# Photon spin

1. Sep 1, 2004

### Marjan

As i understand spin of a photon can be -1 or +1. In a classical view we can say that one is spinning right and other to the left ?

What would be the difference between light with all photons -1 and light with all photons 1 ?

PS: Can be spin of a photon also 0 ?

2. Sep 1, 2004

### ZapperZ

Staff Emeritus
There is a slight problem in terms of terminology here. I will give you a general description, and hopefully, you can adapt that to answer your own question.

There is a spin quantum number s, and the PROJECTION of that spin along an axis of measurement (typically designated as the z-axis), m_s. You are confusing the two. If a particle has a spin of s, then a measurement of m_s can yield values of

m_s = s, s-1, s-2,....., -s+2, -s+1, -s.

An electron has a spin s = 1/2.

A measurement of the electron spin will give m_s = 1/2 or -1/2.

Zz.

3. Sep 1, 2004

### zefram_c

A photon's spin is described by a polarization 4-vector. Although this makes it seem like a photon can have four polarization states, it is in fact not so: two of those degrees of freedom are present due to the gauge invariance of the photon wave equation. Once that is fixed, the physical photon states can be described with only two linearly independent vectors - hence a photon only has two spin states (|1 [+/-]1>). These states can be associated with classical transverse polarization states for light, but I don't know offhand the exact correspondence. If a photon were not massless, it would admit a third state (|1 0>).

4. Sep 1, 2004

### vanesch

Staff Emeritus
Light that is right circularly polarized, or left circularly polarised!

No. This is a consequence of the fact that the photon is a massless particle. It's a quite tricky issue, actually.

cheers,
Patrick.

5. Sep 1, 2004

### humanino

Actually.. yes. A photon can have spin zero, but then it needs (rest) mass, that is a virtual photon.
The four-momentum $$q$$ of a virtual photon has non-vanishing invariant square : $$q^2$$ is non-vanishing and represent a squared rest mass (maybe up to a sign).

Partrick knows that, and also knows that it is useless confusing somebody learning general rules with specific exceptions to the rules.

6. Sep 1, 2004

### zefram_c

Granted, a virtual photon can have spin zero. But I'm quite sure the original question dealt with real photons, and there's no need to confuse the issue by introducing virtual photons.

Edit: to further clarify for the original poster, in this thread "spin 0" or "spin 1 / -1" means the spin component sz along the direction of motion; the total spin s of a photon is always +1.

Last edited: Sep 1, 2004
7. Sep 1, 2004

### jnorman

not to change the topic, but i still believe that virtual photons are a myth - nothing more than an abject admission that we really have no clue how fields interact.

8. Sep 1, 2004

### ZapperZ

Staff Emeritus
Do you think the concept of virtual photons (or any virtual particles) is simply a handwaving argument that is purely qualitative? Have you looked at all the RULES and requirement for them, i.e. they are not simply made up as one goes along? And more importantly, have you considered all the QUANTITATIVE predictions that are made using them that are in agreement with a gazillion experimental observations?

I suggest you look at the physics description of the very materials that you are using in your modern electronics and see if virtual particles of some kind (especially those pesky virtual phonons) are responsible for how we know so much about those materials.

Zz.

9. Sep 3, 2004

### Eye_in_the_Sky

We talk about a photon as having "spin 1". Similarly, we talk about an electron as having "spin 1/2". In general, for a particle with "spin s" (note: according to Quantum Mechanics, s must be restricted to an integer or half-integer), the associated angular momentum is given by
hbarsqrt{s(s+1)}. Thus, the total spin angular momentum for a photon
(s = 1) is hbar√2.

Now, when you say that the spin of a photon can be ±1 in connection with "spinning" right (+1) or left (-1), what you are talking about are two specific quantum-mechanical photon spin-states. These states are called "helicity" sates.

Say, for example, that we have a photon which is traveling in the +z direction. If the photon is in the +1 helicity state, then this means that the z-component of the spin vector is positive (and equal to +hbar). On the other hand, if the photon is in the -1 helicity state, then this means that the z-component of spin vector is negative (and equal to -hbar). Observe that the z-component of spin is quantized, which is to say that, in any measurement of that component, it can be found to have a value of ±hbar, and nothing else.

