# Photon spin

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Hello...

I have a question about the spin of the photon.

I have heard that the photon is considered a spin=1 particle, and that is the electromagnetic force-carrier.

My question is, why does the energy of a photon have a two-fold degeneracy instead of a three-fold given by the projections of the spin=1? Does this have to do with the fact that the photon is a massless particle?

And another question: Do all force-carriers particles have to be bosons?

Greetings...

Nicolas.

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My question is' date=' why does the energy of a photon have a two-fold degeneracy instead of a three-fold given by the projections of the spin=1? Does this have to do with the fact that the photon is a massless particle?

[/quote']

Bingo! Give the man a prize.

Yes, the fact that the J=0 state is supressed for photons is a direct consequence of the fact that the photon doesn't have a rest frame.

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Ok, ok... do not applause anymore

That was just a question I kept in mind before a modern physics class.

Thanks a lot Tom, but could you explain why exactly does that imply the supression of the J=0 state?

Nicolas.

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It is not one of these really obvious things unfortunately. Let me try to explain as simply as I can, and you can let me know if you understood.

Maxwell's equations are:

$\begin{array}{ll} \displaystyle \vec{\nabla} \cdot \vec{E} = \rho \hspace*{5cm}& \vec{\nabla} \cdot \vec{B}=0 \\ \displaystyle \vec{\nabla} \times \vec{E} + \frac{\partial \vec{B}}{\partial t} = 0 & \displaystyle \vec{\nabla} \times \vec{B} - \frac{\partial \vec{E}}{\partial t} = \vec{J} \end{array}$

These are more nicely written in terms of a 4x4 matrix (a tensor) $F^{\mu \nu}$ defined by

$F^{\mu \nu} \equiv \left( \begin{array}{cccc} 0 & -E_1 & -E_2 & -E_3 \\ E_1 & 0 & -B_3 & B_2 \\ E_2 & B_3 & 0 & -B_1 \\ E_3 & -B_2 & B_1 & 0 \end{array} \right)$

$\mu$ and $\nu$ are numbers which run from 0 to 3, with entry 0 corresponding to the time component, and 1 to3 being the usual space components.

Then, the free (no source) Maxwell equations become:

$\partial_{\mu} F^{\mu \nu} = 0$

We can write these equations in terms of the photon field $A^{\mu}$ instead

$F^{\mu \nu} = \partial^{\mu} A^{\nu} - \partial^{\nu} A^{\mu}$

So

$\partial_{\mu} F^{\mu \nu} = \partial_{\mu} \partial^{\mu} A^{\nu} - \partial_{\mu} \partial^{\nu} A^{\mu}=0$

Notice that I can change the field by an amount without changing F

$A^{\mu} \to A^{\mu} + \lambda \partial^{\mu} \phi \qquad \Rightarrow \qquad F^{\mu \nu} \to F^{\mu\nu} + \lambda \partial^{\mu} \partial^{\nu} \phi - \lambda \partial^{\nu} \partial^{\mu} \phi = F^{\mu \nu}$

This is called a gauge transformation. I can change the photon field without changing the physics.

This lets me make an extra equation (called fixing the gauge) to make life simpler. I usually choose:

$\partial_{\mu} A^{\mu} =0$

which is the Lorentz gauge. Then, the (free) maxwell equations become

$\partial^2 A^{\mu}=0$

This has solutions $A^\mu = \epsilon^\mu e^{i q \cdot x}$ where $\epsilon^\mu$ is the polarization 4 vector (with 4 degrees of freedom), and $q^2=0$ (which is the statement that the photon is massless).

The Lorentz gauge condition on this solution implies $q \cdot \epsilon = 0$. This removes one degree of freedom (the time-like polarization).

Also notice that I can change the photon field further $A^{\mu} \to A^{\mu} + \partial^{\mu} \chi$ and everything remains the same as long as $\partial^2 \chi=0$. In other words, I have not completely 'fixed the gauge' yet. This allows me to impose another condition

$A^0=0$

This is called the Coulomb gauge. but,

$A^{0} = 0 \quad \Rightarrow \quad \epsilon^{0} =0 \quad \Rightarrow \quad\vec{q} \cdot \vec{\epsilon} = 0$

The Coulomb gauge has removed another degree of freedom from the polarization vector, removing the longitudinal polarization.

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Thanks Severian! I understood so far... I've got some notions on relativity, four-vectors and all that. Although I didn't understood the whole equations I do see the physics behind.

But your demonstration doesn't finish right there, does it? If so I will have to read it several more times to get the point. If not, I will be waiting anxiously the last part of it.

Thanks again!

Greetings.

Nicolas.

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That is it I'm afraid. The statement $\vec{q} \cdot \vec{\epsilon}=0$ is saying that the polarizations are only transverse.

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Yes! I got it!

I briefly discussed your explanation, Severian, with a friend (he is studying fields theory) and I understood.

