# De Broglie wavelength and direction

1. Jun 11, 2012

### edpell

We can calculate a de Broglie wavelength for a particle with momentum p. My question is does this apply to the lateral size of the particle? Perpendicular to the motion the momentum is zero. What can we say about the size of the particle in the perpendicular direction?

2. Jun 11, 2012

### Staff: Mentor

Wavelength ≠ size.

3. Jun 11, 2012

### Darwin123

The wavelength has nothing to do with the size of the particle. The wavelength of an extended object can be far smaller than the size of the extended object.
A wave can extend in all directions forever, even if its wavelength is very small. The entire wave can extend over many cycles of the wave. So the wave can extend far past the boundaries of the extended object, even if the wavelength is less than the diameter of the extended object.
De Broglie did not consider extended objects. The particle that he was thinking of is probably infinitesimal in size. However, his ideas can be extrapolated to extended objects as long as certain conditions are satisfied. One condition that has to be satisfied is that the individual particles making up the composite particle have to be entangled in order to use the De Broglie relations.
Suppose we have a system that consists of several particles bound together. Each particle will act independently. One can't treat the bound system as a composite particle unless these separate particles are constrained to act together. Therefore, you can't use the De Broglie relations on the bound system unless the waves of separate particles are in phase.
One can't use the De Broglie relations on a composite particle unless the component particles are entangled. Unless the separate particles are entangled, they won't act as a quantum mechanical entity.
Consider as an example biphotons. Biphotons are two-photon states that act as a unit. The energy of the biphoton is the sum of the energies of the individual photons. The momenta of the biphoton is the sum of the momenta of the biphoton. The result is that the biphoton diffracts off a diffraction grating as though it has the wavelength given by the De Broglie relations for a bound two-photon state. The biphoton has beats corresponding to the sum of the frequencies of the two photons. The biphoton can thus be treated as a composite particle.
One can't construct a biphoton unless the individual photons are entangled. Thus, the De Broglie relations in this case can only be applied to an entangled state of the individual relations.

4. Jun 11, 2012

### Darwin123

A particle with a finite size has to be a composite particle, made of several component particles. However, composite particles only satisfy the de Broglie relations if the two particles are entangled.
If the two particles are entangled, their diffraction properties still have nothing to do with size. For example, a Cooper pair of electrons in a semiconductor can be very big. The two electrons are far apart. However, the wavelength of the Cooper pair can be very small. The single particle behavior of the Cooper pair does not make sense in classical physics. This is because the individual particles in the Cooper pair are not entangled.
The De Broglie relations can't be applied to a classical object like a baseball because the classical object is comprised of particles that are not entangled. However, there are composite particles in nature that are comprised of particles that are entangled. Recent experimental work with these objects has shown that they satisfy the De Broglie relations.
The de Broglie theory can only be applied to a composite particle when the component particles are entangled. This is true for molecules, which are comprised of separate atoms. The de Broglie theory is applicable to the molecule only when its atoms are entangled. Similar conditions apply to the biphoton, which is a composite particle comprised of two photons.
The following link describes how the de Broglie relations apply to biphotons.
http://www.aip.org/pnu/2002/split/613-1.html [Broken]
“A bi-photon de Broglie wavelength has been directly measured in an interference experiment for the first time. In the early days of quantum mechanics, Louis de Broglie argued that if waves could act like particles (photoelectric effect) then why couldn't particles act like waves?
They could, as was borne out in numerous experiments (the double-slit experiment for electrons was voted the "most beautiful" experiment in a recent poll—see Physics World, Sept 2002).
In fact, intact atoms in motion and even molecules can be thought of as "de Broglie waves." Molecules as large as buckyballs (carbon-60) have been sent through an interferometer, creating a characteristic interference pattern (see Update 579).
The measured wavelength for a composite object like C-60 will in part depend on the internal bonds of the molecule. What then if the corporate object is a pair of entangled photons?
One of the more fascinating predictions made regarding quantum entanglement (Jacobson et al., Physical Review Letters, 12 Jun 1995) was the suggestion that the de Broglie wavelength for an ensemble consisting of N entangled photons (each with a wavelength of L) would be L/N.
This proposition has been verified now by physicists at Osaka University (Keiichi Edamatsu, 81-6-6850-6507, eda@mp.es.osaka-u.ac.jp) for the case of two entangled photons. The daughter photons were created by the process of parametric down-conversion, in which an incident photon entering a special crystal will split into two correlated photons. These photons are then sent through an interferometer (see figure).”

I notice that they use the word "corporate" instead of "composite". It is the same thing. The corporate object has to be made of entangled particles before one uses the De Broglie relations.

Last edited by a moderator: May 6, 2017
5. Jun 11, 2012

### Ken G

Interesting post!

6. Jun 11, 2012

### edpell

humm this has gone in directions I did not anticipate.

If the particle is an electron (a simple non composite object) it has a wavelength in the direction of motion. Does this wavelength also apply to the two axis perpendicular to the motion?

