# Light polarization vs photons spin

1. Nov 5, 2011

### maxverywell

If I'm not mistaken, right (or left) circularly polarized light consists of photons with right-handed helicity (or left-handed).
But what are the states of the photons in the (horizontally or vertically) linearly polarized light? Are all the photons in a superposition of two helicity eigenstates or/and is a mixture of both right-handed and left-handed photons?

Last edited: Nov 5, 2011
2. Nov 5, 2011

### clem

R circular polarization corresponds to negative (L) photon helicity. This is because you look at light as it comes toward you, but the photon helicity is its spin in its direction of motion. Photons in plane polarization are a coherent mixture of R+L or R-L helicity.

3. Nov 5, 2011

### maxverywell

Ok, it's just a matter of convention.

I think that the mixture corresponds to the unpolarized light.

4. Nov 5, 2011

### maxverywell

The linearly polarized light consists of photons that are in a superposition of the form $a|R\rangle+b|L\rangle$ where the helicity eigenstates $|R\rangle$ and $|L\rangle$ are equally weighted. However the two helicity eigenstates can have different phases. If the phases are equal then we get horizontally linearly polarized light. If the phase difference is $\pi$, we get vertically linearly polarized light etc. For any linear polarization angle $\theta$: $\frac{1}{\sqrt{2}}(e^{-i\theta}|R\rangle+e^{i\theta}|L\rangle)$, the polarization angle determines the phases. Right?

Last edited: Nov 5, 2011
5. Nov 6, 2011

### bbbeard

Unpolarized light is a quantum mechanical phenomenon. Classically, any superposition of polarizations gives light which has a definite polarization. The fact that light can be unpolarized tells us something non-classical is going on: that the light is not in a polarization eigenstate.

6. Nov 6, 2011

### kith

I think this is misleading. Incoherent mixtures are nothing inherently quantum mechanical. They occur all over the place in classical statistical mechanics.

Also, in the Copenhagen interpretation, an unpolarized beam is considered as a mixture of individual particles, each of them beeing in an eigenstate.

7. Nov 6, 2011

### maxverywell

Superpositions of polarizations is a quantum mechanical phenomenon. This leads to a polarized light. Unpolarized light is a statistical mixture (ensemble) of different polarizations (superpositions included).

8. Nov 6, 2011

### bbbeard

No, this is not correct. Sunlight, for example, is unpolarized. This does not mean that it arrives in little packets of definite but random polarization; each individual photon is unpolarized. In other words, there is no requirement that a photon state be a polarization eigenstate. If you pass the photons one by one through a polarizing filter, half will be blocked and half will pass through, regardless of the orientation of the filter. And if you pass the unpolarized beam through a quarter-wave plate (which changes circularly polarized light to plane-polarized) the beam remains unpolarized!

BBB

9. Nov 6, 2011

### maxverywell

10. Nov 6, 2011

### y33t

Polarization of light and spin of an individual photon are different phenomena. You can polarize the wave not the photon. You can spin a photon but not the wave.

Consider a generic wave and evaluate both from classical and quantum approaches. While it is possible to express the wave behavior using quantum approach, it doesn't provide more information than classical approach. Maxwell equations precisely determine the behavior of waves and fields. But when you quantize the wave components, you get much much more information than classical approach. Lowest level is the level of an individual photon where polarization doesn't mean anything anymore. They are different phenomena and not linked in any mean.

11. Nov 6, 2011

### fisico30

Hello,

unpolarized light is perfectly polarized for a "super short" interval of time. It changes rapidly to another type of polarization and so on....Unpolarized light is instantaneously perfectly polarized...

Perfectly polarized light maintains instead the same polarization for a looong interval of time...
But how long?

And finally, there is partially polarized light. It stays polarized in on specific polarization for an "intermediate" interval of time, it changes then to a possibly different polarization state and so on...But how long is this intermediate interval of time? How much longer than the interval for unpolarized light?

thanks
fisico30

12. Nov 7, 2011

### kith

This is a question of interpretation. As I wrote, in the Copenhagen interpretation, each photon is in polarization eigenstate, because the density operator of an unpolarized beam is interpreted as a statistical mixture of particles with definite states.

