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Self-organization of Cold Atomsin Optical Resonators

Claus Zimmermann

University of Tübingen, Germany

macroscopic self organization

biology: crickets, fire flies

architecture: millennium bridge

medicine: pace maker cells

economy: euro crisis , stocks

...

common feature: global couplingA.-L. Barabási, Nature 403, 849 (2000)

fire flies

chirping crickets

Strogatz, et. al, Nature, 438, 43-44 (2005)

self organization of a Bose-Einstein condensate

experiment

theorypump

light atoms

Superradiant Rayleigh scattering and collective atomic recoil lasing in a ring cavityS. Slama, S. Bux, G. Krenz, C. Zimmermann, Ph.W. Courteille, Phys. Rev. Lett. 98, 053603 (2007)

self organization of a Bose-Einstein condensate

self organization of a Bose-Einstein condensate

scattering opt. standing wave periodic potential density grating Bragg-reflection

self organization of a Bose-Einstein condensate

scattering opt. standing wave periodic potential density grating Bragg-reflection

with cavity: "collective atomic recoil lasing (CARL)"

without cavity: "matter wave superradiance"

self organization of a Bose-Einstein condensate

scattering opt. standing wave periodic potential density grating Bragg-reflection

with cavity: "collective atomic recoil lasing (CARL)"

without cavity: "matter wave superradiance"

role of cavity: • Purcell factor (1D)• new control parameter:

cavity pump detuning

observations

threshold behavior stability diagram

Dynamical instability of a Bose-Einstein Condensate in an Optical Ring ResonatorD. Schmidt, H. Tomczyk, S. Slama, C. Zimmermann, Phys. Rev. Lett. 112, 115302 (2014)

experimental parameters:

cavity-pump detuning

Pp pump power

apparatus

regime:

r

cavity decay > coupling > recoil

1.2MHz 14kHz

semi classical description

atoms: • no temperature • GP equation without interaction• no "particles"

light: • classical light field• no photons

photon recoil: discrete momentum states

cavity assisted spontaneous Raman decay between momentum states

CARL equation

equations of motion for the mean fields

population of the momentum states

initial condensate

positive momentum states

negativemomentum states

simulation for n=-5...5

onset shifts withpump powerand detuning

initial condensate

positive momentum states

negativemomentum states

phase diagram

population population of n=0 after 50 s

standing wave scenario

Klinder et al. PNAS 112, 3290 (2015)Baumann et al. Nature 464, 1301 (2010)

Zürich, Esslinger group Hamburg, Hemmerich group

Dicke phase transition

coupling: cavity assisted spontaneous Raman decay

"Dicke (matter wave) superradiance"

excited matter state ctwo fields:

interaction:

light mode in cavity a

Dicke model in the Holstein Primakoff approximation

Dicke model in the Holstein Primakoff approximation

excited matter state ctwo fields:

interaction:

causes instability

light mode in cavity a

works for the standing wave scenario

counter rotating terms in standing wave scenario

scattering out of the mode and destroy lattice

scattering into the mode and create lattice

scattering out of the mode and create lattice

scattering into the mode and destroy lattice

terms in a ring scenario

exchanging energy between scattered particles

exchanging energy between scattered particles

scattering into the pump and condensate

scattering out of the pump and condensate

ring cavity

ring cavity

standing wave vs. ring

two modes(only counter rotating terms)

three modes(ring cavity)

two modes(standing wave cavity)

... + ... +

three mode approximation

light mode a

momentum state c1momentum state c-1

= coupling constant (contains pump power)r = recoil energy

exceptional points

equations of :

see also Carmichael group, PRA 75, 013804 (2007) and Meystre group, PRA 60, 1491 (1999)

is solved by

exceptional points

W.D. Heiss, J. Phys. A: Math. Theor. 45 444016 (2012)

some of our favorite Riemann-surfaces

future

quantum effects: correlations, entanglement

excpetional points

Hannah Tomzcyk

Dag Schmidt

thanks to

Hefei-Lectures 2015First Lesson: Optical Resonators

Claus Zimmermann, Eberhard-Karls-Universität Tübingen, Germany

October 11, 2015

1.1 Power and Field in an Optical Resonator (steady state solu-tions)

The left mirror is the input coupler with a given …eld re‡ectivity r1, the right mirror is theoutput coupler with a re‡ectivity r2. The third mirror re‡ects the light back to the leftmirror and should be as ideal and lossless as technically possible (r3 = 1) . The resonatormay contain a nonlinear crystal, a cloud of atoms, or something else.

