# Electromagnetic wave in quantum physics?

1. Sep 27, 2005

### pivoxa15

In quantum physics light is considered both (or neither) a probability wave and a photon. But where does the eletromagnetic wave come into the scheme of things in quantum physics?

Is the electomagnetic wave merely an approximation to many different types of discrete photons (ranked by their energy level)? Hence instead of light being a range of frequencies on the electomagnetic wave, it is a range of energy levels of finitely many photons? So photons with lower or higher energy levels than those does not register in our eyes, so they would be other sources of radiation such as the so called microwaves, radiowaves, gamma rays, x rays and so on.

Thanks

2. Sep 28, 2005

Staff Emeritus
No, the single photon can manifest as an EM plane wave, for example. Wave particle duality is serious, not just a statistical phenomenon of little bullets "doing the wave".

3. Sep 28, 2005

### pivoxa15

I thought wave particle duality was between the photon and the probability wave (given by the Schronger eqn)?

Or is wave particle duality referring to how light exists as photons and an EM wave at the same time. So the probability wave merely gives an indication of where one is likely to detect the photon when measured?

4. Sep 28, 2005

### MalleusScientiarum

Wave-particle duality states that everything "particulate" in nature can diffract, self-interfere, and on much like a wave. This means electrons, protons, nuclei, you, me, my dog, etc.

5. Sep 28, 2005

### pivoxa15

So what kind of wave is this? In the case of photons, is it the electromagnetic wave?

How does the probability wave (given by Schrondger) come into it?

6. Sep 28, 2005

### Spin_Network

The concept of "Duality" is based on detection of waves and particles, Photons and E-M waves.

The Photon, is the leading edge detected by a detector that is tuned specifically for the detection of..Photons!

E-M waves travel in unison with Photons, they are one and the same. Think of Photons as being non-existing until you place some sort of detecting devise in their path, then hey presto you have Photons. Likewise, you can detect E-M waves using a devise that is tuned likewise.

What you cannot do is detect both E-M waves, and Photons with a "single" detector, although they travel in unison, they cannot arrive at the single detector at the same instant, in this sense both PARTICLES and WAVES are dependant upon Detectors for their existence.

Waves are confirmed by 'wave-detectors', and likewise Particles are confirmed by 'particle-detectors'.

The "wave-particle-duality" and the collapsed wavefunction, are nothing more than a take on the "detector-paradox" stated above. Restate the concept of duality into the notion of source of observation/detector, and you can think this:To collapse a wavefunction you need a Particle-Detector, Observer?

7. Sep 29, 2005

### pivoxa15

Thanks Spin_Network, that clears a few things up but I have more questions. What do you mean by a collapsed wavefunction?

So the probability wave function is just a prediction of where a photon or the EM wave is likely to be at a given instant?

What about the explanation where if there are no observers or detectors, the EM wave and photon is everywhere and only when you detect it will it be at a specific place. At the stage where it is everywhere (hence no one watching), what would you call it? i.e a photon is a particle. An EM wave is a wave. When something (photon, EM wave, electron) is everywhere it is a ...

8. Sep 29, 2005

### the blob inc

although wave-partical duality was thought up be people that are far more inteligent than i, logically how can you have one thing that appears to be two. and in the case of waves in general do you not require a or an "aether" or medium for waves to traval.

9. Sep 29, 2005

Staff Emeritus

The "thing" is richer than the categories particle and wave would suggest. Particle and wave are the two ways we can observe it, but when it's at home it's an infinite dimensional amplitude function, that doesn't live in spacetime. The function has to be "reduced" or "projected" down by observation to become a probability of observing the wave properties or the particle ones.

The photon wave is a variation in the quantum electrodynamic field, which is present in all spacetime. It is not an ether; the ether has been dropped from modern physics as unnecessary.

10. Sep 29, 2005

### Ratzinger

EM waves are classic beasts that cary energy. The total energy of an EM wave that occupies some volume equals nhf. n- number of photons, h- planck constant diveded by 2pi, f- frequency of the classic wave. Probability waves carry statisical information about photons.
So the duality is between the photon and the probability wave.

Photons are bosonic particles that can stay close together that's why you get classic fields and waves.

But don't take my word for it, it's just a layman talking.

