# Homework Help: Photon polarization question

1. Dec 19, 2008

### Sam_Goldberg

1. The problem statement, all variables and given/known data

Hi, I have a question regarding a set of equations in Feynmann volume 3. On chapter 11, page 11, Feynmann discusses the right-hand and left-hand circulation for the polarization of the photon. He states: "In the classical theory, right-hand circular polarization has equal components in x and y which are 90 out of phase. In the quantum theory, a right-hand circularly polarized (RHC) photon has equal amplitudes to be polarized |x> or |y>, and the amplitudes are 90 out of phase. Calling a RHC photon a state |R> and a LHC photon a state |L>, we can write:

2. Relevant equations

|R> = (|x> + i|y>) / sqrt(2)
|L> = -(|x> - i|y>) / sqrt(2)."

3. The attempt at a solution

This is more of a question, but isn't it true that for RHC the x-component is 90 ahead rather than behind? Thus, shouldn't we place the i (which is a 90 phase factor) in front of the |x> instead of the |y>? Same question goes for the LHC equation, but with phases reversed. Thanks.

2. Dec 21, 2008

### Sam_Goldberg

Sorry if the question was not made clear; let's rephrase it. In RHC photon polarization, it is the x-component of the electric field vector that is 90 degrees, or exp(i * pi / 2) = i phase ahead. The complete vector (from the sum of the components) rotates in a counterclockwise fashion, hence the right hand orientation with the thumb pointing out of the screen.
However, when we look at the equation for the state |R> in Feynmann, we get, as before,

|R> = (|x> + i|y>) / sqrt(2).

Here is the problem. Take the amplitude to be x-polarized and the amplitude to be y-polarized:

<x|R> = 1 / sqrt(2)
<y|R> = i / sqrt(2)
Therefore, <y|R> = i<x|R>.

This equation shows the y-component to be 90 degrees ahead rather than the x-component. So, the question is, is the equation wrong and should we instead put the i in front of the |x> in the equation for |R>? Or is it really the case that I am overlooking something? Given that this is the Feynmann lectures on physics, the latter is probably the case...

Finally, I have to ask as I'm new here, is this the right place to post a question such as this one, or should I ask in a different area? Thanks for all your time.

3. Dec 23, 2008

### Redbelly98

Staff Emeritus
Welcome to PF

I've been thinking about your question, and finally this morning wrote out the equations for Feynman's |R> state,

|x> + i|y>​

in terms of sines and cosines. I get a left-handed polarized wave! I wonder if Feynman was defining RH and LH polarizations differently than us.

I have to leave for work soon, and after today I'll be traveling for the holidays. I'll try to post back here on Saturday, with more details. Feel free to post a VM* as a reminder if I haven't done that by Saturday evening (Eastern USA time).

VM = Visitor Message, on my profile page:
https://www.physicsforums.com/member.php?u=122961

Regards,

Mark

EDIT:
p.s., sorry about the delayed response. This one is a little more involved than most questions in here.

4. Dec 24, 2008

### Redbelly98

Staff Emeritus
Okay, here is what I have done.

I write the electric fields of the |x> and |y> states as follows. I'm leaving out the normalization factors and E-field amplitude, since those don't affect the orientation or shape of the fields.

|x> → Ex = Re { exp[i(kz-ωt)] } = cos(kz-ωt)
|y> → Ey = Re { exp[i(kz-ωt)] } = cos(kz-ωt)​
and
i|y> → Ey = Re { i exp[i(kz-ωt)] } = -sin(kz-ωt)

Combining
|R> = |x> + i|y>​
gives
Ex = cos(kz-ωt)
Ey = -sin(kz-ωt)​

This describes a left-hand circularly polarized wave, i.e. the E-field resembles a left-handed screw.

However, if you point your right thumb in the direction of propagation (+z), your right-hand fingers will curl in the direction that the E-field rotates at a stationary point. I am wondering if Feynman had this in mind for right-handed circular polarization?

5. Dec 25, 2008

### Sam_Goldberg

Here is how RHC is defined in volume 1, chapter 33, page 2: "If the end of the electric vector, when we look at it as the light comes straight toward us, goes around in a counterclockwise direction, we call it right-hand circular polarization." So apparently we are not viewing the light from the origin as it travels away from us in the positive-z direction; rather, we are viewing the light coming towards us.

Even Feynman himself admits that in RHC, the y-component of E is -i ahead, or, in other words, i behind the x-component. So we are stuck with the definition we first were thinking of. Let's then go back to the equation in my second post:

<y|R> = i<x|R>.

Here's a wild guess: We would think that <y|R> corresponds to Ey. Well, perhaps not. Maybe <R|y> is really what works. This is a possibility, because if we take the complex conjugate of each side, we get:

<R|y> = -i<R|x>,

which is what we are hoping for. I doubt <R|y> really gives the y-component, but it's a possibility...

Well redbelly, let me know what you think about this way of looking at it. If, in fact, <R|y> is the correct representation, then it would be great to know why this is the case.

6. Dec 28, 2008

### Sam_Goldberg

Okay guys, I talked with a friend of a friend (he's actually a PhD in physics) and have come up with the answer, so here we are. If we look at an RHC photon (using Feynmann's convention) coming towards us, then the x-component will be 90 degrees ahead, given our x-y plane. But, if we look at it coming away from us, then, given the mirror image of this x-y plane, the y-component will be 90 degrees ahead. So we may use either coordinate system and put the i either in front of the |x> or the |y>.

Ultimately, however, it does not matter. Physics is not described by amplitudes; rather, it is described by probabilities, which are the absolute squares of the amplitudes. In fact:

P(x-polarization) = |<x|R>|^2 = (1 / sqrt(2))^2 = 1 / 2,
P(y-polarization) = |<y|R>|^2 = (i / sqrt(2)) * (-i / sqrt(2)) = 1 / 2.

So really, we have a 50-50 chance of polarization in both directions and it ultimately does not matter which coordinate system we use, whether it be viewing an x-y plane as the light comes towards us or an x-y plane as the light moves away from us.

7. Dec 29, 2008

### Redbelly98

Staff Emeritus
Ah, there it is. RHC is defined oppositely in the well-known textbook Optics by Hecht and Zajak (or by just Hecht, depending on what edition it is). I just checked there, and they say E rotates clockwise for RHC coming towards the observer.

I've always used the Hecht and Zajak definition, and must say I prefer it because the spiraling E-field vector in an RHC wave has the same sense as the screw threads in a right-handed screw.

But if somebody like Feynman wants to define it differently, who am I to argue?

Regards,

Mark