesn't make this nonsense true.

Now let's look at the issue of Galileo transformations, and I'll keep the usual setting of the energy of a free particle in non-relativistic physics, because there's no reason to deal with unnecessary phase factors, but it's important to keep the necessary ones. A general single-particle wave function transforms under Galilei boosts ##\vec{x}'=\vec{x}-\vec{w} t##, ##\vec{p}'=\vec{p}-m \vec{w}## as
$$\psi'(t,\vec{x}')=\exp(-\mathrm{i} m \vec{w} \cdot \vec{x}'-\mathrm{i}m \vec{w}^2/2)\psi(t,\vec{x}'+\vec{w} t).$$
Now for the plane wave we have
$$psi(t,\vec{x})=\exp[-\mathrm{i} t\vec{p}^2/(2m)+\mathrm{i} \vec{p} \cdot \vec{x}]$$
and thus
$$\psi'(t,\vec{x}')=\exp[-\mathrm{i} t (\vec{p}^2/(2m)+m \vec{w}^2/2)+\mathrm{i} (\vec{x}'+\vec{w} t)(\vec{p}-m \vec{w})].$$
Plugging in ##\vec{p}'=\vec{p}-m \vec{w}##, you get
$$\psi'(t,\vec{x}')=\exp[-\mathrm{i} t \vec{p}'^2/(2m) + \mathrm{i} \vec{x}' \cdot \vec{p}'],$$
and thus the phase speed is precisely the value that you expect in the "new" reference frame, namey ##\vec{v}'=\vec{p}'/m=\vec{p}/m-\vec{w}=\vec{v}-\vec{w}.##

Of course, the non-trivial phase factor is crucial in this calculation, and it's a mathematical fact that non-relativistic QT works only with such a ray representations of the classical Galilei group, which is a unitary representation of the quantum extension of the Galilei group with the mass as a non-trivial central charge. For details, see Ballentines textbook.

kith
The phase speed was deduced by de Broglie in his special way with the result $$c_{\text{ph}}=\frac{h}{p}.$$
This quantity has the dimension of a length. It is not the phase speed but the de Broglie wavelength.

• vanhees71
Of course you are right. This is the de Broglie wavelength: lambda = h / p , which is infinite for p = 0.
The de Broglie phase speed is
cph = c2 / v
if v is the speed of the particle.
Sorry!

The "rest energy" is just an additive constant to the total energy and thus physically irrelevant (except in GR, but we are talking about non-relativistic QT here). One finds recently a lot of nonsense concerning such additive constants to the total energy (even in peer reviewed journals like the current issue of EJP), but this doesn't make this nonsense true.

Now let's look at the issue of Galileo transformations, and I'll keep the usual setting of the energy of a free particle in non-relativistic physics, because there's no reason to deal with unnecessary phase factors, but it's important to keep the necessary ones. A general single-particle wave function transforms under Galilei boosts ##\vec{x}'=\vec{x}-\vec{w} t##, ##\vec{p}'=\vec{p}-m \vec{w}## as
$$\psi'(t,\vec{x}')=\exp(-\mathrm{i} m \vec{w} \cdot \vec{x}'-\mathrm{i}m \vec{w}^2/2)\psi(t,\vec{x}'+\vec{w} t).$$
Now for the plane wave we have
$$psi(t,\vec{x})=\exp[-\mathrm{i} t\vec{p}^2/(2m)+\mathrm{i} \vec{p} \cdot \vec{x}]$$
and thus
$$\psi'(t,\vec{x}')=\exp[-\mathrm{i} t (\vec{p}^2/(2m)+m \vec{w}^2/2)+\mathrm{i} (\vec{x}'+\vec{w} t)(\vec{p}-m \vec{w})].$$
Plugging in ##\vec{p}'=\vec{p}-m \vec{w}##, you get
$$\psi'(t,\vec{x}')=\exp[-\mathrm{i} t \vec{p}'^2/(2m) + \mathrm{i} \vec{x}' \cdot \vec{p}'],$$
and thus the phase speed is precisely the value that you expect in the "new" reference frame, namey ##\vec{v}'=\vec{p}'/m=\vec{p}/m-\vec{w}=\vec{v}-\vec{w}.##

Of course, the non-trivial phase factor is crucial in this calculation, and it's a mathematical fact that non-relativistic QT works only with such a ray representations of the classical Galilei group, which is a unitary representation of the quantum extension of the Galilei group with the mass as a non-trivial central charge. For details, see Ballentines textbook.

The frequency of a particle is given by its entire energy and not by its kinetic energy. So the difference of both is indeed of physical relevance.

To the equations shown here I have a question. An exponent should only contain dimension-less numbers. But the product p*x is not dimension-less. So how shall we understand the exponents of the equations?