² electric …eld in the resonator

If the resonator is in a steady state, the …eld inside the cavity, right behind the inputcoupling mirror, Ec, should be the same after one round trip

En+1 = En

During a round trip of length l the …eld accumulates a phase

' = kl;

with k being the wave number. Furthermore the …eld is reduced by a factor

rm := r1p1¡ L;

with the …eld-re‡ectivity of the input coupler r1 and the power losses L of one round tripincluding the losses at the output coupler and the third mirror (but without the lossesdue to transmission back through the input coupler). Finally there is …eld coupled inthrough the input coupler

t1Ei =q1¡ r21Ei:

In total one obtainsEn+1 = rme

i'En + t1Ei = En := Ec;

orEc

Ei

=t1

1¡ rmei'

1

² intensity circulating in the resonator

The power time averaged over one oscillation of the light …eld is obtained from themodulus of the …eld:

Pc

Pi=jEcj2jEij2

=t21

(1¡ rmei') (1¡ rme¡i')=

t211¡ 2rm cos'+ r2m

There is a maximum for' = q ¢ 2¼

with q being a natural number. One obtains a series of resonances

' = q ¢ 2¼ = kl

k =q ¢ 2¼

l

º =!

2¼=

ck

2¼=

c

2¼

q ¢ 2¼l

= q ¢ cl= q ¢ ºFSR

ºFSR : =c

l

Close to a resonance one can expand cos(x) ' 1¡ x2=2 and obtains

Pc

Pi

' t21=r2m

(1¡ rm)2 =r2m + '2

:

This is a Lorentzian line shape function in '.

² impedance matching

On resonance, ' = 0, one obtains

Pc

Pi

=t21

(1¡ rm)2 =

1¡ r21¡1¡ r1

p1¡ L

¢2With the de…nition of the power transmission T and the power re‡ectivity R

R1 : = r21T1 : = t21 = 1¡R1

Rm : = r2m = R1 (1¡ L)

we obtainPc

Pi

=1¡R1¡

1¡pRm

¢2 :

2

Let’s calculate the optimal input coupler i.e. the R1 for which the power in the cavityis largest:

d

dR1

1¡R1³1¡pR1 (1¡ L)

´2 =

p¡R1 (L¡ 1) + L ¡ 1³¡1 +p¡R1 (L¡ 1)

´3p¡R1 (L ¡ 1)= 0

p¡R1 (L¡ 1) = 1¡ L

R1 (1¡ L) = (1¡ L)2

R1 = 1¡ L

This is the impedance matched case. Lets calculate the enhancement A :

A :=Pmax

Pi

=1

1¡R1

=1

L:

With 1% losses and perfect impedance matching the power in the cavity is 100 timesthe power incident on the input coupler.

² re‡ected intensity

The re‡ected …eld consists of a part which is directly at the input coupler and a partwhich comes from inside the cavity:

Er = ¡r1Ei + Ec

p1¡ Leiklt1

= ¡r1Ei + Ei

t1

1¡ r1p1¡ Lei'

p1¡ Lt1e

i'

The negative sign of the …rst term takes care of the time reversal symmetry of a beamsplitter: Any energy conserving surface re‡ects the …eld on one side with a ¼ phaseshift relative to the re‡ection from the opposite side.

On resonance on gets

Er

Ei

= ¡r1 + t21rl1¡ r1rl

= ¡r1 + (1¡ r21) rl1¡ r1rl

=(1¡ r21) rl ¡ r1 (1¡ r1rl)

1¡ r1rl

=rl ¡ r11¡ r1rl

with the abbreviationrl :=

p1¡ L

The power may be calculated form the …eld modulus as above. For perfect impedancematching ,i.e. for r1 = rl, the re‡ection vanishes and all the incident light is coupledinto the resonator.

3

² Question to think about

Assume that the only round trip losses are given by the transmission of the outputcoupler:

L = 1¡R2 = T2

In the impedance matched case one obtains for the intensity behind the output coupler

Ptrans = Pc ¢ T2 = Pin ¢ 1T2¢ T2 = Pin:

This holds independently of the value of T2 , i.e. also for T2 = 0 . Impedance matchingthen requires that also T1 = 0. In other words: If you build a resonator with two losslessmirrors with 100% re‡ectivity the transmission through the cavity on resonance willbe 100% . How can that be?

