Photon Physics: EIT and Slow lightstrat102/photonphysics/transparancies/week… · 3 The two-level...

1

Photon Physics: EIT and Slow light

Dries van Oosten

Nanophotonics Section

Debye Institute for NanoMaterials Science

3

The two-level system

I Start with atomic levels |g〉 and |e〉I They have energies ~ωg and ~ωe , with ωeg = ωe − ωg

I The Hamiltonian of the system is thus

H0 = ~ωg |g〉〈g |+ ~ωe |e〉〈e|

I We want to know the response to an external field

E = εE0 cosωt

I The polarisation of the atom is

p = 〈qx〉

How do we calculate the polarization?

4


I We first look at the coupling

p = 〈e|qx · E|g〉

I We rewrite the operator qx · E as

H1 = (|g〉〈g |+ |e〉〈e|) qx · E (|g〉〈g |+ |e〉〈e|)

I This yields

H1 = |g〉〈g |qx · E|e〉〈e|+ |e〉〈e|qx · E|g〉〈g |

I We can write

〈e|qx · E|g〉 = −µegE0 cosωt

5


I We define the Rabi Ω frequency as in week 12

Ω =1

~µegE0

I The total Hamiltonian then becomes

H = ~ωg |g〉〈g |+ ~ωe |e〉〈e| − ~Ω (|e〉〈g |+ |g〉〈e|) cosωt

I First, we take the RWA

H = ~ωg |g〉〈g |+~ωe |e〉〈e|−1

2~Ω(|e〉〈g |e iωt + |g〉〈e|e−iωt

)

6


I Next, we absorb e iωt into |e〉

H = ~ωg |g〉〈g |+ ~(ωe − ω)|e〉〈e| − 1

2~Ω (|e〉〈g |+ |g〉〈e|)

I Finally, we shift the energy down by ~(ωg − ω)

H = ~ω|g〉〈g |+ ~ωeg|e〉〈e| −1

2~Ω (|e〉〈g |+ |g〉〈e|)

I If we define basis vectors (1, 0) = |g〉 and (0, 1) = |e〉

H =

(~ω −~Ω/2−~Ω/2 ~ωeg

)We call this the dressed state picture

7


I Make the matrix look more symmetric, define δ = ω − ωeg

H =~2

(δ −Ω−Ω −δ

)

I Let us denote an eigenstates of H by |ψ〉 = cg |g〉+ ce |e〉I What is the polarization of this state?

p = 〈ψ|qx|ψ〉 = cgc∗e 〈e|qx|g〉+ c∗gce〈g |qx|e〉

I Which, in the RWA, can be written as

p = µeg(cgc

∗e + c∗gce

)

8


I What is the density matrix for |ψ〉 = cg |g〉+ ce |e〉

ρ =

(ρgg ρgeρeg ρee

)=

(|cg |2 c∗gcecgc

∗e |ce |2

)

I Thus the polarization is

p = µeg(cgc

∗e + c∗gce

)= µeg (ρeg + ρge)

How do we determine the density matrix?

9

Bloch equations revisited

Write the Hamiltonian using Pauli spin matrices

H =~2

(δσz − Ωσx)

The Liouville equation

ρ =i

~[ρ,H] + L

where L describes the damping (Lindblad operator)

L =γ

2(2σ+ρσ− − σ−σ+ρ− ρσ−σ+)

σz =

(1 00 −1

), σx =

(0 11 0

), σ+ =

(0 01 0

), σ− =

(0 10 0

),

10

Bloch equations revisited in MathematicaFirst , we define the Pauli spin matrices:they operator on vectors |g>=(1,0) and |e>=(0,1)

In[1]:= g := 81, 0<e := 80, 1<Σz := 881, 0<, 80, -1<<Σp := 880, 0<, 81, 0<<Σm := 880, 1<, 80, 0<<Σx := Σp + ΣmMatrixForm@ΣzDMatrixForm@ΣpDMatrixForm@ΣmDMatrixForm@ΣxD

Out[7]//MatrixForm=

K1 0

0 -1O

Out[8]//MatrixForm=

K0 0

1 0O

Out[9]//MatrixForm=

K0 1

0 0O

Out[10]//MatrixForm=

K0 1

1 0O

Do the lowering and raising operators work correctly?

