4.2 Effect of the SLM
Certainly, to understand what phase is accumulated to the Electric field, it is necessary
to correctly quantify the evolution of the Electric field while passing through the SLM.
For this reason, we will construct the Jones matrix of the SLM [13]. And calculate the
phase it imparts on two different circular polarization states of light.
We know TNSLM is made up of many layers of birefringent layers. First, lets assume
the twisting is linear, i.e. the twist φ can be interpreted as
ψ(z) = αz
(4.4)
where z is the distance along the propagation direction and α is contant.For our
TNSLM total twist angle, ψ(d ) = π2 ; d being the thickness of the SLM. And, we denote
the total birefringence of the SLM to be
δt =
2π
(n e − n 0 )d
λ
(4.5)
Now let there be N number of layers in the SLM. With these properties, each of the
layer will be twisted at an angle
φ
N
and the birefringence of that layer is given by
δt
N.
So
the full Jones matrix of the SLM is given as:
 2πn0 d
e i λN
J = R(ψ) 
0
N
ψ
 R(− )
2πn e d
N
e i λN
0

(4.6)
Which can be further simplified as:
h
ψ iN
J = R(ψ) ψ0 R(− )
N
(4.7)
where
ψ0 = e i
⇒ J =e
2πn d δ
i ( λ0 + 2t
2πn 0 d
λN

δt
ψ
cos N e −i 2N
)
R(ψ) 
δt
ψ
si n N e i 2N
ei
δt
2N

δt
e −i 2N

0
δt
ψ
si n N e −i 2N
δt
ψ
cos N e i 2N
N
0
δt
e i 2N
 =e

(4.8)

2πn d δ
i ( λ0 + 2t
)

R(ψ) 
X −iY
Z
−Z
X +iY
(4.9)
with
X
Y
Z
s
= cos 
ψ2 + (


δt 2 
)
2

s



δt /2
 ψ2 + ( δt )2 
¶
= 
 µq
 si n
2
ψ2 + ( δ2t )2



s


ψ
 ψ2 + ( δt )2 
¶
= 
 µq
 si n
2
ψ2 + ( δ2t )2
(4.10)
Here, the phase factor in front of the rotation matrix gives the dynamical phase due to
the SLM. So, dynamical phase
=
µ
2πn 0 d δt
+
λ
2
¶
(4.11)
Now the geometrical phase associated to a particular polarization state due to J’ is
given as ar g ⟨ψ|J ′ |ψ⟩ for |ψ⟩ ∈ {|LC P ⟩, |RC P ⟩}. where the matrix J’ is given as:

J ′ = R(φ) 
X −iY
Z
−Z
X +iY

 = R(φ)U (δt , ψ)
(4.12)
Here, at this point we face an experimental difficulty that the parameter δt i.e. the
total birefringence present in J’ that will be present in thus calculated geometrical


phases, cannot be measured directly. For this reason, we record the Mueller matrices
of the SLM extract the "effective" birefringence and rotation and calculate back the
total birefringence[14].
4.2.1 Rotator Retarder Model
We employ the fact that any unitary matrix U has an optically equivalent system
consisting of one retarder and one rotator [15]. So we write
U (δt , ψ)
= R(ψeq )Ret (δeq , θeq )


cosψeq si nψeq cosθeq

= 
−si nψeq cosψeq si nθeq

δeq
−si nθeq e −i 2

cosθeq
0
0
ei
δeq
2

cosθeq

−si nθeq
(4.13)

si nθeq

cosθeq
From this relation and equation 4.12, we have :
si n
δeq
2
X
= cos
Y
= si n
Z
= cos
δeq
2
δeq
2
δeq
2
cosψeq
cos(2θeq − ψeq )
si nψeq
(4.14)
si n(2θeq − ψeq ) = 0
4.13 we get δ in terms of δ and ψ as:
comparing eqns. 4.10 and 4.15,
t
eq
eq
δt = 2
s
·
µ
µ
¶
¶¸2
δeq
−1
cos
cos
− ψ2
cosψeq
2
(4.15)