1.2 Equation of motion

² di¤erential equation for the …eld

We require that the …eld after one round trip is the …eld at the beginning of the roundtrip corrected for the round trip losses and the round trip phase shift. In addition, onehas to add the …eld which is coupled in during on round trip

E(t+ ¿) = E(t)rmei' + t1Ei

For very short round trip times (what is the typical round trip time?) one can expandthe electric …eld as

E (t+ ¿ ) ' E (t) +dE

dt¿

4

and obtains

E (t) +dE

dt¿ = E(t)rme

i' + t1Ei

dE

dt=

1

¿E(t)

¡rme

i' ¡ 1¢+

1

¿t1Ei

Near resonance, '¿ 1, we expand the exponential

ei' ' 1 + i'

and obtaindE

dt= E

(rmi'¡ (1¡ rm))

¿+

t1¿Ei:

² resonators with low losses

We …rst write down the round trip transmission

tm :=p1¡ r2m =

p1 + rm

p1¡ rm

then solve for

1¡ rm =t2m

1 + rm:

For small losses rm ' 1 :t2m

1 + rm' t2m

2

with this one obtains

dE

dt= (i¢¡ ·)E + ´

¢ : = rm'

¿' '

¿, cavity detuning

· : =1¡ rm

¿' t2m

2¿=

L

2¿, …eld decay rate

´ : =t1¿Ei, …eld pump rate

We look for the solution on resonance of a …lled cavity after the pump …eld has turnedof:

dE

dt= ¡·E

E(t) = E0e¡·t:

measuring the decay rate is one way to determine the losses of the resonator (worksonly for very good cavities and fast detectors). Which decay rates do you expect fortypical resonators?

5

² general solution

isE (t) =

´

·¡ i¢+ Ce(i¢¡·)t

C is given by the initial conditions. After several decay times the second term decaysand one obtains (compare to the steady state solution above):

Ec =´

·¡ i¢

= ´i¢+ ·

(·¡ i¢) (· + i¢)

= ´i¢

¢2 + ·2+ ´

·

¢2 + ·2

The …eld oscillates with a dispersive and an absorptive amplitude (plot the two terms,does it remind you to something ?). In fact, the cavity behaves like a driven, dampedmechanical harmonic oscillator only that the phase µ between the driver and the os-cillator is di¤erent. If the phase of the driving …eld is taken as reference, the relativephase is given by the phase of the …eld inside the cavity:

µ = arctanIm (Ec)

Re (Ec)= arctan

´ i¢¢2+·2

´ ·¢2+·2

= arctan¢

·:

For ¢À · the phase shift is ¼ relative to the limiting case of ¢¿ ·. This is the samefor a mechanical resonator. What di¤ers is the phase at resonance which is 90± for themechanical oscillator and 0± for the optical resonator.

1.3 Line width, …nesse, decay rate

² line width

is de…ned as full width at half maximum (FWHM) of the power resonance function.

Pc

Pi=

t211¡ 2rm cos'+ r2m

close to resonance ('¿ 1) the expansion yields

Pc

Pi

' t211¡ 2rm

¡1¡ 1

2'2¢+ r2m

=t21

(1¡ rm)2 + rm'2

half maximum is obtained for

prm'1=2 = 1¡ rm

6

which corresponds to a detuning ±1=2

º02¼

=±1=2'1=2

so±FWHM = 2±1=2 =

º0¼'1=2 =

º0¼

1¡ rmprm

:

² …nesse

is de…ned as the ratio

F =º0

2±1=2= ¼

prm

1¡ rm

withrm = r1

p1¡ L =

p(1¡ T1) (1¡ L)

and in the impedance matched case

r1 =p1¡ L;

one obtains

F = ¼

pr1p1¡ L

1¡ r1p1¡ L

= ¼

p1¡ L

L:

and for small losses, L¿ 1, one …nally gets

F ' ¼

L:

² decay rate

the power decay rate °P is twice the …eld decay rate

°P := 2· = 21¡ rm

¿:

7

with this, one obtains

±FWHM =º0¿

¼

·

rm=

1

¼

·

rm' ·

¼=

°P2¼

=1

2¼

1

Tres;

where Tres is the 1=e-lifetime of the power in the resonator. The …nesse and the lifetime is connected according to

F =º0

±FWHM= 2¼º0Tres = !0Tres:

1.4 Optical Beams (Kogelnik, Li, Applied Optics 5, 1550, (1966))

To understand laser beams in resonators we …rst need to talk about geometric optics.