13. Nov 8, 2011

### bbbeard

I've been looking into this over the past few days, and I will admit that the situation is murkier than I thought. A few things:

(1) I'm pretty sure the Copenhagen interpretation is inconsistent with "hidden variables" theory -- which incorporates the idea that quantum systems have definite values of all physical parameters before and after measurement. In other words, in the Copenhagen interpretation, a system does not acquire a definite value of momentum, polarization, etc. until you measure it.

(2) I'm pretty sure that there is no requirement in the postulates of quantum mechanics that a quantum system be in a "pure" state. If the system is in a pure state, then you can make certain calculations, but there is no law of nature that says that the system must be in a pure state.

(3) To that end, in 1927 John von Neumann invented the concept of quantum density matrices and the density operator. This generalizes the computations of quantum mechanics to states which are mixed, i.e. can be represented by the density matrix

ρ=Ʃpii><ψi|

(4) Many texts discuss the density matrix in terms of the classical analogue, although you have to read carefully not to be deceived. Quantum systems described by a density matrix are quantum, not classical. In particular there is no "law of large numbers" for the quantum density matrix formalism. A single photon can be in a mixed state.

(5) For unpolarized light in particular, the idea that the density matrix implies that each photon has a definite polarization is incoherent. For example, in the {x,y} eigenbasis (={x-polarization, y-polarization}), one way to represent unpolarized photons is with the density matrix

$\rho=\frac{1}{2}\left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \\ \end{array} \right)$

Classically, this describes a system with a statistically large number of particles, half of which are in state x, and half in state y. Quantum mechanically, we might be tempted to use the same words, but it is more correct to say that the state is such that there is a 50% probability that a measurement will find the polarization is x, and 50% for y.

Here's the rub: if we change the basis to {R, L} (={right circularly polarized, left circularly polarized}), then the density matrix doesn't change. The state is such that there is a 50% probability that a measurement will find the polarization is R, and 50% for L. But quantum mechanically, this is the identical mixed state.

So you see the problem with the "semi-classical" interpretation? Classically, we think that the collection of photons is half x, half y, and that if we examine each one, we can find which is which. In the classical view, the photon actually has an x-polarization, and is not circularly polarized. Quantum mechanically, though, we are agnostic about the state prior to measurement. We only claim knowledge of the state after the polarization has been measured.

(6) In the case of photons, it's difficult to say why the Copenhagen interpretation (my version, not yours) is forced on us. Classically, because the beam has no net polarization, one is tempted to conclude what fisico30 said: that the beam has a definite polarization at any time, but it switches from one state to another very rapidly. Quantum mechanically, the beam is just a stream of photons in the unpolarized mixed state. But it takes a subtle experiment to see why the classical view is untenable.

(7) At the risk of playing bait-and-switch, this is easier to discuss in terms of systems with polarized atoms, like the paradigmatic Stern-Gerlach (SG) experiment. In the atomic polarization case, we can have x, y, and z polarizations, each with +,- states. The SGx device is designed to sort the beam into x+ or x- atoms, ditto SGy, SGz. Now, classically, an unpolarized beam of atoms is composed of atoms of definite but random polarization. Suppose we send an unpolarized beam of atoms into an SGz detector. Half come out z+, half z-. Classically, half of the z+ atoms are in the x+ state as well, half x-, and if we send the z+ beam into an SGx device, this what we observe. But the crucial difference between classical and quantum measurements is that if we send the doubly-filtered beam into an SGz device again, we see z- atoms coming out -- even though we selected only z+ atoms in the first device. It is this peculiar behavior of quantum mechanical systems that leads us down the path of defining eigenstates, operators, amplitudes, expectation values, and the probabilistic Copenhagen interpretation, which is agnostic about states prior to measurement.