Another question: what does this Galilean transformation have to do with the de Broglie wavelength? That wavelength is defined as lambda = h / p; it has pole for p = 0. My question was, how this pole is transformed into another frame so that the physical content is still valid. Is this threat really an answer to my question?

I am afraid, not.

vanhees71
Gold Member
2021 Award
Again, this is wrong. If you were right, nearly all calculations in non-relativistic quantum theory were wrong, because there one uses ##E_{\text{kin}}=\vec{p}^2/(2m)## as the dispersion relation and not the non-relativistic approximation of relativistic energy ##E=\sqrt{\vec{p}^2+m^2} \simeq m + \vec{p}^2/(2m)##. I've given the reason for why this constant energy shift doesn't do anything to any prediction of physical observables. In this connection one should keep in mind that the representation theory of the Galileo group leads to a superselection rule forbidding the superposition of states of different mass!

Now concerning you question about the de Broglie wavelength. I don't know, where you have a problem with the de Broglie wave length diverges for ##p \rightarrow 0## and also I don't know what you think the physical meaning of this wavelength is. I personally have very little use for it concerning my physical intuition.

Again, this is wrong. If you were right, nearly all calculations in non-relativistic quantum theory were wrong, because there one uses ##E_{\text{kin}}=\vec{p}^2/(2m)## as the dispersion relation and not the non-relativistic approximation of relativistic energy ##E=\sqrt{\vec{p}^2+m^2} \simeq m + \vec{p}^2/(2m)##. I've given the reason for why this constant energy shift doesn't do anything to any prediction of physical observables. In this connection one should keep in mind that the representation theory of the Galileo group leads to a superselection rule forbidding the superposition of states of different mass!

Now concerning you question about the de Broglie wavelength. I don't know, where you have a problem with the de Broglie wave length diverges for ##p \rightarrow 0## and also I don't know what you think the physical meaning of this wavelength is. I personally have very little use for it concerning my physical intuition.

For all calculations of a particle’s wavelengths the internal frequency of that particle is an essential quantity. This frequency depends on the entire energy of a particle, not on the kinetic energy. That was my point.

What is my problem with de Broglie's wavelength? When de Broglie introduced matter waves he concluded that the wavelength is given by lambda = h/p. This was proven correct in experiments where electrons are scattered at a double slit, if investigated in a frame in which the double slit is at rest. The interference pattern behind the double slit could be correctly determined by use of this wavelength. However, if such experiment is seen from the frame of the moving electron, the situation changes considerably. In its own frame the electron has momentum p=0 and so its wavelength is infinite. The double slit moves now towards the electron. Which interference pattern can an observer, residing in the frame of the electron, expect if the wavelength is infinite?

It is a fundamental rule in physics (I think since Galileo) that we should observe the same physics in every inertial frame. De Broglie’s definition of the wavelength of an electron seems to be in conflict with this fundamental rule. – That was my original question.

vanhees71
Gold Member
2021 Award
For all calculations of a particle’s wavelengths the internal frequency of that particle is an essential quantity. This frequency depends on the entire energy of a particle, not on the kinetic energy. That was my point.
Can you give an example within non-relativistic physics, where the choice of the absolute energy-zero point is of any relevance? I've no clue what that might be.

Then there is a predecessor theory to modern quantum mechanics, which in a sense was a crucial transition stage between the old Einstein-Bohr to the modern Heisenberg-Schrödinger-Dirac QT, and that was de Broglies idea of "wave-particle duality", but from the modern point of view it's part of "old quantum theory" and thus obsolete. In this theory the wavelength is not related to energy but to momentum via ##\lambda=h/p##. So there the choice of the energy-zero doesn't play any role anyway.

What is my problem with de Broglie's wavelength? When de Broglie introduced matter waves he concluded that the wavelength is given by lambda = h/p. This was proven correct in experiments where electrons are scattered at a double slit, if investigated in a frame in which the double slit is at rest. The interference pattern behind the double slit could be correctly determined by use of this wavelength. However, if such experiment is seen from the frame of the moving electron, the situation changes considerably. In its own frame the electron has momentum p=0 and so its wavelength is infinite. The double slit moves now towards the electron. Which interference pattern can an observer, residing in the frame of the electron, expect if the wavelength is infinite?
As I've demonstrated in an earlier posting, the Schrödinger equation is Galilei invariant. If you want the interference pattern for the case of moving slits you just Lorentz boost the usual wave function for the slits at rest. There cannot be any inconsistency in the math, because of the Galilei invariance of the Schrödinger equation. Of course, if the slit runs as fast as the electrons, they'll never reach the slits at all and there's no interference pattern. Maybe again I don't understand your setup.