² beam vector

we introduce a vector

~x :=

µxx0

¶=

µdistance to optical axisslope of the beam

¶

² ABCD-matrices

optical elements may be described by matrices

1) path of length d

Md =

µ1 d0 1

¶2) lens

Mf =

µ1 0¡ 1

f1

¶(convex lenses f > 0, concave lenses f < 0)

3) path within a material with index of refraction

M_d=

µ1 d=n0 1

¶8

In general, the matrix

M =

µA BC D

¶is called ABCD-matrix.

It transforms an optical beam into a new one

~x2 = M ~x1

² example 1: propagation along a distance dµx2x02

¶=

µ1 d0 1

¶µx1x01

¶=

µx1 + d ¢ x01x1

¶changes distance to optical axis but keeps the slope constant

² lens with focal length f :µ1 0¡ 1

f1

¶µx1xp1

¶=

µx1x01 ¡ x1

f

¶=

µx2x02

¶changes slope but keeps the distanceµ

1 0¡ 1

f1

¶µx10

¶=

µx1¡x1

f

¶The new beam intersects the optical axis at a distance f from the lens.

² optical systems

composite systems are described by the product of the corresponding matrices. Exam-ple: path d1, lens f , path d2

M =

µ1 d20 1

¶µ1 0¡ 1

f1

¶µ1 d10 1

¶=

µA BC D

¶

9

² lens conductor

The possible trajectories between two mirrors

with radius of curvature r corresponds to a periodically repeating series of lenses withfocal length

f =r

2

One round trip is described by the Matrix

M = Md ¢Mf2 ¢Md ¢Mf1

10

A number of n round trips are described by

Mn = (Md ¢Mf2 ¢Md ¢Mf1)n

² Sylvester TheoremµA BC D

¶n

=1

sin µ

µA sin(n ¢ µ)¡ sin ((n¡ 1) ¢ µ) B (sinn ¢ µ)C sin (n ¢ µ) D sin (n ¢ µ)¡ sin ((n¡ 1) ¢ µ)

¶with

cos µ :=1

2(A+D)

periodic focusing and defocussing with the ”frequency” µ.

² stability

unstable solutions are obtained if µ is unde…ned i.e.¯̄12(A+D)

¯̄> 1. Stability range

is thus¡1 <

1

2(A+D) < 1

standing wave resonator:

0 <

µ1¡ d

r1

¶| {z }

:=g1

µ1¡ d

r2

¶| {z }

:=g2

< 1

0 < g1g2 < 1

² stability diagram

Since g1g2 > 0 the point (g1g2) is in the …rst and third quadrant. The stabilityboundaries ful…ll g1g2 = 1, i.e. they form the functions g2 = 1

g1:

11

² boundary at g1g2 = 1 for resonators with equal mirrors

for r1 = r2 = r; i.e.

g1g2 = 1µ1¡ d

r

¶µ1¡ d

r

¶= 1

1¡ d

r= §1

d

r= 0 or

d

r= 2

² plane resonatord

r= 0

r is in…nite and the resonator consists of two plane mirrors. The resonator is unstableagainst an in…nitesimal tilt of the beam.

² concentric resonatord

r= 2

The radius is half the distance and the mirror planes lie on a sphere. The resonator isunstable against an in…nitesimal transversal shift of the beam.

² confocal resonator, boundary at g1g2 = 0

for equal mirrors one obtains

1¡ d

r= 0

d = r:

The mirrors have their focal point at the same position. The resonator is stable.However an in…nitesimal di¤erence in the radius makes the resonator unstable.

² Questions to think about

After how many round trips does the beam comes back to the starting point in aconfocal resonator if the beam is injected not along the optical axis? What is the freespectral range in this case? If the cavity has two equal mirrors and no losses, is thecavity still impedance matched?

12

1.5 Gaussian beams (Kogelnik, Li, Applied Optics 5, 1550, (1966))

² paraxial wave equation

In vacuum, Maxwells equations can be used to derive the Helmholtz equation.

r2u(~r) + k2u(~r) = 0

withk =

2¼

¸:

Ansatz for the solution.u(~r) = ª(x; y; z) ¢ e¡ikz

ª is slowly varying in space. The Helmholtz equation then becomes the ”paraxial waveequation”.

@2

@x2ª+

@

@y2ª¡ 2ik

@ª

@z= 0:

The term » @2ª@2z

has been neglected. It would by responsible for a fast spatial modula-tion which is already taken care of in the Ansatz. The remaining changes of ª in thez-direction should stay linear. This is the paraxial approximation.

