Atomic and Molecular Physics for Physicists Ben-Gurion University of the Negev

Ben-Gurion University of the Negev
Atomic
Atomic and
and Molecular
Molecular Physics
Physics
for
for Physicists
Physicists
Ron Folman
Chapter 2:
Linear and angular momentum of radiation, helicity and the spin of a photon
(half wave plates, quarter wave plates, filters, polarizing beam splitters)
Main References: Corney A. Atomic and Laser
Spectroscopy, Oxford UP, 1987; QC 688.C67
1987; Any good optics and lasers book.
Exercises:
Dudi Moravchik.
www.bgu.ac.il/atomchip
www.bgu.ac.il/nanofabrication
www.bgu.ac.il/nanocenter
We ended the last class with the following picture:
But what does polarization really mean?
E(r,t)= (ε1Ε1+ε2Ε2) exp{i(k*r-ωt)}
where ε are two unit vectors perpendicular to k
and E1 and E2 are corresponding complex amplitudes.
When the latter have the same phase we get a linear polarization and when a π/2
phase difference we get a circularly polarized wave.
(Namely, the above mathematics is in the Vertical/Horizontal linear basis.
We could just as well do all the calculations in the Right/Left handed circular basis)
And this is how the 3 general solutions look like: Linear, circular, elliptical.
The photon is a Boson with J=1.
This J is due to spin and not to orbital angular momentum.
It has a J of -1,0,1 in the direction of its motion if its circularly polarized (left handed),
linearly polarized, and circularly polarized (right handed), respectively.
These 3 states are called: σ+, π, and σ-.
Angular Momentum conservation examples:
Classical:
1. When circularly polarized EM radiation hits an antenna it applies a torque.
2. When a λ/4 plate turns π into σ it feels a torque.
Quantum Mechanics:
1. This gives us another degree of freedom through which we can manipulate
the atom which the light interacts with. For example, through a process called
Optical pumping (which we will discuss later in the course) we can decide what
will be the J of the atomic (electronic) state, as angular momentum conservation
means that when a photon is absorbed by the atom, the atom must also absorb its J.
2. When an excited electronic state doubly decays (i.e. a 2-photon cascade) e.g. as a
source for an EPR experiment, the two photons will have opposite spin.
Relevant Optics:
Mirror: learn when it maintains the input polarization. Learn about critical angles.
(read Feynman’s “light and matter” book).
BS: always 50/50 splitting.
PBS: what input light do you need in order to get 50/50 splitting?
λ/2 half wave plate:
A Fast and Slow axis tuned such that
The Slow axis gets a π phase shift
relative to the Fast axis.
λ/4 quarter wave plate: there is also a Fast and Slow axis. They are tuned such
That with the width of the plate and with an input linear light at 45 degrees to the
wave plate axis, the outgoing light with be perfectly circularly polarized.
That means, the plate introduces a π/2 phase difference!
E(r,t)= (ε1Ε1+ε2Ε2) exp{i(k*r-wt)} where |Ε1|=|Ε2| and Ε1=Ε2 exp{i π/2}
Think and develop your intuition!
Some technological applications:
Linear polarization enables very simple filters for visible light (e.g. sun glasses),
Radio waves (just an array of wires), and so on.
Two way isolator:
One way isolator:
Nano second resolution on/off switch:
Lets calculate the outgoing helicity for a λ/4 + mirror for a +1 incoming helicity.
Helicity is defined as H= J*k / (|J| |k|) i.e. the normalizes scalar product between the
Direction of the photon angular momentum and the direction of it propagation.
H can be zero or +1 or -1.
We start with a circular wave E(r,t)= (e1E1+e2E2) exp{i(k*r-wt)}
where |E1|=|E2| and E1=E2 exp{i π/2} and H=+1
Lets assume that there is an angle α between the above e1,e2 basis and the basis of
the wave plate. We now decompose the above E(r,t) in the basis of the wave plate.
E1’=E2’ exp{i π/2} and E2’=E2 (up to a common constant).
Namely, the same phase relation even in the new basis! Prove it at home!!!!!
After going through the wave plate, these amplitudes will now be
E1’=E2’ exp{i (π/2+π/2)} and E2’=E2.
The way to the mirror and back does not add any differential phase.
Now the two waves go again through the wave plate and hence:
E1’=E2’ exp{i (-π/2)} and E2’=E2.
We now find that the right handed polarization turned into a left handed. As k has
also flipped sign, the helicity remains the same!
Homework: calculate where the light will go after a PBS → λ/4 → mirror.
(the PBS axis is 45 degrees from the wave plate axis)
Lets calculate the amount of angular momentum (Corney):
We start with the Poynting vector N=Re(HxE)/2 and we remember from the previous
lecture that N= c * Energy Density. We note the energy density: U=N/c.
The relativistic energy is
U 2 = G 2c 2 + mo 2c 4
Where G is the momentum density. As m=0, G=N/c2.
The torque applied on a surface dS is therefore c dS (r x N/c2) (the c factor is always
there to turn density into flux i.e. into something per unit time.
And we get an angular momentum L:
1
L = ∫ ( r ∧ N )dS
c
We now expand according to one of the identities from the previous lecture and get:
L=
1
*
*
⋅
−
⋅
{
r
H
E
r
E
H
}dS
(
)
(
)
∫
2c
We note that H is perpendicular to r (see previous lecture) and the first term is zero.
However, in the far field E is also perpendicular to r, and so also the second
term should be zero. We therefore use the 1/r2 E component and get:
L=
ik
4πε 0c
*
r
⋅
p
H
∫ ( ) dΩ
Prove at home!!! (incl. the ¼ factor)
We now use the far field result for H and find:
ik 3
L=
16π 2ε 0
*
ˆ
ˆ
r
⋅
p
r
∧
p
)d Ω
∫ ( )(
Strange result as we used Far field H and intermediate field E.
Without proving, I tell you that
ik 3
*
*
Lz =
p
p
−
p
p
(
x y
y x )
12πε 0
This equals ZERO is px and py are real
or if they have the same phase (convince yourself!).
Now, THINK why Lz is zero for this above case (which also describes a linear dipole).
For py= +/- i px
(circularly polarized)
2
3
we get:
k p
Lz
1
and we get:
Lz = ±
=± =±
12πε 0
pEl
ω
ω
Lets use the polarization as a tagging device:
The experiment is the famous double slit experiment and the question is:
“is complementarity more fundamental than the uncertainty principle…?”
You don’t have to measure the photons (or kick them)… its enough that you make such
“which path knowledge” available in principle i.e. tag the photons (e.g. polarizers)
(or their environment e.g. emit a photon in a cavity / scully).
No
pattern
Orthogonal
polarizers
But then you can also erase the tagging!
(Raymond Chiao, Paul Kwiat, Aephraim Steinberg)
Polarization
eraser
Pattern
re-appears
How does the photon know at the slit that in its future there is going to be an eraser?
Final note: even if people will tell you photon spin is trivial, don’t believe them.
Mind boggling papers are still coming out on this issue…….
Also look up photon orbital angular momentum…..