What is the Hamiltonian?

Define the density matrix

And write down the Liouville equation

In components

11



In[11]:= Σp.g eΣm.e g

Out[11]= True

Out[12]= True




In components

12




In[13]:= H :=

1

2H∆ Σz - W ΣxL

MatrixForm@HD

Out[14]//MatrixForm=∆

2-

W

2

-

W

2-

∆

2


In[33]:= Ρ@t_D := 88Ρgg@tD, Ρeg@tD<, 8Ρge@tD, Ρee@tD<<MatrixForm@Ρ@tDD


KΡgg@tD Ρeg@tD

Ρge@tD Ρee@tDO


In components

13






In[35]:= Ρdot := FullSimplifyB-ä HH.Ρ@tD - Ρ@tD.HL +

Γ

2H2 Σm.Ρ@tD.Σp - Ρ@tD.Σp.Σm - Σp.Σm.Ρ@tDLF

MatrixForm@ΡdotD


Γ Ρee@tD -

1

2ä W HΡeg@tD - Ρge@tDL

1

2ä HW Ρee@tD + ä HΓ + 2 ä ∆L Ρeg@tD - W Ρgg@tDL

-

1

2ä HW Ρee@tD + H-ä Γ - 2 ∆L Ρge@tD - W Ρgg@tDL

1

2H-2 Γ Ρee@tD + ä W HΡeg@tD - Ρge@tDLL

In components

14






In components

In[41]:= D@Ρ@tD, tD@@1, 1DD == Ρdot@@1, 1DDD@Ρ@tD, tD@@2, 2DD == Ρdot@@2, 2DDD@Ρ@tD, tD@@1, 2DD == Ρdot@@1, 2DDD@Ρ@tD, tD@@2, 1DD == Ρdot@@2, 1DD

Out[41]= Ρgg¢@tD Γ Ρee@tD -

1

2ä W HΡeg@tD - Ρge@tDL

Out[42]= Ρee¢@tD

1

2H-2 Γ Ρee@tD + ä W HΡeg@tD - Ρge@tDLL

Out[43]= Ρeg¢@tD

1

2ä HW Ρee@tD + ä HΓ + 2 ä ∆L Ρeg@tD - W Ρgg@tDL

Out[44]= Ρge¢@tD -

1

2ä HW Ρee@tD + H-ä Γ - 2 ∆L Ρge@tD - W Ρgg@tDL

15

Bloch equations revisited in Mathematica

Steady state solutions

SolveA9Ρdot@@1, 1DD 0, Ρdot@@2, 2DD 0,

Ρdot@@1, 2DD 0, Ρdot@@2, 1DD 0=,

8Ρee@tD<, 8Ρgg@tD, Ρeg@tD, Ρge@tD<E

Out[60]= 88<<

Hmm, no solution because it is overdetermined!

SolveA9Ρgg@tD + Ρee@tD 1, Ρdot@@2, 2DD 0,

Ρdot@@1, 2DD 0, Ρdot@@2, 1DD 0=,

8Ρee@tD<, 8Ρgg@tD, Ρeg@tD, Ρge@tD<E

Out[61]= ::Ρee@tD ®

W

2

Γ

2+ 4 ∆

2+ 2 W

2>>

I That is a Lorentzian with γ →√γ2 + 2Ω2

I Power broadening!

I Before we look at the result, an alternative approach!

16


I We use a ”trick” to put the damping in the Hamiltonian.

