# I What does it mean -- polarization of a single photon?

Tags:
1. Nov 24, 2016

### Uriel

Hi.

I have a rather silly question. When speaking about a single photon? What do people mean when they speak of the polarization of a single photon.

For instance, in classic electromagnetic theory, this would be the direction in which the electric field of the wave is oscillating . But does this means that the photon have a electric field?

I'm kind of confused here

2. Nov 24, 2016

### Orodruin

Staff Emeritus
As a massless spin 1 particle, the photon has two helicity states corresponding to the two possible spin directions. Those (or a linear combination) are the photon polarisations.

A circular polarised classical EM field is essentially a coherent state of photons in one of those spin states.

3. Nov 25, 2016

### A. Neumaier

Of course.

4. Nov 25, 2016

5. Dec 4, 2016

### jeremyfiennes

Related: when a light wave of magnitude A, polarized at angle α to the vertical, is shone on a vertical polarizing filter, a vertically polarized wave of magnitude A.cos(α) is observed on the far side (?). If a corresponding single photon is fired at the filter, the probability of a photon being detected on the far side is then cos(α)?

6. Dec 4, 2016

### A. Neumaier

yes, since the detection rate is proportional to the intensity. In fact, this is Malus' law from 1809, recast in terms of quantum mechanics, where it is just the Born rule applied to polarization states.

7. Dec 4, 2016

### jeremyfiennes

Thanks. So the probability of detecting a single photon after the polarizer should have been cos2(α), rather than cos(α)?

Continuing, if I measure the photon spin along the axis of polarization, I have a 100% chance of getting spin up; perpendicularly to it a 100% chance of getting spin down; and at 45o to the axis a 50% chance of each?

8. Dec 4, 2016

### FilipLand

Yes, a photon is an electromagnetic wave, with an oscillating electric field which "drive" the oschillating magnetic field, which drive the oschillating electric field and so on and so on (and will do so til the end of time if nothing absorbes it). And a really spooky question is: what gives rise to the "independent oschillating electric wave" which trigger the cycle of the photon?

9. Dec 4, 2016

### A. Neumaier

Yes, i hadn't noticed the missing square.
No. With which polarization are you starting?

10. Dec 4, 2016

### Staff: Mentor

Small point: Spin-up and spin-down applies to spin-1/2 particles, not spin-1 photons. The two basis states that describe photon polarization can't be sensibly described as "up" and "down".

Bigger point:
You are assuming that the photon has an axis of polarization before you measure it. That's a bad assumption (and you should know that it's a bad assumption, because you've recently started this Bell's theorem thread).

Let's look at what actually happens when we measure the polarization of a photon by passing it through polarizing filter at some angle $\theta$: Either the photon pass through or the photon is absorbed. If it goes through, we can make predictions about how it will interact with subsequent polarizing filters: 100% chance of passing through a filter at the same angle, no chance of passing through a filter at right angles to that, probability $\cos^2(\theta-\phi)$ of passing through a filter at angle $\phi$.

But what did we actually measure? Certainly not the polarization axis of the photon before it encountered the filter - we still don't know what that was. For all we know, the photon was circularly polarized and would have interacted similarly with our polarizing filter no matter what angle it was at.

11. Dec 4, 2016

### jeremyfiennes

(Good thing this is a basic thread!)
So what are those directions, in this case?

No. I was assuming that if I make a measurement I will get a result, which will then put the photon in that state (?).

12. Dec 7, 2016

### jeremyfiennes

I'm still unclear on the relation between polarization along a given axis and photon-particle spin.

13. Dec 7, 2016

### Staff: Mentor

Clockwise and counterclockwise, as "viewed" along the direction of propagation.

The corresponding polarizations of a classical electromagnetic wave are the two forms of circular polarization, in which the electric field vector "spins" or "pinwheels" around the direction of propagation (in a clockwise or counterclockwise sense) instead of oscillating back and forth along a single direction perpendicular to the direction of propagation.

Last edited: Dec 7, 2016
14. Dec 7, 2016

### jeremyfiennes

Still not there. A photon-wave can be polarized vertically or horizontally. (Or circularly, clockwise or counterclockwise, but these cases can be represented by super-positions of two separate waves with a +/- 90o phase lag.) Two variables are thus needed to define the wave: a vertical and a horizontal component. And in photon-particle spin terms? Or is my thinking naive and/or muddled?

15. Dec 7, 2016

### Staff: Mentor

A photon can be described using two orthogonal spin eigenstates, $|R \rangle$ and $|L \rangle$, which have values of the projection of the angular momentum on the axis of propagation of $+\hbar$ and $-\hbar$. Linearly polarized photons are in a state that is a superposition of these two spin eigenstates,
$$\frac{1}{\sqrt{2}} \left( |R \rangle + e^{i \phi} |L \rangle \right)$$
with the orientation of the polarization axis depending on $\phi$.

16. Dec 7, 2016

### Staff: Mentor

More explicitly, for $\phi = 0$ and $\phi = \frac {\pi} 2~\rm{rad} = 90^\circ$ you get the horizontally and vertically polarized states: $$\frac 1 {\sqrt 2} (|R\rangle + |L\rangle) = |H\rangle \\ \frac 1 {\sqrt 2} (|R\rangle + i |L\rangle) = |V\rangle$$

17. Dec 7, 2016

### Staff: Mentor

That depends on the orientation of the coordinate system (which is why I had refrained to give specific values).

18. Dec 7, 2016

### Staff: Mentor

Sure. I should have written "if you define 'horizontal' as $\phi = 0$ and 'vertical' as..."

19. Dec 8, 2016

### vanhees71

This is a bit misleading. The "spin" of the photon refers to the corresponding massless representation and is the casimir operator of the rotation group, $s=1$. Now photons are massless quanta and as such do not have 3 spin components in any "quantization direction" but only two. The natural choice is the helicity, i.e., the projection of the total angular momentum (there's no gauge independent way to split the total angular momentum of a photon in spin and orbitral parts; so it's also important to keep in mind that it's the well-definied total angular momentum here!) to the direction of the photon's momentum, and the two possible values are $\pm 1$. These plane-wave helicity eigenmodes correspond to the left- and right-circular polarized eigenmodes in the more conventional language of classical electrodynamics.

20. Dec 8, 2016

### A. Neumaier

This is indeed the only coordinate-independent choice. But a linear polarizer (unlike a circular polarizer) produces linearly polarized light, and this is more naturally written as a superposition of horizontal and vertical polarizations, well-defined in the lab coordinate system. The latter is also easier to visualize. Thus horizontal and vertical are also quite natural light analogues of spin up and spin down for electrons.