Polarization of Light: Angular Momentum

[BeauGeste]

So from what I understand, right handed circularly polarized light has angular momentum +1,
left handed circularly polarized light has angular momentum -1.
What is the angular momentum of linear polarized light?

 

[Claude Bile]

Are you referring to the angular momentum of a photon with circular polarisation? Or are you referring to the angular momentum of a classical wave?

Also, when you say +1, what quantity are you referring to, the angular momentum itself (in which case you should denote units) or the quantum number associated with angular momentum?

Claude.

 

[BeauGeste]

i guess I'm referring to the quantum number of the photon. but I'm interested in the classical answer too if there is one.

 

[Claude Bile]

For a single photon, the quantum number for angular momentum must be +/- 1, sometimes we say that the photon has spin +/- 1.

For a collection of photon with linear polarisation, we can regard half has having spin +1 and half as having spin -1 (with spin +1 corresponding to left hand and spin -1 corresponding to right hand circular polarisation). The average angular momentum from a group of linearly polarised photons is therefore 0.

Claude.

 

[Meir Achuz]

So from what I understand, right handed circularly polarized light has angular momentum +1,
left handed circularly polarized light has angular momentum -1.

There is actually a mismatch between the definition of circular polalrization
(R and L) and the photon spin direction. The photons of R polarized light are actually left handed. That is they have negative helicity, or spin component
of -hbar (or -1 in units where hbar is absent). The reason for this difference is that opticians define the rotation of the E vector as you look toward the light. Particlelists define the spin of the photon in the direction of advance of the light.

Classical light is the case when the number of photons is so large that the amplitude seems continuous. The angular momentum of classical L polarized light of a pure frequency would be n hbar, where n is the number of photons.

 

[Voltage]

Sorry if I ought to start a new thread with this, but can anybody tell me:

Can a single photon exhibit linear polarisation?

What's the difference between polarization plane and polarization?

 

[BeauGeste]

For a collection of photon with linear polarisation, we can regard half has having spin +1 and half as having spin -1 (with spin +1 corresponding to left hand and spin -1 corresponding to right hand circular polarisation). The average angular momentum from a group of linearly polarised photons is therefore 0.

Claude.

I thought that non polarized light would have avg. ang. mom. of 0, because it is have left handed and half right handed. It seems that linear polarized light is a different situation. You see what I'm saying?

 

[Claude Bile]

Voltage - The polarisation of an EM wave is typically defined as the direction of the E-field vector, with the polarisation plane being the plane defined by the E-field vector and the wave-vector, k.

Voltage, BeauGeste - The question of whether a single photon can possesses a linear polarisation pushes the edge of my knowledge to be honest. While a single photon must possesses angular momentum of +/- h-bar units, a photon can exist in a superposition of two states with the photon having a 50% probability of becoming either RH or LH polarised when a "measurement" is performed. You could perhaps call a photon in such a superposition of states a single photon with linear polarisation. You could similarly regard an unpolarised photon as being in a superposition of two orthogonal linearly polarised states. (I would recommend asking this question again in the QM forums, you will most likely get a more accurate answer there.)

The bottom line though is that a photon MUST have a momentum of either +/- h-bar units.

Claude.

 

[Creator]

I thought that non polarized light would have avg. ang. mom. of 0, because it is have left handed and half right handed. It seems that linear polarized light is a different situation. You see what I'm saying?

Classically, BeauGeste, it is different because you are not dealing with discrete particles but with continuous electric field amplitudes. Classically, circular polarization is composed of two linear polarized waves (of equal amplitude) oriented orthogonal to each other which are simply out of phase by 90 degrees.
The resulting superposition of the electric fields creates a continuosly rotating electric field vector thru space.

This can be seen experimentally simply by sending linear polarized light through a 'quarter wave' plate, creating circular polarized light.

Of course, classically, I don't think linear polarized waves have a net angular moment, but rather have a transverse oscillating field which can been seen in the effect on charges in matter as the wave passes through.
And they do have linear momentum (as per the Poyting vector) in the direction of propagation.

Creator

"Never eat more than you can lift" - Ms. Piggy

 

[Voltage]

Many thanks Claude and Creator.

 

[Milind_shyani]

With you

So from what I understand, right handed circularly polarized light has angular momentum +1,
left handed circularly polarized light has angular momentum -1.
What is the angular momentum of linear polarized light?

I would like to know first of alll what is right handed circularly polarized light and left handed circularly polarized light ??

 

[Voltage]

..Classically, circular polarization is composed of two linear polarized waves (of equal amplitude) oriented orthogonal to each other which are simply out of phase by 90 degrees...

Are we talking about them being out of phase by 90 degrees in a left handed direction or a right handed direction? Like by +90 or -90 degrees, perhaps better expressed as 90 degrees or 270 degrees?