² solution

The most simple solution (fundamental mode) has the form

ª(r; z) = e¡i(P (z)+ k2q(z)

¢r2)

with the functions q(z) and P (z) obeying the di¤erential equations

dq(z)

dz= 1

anddP (z)

dz= ¡ i

q(z)

alsor2 := x2 + y2:

Fromdq

dz= 1

an getsq = q0 + z

with q0 can be complex.

13

We choose the origin such that the real part of q0 vanishes. q0 is then complex and wecan write

q0 = iz0

with the real valued ”Rayleigh-length” z0. This is the most important parameter of alaser beam

² beam waist, Rayleigh length, confocal parameter

with q = iz0 + z the solution ª writes

ª = e¡iP (z) ¢ e¡i k¢q¤

2q¢q¤ r2

ª = e¡iP (z) ¢ e¡ik2¢ z+iz0z2+z20

r2

ª = e¡iP (z)¡i k

2r2z

z2+z20 ¢ e¡ r2

w2

with

w2 :=2

k¢ z

2 + z20z0

=2z0k¢Ã1 +

µz

z0

¶2!:

The …eld amplitude has an envelope according to a Gaussian e¡r2

w2 with a 1e-beam

radius w of

w = w0 ¢s1 +

µz

z0

¶2

:

The beam waist w0 is the radius at z = 0:

w20 :=

2z0k

=b

k

The ”confocal parameter” is de…ned as twice the Rayleigh length

b := 2z0:

14

² far …eld angle

in the limit jzj À jz0j one getsw ' w0 ¢ z

z0w

z=

w0

z0= tan µ

µ is the far …eld angle

tan µ =w0

z0=

r2

kz0=

r¸

¼ ¢ z0 =¸

¼ ¢ w0

² wave fronts

The solution ofdP

dz= ¡ i

q= ¡ i

z + iz0is

iP (z) = ln(1¡ iz

z0);

(check it by taking the derivative). Decomposition it in real and imaginary part yields

Re (iP (z)) =1

2

µln(1¡ i

z

z0) + ln(1 + i

z

z0)

¶=

1

2ln

µ(1¡ i

z

z0)(1 + i

z

z0)

¶=

1

2ln

µ1 + (

z

z0)2¶= ln

r1 + (

z

z0)2

and

Im(iP (z)) =1

2i

µln(1¡ i

z

z0)¡ ln(1 + i

z

z0)

¶=

1

2iln

Ã1¡ i z

z0

1 + i zz0

!

= ¡i lnÃs

1¡ i zz0

1 + i zz0

!= ¡ arctan

z

z0

(note that arctan x = i lnq

1¡ix1+ix

). As result one obtains

iP (z) = ln

r1 + (

z

z0)2 ¡ i arctan(

z

z0):

15

With

¡iµP (z) +

k

2

r2z

z2 + z20

¶= ¡ ln

r1 + (

z

z0)2

+i

µarctan(

z

z0) ¡ k

2

r2z

z2 + z20

¶:

the total …eldu(~r) = ª(x; y; z) ¢ e¡ikz

can now be written

u(~r) = ª(x; y; z) ¢ e¡ikz

= ª0 ¢ 1q1 + ( z

z0)2| {z }

conservation of total power of u2

¢ e¡(r

w(z))2| {z }

envelope

¢ e¡i³kz¡arctan( z

z0)´| {z }

phase factor

¢ e¡ik2¢ r2

R(z)| {z }curved wave fronts

WithR(z) := z ¢

³1 + (

z0z)2´

We now discuss the di¤erent terms:

² wave front curvature

R(z) is the radius of the wave fronts which can be seen by looking at the position ofconstant phase

k

2¢ r2

R(z)+ kz + arctan(

z

z0) = const:

arctan( zz0) depends only slowly on z and can be neglected. One gets

z(r) =const

k¡ 1

2¢ r2

R(z):

The wave front runs along a parabola z(r) » r2 with a curvature

d2

dr2z(r) =

1

R(z):

Plotting 1=R (z) yields:

16

Strongest curvature is at §z0.² Gouy phase

…nally the phasearctan(

z

z0)

is called Gouy phase. While going through the focus there is an overall phase shift of¼.

This phase is fundamentally responsible for the absorption of light by a point likescatterer as for instance an atom. (see also optical theorem)

² transverse modes

Other solutions are obtained with the Ansatz

ª = g(x

w) ¢ h( y

w)e¡i(P (z)+ k

2q(z)(x2+y2))

g and h are real functions with real variables. The Helmholtz equation is solved if g isone of the functions

Nm ¢Hm

³p2x

w

´17

withd2

dx2Hm ¡ 2x

d

dxHm + 2mHm = 0:

andNm =

1rq2¼w ¢ 2mm!

