# Quantum field theory, photons, and polarization.

1. Feb 18, 2013

### Spinnor

A circularly polarized electromagnetic wave can be thought of proper combinations of orthogonal linear polarized waves, and a linear polarized wave can be thought of proper combinations of left and right circularly polarized waves. It seems one type of wave is no more fundamental then the other, circularly polarized verses linear polarized. In the quantum field theory of photons is this still the case or not?

Can I think of both linear and circular polarized classical electromagnetic waves as proper combinations of many spin 1 particles? Is there another way of thinking of linear and circular polarized classical electromagnetic waves? Is it the spin 1 that is fundamental?

Hope I'm clear, thanks for any help!

2. Feb 19, 2013

### tom.stoer

In QED it's nothing else but the choice of basis vectors in Fock space

3. Feb 19, 2013

### Zarqon

From a mathematical point of view yes. But from a more "intuitive" point of view, would you say that it's more natural to consider the fundamental state of a particle with spin = 1 to be one with some form of inherent rotation, i.e. being circularly polarized?

4. Feb 19, 2013

### andrien

it is possible to prove that circularly polarized em wave has spin value of 1.However in the case of linear polarization,it is a linear combination of +1 and -1 spin state.In quantum theory of photons,however all photons are either spin +1 or -1.

5. Feb 19, 2013

### Spinnor

Thank you, that answers my question, I'm guessing Tom answered my question as well but was not sure how to interpret his answer. Thanks to all!

6. Feb 19, 2013

### Avodyne

Not so. There can be arbitrary linear combinations of the +1 and -1 states, including states that correspond to linear polarization.

7. Feb 20, 2013

### andrien

can you exactly point a reference for photons,not light wave because there is something which is preventing me from that.

8. Feb 20, 2013

### Avodyne

See section 10.6.3 of Mandel and Wolf, Optical Coherence and Quantum Optics.

9. Feb 20, 2013

### Spinnor

10. Feb 20, 2013

### Cthugha

Let me respond with a (deliberately) ridiculous question in return: You have a 2d coordinate system and two vectors along the x- and y-direction, respectively. Is a vector going at 45 degrees to both less fundamental than these two are?

To get a bit less ridiculous: It is important to understand what "spin 1 particle" or s=1 particle means in this case and what the number really means. It is not the total spin angular momentum of the photon (which is $$\hbar \sqrt{(s (s+1))}$$, but it rather defines the range of results you can get in an experiment if you choose to measure the spin along one certain axis (typically the z-axis is chosen). Now if you perform a measurement along this axis, you will measure a value between -s and s with only integer steps allowed. For spin-1 particles, the allowed values are therefore -1,0,+1.

Photons as massless particles are special and for free photons the value of zero is not allowed (and one should rather talk about helicity than about spin). Therefore, you will only get a value of -1 or +1 if you measure along this certain axis, so these are the helicity states. Note that the direction of the spin in the plane perpendicular to this axis is undefined!

However, any normalized superposition of these two helicity states is allowed, too. These states then correspond to elliptically or linearly polarized light. If you have an equally weighted superposition of the two helicities, you will get both -1 and +1 when measuring along the z-axis with equal probability and the expectation value of that measurement along that axis will be zero. I do not see any reason to consider this state less fundamental than the helicity states just because they are not eigenstates(*) of the spin measurement along the z-axis. For example they are instead eigenstates of measurements along the x-axis and y-axis, respectively. This is something the helicity states are not - and nobody would consider helicity states less fundamental than the linearly polarized states because of that.

Also note that states in qm are (in most interpretations) a statistical concept and a single photon state does not mean a single measurement, but a series of measurements on an identically prepared ensemble of single photons.

(*) Using the term eigenstate in connection with photons is a bit complicated, but it is ok at the initial level to get a rough understanding.