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

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1. Jun 9, 2016

### Albrecht

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?

2. Jun 9, 2016

### Heinera

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.

3. Jun 9, 2016

### Jilang

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

4. Jun 9, 2016

### Heinera

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).

5. Jun 9, 2016

### Albrecht

"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.

6. Jun 9, 2016

### Heinera

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).

7. Jun 10, 2016

### Albrecht

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?

8. Jun 10, 2016

### Jilang

Don't t and x also transform?

9. Jun 10, 2016

### Heinera

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: May 7, 2017
10. Jun 11, 2016

### 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.

11. Jun 11, 2016

### vanhees71

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

12. Jun 11, 2016

### haushofer

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

13. Jun 15, 2016

### Albrecht

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: May 8, 2017
14. Jun 15, 2016

### Jilang

I don't like your phase speed. It tends to infinity as v1 tends to zero.

15. Jun 16, 2016

### vanhees71

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).

16. Jun 16, 2016

### Albrecht

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.

17. Jun 16, 2016

### Albrecht

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.

18. Jun 17, 2016

### vanhees71

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.

19. Jun 17, 2016

### kith

This quantity has the dimension of a length. It is not the phase speed but the de Broglie wavelength.

20. Jun 18, 2016

### Albrecht

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!