# Wave particle duality again(photons)

1. Aug 21, 2007

I'm getting confused with the idea of probability(Schrodinger) waves and photons. Is the wave side of a photon a probability wave or an electromagnetic wave or both? In the 2 slit experiment with 1 photon at a time, it seems like a photon is described by a probability wave with a probability of going through either slit. However we all know that light is made of electromagnetic radiation. How do these fit together in the case of photons?

2. Aug 21, 2007

### ZapperZ

Staff Emeritus
You might want to start with our FAQ, and then get a copy of this paper:

T.V. Marcella Eur. J. Phys. v.23 p.615 (2002).

He shows in painful detail how you get all of those interference patterns using purely quantum mechanical derivation without invoking any classical wave description. It is also the reason why we don't torture undergraduate physics students to do it this way, because the classical wave picture is so much easier and gives the same results for many practical purposes. But it doesn't mean that such a method in QM doesn't exist.

Zz.

3. Aug 21, 2007

So you are saying it is possible to explain the interference pattern by using probability waves to describe where the photons will be detected. How do probability waves that describe the position of particles correlate to the electromagnetic description of light? You have explained a probability wave/particle duality but not an electromagnetic wave/particle duality. It still doesnt make sense to me.

4. Aug 21, 2007

### Staff: Mentor

You make the correspondence by way of energy, not position. Suppose you have a classical electromagnetic wave with the following electric field:

$$\vec E = {\vec E}_0 \cos (kx - \omega t) = {\vec E}_0 \cos \left[ \left( \frac{2 \pi}{\lambda} \right) x - 2 \pi f t \right]$$

That wave carries a certain amount of energy per second, per square meter of surface perpendicular to the direction the wave is traveling. I'll call it I (intensity or irradiance):

$$I = \frac{1}{2} \epsilon_0 c |{\vec E}_0|^2$$

whose units are joules / sec / m^2.

A photon with frequency f has energy E = hf, so this corresponds to a certain number of photons per second per square meter:

$$n = \frac {\epsilon_0 c |{\vec E}_0|^2} {2 h f}$$

But you have to interpret this number as an average. The actual number of photons fluctuates randomly from one second to the next and from one m^2 to the next, like the number of decays in a radioactive sample fluctuates randomly from one second to the next. If you have a very large number of photons, the fluctuations are so small in percentage terms that you can't detect them.

Beyond that, I don't think you can correlate individual photons with specific configurations of electric and magnetic fields. Photons are not tiny localized "bundles" of $\vec E$ and $\vec B$ fields that propagate independently through space. Neither can you take a classical electromagnetic wave, from a light bulb for example, chop it up into lots of little pieces, and call those pieces "photons."

5. Aug 21, 2007

### meopemuk

This is an excellent question! Indeed, modern physics offers two diferent explanations of the light interference. One (classical) explanation is based on electromagnetic waves satisfying Maxwell's equations. Another (quantum) explanation is based on wavefunctions satisfying Schroedinger equation. A consistent and logical theory cannot have two explanations for the same effect. Which explanation is more fundamental?

I think that quantum-mechanical explanation is more fundamental. In the limit of very low intensity light (when photons can be counted one-by-one) the quantum theory still works well. However, the classical description of light by continuous electromagnetic waves fails.

From this perspective Maxwell's equations in free space can be regarded as surrogate quantum equations for photon's wavefunctions. It also follows that wave properties of light (diffraction, interference, Newton rings, etc.) are, in fact, quantum properties of photons. Grimaldi, Huygens, Newton, Young, Fresnel, Maxwell, et al. were doing quantum mechanics without knowing it.

Eugene.

6. Aug 21, 2007

Thanks for the answers they have been very helpful. One thing i dont understand is this - what is it about a photon that is electromagnetic? If a photon is not a tiny bundle of E and B fields as jtbell states then what is it? A photon has no mass and no charge, so if it was not a bundle of electromagnetic radiation it would have no detectable properties at all - there would be nothing there to detect. It seems to me that one single photon must be a bundle of E and B fields. Did you simply mean that photons ARE tiny bundles of E and B fields but that they dont propagate independently?