EDIT: Well, there is of course also a duality between photons and EM waves. Both got energy. So there are two dualism. Or what? help here, please

Last edited: Sep 29, 2005
11. Sep 29, 2005

### masudr

Surely EM waves are the result of classical EM theory and are solutions of the equations provided by the Lagrangian density

$${\cal L}=-\frac{1}{4c\mu_0}F^{\mu\nu}F_{\mu\nu}$$

in the language of classical field theory?

And photons are the result of QFT of which I have no knowledge; so I hope someone can fill that in, but I assume it involves quantising the field variable $A^\mu$ and it's conjugate variable, and ensuring they satisfy the commutation relations.

In any case, my point is that EM waves and photons are results from two very different theoretical machines. All we can do is draw some similarities from them, but surely hoping for any real correspondence is as futile as expecting momentum to be given by $\vec{p}=m\vec{v}$ in quantum mechanics?

EDIT: Corrected LaTeX

Last edited: Sep 29, 2005
12. Sep 29, 2005

### pivoxa15

I haven't formly studied any of this but the energy of a photon is given by E = hf.

It was Einstein who suggested this in 1905 where the electomagnetic radiation is quantised and exists in elementray amounts we call photons.

the f in the equation corresponds to the frequency of the electomagnetic wave so a particular photon with a particular energy corresponds to an EM wave with a given frequency (that was used to calculate the photon energy). So from this they look to be the same thing. According to Spin_Network, this theory has been experimentally verified as well.

13. Sep 30, 2005

Staff Emeritus
The Lagrangian for Quantum Electrodynamics (QED) contains the EM Lagrangian you cite as one of its terms. A second term describes Dirac's relativistic electron, and the third term describes their interaction. From Peskin & Schroeder equation (4.3):

$$\mathcal{L}_{QED} = \bar{\psi}(i\kern+0.1em /\kern-0.55em \partial - m)\psi - \frac{1}{4}(F_{\mu\nu})^2 - \epsilon \bar{\psi}\gamma^{\mu}\psi A_{\mu}$$

(Edit: Thanks to Tom Mattson for the LaTeX to put a stroke through the del.)

This is before quantization of course, but it shows that QED is not a contradiction of EM but an enlargement of it.

Last edited: Sep 30, 2005
14. Sep 30, 2005

### Tom Mattson

Staff Emeritus
$$\kern+0.1em /\kern-0.55em \partial$$

Don't ask me what those "kern" terms mean though!

15. Sep 30, 2005

### masudr

Is that literally all there is to it?

And can someone explain exactly what $/\kern-0.55em \partial$ means?

EDIT: Wikipedia says the Lagrangian is

$$\mathcal{L}=\bar\psi(i\gamma_\mu D^\mu-m)\psi -\frac{1}{4}F_{\mu\nu}F^{\mu\nu}$$

with $D^\mu$ being the gauge covariant derivative $D_\mu = \partial_\mu+ieA_\mu$. The interaction term appears to be missing, though. I also wondered if someone could explain how this covariant derivative is related to the covariant derivative from general relativity whereby it's just the partial derivative plus the connection coefficient that makes the metric tensor vanish. I don't really get how the space for the Dirac spinors can be curved, or maybe I'm getting confused with something I don't understand.

Last edited: Sep 30, 2005
16. Sep 30, 2005

### Staff: Mentor

Substitute the expression for the gauge covariant derivative into Wikipedia's Lagrangian, expand things a bit to get rid of some of the parentheses, and you should see the interaction term appear. You'll also get the answer to your first question.

17. Oct 1, 2005

### Blackforest

As far as I can understand the work of Cartan and of Dirac, we don't have a theory for a quantum general relativity. The definition of a spinor given by Cartan is unsufficient to push the theory of spinors in the world of the GR and Dirac did "only" (and that's enorm) make the junction between the special relativity and the QT. One could perhaps extend the relation concerning the matrices of Dirac in arguing that it holds "in average", writing: $\gamma_\alpha.\gamma_\beta + \gamma_\beta.\gamma_\alpha = g_{\alpha\beta}$ where $g_{\alpha\beta}$ is any value of the metric tensor and <$g_{\alpha\beta}> = \eta_{\alpha\beta}$. A subsidiary remark can be done within my personal approach of the $F_{\mu\nu}$ tensor that allows the introduction of the Christoffel's symbols of second kind in the continuation of the theory of Cartan; but personal theories are not wellcome on these forums. Best regards

18. Oct 1, 2005

### masudr

It does indeed. Cheers.