Amount of energy required to change the spin of a photon

[San K]

Not sure if the question makes sense, nevertheless it can help clarify some concepts, I guess.

What is the amount of energy required to change the spin (intrinsic angular momentum) of a photon?

 

[The_Duck]

Zero. You can simply pass the photon through a half wave plate. Afterwards the photon still has the same energy; you don't need to do any work. However, there is a transfer of angular momentum between the wave plate and the photon.

 

[San K]

Zero. You can simply pass the photon through a half wave plate. Afterwards the photon still has the same energy; you don't need to do any work. However, there is a transfer of angular momentum between the wave plate and the photon.

Thus ...is the transfer of angular momentum, in this case, friction-less?

 

[andrien]

well if you are talking about making a spin 1 into spin 1/2 ,then it is not possible.

 

[San K]

well if you are talking about making a spin 1 into spin 1/2 ,then it is not possible.

Hi Andrien, I am talking about changing the spin (along a particular axis) from, say, spin up to spin down.

 

[Bill_K]

There are only two things you can do to a photon:
a) You can emit it.
b) You can absorb it.
You cannot change its spin. The comments above apply to a light ray, in the classical sense, but not to an individual photon. You cannot pass a photon through a half wave plate, or focus it with a lens, or bounce it off a mirror. These are collective interactions with the atoms of a solid object. A photon is simply absorbed by the first thing it encounters.

 

[Cthugha]

You cannot change its spin. The comments above apply to a light ray, in the classical sense, but not to an individual photon. You cannot pass a photon through a half wave plate, or focus it with a lens, or bounce it off a mirror. These are collective interactions with the atoms of a solid object. A photon is simply absorbed by the first thing it encounters.

Hmm, I suppose all the people working on polarization entangled photons and polarization properties of single photon sources should retract all their papers then because they pass single photon states through quarter and half wave plates all the time. They focus them and use mirrors on them all the time. Of course one can do all of that with single photon states. See e.g. Phys. Rev. Lett. 86, 1502–1505 (2001) where all of this is done with single photons.

 

[Bill_K]

Interesting. Please draw me the Feynman diagram of a photon interacting with a quarter wave plate.

 

[The_Duck]

<accidental double post, see below>

 

[The_Duck]

Surely this is just a disagreement over the word "photon." I'm sure everyone agrees on the physics if not the terminology. If by photon we mean "a single wavy line in a Feynman diagram" then we can only emit it or absorb it and the spin of a single photon can't be changed. If by photon we mean something like "a state that, when it interacts with a photon detector, causes one count" [what's a better way of stating this?] so that we include collective excitations within transparent materials, then we can speak of sending a photon through a wave plate or focusing it with a mirror.

In any case, to answer the original question, everyone should agree that we can go from a state with a positive helicity photon to a state with a negative helicity photon at no energy cost.

Thus ...is the transfer of angular momentum, in this case, friction-less?

Well, I imagine we can't get perfect efficiency. Some of our photons will be absorbed or reflected by the wave plate instead of being transmitted. But not very many (for a well-constructed wave plate) and I think you could get arbitrarily close to perfect efficiency by improving the quality of your wave plate.

 

[andrien]

In any case, to answer the original question, everyone should agree that we can go from a state with a positive helicity photon to a state with a negative helicity photon at no energy cost.

Do you have a reference for that,because it is really silly.

 

[The_Duck]

Do you have a reference for that,because it is really silly.

This is what a half-wave plate does, as I mentioned in my first post above. Wikipedia explains

the effect of the half-wave plate is to mirror the wave's polarization vector through [a certain plane]. For linearly polarized light, this is equivalent to saying that the effect of the half-wave plate is to rotate the polarization vector through an angle 2θ; however, for elliptically polarized light the half-wave plate also has the effect of inverting the light's handedness.

As you mentioned above, the photon does remain a spin 1 particle; it's the direction of the spin that changes. A half-wave plate changes left circularly polarized photons to right circularly polarized photons and vice versa. Circularly polarized photons are in states of definite spin/helicity, and inverting the handedness of the polarization switches the spin/helicity.

 

[andrien]

Well I was really thinking in terms of some quantum mechanical situation because spin is nevertheless a quantum phenomenon,however it is possible to prove that light does have a spin 1 character but that really goes with circular polarization and not plane polarized .So I was thinking about some quantum mechanical way of doing it ,if it is there.

 

[San K]

There are only two things you can do to a photon:
a) You can emit it.
b) You can absorb it.
A photon is simply absorbed by the first thing it encounters.

thanks Bill_K. Is this process (emittance, absorption, re-emittance, re-absorption) friction-less?

 

[San K]

In any case, to answer the original question, everyone should agree that we can go from a state with a positive helicity photon to a state with a negative helicity photon at no energy cost.

Thanks for the information Duck.

In classical physics, we are used to the idea that any kind of change must involve energy (transfer/use).

So the above, like many other quantum mechanical phenomena, is difficult to comprehend in the Newtonian-Einsteinian mindset.

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[Cthugha]

Interesting. Please draw me the Feynman diagram of a photon interacting with a quarter wave plate.

You were explicitly talking about "individual" photons. This implies a single photon Fock state in terms of quantum optics. Nobody would draw Feynman diagrams for such a situation as it is quite cumbersome and the issue can be easily handled in terms of quantum optics.

If instead you want a particle-physics like approach calling the "basis modes" of the em field photons, it is obviously pointless trying to change its spin. However, this is not the kind of photon you get in the lab or can realize and it is also not what people typically have in mind (unless one is discussing elementary particle physics topics of course) when talking about photons. As the whole discussion would be pointless if one assumes such a meaning of the term "individual photon", it seems ok to me to discuss the quantum optics definition as in this case you definitely can change the spin of a single photon and the question carries some meaning.