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

[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

 

[Orodruin]

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.

 

[A. Neumaier]

the photon have a electric field?

Of course.

 

[Uriel]

Thanks to both of you. I will read more about this topic.

 

[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)?

 

[A. Neumaier]

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)?

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.

 

[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 45^o to the axis a 50% chance of each?

 

[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?

 

[A. Neumaier]

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

Yes, i hadn't noticed the missing square.

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 45^o to the axis a 50% chance of each?

No. With which polarization are you starting?

 

[Nugatory]

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 45^o to the axis a 50% chance of each?

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.

 

[jeremyfiennes]

(Good thing this is a basic thread!)

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.

So what are those directions, in this case?

You are assuming that the photon has an axis of polarization before you measure it.

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

 

[jeremyfiennes]

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

 

[jtbell]

So what are those directions, in this case?

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.

 

[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 +/- 90^o 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?

 

[DrClaude]

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 $\phi$.

 

[jtbell]

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

 

[DrClaude]

More explicitly, for $\phi = 0$ and $\phi = \frac {\pi} 2~\rm{rad} = 90^\circ$ you get the horizontally and vertically polarized states

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

 

[jtbell]

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

 

[vanhees71]

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 $\phi$.

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.

 

[A. Neumaier]

. The natural choice is the helicity, [...] the two possible values are ±1±1\pm 1. These plane-wave helicity eigenmodes correspond to the left- and right-circular polarized eigenmodes

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.

 

[vanhees71]

Sure, there the directions are given by the measurement/preparation apparatus. However usually the aim is to define properties of entities like particles or massless quanta in "natural" frames, defined by the entities themselves. That's also why one defines the thermodynamic quantities of a condensed-matter system like temperature, chemical potentials, energy and charge densities etc. in the (local) restframe(s) of the fluid (cells) and thus as scalar (fields).

 

[forcefield]

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

I'm puzzled because the Feynman Lectures on Physics gives: $$\frac 1 {\sqrt 2} (|R\rangle + |L\rangle) = |x\rangle \\ \frac {-i} {\sqrt 2} (|R\rangle - |L\rangle) = |y\rangle$$ which I can't make equal with your superpositions.

 

[jeremyfiennes]

Thanks all. My question is more than answered.