Along these lines, imagine a Stern-Gerlach device for photons. An SGxy device sorts the photons into x and y polarizations. An SGRL device sorts the stream into R and L polarizations. If we send an unpolarized stream of photons into an SGxy device, half come out x, half y. Now send the x stream into an SGRL device. Half of those come out R, half L. Now send the x,R stream into an SGxy device -- and we will find y photons coming out. Why does this not surprise us? I would argue that this result does not surprise us because we already had a workable wave theory of light long before we were confronted with the quantum mechanical properties of photons. And in the case of EM wave propagation, we already have the idea that polarizing filters project the wave into a different state. In the case of atoms, the idea of discreteness was around from the inception. It took awhile before we got the point about the wave nature of atoms.

But here's the point: an unpolarized stream of atoms is represented by a density matrix, and it is not just "an unpolarized stream of atoms" but literally "a stream of unpolarized atoms". It's not that half the atoms are in state |z+> and half in |z->. The state of each atom is not determined until we measure it. In the same way, "an unpolarized stream of photons" is really "a stream of unpolarized photons". The polarization is not set until we measure it.

At least, that's what makes sense to me. I'm open to other interpretations or insights.

BBB

Last edited: Nov 8, 2011
14. Nov 8, 2011

### maxverywell

@bbbeard

Photons are always completely polarized. A statistical mixture is a concept from statistical interpretation of quantum mechanics, so when we talk about mixed states we are talking about infinitely big ensembles of different particles (quantum systems). Same for a pure state - infinitely big ensemble of identical particles (quantum systems).

Unpolarized beam consists of photons that are described by the density matrix $\rho=\frac{1}{2}(|R\rangle\langle R|+|L\rangle\langle L|)=\frac{1}{2}(|H\rangle\langle H|+|V\rangle\langle V|)$. The choice of the base has to do with the measuring device (polarizer).

Last edited: Nov 8, 2011
15. Nov 8, 2011

### bbbeard

Well, that's the rub, isn't it? You seem to be claiming that any given photon is in a completely determined polarization state -- but then you turn around and say that even what kind of state it is in (e.g. linearly or circularly polarized), not even to mention R or L, H or V, depends on what kind of detector you pull out of the cabinet that day. Which is it?

BBB

16. Nov 8, 2011

### maxverywell

Because the connection between the states states $|R\rangle$, $|L\rangle$ and $|H\rangle$, $|V\rangle$ is similar to the connection between position state $|x\rangle$ and momentum state $|p\rangle$. Not exactly but here is the point:
When you measure R and L of unpolarized light, you get 50-50 R and L. Then measure H and V. Because R and L are equally weighted superpositions of H and V, you will get 50-50 H and V.
If you measure directly H and V of unpolarized light, you will get 50-50 H and V.
This leads to the same final result.

You could use another basis. For example $|A\rangle$ and $|D\rangle$ for ±45° linear polarizations. The density matrix is the same:

$\rho=\frac{1}{2}(|A\rangle\langle A|+|D\rangle\langle D|)=\frac{1}{2}(|R\rangle\langle R|+|L\rangle\langle L|)=\frac{1}{2}(|H\rangle\langle H|+|V\rangle\langle V|)$.

Last edited: Nov 8, 2011
17. Nov 8, 2011

### bbbeard

But you would not claim that the photon has a specific position and momentum prior to measurement, would you? That would be the analog of what you are claiming with respect to polarization.

BBB

18. Nov 8, 2011

### maxverywell

I think that before the measurement you don't know the exact states (Copenhagen interpretation). So it (unpolarized light) can be a mixed state of R,L or H,V or any two mutually orthogonal polarized states. However after the measurement you get the same result. This is the reason why the density matrix is the same for all this cases.

Last edited: Nov 8, 2011
19. Nov 8, 2011

### bbbeard

In the Copenhagen interpretation it's not just that you don't know the value of the parameter being measured (such as polarization in the {R,L} basis), it's that the parameter doesn't have a value until you measure it. I'm okay with that, that's how quantum mechanics works, but apparently some folks really want to hold on to the semi-classical picture.

BBB

20. Nov 8, 2011

### maxverywell

Yeah, I think we can say that the measurement chooses (determines) the base. There is some kind of symmetry, the density matrix is same for any of this bases.

Classically we interpret it as that a unpolarized light have all the possible polarizations.