The same holds for h accordingly.

Thusgm ¢ hn = Nm ¢Hm

³p2x

w

´¢Nn ¢Hn

³p2y

w

´:

The …rst few Hermitian polynomials are

H0(x) = 1

H1(x) = 2x

H2(x) = 4x2 ¡ 2

H3(x) = 8x3 ¡ 12x

The Gouy phase now reads:

Á(m;n; z) = (m+ n+ 1) arctan(z

z0):

Theses solutions are called ”transverse electrical modes (TEMnm)”.

² intensity pro…les

18

² transformation of a Gaussian beam and ABCD-law

The q-parameters of a Gaussian beam before (q1) and after (q2) an optical system isconnected by the ABCD-law

q2 =Aq1 +B

Cq1 +D

A, B, C, D are the Elements of the above ray matrix M of geometrical optics. A proofis found in A. Siegman: ”Laser”, University Science Books, 1986, chapter 20.

1.6 Modes of a standing wave resonator

² self-consistency of q.

After one round trip q should reproduce:

q =Aq +B

Cq +D

with

M =

µA BD C

¶describing the optical system of one round trip. In general q is a complex number

q = z + iz0

The real part gives the position of the beam waist, the imaginary part is the Rayleighlength of the beam.

z0 = b=2 = z0 =1

2kw2

0

19

The condition for self-consistency yield

z =1

2

A¡D

C

z0 =

s¡µ1

2

A¡D

C

¶2

¡ B

C

z0 =

r¡z2 ¡ B

C

with A, B, C, D containing the geometrical variables such as r = 2f and d.

² Standing wave resonator with two curved mirrors

The matrix is

M =

µA BC D

¶=

µ1 d

2

0 1

¶µ1 0

¡1=f1 1

¶µ1 d0 1

¶µ1 0

¡1=f2 1

¶µ1 d

2

0 1

¶:

From this one can calculate the matrix elements which are plucked in the generalsolution above. For two equal mirrors (f1 = f2) one obtains:

z0 =1

2

pd(2r ¡ d) Rayleigh-length

z = 0 position of the waist

² stability range

with the de…nition

g := 1¡ d

r

one may write

g2 =

µ1¡ d

r

¶2

= 1 +

µd

r

¶2

¡ 2d

r

= 1¡ 1

r2¡2rd¡ d2

¢= 1¡ (2z0)

2

r2

or

g2 +

µ2z0r

¶2

= 1

g the Rayleigh-length z0 normalized to the only length scale of the system, f = r=2,form half a circle.

20

² spectrum

The resonance frequencies are given by the phase that is collected by propagating fromone mirror to the other including the Gouy phase

' = k ¢ d¡ 2(m+ n+ 1) arctan(d=2

z0)| {z }

Gouy-phase

The phase of a complete round trip, 2', must be a multiple of 2¼.

2' = k ¢ 2d¡ 4(m+ n+ 1) arctan(d=2

12

pd(2r ¡ d)

) = 2¼q

k =¼

dq +

2

d(m+ n+ 1) arctan(

d=212

pd(2r ¡ d)

)

The corresponding frequency in units of the free spectral range º0 =c2d

then reads

º(q;m; n)

º0= q +

1

¼(m+ n+ 1) ¢ 2 arctan( 1q

2rd¡ 1

):

and sincearccos(1¡ x) = 2 arctan(

1q2x¡ 1

)

one …nally gets:

º(q;m; n)

º0= (q + 1) +

1

¼(m+ n+ 1) arccos(1¡ d

r):

The natural number q counts the longitudinal modes. The numbers m and n count thetransverse modes. The additional 1 in the brackets, (q + 1), is a matter of convention:The mode with the lowest frequency is here counted as the zeroth longitudinal mode.

21

By introducing the transverse oscillation frequency

ºt :=1

¼arccos(1¡ d

r)

the spectrum has the very simple form

º (q;m; n) = (q + 1) º0 + (m+ n+ 1)ºt:

² plane cavity

d

r= 0

arccos(1) = 0

º(q;m; n)

º0= q + 1

ºt = 0

The mode consists only of the near …eld part of the Gaussian beam. The Gouy phasecan be neglected. The transverse frequency is zero. All transverse modes are degener-ate.

² concentric resonator

d

r= 2

arccos(¡1) = ¼

º(q;m; n)

º0= (q + 1) + (m+ n+ 1)

ºt = º0

Each transverse mode is degenerate with a longitudinal mode. As in the plane cavity,all modes are degenerate.

