what is the spin of photons ?

[Widdekind]

Photons have an intrinsic angular momentum, pointing parallel or anti-parallel to their path of propagation, with a magnitude of exactly one unit of h-bar... yes?

 

Thus, when electrons emit or absorb photons, they must spin-flip, which represents a change of -1/2 hbar <----> +1/2 hbar = one unit of h-bar... yes?

[John Cuthber]

Something like that

http://en.wikipedia.org/wiki/Selection_rule#Angular_momentum

[Enthalpy]

Parallel or anti- to the propagation are not the only possiblities!

 

A photon can have a straight polarization also, like vertical, or horizontal, or biassed. This is just as fundamental a base as circular polarization, including for one single photon.

 

Or the polarization can be elliptic. It's just that elliptic polarizations are less simple to make orthogonal, and one would have to tell how much elliptic.

 

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A linear photon can be detected by a circular receiver, either left or right. Similarly, a circular photon fits a linear detector, either 17° or 107°. The photon has just 50% chances when it "decides" at the detector, instead of 0% or 100% if the nature of the polarization matches.

 

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Same picture with atomic orbitals. A 2p orbital can be doughnut-shaped, with the wave's phase turning 360° per geometric turn (360*1=360, quantum number=1), hence a circular polarization. Or it can be peacock-shaped, with the wave positive at one side and negative at the other, hence a linear polarization.

 

This is just for the phase of the time-independent part. Multiply by some exp(j*w*t) depending on the energy (...which is not absolute).

 

Just like photons, a peacock 2p is a 50-50 sum of right and left turning doughnut 2p, AND a doughnut is a (phased) sum of N-S and W-E peacocks. Both are equally "fundamental".

 

Better: if the transition is from or to an s orbital, the superposition of p and s has the corresponding polarization, and so does the photon.

 

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To conserve the spin, an electron can flip - or rather it can change its orbital. A sum of orbitals is a strong electric current that couples well with a photon.

 

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This story of straight being as fundamental as circular gets fun if you consider entanglement, because entanglement working equally with both is the kind of arguments that defeats theories of hidden variables. If photons decided their polarization when emitted, and not when detected, then the correlation would work with one kind of polarization only.

Edited September 2, 2013 by Enthalpy

[swansont]

Photons have an intrinsic angular momentum, pointing parallel or anti-parallel to their path of propagation, with a magnitude of exactly one unit of h-bar... yes?

 

Thus, when electrons emit or absorb photons, they must spin-flip, which represents a change of -1/2 hbar <----> +1/2 hbar = one unit of h-bar... yes?

 

Bare electrons do not absorb photons. They must be in a composite system. Once you have a composite system like an atom, there is not only the spin orientation but also orbital angular momentum. Since orbital angular momentum changes by whole units of hbar, a spin flip is not necessary when absorbing a photon, though it is possible.

[Widdekind]

if photons have angular spin momentum (which i understand to be distinct from plane of polarization)...

 

then do photons also have a corresponding magnetic moment ?

 

(the only way i know of, to get angular momentum out of an EM field, is to combine an electrostatic field, with a magnetic dipole field... then the resulting field has angular momentum, in the magnetic-axis direction,

 

Lz ~ z * (r x (E x B)) ...

 

so, does a photon have some small magnetic moment, whose propagation with the super-imposed plane-wave EM field, accounts for the angular momentum of photons ?

 

what generates Lz is the axially-directed field lines of B, crossed with radial E... so the angular momentum, of an oscillating E, varying perpendicular to the magnetic axis, might be possible to make to have axial angular momentum)

[swansont]

if photons have angular spin momentum (which i understand to be distinct from plane of polarization)...

Plane polarization is simply a superposition of the opposite circular polarization states.

 

 

then do photons also have a corresponding magnetic moment ?

No

(the only way i know of, to get angular momentum out of an EM field, is to combine an electrostatic field, with a magnetic dipole field... then the resulting field has angular momentum, in the magnetic-axis direction,

But EM radiation is not the result of an electrostatic field. It's oscillating E and B fields. Circular polarization occurs when the E and B are 90º out of phase. The field rotates about the propagation axis.

 

http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/polclas.html

[Enthalpy]

if photons have angular spin momentum [...] then do photons also have a corresponding magnetic moment ?

No, because they have no charge.

Though, neutrons do have a magnetic momentum, but they're composite particles.

[Widdekind]

magnetic moment is, ultimately, purely magnetic fields...

 

and EM fields can possess "mechanical" spin angular momentum...

 

perhaps "swirling vector potential" could account, for said spin and also imply a magnetic moment (??)

[swansont]

No. A photon has no magnetic moment.

[Widdekind]

 

But EM radiation is not the result of an electrostatic field. It's oscillating E and B fields. Circular polarization occurs when the E and B are 90º out of phase. The field rotates about the propagation axis.

 

http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/polclas.html

 

not what i suggested... perhaps a photon carries a magnetic dipole field, associated w/ its spin, along w/ that oscillating field (B = B_oscillate + B_dipole)

 

what about other bosons ? What about charged Weak-bosons... w/ charge & spin, they plausibly possess magnetic moments (?)

[swansont]

 

not what i suggested... perhaps a photon carries a magnetic dipole field, associated w/ its spin, along w/ that oscillating field (B = B_oscillate + B_dipole)

 

what about other bosons ? What about charged Weak-bosons... w/ charge & spin, they plausibly possess magnetic moments (?)

 

I know that's not what you suggested. I'm telling you what the current, well-tested physics says.

 

The photon does not have a dipole moment.

[Widdekind]

according to vector calculus, the photon's momentum density is [math]\propto \vec{E} \times \vec{B}[/math]... e.g. [math]\vec{A} = \hat{\phi} \frac{A_{\phi}{r} e^{\imath \left( k z - \omega t \right) }[/math]

 

so any angular momentum would require [math]\vec{r} \times \left( \vec{E} \times \vec{B} \right)[/math]... which would require photons' fields to be (partially) longitudinal...

Edited September 13, 2013 by Widdekind

[swansont]

according to vector calculus, the photon's momentum density is [math]\propto \vec{E} \times \vec{B}[/math]... e.g. [math]\vec{A} = \hat{\phi} \frac{A_{\phi}{r} e^{\imath \left( k z - \omega t \right) }[/math]

 

so any angular momentum would require [math]\vec{r} \times \left( \vec{E} \times \vec{B} \right)[/math]... which would require photons' fields to be (partially) longitudinal...

 

Yes, plane-wave EM radiation has no orbital angular momentum. Spin is intrinsic and not dependent upon the linear momentum.