7. Aug 21, 2007

### meopemuk

In my opinion, it makes more sense to imagine the photon as a point particle. This particle has zero mass and (generally) non-zero values of such observables as momentum, position, energy, and polarization. The propagation of photons is described by quantum-mechanical wavefunctions. This is the reason for their "wave properties", like diffraction and interference. I think that E and B fields (which are defined for large ensembles of photons with continuously distributed properties) should have definite expressions in terms of photon's wavefunction. However, I don't know the exact nature of this connection.

I don't think it is useful to imagine photons as "bundles of E and B fields". This would be the same mistake as representing electrons as "bundles of wavefunctions".

Eugene.

8. Aug 21, 2007

Now I am even more confused. Why is the momentum of a photon defined using the wavelength of the electromagnetic wave whereas the momentum of an electron (for example) is defined using the wavelength of its probability wave? How do point particles with no electromagnetic fields combine to produce electromagnetic radiation? Sorry to keep asking more questions but this really doesn't make sense to me.

9. Aug 21, 2007

### nrqed

One thing that is important to realize is that a "particle" may have other properties even if it has no mass or charge: momentum, energy, spin (and color and other quantum numbers if we go beyond photons and electrons). In classical, non-relativistic physics having no mass implies no particle (no momentum, no energy, etc). This is not so in relativity.
The way we define an electric of a magnetic field in classical physics is through the *interaction* they have with charged particles. Nobody has ever seen an E field or a B field. what we see are charges reacting to E and B fields.
In quantum physics, the photon is a massless spin 1 particle and it turns out that the way it interacts with charged particles (like an elctron) it leads to an interaction which reproduces the effect classical E and B fields have on charged particles.

10. Aug 21, 2007

### nrqed

Very roughly speaking, the electromagnetic wave of classical physics is the probability wave of photons.

11. Aug 21, 2007

Ok thanks. I guess I just haven't learned enough QM yet to fully understand it. Maybe I'll find out next year of uni. I definitely think I understand it a bit better than before, so thanks for your help.

12. Aug 21, 2007

### meopemuk

This is a good way to put it, in my opinion. However, I think it is also fair to say that the interpretation of Maxwell's equations as analogs of the Schroedinger equation for photons (with probability interpretation and all that) is not well-understood yet.

Don't be surprised if you still don't understand it after learning QM and even QED. These are not easy questions, and I haven't seen their clear explanation in the literature. There is no "canonical" answer, and different people may have completely different views.

Eugene.

13. Aug 21, 2007

### JDługosz

But in radio, we find real E fields. Where do they come from? (I call it real because some potential can be seen driving charges back and forth along a wire)

I was under the impression that E fields (with so many low-energy photons that it can be treated as continuous) do exhibit classical interference. For example, two different radio stations will cause the antenna to receive the sum of the radio waves. That is how jammers can work. Different photons contribute.

Meanwhile, QM wavefunctions only work with a single photon: each photon only interferes with itself.

I'm also interested in understanding this.

—John

14. Aug 21, 2007

### meopemuk

But individual photons also interact with electrons (e.g., Compton effect). I think it is entirely possible that the radio "wave" is just a bunch of photons. Each photon is described by a plane-wave wavefunction, and phases of different photons are correlated. They interact with electrons in the wire forcing them to move back and forth. In a good approximation we can describe this action of billions of photons by two vectors E and B, but it is just an approximation.

I remember watching one of Feynman lectures online (it should be somewhere on YouTube) where he said that quantum interference of two photons from different sources is also possible. He also mentioned an experiment, but I don't recall the details. Can somebody help?

Eugene.

15. Aug 24, 2007

### S1nG

I have this proposal. I was wondering, in theory wise, whether its possible to conclude that an electron particle superpositions itself in an area before passing through a double slit. But only appearing as single when it finally hits the screen/target. I'm still confused with the experiment behind it.