I Introduce δ = δ + iγ/2

H =~2

(δ −Ω

−Ω −δ

)

I Now, we just determine the eigenvectors of this matrix

17


I With γ 6= 0, define tan 2θ = −Ω/δ

I ~ω± = ±~√

Ω2 + δ2/2

I Eigenvectors e− = (cos θ, sin θ) and e+ = (− sin θ, cos θ)

I These are not truly eigenstates and eigenenergies, as theycontain damping

I As in Week 12, the polarization is now

p = 〈qx〉 = 2µegρge = 2µeg cos θ sin θ

18


I Can also be written p = ε0αE, thus

α =2µegε0E0

cos θ sin θ

I α has the unit of volume

I The susceptibility of a dilute gas of these atoms is

χ = Nα

with N the density of atoms

19


-6 -4 -2 0 2 4 6detuning (°)

-2.0

-1.5

-1.0

-0.5

0.0

0.5

1.0

1.5

2.0

pol

ariz

ability (m3)

1e-20

0.0

0.5

1.0

1.5

2.0

2.5

3.0

cros

s se

ctio

n (m2)

1e-13

I polarizability α andscattering cross section σsc

I green lines: numericaleigenvalues

I red lines: optical blochequations

I Analytical eigenvalues alsogive the same result

20

The three-level system

I Now, we have levels |g〉, |e〉 and |m〉I ~ωe > ~ωm, ~ωg

I We want to know χ for this system

I with a strong coupling between |e〉 and |m〉

H = ~ωg |g〉〈g |+ ~ωe |e〉〈e|+ ~ωm|m〉〈m|

− 1

2~Ωp

(|e〉〈g |e iωpt + |g〉〈e|e−iωpt

)− 1

2~Ωc

(|e〉〈m|e iωc t + |m〉〈e|e−iωc t

)

21


I |g〉e−iωpt → |g〉, ωg = ωg + ωp

I |m〉e−iωc t → |m〉, ωm = ωm + ωc

I ωeg = ωe − ωg , ωem = ωe − ωm

H = ~(ωp − ωeg)|g〉〈g |+ ~(ωc − ωem)|m〉〈m|

− 1

2~Ωp (|e〉〈g |+ |g〉〈e|)− 1

2~Ωc (|e〉〈m|+ |m〉〈e|)

I Introduce two detunings δp = ωp − ωeg and δc = ωc − ωem

H = ~δp|g〉〈g |+ ~δc |m〉〈m|

− 1

2~Ωp (|e〉〈g |+ |g〉〈e|)− 1

2~Ωc (|e〉〈m|+ |m〉〈e|)

22


I Basis |g〉 = (1, 0, 0), |e〉 = (0, 1, 0) and |m〉 = (0, 0, 1)

H = ~

δp −Ωp/2 0−Ωp/2 −iγ/2 −Ωc/2

0 −Ωc/2 δc

I Where we assume that only level |e〉 decays

I An analytical solution is clearly cumbersome, but finding anumerical solution is easy

23

The three-level systemElectro-magnetically induced transparency

I Ωp is weak, δc γ

-8 -6 -4 -2 0 2 4 6 8detuning (°)

0.0

0.2

0.4

0.6

0.8

1.0

abso

rption

cro

ss-s

ection( ¾0)

c =0

c =°=4

c =°=2

c =°

c =4°

I Vertical dashed lines indicate ±Ωc/2

I Transparancy window is Ωc wide

24


I Ωp is weak, δc γ, Ωc = γ

I What about dispersion?

-10 -5 0 5 10detuning (°)

-10000

-5000

0

5000

10000

15000

¢n(10¡6)

I Index of refractionn(ω) =

√1 + χ(ω)

I Ωc = γ

I Highly dispersive in theEIT window!

25


I Group velocity vg = ∂ω/∂k

I Group index

ng =c

vg≈ 1 +

ω

2

∂χ

∂ω

-10 -5 0 5 10detuning (°)

-600

-500

-400

-300

-200

-100

0

100

200

300

ng(103)

0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

abso

rption

cro

ss-s

ection

( ¾0)

I ng ∼ 250000

I Inside the transparencywindow!

I Quite impressive (1 km/s)

I 1 µs pulse → 1.2 mm long

I But there is more nextweek!