22

² confocal resonator

d

r= 1

arccos(0) =¼

2º(q;m; n)

º0= (q + 1) +

1

2(m+ n+ 1)

ºt =º02

Between two longitudinal modes there is always a transverse mode. In addition, halfof the transverse modes are degenerate with a longitudinal mode.

² Interpretation

For a …xed q, the transverse modes for the spectrum of a harmonic oscillator of fre-quency ºt. The light swings around the optical axis with a frequency that depends on

23

the curvature of the mirrors. Small curvature results in a low frequency, no curvaturein a vanishing frequency of the degenerate plane cavity. In the confocal cavity the lightbounces twice, i.e. it needs two round trips before it comes back to the initial position:the transverse frequency is twice as large as the free spectral range which gives raise toa semi degenerate spectrum. The concentric resonator both frequencies are the sameand the spectrum is fully degenerate.

The mode pro…le is given by Hermitian polynomials since this is the solution of thequantum harmonic oscillator. For the photons in the resonator Helmholtz equationplays the same role as Schrödinger’s equation for a massive particle in a harmonicpotential. The resonator is a harmonic trap for the transverse motion of the photon.

1.7 Mode Matching

² mode matching

How can the output mode of a laser (or other source) be transformed such that ismatches with the mode of the resonator (or other device)?

² Lens

Transformation of a Gaussian beam by a lens.

The matrix of the lens

M =

µA BC D

¶=

µ1 0¡ 1

f1

¶the beam parameter of the input beam

q1 = z + iz0 = z + ib12

24

and that of the output beam

q2 = z2 + iz0;2 = z2 + ib22

are connected by the ABCD-law

q2 = z2 + ib22=

Aq1 +B

Cq1 +D

=1 ¢ q1 + 0

¡ 1fq1 + 1

=z + iz0

¡ zf¡ iz0

f+ 1

=(z + iz0)(1¡ z

f+ i z0

f)

(1¡ zf)2 + ( z0

f)2

=z(1¡ z

f)¡ z20

f

(1¡ zf)2 + ( z0

f)2

+ iz0(1¡ z

f) + z0z

f

(1¡ zf)2 + ( z0

f)2:

This yields an equation for the real and the imaginary part.

z2 =z(1¡ z

f)¡ z20

f

(1¡ zf)2 + ( z0

f)2:

b22=

z0(1¡ zf) + z0z

f

(1¡ zf)2 + ( z0

f)2:

² position of the new waist

We use dimensionless quantities.

~z : =z

f;

~q : =q

f;

~z0 : =z0f

~d1 : =d1f

~d2 : =d2f

and obtain

~z2 =~z(1¡ ~z)¡ ~z20 + 1¡ 1 + ~z ¡ ~z

(1¡ ~z)2 + ez20=¡(1¡ ~z)2 ¡ ~z20 + 1¡ ~z

(1¡ ~z)2 + ~z20

= ¡1 + 1¡ ~z

(1¡ ~z)2 + ~z20:

25

² diagram

As seen from the input beam the lens is at a distance d1 from the waist.

z = d1

As seen from the output beam the lens is at a distance ¡d2 from the waist.

z2 = ¡d2This yields

1¡ ~d2 =1¡ ~d1

(1¡ ~d1)2 + ~z20:

Introducing

p1 : = 1¡ ~d1

p2 : = 1¡ ~d2

one obtains a dispersive Lorentzian curve with a width ~b1

p2 =p1

p21 + ~z20:

At the origin of the diagram p1 = 0 and p2 = 0 , i.e. d1 = f and d2 = f : a waist ofthe input beam at a distance f from a lens is mapped to a waist of the output beamat the same distance f from the lens

26

In contrary, geometric optics images an object at f to an image at in…nite distance(dashed line)!

The largest possible distance from the output waist to the lens is 1eb1 .For a collimated input beam (p1 = 1) the output is p2 = 0 i.e. d2 = f . This agreeswith geometric optics.

For small waists the transformation approaches the classical limit.

