# A question that bothers me - Photon spin up/down state?

1. Jul 19, 2007

### joechien0218

Hi there,

There has been 2 questions that bother me and couldn't find a good answer to them. Maybe some experts out there could help?

Question 1:
It is said that the representation of photon spin-up is
-(1/Sqrt(2)) (1, i, 0)
And the representation of photon spin-down is
(1/Sqrt(2)) (1, -i, 0)
Why? And where did these come from?

The place where I saw the above being used is in Griffiths's text Elementary Particles PG 244 EQ 7.156. That's where he uses these to compute the amplitude with rules derived from Feynmann calculus.

Question 2:
Photon is known to be a spin 1 particle, but why does it have just two states? Shouldn't there be 3? I consulted a book and it says that it has to do with the fact that photon has zero mass. What does mass have to do with this?

Thank you for your response! I appreciate it!
Joe

2. Jul 20, 2007

### meopemuk

The theory of wavefunctions of elementary particles was developed by E.P. Wigner in 1939. He proved that elementary particles should be described by unitary irreducible representations of the Poincare group.

Poincare group = the group of transformations between inertial observers (time and space translations, rotations, and boosts)

"representation" means that these transformations are mapped to transformations of wavefunctions such that essential relationships between transformations (the multiplication law) are preserved.

unitary = probabilities are preserved

irreducible = (roughly) any state of the particle can be reached from any other state by applying Poincare transformations. In other words, elementary particles do not have "internal structure".;

Wigner was able to classify all unitary irreducible representations of the Poincare group and found that they are parametrized by two numbers $m > 0$ and s =0, 1/2, 1, 3/2, ... which can be identified with particle's mass and spin, respectively. For massless particles, parameter s is also called helicity.

Photon is not a truly elementary particle. Its Hilbert space is a direct sum (superposition) of spaces carrying two irreducible representations (m,s) = (0,1) and (0,-1). In classical physics, they correspond to electromagnetic waves with left and right circular polarization.

You can read more about Wigner's theory in the beginning of Weinberg's "The quantum theory of fields".

Eugene.

3. Jul 20, 2007

### Sojourner01

This is a curious statement. Is it possible, in principle, to separate these left- and right-circular polarisations? I know that you obviously can produce circularly-polarised light. Is this a direct manifestation of photon helicity or is it some other, compounded effect?

The thought that one photon is intrinsically two entities boggles my mind.

4. Jul 20, 2007

### marlon

Hey

We discussed this topic before :

You need to know some grouptheory in QFT context though.

QFT shows us that the $$s_{z} = 0$$-state corresponds to a non-physical degree of freedom. Such photons will exhibit a negative probability-distribution. The reason for these troubles are the fact that a photon has zero-restmass.

regards
marlon

5. Jul 20, 2007

### meopemuk

I think that left (or right) circularly polarized light consists of photons of one helicity type. It is easy to understand why definite circular polarization corresponds to an irreducible representation of the Poincare group. The photon will keep its polarization (left or right) no matter from which reference frame one is looking at it.

Eugene.