1

Photon Physics: EIT and Slow light

Dries van Oosten

Nanophotonics Section

Debye Institute for NanoMaterials Science

2

This weekSlow light in atomic and nanophotonic structures

What are we going to discuss today?

I Light stopping/storage in a three-level system

I Photonic bandstructures

I Slow light in photonic crystal waveguides

3

Last weekSlow light in the three-level system

I Group velocity vg = ∂ω/∂k

I Group index

ng =c

vg≈ 1 +

ω

2

∂χ

∂ω

-10 -5 0 5 10detuning (°)

-600

-500

-400

-300

-200

-100

0

100

200

300

ng(103)

0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

abso

rption

cro

ss-s

ection

( ¾0)

I ng ∼ 250000

I Inside the transparencywindow!

I Quite impressive (1 km/s)

I 1 µs pulse → 1.2 mm long

4

Light stopping/storage in a three-level systemExperiments

laser

Pockels

cell

λ/ 4

plate

87Rb cell in oven,

solenoid & shields

λ/ 4

platephoto-

detector

AOM

photo-

detectoriris

polarizing

beam splitter

Ωc

|+>|->

Ωs

|e>5P1/ 2 , F=1

5S1/ 2 , F=2

70

-100 -50 0 50 100

tra

nsm

issio

n (

%)

applied magnetic field (mG)

50

30

10

5

Light stopping/storage in a three-level systemExperiments

50

0

50

0

50

0

250200150100500-50-100

time (µs)

b)

c)

a)I

I

I

II

II

II

τ = 50 µs

τ = 100 µs

τ = 200 µsnorm

aliz

ed s

igna

l inte

nsi

ty (

%)

Experimental cycle:

I First, zero the magneticfield

I Switch on AOM

I Use Pockels cell to createprobe pulse

I Switch off AOM

I Wait

I Switch on AOM

6

Photonic crystal waveguidesThe wave equation

Dielectric function varies in space, the wave equation reads

∇2E(x) + ε(x)k20E(x) = 0

Now assume the dielectric function is periodic in space

ε(x) = ε (1 + ∆ε cos kLz)

And write the electric field as

E = xE0f (z)

Then the wave equation becomes

∇2f (z) + εk20 (1 + ∆ε cos kLz) f (z) = 0

7

Photonic crystal waveguidesBloch theorem

Bloch theoremfk(z) = e ikzuk(z)

Withuk(z) = uk(z + L)

In other words

uk(z) =∞∑

n=−∞unke

inkLz

and thus

fk(z) =∞∑

n=−∞unke

i(k+nkL)z

8

Photonic crystal waveguidesThe mode equation

Plug this into the wave equation

∇2f (z) + εk20 (1 + ∆ε cos kLz) f (z) = 0

To find, after some algebra that for each n and k

unk

εk20 − [k + nkL]2

+

1

2ε∆εk20 un−1k + un+1k = 0

Diagonalizing this band diagonal matrix will yield k0 as functionof n and k.

Problem, the matrix depends on the eigenvalue!

9

Photonic crystal waveguidesThe mode equation

Start from the wave equation for H

∇× 1

ε(x)∇×H = k20H

Has the eigenvalues k20 isolated on the right-hand side. Great!Using the same trick as before, we find

k20unk =∑n′

[εn−n′ ]−1 (k + nkL)

(k + n′kL

)un′k

Whereεn−n′ = εδn,n′ + ∆

(δn+1,n′ + δn,n′+1

)

10

Photonic crystal waveguidesSolving the mode equation in Python/Numpy

impor t numpy

de f i n v e p s i l o n (em , ed ,N) :e p s i l o n=em∗numpy . i d e n t i t y (2∗N+1, dtype=f l o a t )+\

ed∗numpy . i d e n t i t y (2∗N+2, dtype=f l o a t )[1: , :−1]+\ed∗numpy . i d e n t i t y (2∗N+2, dtype=f l o a t ) [ :−1 ,1 : ]

r e t u r n numpy . l i n a l g . i n v ( e p s i l o n )

