# Polarization of individual photons.

1. Mar 17, 2010

### G01

Hi Everybody,

I'm currently working my way through Sakurai's second book, "Advanced Quantum Mechanics."

I'm getting held up on a short several paragraphs on the connection between photon spin and polarization.

On pages 42-43, Sakurai states:

"Since the polarization transforms like a vector, the general theory of angular momentum encourages us to associate it with one unit of angular momentum. This is what is meant by the statement that they photon has one unit of spin angular momentum. [He then goes on to describe the circular polarization basis, and how individual photons are have spin parallel or antiparallel to the propagation direction.]"

So does this mean that all photons taken individually, are always circularly polarized? I want to say that I'm missing something, but this seems to be what he is saying. If someone could clarify this discussion on pages 42-43 of Sakurai's AQM, I'd really appreciate the help.

2. Mar 18, 2010

### ytuab

All I know about the photon spin is as follows; (Though I don't have that Sakurai's book. )

The photon spin $$\pm$$ 1 correspond to the electromagnetic(E-M) waves with left and right circular polarization.
So the "linear" polarization corresponds to the superposition of these two states. (Though it's a little difficult to imagine.)

The QED shows the photon has the $$\pm\hbar$$ angular momentum. (To put it simply, the fact that the light polarization is the "vector" not spinor, leads to its "spin 1")

The "experimental evidence" of the photon spin 1 is as follows;
The fine structure of the hydrogen atom first indicated the relativistic energy difference between 2S and 2P in the Sommerfeld model.
But after the "electron spin" appeared, the interpretation of the fine structure was changed to the spin-orbital interaction (like the energy difference between 2P1/2 and 2P3/2 of the hydrogen atom).
So the "unnecessary" transition like 2S to 1S need to be inhibited to explain the hydrogen spectrum results.
The selection rule(2S --x--> 1S) means the existence of the photon spin 1.
[But, the energy levels of 2S1/2 and 2P1/2 are the same. So I wonder if this selection rule is really necessary or not. A little complex Lamb shift seems to be caused by the two photons...]

The polarization axis of one photon is important when we consider its interaction with the polarizing filter like in the Bell test experiment.
For example, when the angle difference between the photon polarization axis and the polarizing filter is 6 degrees,
The probability that the photon can pass through this filter is $$(E^2) \cos^2 6 = 0.99$$ = 99%.

(But personally I wonder why the photon particle (not wave) with the polarization axis different from the filter can pass through this filter so much.
If the photon is a "indivisible" particle, even when the angle difference is as small as 6 degrees, the particle can't pass through the filter at all, as a key doesn't match with a keyhole? )

Last edited: Mar 18, 2010
3. Mar 18, 2010

### G01

OK. So, then the spin basis is |1> |0> |-1>. Then a photon with linear polarization would be in the spin state:

$$|\chi>=\frac{1}{\sqrt{2}}|1> + \frac{1}{\sqrt{2}}|-1>$$

?

Does the m=0 state for the photon play a role in polarization states?

4. Mar 18, 2010

### ytuab

That's "probably" right.
Sorry, personally I wonder what the photon spin , superposition and entanglement really are.
So I also need some other persons' help.

R = right circular polarized wave, L= left circular polarized wave,
H=horizontal linear polarized wave, V=vertical linear polarized wave,
So the H linear wave can be expressed as,

$$|H\rangle = \frac{1}{\sqrt{2}}|R\rangle + \frac{1}{\sqrt{2}}|L\rangle$$

And V linear wave is,

$$|V\rangle = -\frac{i}{\sqrt{2}}|R\rangle + \frac{i}{\sqrt{2}}|L\rangle$$

These are all one photon state which have the superposition of R, L, H and V.

5. Mar 18, 2010

### Ben Niehoff

No. m=0 would correspond to longitudinal polarization, which photons cannot have because they are massless (think back to freshman E&M: EM waves are always transverse).