Notice that, in the above, I have mentioned only a single component of spin and not the total spin vector itself. No doubt, you have heard of an "uncertainty principle" regarding the position and the momentum of a particle. Well, as it turns out, such a principle also applies to the spin of a particle. More precisely, a pair of spin components along two nonparallel axes are subject to an "uncertainty principle". Even more precisely, and in particular, if the particle is in a spin state such that, say, the z-component of spin is completely definite (as is the case for our two helicity states), then the direction of the spin component in the xy-plane will be completely uncertain (only the magnitude will be certain).

From these remarks, you will understand that helicity states refer to the direction (+ or -) of the spin component (parallel to the direction of propagation), and not to the total spin vector itself. This is what ZapperZ was alluding to in post #2. Of course, from the description given there and the fact that s = 1 for a photon, we would expect that the allowable eigenvalues for Sz are m = -1, 0, and 1. If so, we would then also have a zero helicity state (Sz eigenstate with m = 0) for the photon ... but as pointed out by zefram_c (post #3) and vanesch (post #4), for a physical photon, this m = 0 case is excluded.
______________
Yes, viewing the photon in a 'classical' way (note: this is a 'simplification' to be used only with caution!), we can say that the +1 helicity state corresponds to the rotation of a "right-handed" screw (again, this is said with regard to the z-component of spin only, and not the total spin vector), and -1 to a "left-handed" screw.
______________
Let us denote the +1 helicity state as |R>, and the -1 helicity state as |L>. Then, as vanesch (post #4) has pointed out, |R> (|L>) corresponds to right- (left-) circularly polarized light.
______________
As already indicated, the m = 0 eigenstate of Sz is excluded. However, this does not mean that the only possible photon states are eigenstates of Sz with m = ±1. In general, any (normalized) linear combination of |R> (eigenstate of Sz with m = +1) and |L> (eigenstate of Sz with m = -1) is also a possible state. Each such (nontrivial) linear combination will correspond to either elliptical or linear polarization. For example,

|x> ≡ (1/√2)(|R> + |L>)

and

|y> ≡ (-i/√2)(|R> - |L>)

are suitable choices for linearly polarized photons along the x- and y- axes, respectively. In both cases <Sz> = 0.

Note that |x> is an eigenvector of Sy with eigenvalue m = 0, and |y> is an eigenvector of Sx with eigenvalue m = 0.

Last edited: Sep 3, 2004
10. Oct 6, 2004

### wm

Photon spin & photon polarization?

1. Should this be a new thread? (I'm not sure of the correct procedure here.)

2. What is the relation of photon spin to photon polarization? (a) I understand that a photon always has spin s = 1. (b) I also understand that a photon can have any degree of polarization from 0 to 1.

3. When discussing this matter I've been told (authoritatively) that photon polarization is just another term for photon spin and that photon spin is exactly analogous to electron spin.

4. I cannot see this (point 3), given #2(b) above and those degrees from 0 to 1? I'm not aware that electrons have such degrees associated with them.

11. Oct 7, 2004

### Gonzolo

1. Usually, a new question is a new thread, but small related questions that don't disrupt the flow of the original discussion are common, it's a matter of respect to the thread starter.

2. A bunch of photons with either -1 or +1 spin correspond to a left or right circularly polarized light (or the opposite). If you have 50% of photons in one state and 50% in the other, you have linearly polarized light. Other relative percentages corespond to the range of elliptically polarized light.

3. "Just another term" is correct in the same sense as wave-particle duality.

4. As for electrons, as far as I know, you could define the polarization of a stream of electrons in the same way as I said in 2. Your question makes me realize that the Stern-Gerlach experiment (which seperate + and - 1/2 spin particles) seems to be somekind of an equivalent to a birefringent crystal (except birefringent crystals separate the 2 linear polarizations instead of the 2 circular polarizations) in this respect.

Last edited by a moderator: Oct 7, 2004
12. Oct 7, 2004

### wm

photon spin vs polarization

Thank you.

