# Do particles behave like electromagnetic waves?

From double-slit experiments we know particles have wave-like behavior: they statistically form an interference pattern.

My question is: Is this wave-like behavior similar to the photons' behavior?

More specifically:

• Do they behave like transverse waves? For example: can you polarize an electron beam?
• Can you (internally) reflect and refract a particle beam? For example can you make a lens or prism the refract electron beams?
• Although photons are neutral they can cause current in a receiver rod, so radio works. Can other particles do this? For example can you tune a radio to receive very slow neutrons or electrons (whose de Broglie wavelength is sufficiently large)
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Electron microscope exploits wave-like properties of electrons, and has lenses for electrons. – yuggib Jul 10 '14 at 9:12
For the first question see What does a de Broglie wave look like? – John Rennie Jul 10 '14 at 9:45
To the last point: radio "works" because of a stimulated E-M field within the antenna. to the extent you could create a stream of electrons w/ some sort of modulation, the EM field they cause might possibly be picked up by some antenna. – Carl Witthoft Jul 10 '14 at 11:57
@CarlWitthoft - and that would be the classic "spark gap" transmitter from the early part of last century, so indeed it works. Of course, that stream of electrons just radiates radio waves... – Jon Custer Jul 10 '14 at 14:09
@yuggib: But it uses electric fields isn't it? I wonder whether slow particles eg. neutrons can be refracted by eg. a glass. – Calmarius Jul 10 '14 at 17:17

As John Rennie, says, what-does-a-de-broglie-wave-look-like has helpful answers which you should read first, but I don't think they are complete.

Do they behave like transverse waves?

No - the wave function for a single particle with no spin from the Schrodinger equation is just a scalar so there is no direction connected with it.

For example: can you polarize an electron beam?

You can polarise an electron (it has spin 1/2, so two options for spin direction). However, the spin part of an electron's wavefunction is separate from the spatial (wave) part - this is why the Schrodinger equation works for electrons even though it ignores spin. Thus the de Broglie wave itself is unaffected. (I believe this a valid alternative to the answer that a spin 1/2 particle has two de Broglie waves.)

Can you (internally) reflect and refract a particle beam? For example can you make a lens or prism the refract electron beams?

As yuggib mentioned, electron microscopes work by refracting electron beams. However, they use electric and magnetic fields in vacuum - so not much like ordinary refraction. Moreover, the lenses are well-described by classical physics. The problem is that the particles we are familiar with (other than the photon) have very short ranges in ordinary matter.

You can certainly diffract particle beams using crystals, in a similar way to the diffraction of x-rays by crystals, or light by a grating. The question is about refraction, though. See below.

can you tune a radio to receive very slow neutrons or electrons ..?

1. While a single particle has a wave function which can be a simple wave, the wave function of two particles is a function of the the positions of both particles (ie a function of 6 variables), etc. Not so easy to visualise.
2. The wave function is a complex number. In fact, for a simple de Broglie plane wave, the modulus is constant (the particle can be anywhere) - only the argument varies (known as the phase in quantum mechanics).
3. The argument (phase) can be changed without any physical change. For example, this answer by Lubos mentions that you can include or not include the rest energy of a particle ($E=mc^2$) in the formula $E=h\nu$, which changes the frequency, without changing the behaviour of the wavefunction. Clearly this wouldn't work with classical electromagnetic radiation - the tuning dial on the radio shows the frequency.
4. The velocity (phase velocity) of a de Broglie wave is $c^2/v$ - so it equals c for a photon whose velocity $v$ is $c$, the speed of light, but is greater than $c$ otherwise. The group velocity (speed at which a wavepacket travels) is the speed of the particle. This isn't a problem, but shows how significantly massless particles like photons differ.
5. Ron Maiman explains that

If you have many Bosons in a superposition state where they all share the same quantum state, their wavefunction becomes a classical field which obeys the Schrodinger equation.

He doesn't mention it, but I believe this starts to explain why light exists. Photons are bosons - particles that can be in the same quantum state, unlike electrons (fermions) for which the exclusion principle forbids this. In an ordinary beam of light there are many photons in the same or similar states. Somehow their collective wavefunction is manifested as real (not complex) electric and magnetic fields.

I therefore conclude the answer is no - you cannot use fermions such as electrons to produce a signal in an "electron radio".

I would be interested to know if @RonMaimon or @LubošMotl agree.

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you are not stressing that these are probability waves, not matter waves. – anna v Sep 14 '14 at 16:03
For your explanation of light: the "wavefunction" for the photon can be taken to be the vectors $\vec{E}=\left(\left.\left<0\right|\right. \hat{E}^+_j\left.\left|\psi\right>\right.\right)_{j=1}^3$ and $\vec{H}=\left(\left.\left<0\right|\right. \hat{H}^+_j\left.\left|\psi\right>\right.\right)_{j=1}^3$, where $\hat{E}^+_j$ and $\hat{H}^+_j$ are the positive frequency parts of the electric and magnetic field observables and $\psi$ is the photon state (expressed as a superposition of one-photon Fock states). This "wavefunction" is not quite the same as for an electron, as there is no position ... – WetSavannaAnimal aka Rod Vance Dec 17 '14 at 9:09
... but its square magnitude is (1) the probability density to photodetect (destructively detect a photon by absorption), (2) uniquely defines the one-photon state $\psi$ and thus can be taken to be equivalent to the latter and (3) fulfills Maxwell's equations. The means of the obsverables $\hat{E}$ and $\hat{H}$ (i.e. $\left(\left.\left<\psi\right|\right. \hat{E}^+_j\left.\left|\psi\right>\right.\right)_{j=1}^3$ and $\left(\left.\left<\psi\right|\right. \hat{H}^+_j\left.\left|\psi\right>\right.\right)_{j=1}^3$ when the light field is in a coherent state (different from a one-photon ... – WetSavannaAnimal aka Rod Vance Dec 17 '14 at 9:15
.. also fulfill the same Maxwell's equations, and so they both (1) uniquely define the one photon states that can exist for the prevailing boundary conditions (even though they define a coherent state, and not a one-photon state) and (2) if the coherent displacement parameter is big enough (i.e. if the light field is coherently displaced far enough from its ground state), then these means become the same as what we classically measure the electric and magnetic field to be. – WetSavannaAnimal aka Rod Vance Dec 17 '14 at 9:20
I've added to all this in my own answer. +1 by the way, and thanks for referring me to the wonderful Ron Maimon answer, I can't read enough of this guy! – WetSavannaAnimal aka Rod Vance Dec 17 '14 at 9:51