7. Jun 12, 2012

### Darwin123

Three dimensional waves have a wave vector rather than a wavelength. The wave vector of a three dimensional wave has three components. According to the De Broglie relations, Planck's constant times the wave vector is equal to the momentum of the momentum of the electron.
The wave vector is actually a feature of classical physics, applicable to any type of wave. For any wave, the magnitude of the wave vector is equal to 2 pi/wavelength. Diffraction can be described using the wave vector rather than the wavelength. Wavelength is a scalar quantity that doesn't really include direction. However, a wave vector can be determined by direction cosines and a wavelength.
Momentum, like wavevector, is a three dimensional quantity. So to be applicable in three dimensions, the de Broglie relations have to be defined in terms of wave vector, not wave length. De Broglies relation says that the linear momentum is Planck's constant times the wave vector.
One can see that one doesn't really decompose the wavelength into three components. However, one can decompose the wave vector into three components.
The three components of a wave vector relate to the frequency by a dispersion relation. If you want, you can define a wavelength for all three dimensions using the three components of the wave vector. Howe
Please note that the size and shape of objects are not specified by any one wave vector. However, the Heisenberg's uncertainty condition does suggest that the size of an object may be inversely proportional to the bandwidth of the wave packet. One can consider an object of finite size to be a superposition of waves with different wave vectors. Therefore, it is also a "superposition" of "particles" with different linear momenta. The size of the object in any direction times the bandwidth of wave vector components in that direction has to be greater than one.

The dimensions of the object have nothing to do with the components of the wave vector. A corporate object can have a shape and size that is unrelated to the wave vector. However, there is sometimes an inverse relationship between the spatial extent of an object and the wave vector bandwidth. However, the mean wave vector of the object is completely unrelated to the size and shape of the object.
That has a lot to do with wave packets and Ehrenfest's theorem. Fourier transforms in three dimensions determine the "shape" of a wave packet. Wave packets can be used to approximate the behavior of an object. However, that gets into issues more complicated than waves that are single wavelength and unidirectional.
If there is a single wavelength and single direction to your wave, then the uncertainty relations show that the shape of the particle-object is completely undeterminable. So your question regarding "the" wavelength suggests that the object has no shape. If the question were posed in terms of wave packets, then a more satisfying answer could be provided.

8. Jun 12, 2012

### Staff: Mentor

They have both. λ = 2π/k, where k is the magnitude of the wave vector $\vec k$.

9. Jun 12, 2012

### Reptillian

I can see how you'd get a size out of the deBroglie wavelength...after all, the size (expectation value of an electron) of an atom is on the order of the deBroglie wavelength of an electron in the atom, and the size of the nucleus is on the order of the the deBroglie wavelength for a proton/neutron. In the direction perpendicular to motion, the deBroglie wavelength doesn't tell you much. I mean, it could be a wave on the ocean, or a wave around a pebble dropped in a puddle. One has a definitive size in the perpendicular direction, one doesn't. You'd need information about the global conditions in order to place limitations on the particle size.

10. Jun 14, 2012

### Darwin123

The diameter of the atom is the De Broglie wavelength of the electron. The diameter of the atom is not the De Broglie wavelength of the atom.
The De Broglie wavelength of the electron is not the same as the De Broglie wavelength of the atom. The atom is a composite of both the electron and the nucleus.
In the center of mass frame of the atom, the De Broglie wavelength of the atom is infinite. In any frame where the center of mass of the atom moves slowly, the wavelength of the atom will be much bigger than the diameter of the atom.
In the semiclassical picture of the atom, the electron is moving rapidly around the nucleus. If one applies the De Broglie hypothesis to this picture, it is easy to see that the wavelength of the entire atom is much bigger than the wavelength of the electron.
The wavelength of a corporate particle is unlikely to be the same as the size of the corporate particle. The size of the corporate particle has to do with how the separate components of the corporate particle are coupled to each other.
In the case you described, the size of the corporate particle (atom) was equal to the wavelength of one of its components (electron).
Your example just supports my point. The wavelength and size of an object are not the same in any direction. Parallel or perpendicular to momentum, it doesn't matter. The size and shape of the molecule doesn't change with molecule momentum. The wavelength of the molecule does change with molecule momentum.
A real life example would be the diffraction experiments with the Buckball molecule, C60. The diffraction experiment produced data (diffraction angles) consistent with the De Broglie wavelength. This changed with average momentum of molecules. However, the De Broglie wavelength was different than the diameter of the molecule.
This is why many of these entanglement experiments are done on molecules and atoms of low temperature. By keeping the temperature low, the momentum of the molecule is kept low. This ensures that the De Broglie wavelength is very large. The larger the De Broglie wavelength, the easier the diffraction experiment. The experimenter can get to the point where the wavelength is much larger than the corporate particle.
The diameter of the molecule doesn't change greatly with molecule momentum. The wavelength of the molecule does change with momentum. Therefore, the two can't be the same.