de f hami l ( i n v ep s , k , kL ,N) :q=k+kL∗numpy . a range(−N,N+1)q1=numpy . ou t e r (q , numpy . ones (2∗N+1))r e t u r n i n v e p s∗q1∗numpy . t r a n s p o s e ( q1 )

de f e i g e n ( i n v ep s , ks , kL ,N) :r e t u r n numpy . a r r a y ( [ numpy . s o r t ( numpy . l i n a l g . e i g ( hami l ( i n v ep s , k , kL ,N ) ) [ 0 ] ) f o r k i n ks ] )

d e f e i g e n no s c an (em , ed , k , kL ,N) :i n v e p s=i n v e p s i l o n (em , ed ,N)r e t u r n numpy . s o r t ( numpy . l i n a l g . e i g ( hami l ( i n v ep s , k , kL ,N ) ) [ 0 ] )

d e f p e r t u r b (em , ed , k , kL ) :o f f d i a g=−(ed/em∗∗2)∗k∗(k−kL )mat r i x=numpy . a r r a y ( [ [ k∗∗2/em , o f f d i a g ] , [ o f f d i a g , ( k−kL)∗∗2/em ] ] )r e t u r n numpy . s o r t ( numpy . l i n a l g . e i g ( mat r i x ) [ 0 ] )

Try Python, it’s awesome and free!

11

Photonic crystal waveguidesSolving the mode equation in Python/Numpy

0.0 0.1 0.2 0.3 0.4 0.5k (¼=a)

0.0

0.2

0.4

0.6

0.8

1.0

k0 (¼=a)

In practice, we use the MIT Photonic Bands program(does essentially the same thing)

12

Photonic crystal waveguidesSolving the mode equation using perturbation theory

0.0 0.1 0.2 0.3 0.4 0.5k (¼=a)

0.0

0.2

0.4

0.6

0.8

1.0

k0 (¼=a)

Gap seems ok, but mid gap is offset

13

Photonic crystal waveguidesSolving the mode equation using perturbation theory

0.0 0.1 0.2 0.3 0.4 0.5k (¼=a)

0.0

0.2

0.4

0.6

0.8

1.0

k0 (¼=a)

For smaller dielectric constant, the match is great!

14

Photonic crystal waveguidesIn practice

I Photonic band gap causes total reflection

I Waveguide mode is caught between two photonic crystals

I Periodicity also causes band structure of the defect mode

15

Photonic crystal waveguidesGroup velocity

0.0 0.1 0.2 0.3 0.4 0.5k(2¼=a)

0.0

0.1

0.2

0.3

0.4

0.5

!(2¼c=a)

0.25 0.30 0.35 0.40 0.45 0.50k(2¼=a)

0.00

0.05

0.10

0.15

0.20

0.25

v g=c

odd

even

I Only slow modes below the light-line are useful

I Even mode is the slowest

I And has the smallest group velocity dispersion

16

Photonic crystal waveguidesNear-field microscope

to detector

from laser

λ/2

photonic crystal waveguide

400 nm

thickness 200 nmpitch 450 nmhole size 250 nm

I Aperture probe → sub-wavelength resolution

I Interferometer → phase sensitivity

17

Photonic crystal waveguidesReal space images

I Perform a fourier transform on the images yieldswavevector k

I Repeat for many optical frequencies ω yields dispersion

18

Photonic crystal waveguidesDispersion measurements

19

Photonic crystal waveguidesFast-slow interface

I Take a piece of fast waveguide (vg = c/10)

I Connect it to a piece of slow waveguide (vg = c/50)

-10 0 10 20 30y/d

-6

0

6

x/d

-6

0

6x/

d0.0 0.6 1.2 1.8 2.4 3.0 3.6 4.2 4.8 5.4

a

b

λ=1525nm

λ=1533.95nm

I Field enhancement, factor 5

Why the field enhancement?

Photon Physics: EIT and Slow lightstrat102/photonphysics/transparancies/week… · 3 The two-level...

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Transcript of Photon Physics: EIT and Slow lightstrat102/photonphysics/transparancies/week… · 3 The two-level...