I'll need to study this. If you say that the "degree of polarization" applies to an ensemble of photons and not single photons, then you've made my day! (Why?) Because I'm seeking to understand single photons, one by one.

So: Does that "degree" term refer to a classification system for differing ensembles?

Now: I would like to say: Like electrons, photons have intrinsic angular momentum (a vector) which (in pristine photons) may be randomly oriented. If we "measure" circular polarization, then we're "measuring" (projecting actually??) the pristine spin onto the line-of-flight (+1, RHCP; -1, LHCP). Also: if we project the pristine-spin onto a direction orthogonal to the line-of-flight, then we get linear polarization (+1 = LP in measured direction; -1 = LP in orthogonal direction. NB: Both these directions being still orthogonal to the line-of-flight).

Which leads me to say (but is it correct?): When physicists talk of photon polarization they are referring to the spin projection; circular-polarization refers to a line-of-flight projection; linear-polarization refers to an orthogonal projection. When physicists talk of electron spin, they normally refer to the x, y, or z projection. It would not be wrong to use the same language when talking of photon spin. (??Is this close??) Recognising spin s = 1 for photons, s = 1/2 for electrons.

Could you elaborate this bit please?

Yes; you've caught my drift (see PS); it's parallels like this that I'm seeking to understand, to the limit. I am seeking to take the "common" spin features of photons and electrons to the limit (and use a common language/terminology to do so); then to understand their spin differences.

I hope I'm getting close to what a good textbook would tell me; but I've not seen it presented in the way that "I" think would be clearest. Thanks again.

PS: For your info; I claim a simple refutation of Bell's theorem (photons or electrons) and I'd like to express my "spin" findings correctly; ie, in accord with what we definitively know/accept about photons from other studies.

Last edited: Oct 7, 2004
13. Oct 7, 2004

### Gonzolo

I believe it applies to single photons as well, in which case we must think in terms of the probability of finding a photon in either state. A photon has it's wave function, so its state is only -1 or +1 once you measure it. In the meantime, it's in a superposition of both states, such that degrees of polarizations can apply. I would suggest you corroborate this with some other source though, as this is at best an educated understanding from my part.

I understand "degree" [of polarization] refers to whether the light is polarized or not such that :

- 0 => not polarized at all => completely random state for each photon, wavefunction undefined, unmeasured.

- 0<x<1 => not perfectly polarized => +1 vs -1 state distribution works, but with an error $$\Delta$$. Wavefunction defined, measured to a certain accuracy

example 0: unpolarized light that has passed through a polarizer is still not perfectly polarized.
example 1: for linear polarization, if x = 0,5, there's a 50% probability of having 50% photons in the + state and 50% in the - state;
example 2: for right-circular, x = 0,6 means there's a 60% probability of having 100% the photons in the +1 state;
example 3: for a single photon having gone through linear polarizer : there is a x probability of having a 50% chance of it being in the + state.)

This is the general idea I have anyway, you might want to see the "degree of polarization" in terms of a known error $$\Delta$$, i.e. for linear polarization, a photon has a (50 ±$$\Delta$$)% chance of being +1.

- 1 => perfectly polarized => + vs - state distribution exactly known. 100% chance of having all photons in a classical state (perfectly linear, perfectly defined ellipse etc.). I would think that Heisenberg's principle would prevent this however.

A photon doesn't care much which theory we use to describe it. It minds its own business, so yes polarization and photon states are the same. But whether you talk about polarization or photon states largely depends on what experiment you're doing. You don't necessarily have to introduce the other theory to completely understand what's going on in a setup, (although it surely is interesting and worth studying). A complete understanding of quantum optics though should eliminate any contradiction in either wave-particle duality or our subject matter.

If you find or create such a textbook, please tell me about it!

Last edited by a moderator: Oct 7, 2004
14. Oct 27, 2004

### James R

Eye_in_the_Sky,

That's one of the clearest explanations I've seen about the relationship between photon spin and polarisation of light. Thankyou.

Now, if we can just sort out the difference between [tex]\sigma^+[\tex] and [tex]\sigma^-[\tex] polarisations...