It is a fundamental rule in physics (I think since Galileo) that we should observe the same physics in every inertial frame. De Broglie’s definition of the wavelength of an electron seems to be in conflict with this fundamental rule. – That was my original question.
We observe the same physics in each inertial frame within non-relativistic QT since by construction it's Galilei invariant as it must be. I still don't see what problems you have with the de Broglie wavelength and also still do not understand which physical implications this quantity has in your opinion.

Can you give an example within non-relativistic physics, where the choice of the absolute energy-zero point is of any relevance? I've no clue what that might be.

The relation E = h * frequency is fundamental in QM with E = the entire energy (= rest energy + kinetic energy); the relation is not restricted to relativity.
Then there is a predecessor theory to modern quantum mechanics, which in a sense was a crucial transition stage between the old Einstein-Bohr to the modern Heisenberg-Schrödinger-Dirac QT, and that was de Broglies idea of "wave-particle duality", but from the modern point of view it's part of "old quantum theory" and thus obsolete. In this theory the wavelength is not related to energy but to momentum via ##\lambda=h/p##. So there the choice of the energy-zero doesn't play any role anyway.

Not zero energy but rest energy PLUS kinetic energy. The de Broglie relation ##\lambda=h/p## which you mention does not contain the rest energy, and this can cause an infinite wavelengh and an infinite phase speed which both are unphysical in my understanding.
As I've demonstrated in an earlier posting, the Schrödinger equation is Galilei invariant. If you want the interference pattern for the case of moving slits you just Lorentz boost the usual wave function for the slits at rest. There cannot be any inconsistency in the math, because of the Galilei invariance of the Schrödinger equation. Of course, if the slit runs as fast as the electrons, they'll never reach the slits at all and there's no interference pattern. Maybe again I don't understand your setup.

You did not present the Schrödinger equation but something different (in your equation even the dimensions in the exponent are inconsistent as I have mentioned). The correct mathematical description of the interference pattern in the case that the double slit is at rest follows from de Broglie's ansatz for the wavelength. But if seen from the frame of the electron, the de Broglie wavelength is infinite. This case should result in the same interference pattern (in the non-relativistic situation). How can this be argued?
We observe the same physics in each inertial frame within non-relativistic QT since by construction it's Galilei invariant as it must be. I still don't see what problems you have with the de Broglie wavelength and also still do not understand which physical implications this quantity has in your opinion.

The problem is how an infinite wavelength can cause the expected interference pattern.

George Jones
Staff Emeritus
Gold Member
To the equations shown here I have a question. An exponent should only contain dimension-less numbers. But the product p*x is not dimension-less. So how shall we understand the exponents of the equations?

vanhees71 is using "natural" units such that ##c=\hbar =1##. This is quite common.

Albrecht, seen from the frame of the electron, would not the slits which are rapidly approaching seem kind of fuzzy and overlapping? In fact so fuzzy the poor electron is not quite sure which one it went through? On reaching the screen might it conclude that it's likely position was influenced by the slits jostling for position?

vanhees71
Gold Member
2021 Award
The relation E = h * frequency is fundamental in QM with E = the entire energy (= rest energy + kinetic energy); the relation is not restricted to relativity.
Well, the significance is that the Hamiltonian is the generator of time translations. That's a definition valid all over physics. That there's an ##\hbar## as a conversion factor is just due to the choice of unnatural units ;-).

Not zero energy but rest energy PLUS kinetic energy. The de Broglie relation ##\lambda=h/p## which you mention does not contain the rest energy, and this can cause an infinite wavelengh and an infinite phase speed which both are unphysical in my understanding.
Only because you repeat a wrong statement it doesn't become true.

You did not present the Schrödinger equation but something different (in your equation even the dimensions in the exponent are inconsistent as I have mentioned). The correct mathematical description of the interference pattern in the case that the double slit is at rest follows from de Broglie's ansatz for the wavelength. But if seen from the frame of the electron, the de Broglie wavelength is infinite. This case should result in the same interference pattern (in the non-relativistic situation). How can this be argued?
Of course, I used the Schrödinger equation. What else should it be in the case of non-relativistic quantum theory? The interference pattern follows from the solution of the Schrödinger equation with the grating defining boundary conditions. Mathematically it's of course a wave phenomenon and thus has to do with a wavelength. Physically it's of course far from a classical-wave picture, and this dinstinguishes modern quantum theory from old quantum theory, which was a predecessor to the latter, not more but also not less.

The problem is how an infinite wavelength can cause the expected interference pattern.

There is no problem at all in non-relativistic quantum theory. A particle cannot have a sharp momentum (nor a sharp position) as the Heisenberg uncertainty relation tells you. You cannot prepare a particle in a momentum eigenstate, because the corresponding plane-wave solutions of the Schrödinger equation do not represent states, because they are not square integrable functions.

jedishrfu
Mentor
Thank you all for your participation in the discussion of this topic.

The OP has recieved many excellent answers here to consider and now its time to closie this thread.

-- Jedi