Thinking of the photon itself as a single particle, and imagining the building up diffraction patterns with one photon in the apparatus at a time will give you an idea of the correspondence between classical wavelike behaviour and the probability waves of the wavefunction, at least for bosons. The following, I believe, is what Akrasia's answer means when Akrasia says "He doesn't mention it, but I believe this starts to explain why light exists..." and what Ron Maimon in the opening paragraph to his excellent answer means.

The "wavefunction" for the photon can be taken to be the vectors:

$$\vec{E}=\left(\left.\left<0\right|\right.\hat{E}^+_j\left.\left|\psi\right>\right.\right)_{j=1}^3$$ $$\vec{H}=\left(\left.\left<0\right|\right.\hat{H}^+_j\left.\left|\psi\right>\right.\right)_{j=1}^3\tag{1}$$

where $E^+_j$ and $H^+_j$ are the positive frequency parts of the electric and magnetic field observables and $\psi$ is the one-photon state, expressed as a superposition of one-photon Fock states. This "wavefunction" is not quite the same as for an electron, as there is no position observable, at least in the same sense as for the nonrelativistic electron Schrödinger equation. (Note that there are difficulties with defining a position observable for the relativistic Dirac electron too, so you can think of the nonexistence of the true "wavefunction" as an assertion that there is no nonrelativistic description of light). But this six-component wavefunction uniquely defines the one-photon state, so that the two can be taken as equivalent. Now, witness the following interesting physical interpretations and theoretical observations about this six component field:

1. Its normalised "square magnitude" $\frac{1}{2}\,\epsilon\,\|\vec{E}|^2 + \frac{1}{2}\,\mu\,\|\vec{H}|^2$ is the probability density in space and time to photodetect the photon, i.e. destructively detect a photon by absorption with, say, a PMT. So it defines the interference / diffraction pattern that you will build up over time in a one-photon-in-the-apparatus-at-a-time experiment;

2. As already stated, it uniquely defines the one-photon state $\psi$ and contrariwise, so that the two can be taken to be equivalent information and thus we can think of this vector field as the one photon state;

Now we suppose that the light field is in a Coherent State, i.e. the field state is of the form:

$$\psi = \prod\limits_j\,\exp\left(\alpha_j\,a_j^\dagger-\alpha_j^*\,a_j\right)\left.\left|0\right>\right.\tag{2}$$

where $a_j^\dagger$ is the creation operator raising the unique quantum field ground state $\left.\left|0\right>\right.$ to the one photon Fock state in the field mode corresponding to the plane wave $\exp\left(\vec{k}_j\cdot\vec{r}-\omega_j\,t\right)$ and the $\alpha_j$ - the "Displacements" - define the strength of the excitation in each mode. Take heed that in a coherent state, there is superposition of number states in each mode, so that a coherent state the photon number is uncertain and Poisson-distributed, unlike for the one photon Fock state superposition. The "displacements" get their name because they behave like "components" of a displacement "vector" displacing the Wigner quasiprobability distribution away from the origin (ground state) in quantum phase space.

Now withness the means of the field observables $\hat{E}$ and $\hat{H}$:

$$\left(\left.\left<\psi\right|\right.\hat{E}^+_j\left.\left|\psi\right>\right.\right)_{j=1}^3$$ $$\left(\left.\left<\psi\right|\right.\hat{H}^+_j\left.\left|\psi\right>\right.\right)_{j=1}^3\tag{3}$$

(take heed of the subtle difference between (1) and (3)) as well as their following extremely interesting properties for coherent states:

1. The fulfill the same Maxwell's equations as fulfilled by the "wavefunction" in (1), therefore they uniquely define the one photon states that can exist for the prevailing boundary conditions (even though they define a coherent state, and NOT a one-photon state);
2. If the coherent displacement parameters $\alpha_j$ are big enough (i.e. if the light field is coherently displaced far enough from its ground state), then these means become the same as what we classically measure the electric and magnetic field to be.

So, in (1), Maxwell's equations and their boundary conditions define probability waves that determine how our diffraction pattern will be built up one photon at a time. The objects defined in (3) are what our classical measurements of the Maxwellian electromagnetic field vectors would measure, and they are the same Maxwell equations and boundary conditions!

So, if you get a vector voltmeter and magnetometer and measure classically the field distributions in a microwave setup, you are measuring the objects in (3) and, by our discussion above, you are also exactly measuring the one-photon states of the quantum field that can exist for the same boundary conditions.

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