 

[Creator]

Are we talking about them being out of phase by 90 degrees in a left handed direction or a right handed direction? Like by +90 or -90 degrees, perhaps better expressed as 90 degrees or 270 degrees?

Here's a brief discription with diagrams so you can better visualize.

http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/polclas.html

...

 

[Voltage]

Creator: many, many thanks.

 

[Creator]

Creator: many, many thanks.

No problem Voltage.

I guess, in light of BeauGeste's original question, I ought to address the angular momentum question classically also.

Notice in my first post to him I mentioned (as did Claude) it is only the direction of the electric field vector, E, that defines polarization; but this says nothing of the angular momentum.

Not so well known is the fact that, classically, standard Maxwell equations show there should be NO angular momentum vector directed along the direction of propagation ...however, this directly contradicts laser experiments which reveal angular momentum is transferred to objects in the path of circularly polarized light.

Just thought that ought to be mentioned...

Creator

-Honk if you like peace and quiet.-

 

[Weimin]

I'm following the link creator provided, here

http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/polclas.html

According to this picture in the link below, we are looking to the light that is coming toward us. The text says:

If this wave were approaching an observer, it electric vector would appear to be roting counterclockwise. This is called right-circular polarized.

http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/imgpho/polcir.gif

But right after that, it says:

If while looking at the source, the electric vector of the light coming toward you appears to be rotating clockwise, the light is said to be right-circularly polarized

That appears to be contrary!

 

[Creator]

:

:
:


Good catch Weimin; its probably a typo.

Creator:zzz:

 

[Weimin]

So which one is correct as I've seen different definitions of right and left circular polarizations from different books?

 

[Meir Achuz]

If this wave were approaching an observer, it electric vector would appear to be roting counterclockwise. This is called right-circular polarized.

The key words are "If this wave were APPROACHING AN OBSERVER,".
This is like looking at a transparent clock from behind.
HE physicists do look at the wave this way, which is why they call L polarized light, "right handed".

 

[Creator]

So which one is correct as I've seen different definitions of right and left circular polarizations from different books?

Me too; apparently left and right circular polarized light is dependent upon whether it is being observed by an engineer or a physicist...

The point is that neither is 'correct'; it is simply a matter of convention. ( maybe that's why someone invented quantum mechanics..)

The thing to remember is that when the x component trails behind the y component by 90 degrees, the E vector rotates one way; when it preceeds the y component by 90 *, it rotates the other way.

Here is a great animation for each direction, showing it clearly.

http://www.enzim.hu/~szia/cddemo/edemo7.htm

By the way, two circularly polarized waves can also be out of phase by 90 degrees...guess what you get?

See here:
http://www.enzim.hu/~szia/cddemo/edemo8.htm

This is simply a demo re-stating the fact that linearly (plane) polarized light is simply the superposition of two circularly polarized waves.

Creator

 

[Weimin]

The key words are "If this wave were APPROACHING AN OBSERVER,".
This is like looking at a transparent clock from behind.
HE physicists do look at the wave this way, which is why they call L polarized light, "right handed".

I was a bit confused about that expression but look at the picture, the observers says the electric field is roting anticlockwise, that means he is looking against the direction of the light. Otherwise the picture and its caption are contrast themselves.

If it is the matter of convention like Creator said, I will not mind it any more

 

[Creator]

...

The bottom line though is that a photon MUST have a momentum of either +/- h-bar units.

Claude.

Claude, things used to be that simple when we were in college.

It is not commonly known these days that light is not only 'spin' polarized in quantum units, but it can also possesses "orbital" angular momentum which can exist in any number of angular momentum states.

This has created loads of interesting studies of late, especially since the 'orbital' portion can be of any of an 'infinite' number of possible states, making its applicability to communication technology highly desirable.

Here's an update:

The original discovery...by Allen :

http://prola.aps.org/abstract/PRA/v45/i11/p8185_1

I really like this one ...with info on an interferometric/'sorter' used to measure orbital angular momentum:
http://www.physics.gla.ac.uk/Optics/projects/singlePhotonOAM/


Some basics , for teaching / experimental point of view:

http://departments.colgate.edu/phys...al angular momentum in optics instruction.pdf

For early graduates like me (that's another name for old guys), its disconcerting to realize that just after you leave the academic environment, they change all the rules.

Welcome back to college.

Creator

 

[Claude Bile]

I always thought of an EM applying a torque as a "classical" momentum, and never really thought of how it would apply to the quantum world. It always seemed to be an issue artfully dodged by EM textbooks!

Thankyou for the references! Indeed the rules have been "changed", but at least I feel a little bit smarter .

Claude.