So it's indeed a relativistic effect. I first oriented my question to the quantic nature of the photon. Anyways, all is connected

Thanks again!

Nicolas.

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So it's indeed a relativistic effect. I first oriented my question to the quantic nature of the photon. Anyways' date=' all is connected

[/quote']

Yes, that's right. It was the choice of frame (via the Coulomb gauge) which removed the longitudinal polarization. It is definitely not quantum, since there was absolutely nothing quantum in my discussion - this was classical electromagnetism.

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• 2 years later...

Does this mean that the polarization is 2-dimensional, and hence a photon is like a spin 1/2 particle ?

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Does this mean that the polarization is 2-dimensional,

Yes. More precisely, the space spanned by the photon's polarization vectors has dimension 2.

and hence a photon is like a spin 1/2 particle ?

No. First and most basic: 1/2 does not equal 1! The photon carries more spin angular momentum than say, an electron. And second, if the photon were spin 1/2 it would be a fermion. But the photon doesn't obey Fermi-Dirac statistics, it obeys Bose-Einstein statistics.

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I heard a photon was a vector-particle, how do we prove this ?

btw, Do spin 1/3 particle exist ?

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I heard a photon was a vector-particle, how do we prove this ?

Show that the angular momentum changes when it interacts. IIRC, vector means spin 1. Scalar is spin 0.

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I don't really understand the link between angular momentum and spin.

But I heard that "longitudinally" polarized photons were not "seeable", but would "transmit the electromagnetic force". Hence longitudinally polarized photons would make that 2 magnet attract themselves, but of course there is no "light" between the magnets. Is this correct in some sense ?

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I don't really understand the link between angular momentum and spin.

Spin is one form of angular momentum.

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And angular momentum is $m\vec{r}\wedge\vec{\omega}$ or opposite ?

But spin is eigen momentum (spin on oneself, except that with 180 degrees you find yourself as initially) ?

My question was : how to link the spin n (1/2 or 1) to matrices (Pauli or spin 1 matrices)

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Since the photon is a spin-1 particle, how do we recognize it out from a polarization measure, since only +,- are possible (dimension 2), from a spin-1/2 system ?

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I don't really understand the link between angular momentum and spin.

We know that the angular momentum must be conserved, because otherwise the universe would have physical laws which depend on the direction you are facing (this is Noether's theorem). We also know that the Dirac Hamiltonian (which describes the motion of free electrons) does not commute with the traditional angular momentum operator, so traditional angular momentum is not conserved. We also know that the solutions of the Dirac equation are eigenstates of another operator, but at this point we have no physical idea of what this operator represents.

Then we notice something special. If we add the traditional angular momentum operator to this new operator, we find that the sum does commute with the Hamiltonian. The new quantity is conserved, so we interpret this new quantity as an internal angular momentum of the electron which we call spin. In hindsight, we can go back and realise that the spin eigenstates are themselves fundamental representations of the Poincarre algebra, which reinforces out idea.

Since the photon is a spin-1 particle, how do we recognize it out from a polarization measure, since only +,- are possible (dimension 2), from a spin-1/2 system ?

There are lots of reasons. Firstly, Maxwells equations require a vector potential, or in other words a spin-1 object. Then we notice that photons obey Bose Einstein statistics, so no Pauli exclusion principle (which they would have if they were spin-1/2). Also, their couplings to electrons would be entirely different if they were fermions.

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Hi, I've got a question that's bugging me.

Why does light move? Why doesn't it stay in one place? What happens to a photon when it hits an object, is it destroyed?

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Hi, I've got a question that's bugging me.

Why does light move? Why doesn't it stay in one place? What happens to a photon when it hits an object, is it destroyed?

Light is an electromagnetic wave — the solution to the wave equation is something that always moves at c (in a vacuum). Photons can't not move.

Photons can scatter or be absorbed by atoms or nuclei. You get a photon of another energy, or the photon is destroyed. But you can make photons, too.

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The Pauli matrix for a spin-1/2 particle is 2x2

I thought that the dimension of the matrix representation of a spin s was (2s+1)x(2s+1) (spin 0 is a 1x1=1 scalar, 1/2 is 2x2, spin s=1 is 3x3, aso), but then the Dirac matrices are 4x4 which should be a spin 3/2, and if gravitons would exist, they should be by 5x5 matrices represented ?

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I thought that the dimension of the matrix representation of a spin s was (2s+1)x(2s+1) (spin 0 is a 1x1=1 scalar, 1/2 is 2x2, spin s=1 is 3x3, aso), but then the Dirac matrices are 4x4 which should be a spin 3/2, and if gravitons would exist, they should be by 5x5 matrices represented ?

You are forgetting the antiparticles. The Dirac matrices need to be 4 dimensional to have a spin up electron, spin down electron, spin up positron and spin down positron.

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Right, those are block (anti-)diagonal matrices.

Then, how are s=3/2 particles represented (e.g. gravitini, I suppose) ?

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