# Photon spin and polarization

1. Apr 7, 2013

### Alfred Cann

I think I understand this only up to a point.
1. Photon spin is quantized to be +1 or -1 and these represent left- and right-hand circular polarization.
2. A photon can have a superposition of 2 spin states in any proportion.
3. Having probability amplitude of 0.5 of spin 1 and 0.5 of spin -1 yields linear polarization, and the orientation depends on the phase angle between the two.
4. Here is my question:
Suppose I let a photon with vertical linear polarization (obtained from a vertical linear polarizer) pass through another vertical linear polarizer. What is the transmission probability?
Classical reasoning says 1.0.
But if the photon can REALLY have only spin 1 or -1, the transmission probability is only 0.5 in either case and therefore 0.5 overall.
I suspect the problem is with the word 'really' because we are not allowed to ask about objective reality. But please explain and, if possible, without ket signs or matrices.

2. Apr 7, 2013

### Staff: Mentor

Quantum reasoning also says 1.

No, the superposition state is equally valid. It is a question of which basis you decide to use to describe the system and to measure it.

Take the simpler example of a particle with some angular momentum quantum number $l$. By convention, we usually also describe the particle using the projection of the angular momentum on the z axis, $m_l$, related to operator $\hat{l}_z$. But this choice is arbitrary, and we could as well have chosen the x axis instead. Then, if you prepare the particle in an eigenstate of $\hat{l}_x$ and meaure the projection on the x axis, you will always get the same result with probability 1. But if you measure the projection on z, then the result will be probabilistic, because from the point of view of $\hat{l}_z$, the particle is in a superposition of states.

The same thing is going on with the photon. Choosing a basis of linear polarizations is equally valid as choosing a circular polarizations, such that if the photon is in a state of vertical polarization, you will always measure a vertical polarization with probability 1. But if you were to ask if the spin projection is 1 or -1, then you would get a probabilistic measure, and the photon would no longer be vertically polarized.

3. Apr 8, 2013

### Alfred Cann

Thank you for your reply. If I have understood it right, the usual description of photon spin as being either +1 or -1 is only a convention, not fundamental. It is equally legitimate to say that the photon polarization is linear and either vertical or horizontal (or any other orthogonal pair). I take it that this means that, in any experiment to study the transfer of angular momentum from photons to material objects, one must prepare the photons in a (nearly) pure circular polarization state.

In contrast, //mathpages.com/rr/s9-04/9-04.html ( google 'photon angular momentum') states that "...if the intensity of a linearly polarized beam of light is lowered... only one photon is transmitted at a time, it will appear to be circularly polarized (left or right)..."
Are they being simplistic or wrong?

4. Apr 9, 2013

### Staff: Mentor

I had a look at that page, and I think it is incorrect. To be sure, I checked in G. Grynberg, A. Aspect, and C. Fabre, Introduction to Quantum Optics (Cambridge University Press), and section 5C.2.1, Measuring the polarization of a single photon, considers the detection of a single linearly-polarized photon.

I think that the author of the above link confuses two aspects of the angular momentum of the photon, namely the spin and its projection. The spin of the photon is an intrinsic characteristic, $s=1$ is fixed by nature. The corresponding angular momentum must be conserved when the photon is absorbed, whatever its polarization. Then there is the projection of this angular momentum on the z axis, which can take values of $\pm1$, and there the polarization will come into play as to how this projection is conserved.

5. Apr 9, 2013

### Alfred Cann

Please excuse my ignorance. I fail to understand your answer. What do you mean by spin as an intrinsic characteristic? Is it not the same as angular momentum?
As a corollary, doesn't a linearly polarized photon, i.e., an equal superposition of s = 1 and s= -1 have zero angular momentum?

6. Apr 9, 2013

### Staff: Mentor

It is me who should apologize for not being clear...

Spin is a form of angular momentum. It is intrinsic in that it is a property of the particle and cannot change, in contrast to other forms of angular momentum which depend on motion. For instance, an electron has spin 1/2, while a photon has spin 1.

No, it still has a spin of 1, and therefore an angular momentum of $\sqrt{2}\hbar$. What is +/- 1 is what is called helicity, which is the projection of the spin on the direction of propagation.

Imagine that the photon has a vector attached to it that represents its spin angular momentum. According to the rules of quantum mechanics, this vector has a length of $\sqrt{s(s+1)} \hbar=\sqrt{2}\hbar$. Quantum mechanics also tells us that the projection of this vector on an axis is quantized, and in the case of a photon the projection of the spin along the direction of propagation is either $+\hbar$ (the vector is pointing in the direction of propagation, with a slight angle), or $-\hbar$ (the vector is pointing opposite the direction of propagation, with a slight angle).

A linearly-polarized photon is in a superposition of these two helicities, but it still has the same spin.

7. Apr 10, 2013

### Alfred Cann

I'm not catching it. I need to take a course in QM. Can you recommend a website or open courseware?

8. Apr 11, 2013

### Staff: Mentor

I like Leonard Susskind's lecture series on YouTube.