² waist of the output beam

As above one obtains~b2 =

~b1p21 + ~z21

That is an absorptive Lorentzian:

The largest b2 (collimated output)) is obtained for p1 = 0 i.e. d1 = f: The value isb2;max =

4eb1 .1.8 Ring resonators

² Bowtie resonator

Bowtie-resonators are often used for single mode lasers and frequency doublers

27

With the de…nitions

g1 := 1¡ d1r

g2 := 1¡ d2r

the solution may be written as (calculation is a very good exercise and not very di¢cult)µb1r

¶2

= ¡g21 +g1g2

µb2r

¶2

= ¡g22 +g2g1:

Again one obtains a semi circle with a radius

b1max =1

2

r2

d2 ¡ r:

28

A strongly focused beam is obtained for d2 À d1. This is useful for applications wherea high intensity is needed (laser, nonlinear optics ). Similarly focussed beams maybe obtained with standing wave cavities only if they are very small, which is oftenunpractical.

² astigmatism in a ring resonator

Hitting the mirror under an angle leads to di¤erent focal length in x and y direction.

2fx = rx =r

cos(µ):

2fy = ry = r ¢ cos(µ);with the angle of incidence µ. One may write a Gaussian beam as the product of twoGaussian functions

I = I0 exp(¡2x2=w2x) exp(¡2y2=w2

y);

which describe the pro…le for the two transverse directions. The waists wx and wy maybe calculated independently as above taking into account the two di¤erent e¤ectiveradii (see also Jenkins and White, Fundamentals of Optics, McGraw-Hill, New York,1957, page 95).

² Spectrum

By properly taking into account the Gouy phase one obtains (after some lengthy butsimple calculation)

º(q;m; n) = qº0 +

µ1

2+m

¶º tx +

µ1

2+ n

¶º ty

ºtx : = º0 ¢arctan

³d1=2z1x

´+ arctan

³d2=2z2x

´4¼

º ty : = º0 ¢arctan

³d1=2z1y

´+ arctan

³d2=2z2y

´4¼

º0 : =c

L

with the free spectral rangeº0 :=

c

L

² geometric phase for odd number of mirrors: transverse modes

the above spectrum only hold for resonators with a even number of mirrors.

Each mirror creates the mirror image of the function which describes the pro…le in theplane of incidence

Hn(x)! Hn(¡x):

29

Symmetric transverse mode functions have an even index. They are left unchanged bythe mirror.

H2n(x)! H2n(¡x) = H2n(x):

Mode functions with odd index are antisymmetric and thus change sign.

H2n+1(x)! H2n+1(¡x) = ¡H2n+1(x):

The sign change corresponds to an additional phase of ¼ collected during one roundtrip and thus the spectrum of the odd modes is shifted by half a free spectral range.

² geometric phase for odd number of mirrors: polarization

Similarly, light which is linearly polarized in the plane of the resonator su¤ers a signchange at each re‡ection. For cavities with an odd number of mirrors the spectrum forthis polarization is shifted by half a free spectral range. Such cavities are very goodpolarization …lters.

1.9 Locking a laser to a cavity

² Generating an error signal, setup

The method of Pound, Drever and Hall (1983) is based on detecting the phase shift ofthe light re‡ected from the resonator.

A radio frequency generator drives an electro-optical modulator (EOM) which modu-lates the phase of the laser light with the frequency ­ (20¡ 100MHz). The re‡ectedlight is recorded with a photodiode. The electric signal is sent to one of the inputs ofa mixer. The output of mixer is proportional to the product of the two input signals.After time averaging with a low pass …lter one obtains the desired error signal.

30

² Generating an error signal, qualitative explanation

Qualitative explanation: the EOM generates two sidebands frequency shifted from thecarrier frequency ! by §­. If the resonator is replaced by a mirror one cannot observethe modulation with the photo diode. The beat signal between the carrier and the leftside band would be 180 ± out of phase with the beat signal between the carrier and theright side band and both beat signals exactly cancel. If we replace the resonator andtune the carrier close to resonance the phase of the carrier is shifted by the resonator.The two side bands are far detuned from resonance and their phases stay unchanged.Now the two beat signals do not cancel any more and one obtains an electric signal.The phase of the signal is analyzed by …rst mix it with the driving oscillator andthan …lter out harmonics. For a quantitative analysis we …rst have to understand thespectrum of the light behind the EOM.

² spectrum of a phase modulated light …eld

The modulated …eld is given in complex writing by

E(t) = E0 ¢ ei(!t+M cos ­t)

with the modulation index M . We use the so called Jacobi–Anger identity:

ei(!t+M cos­t) = J0 (M) ei!t + 2+1Xn=1

in ¢ Jn(M) ¢ cos (n­t) ¢ ei!t:

with the Bessel- functions of …rst kind Jn(M) (http://mathworld.wolfram.com/BesselFunctionofth

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

J(M)

1 2 3 4 5M

The physical electric …eld is the real part:

Re (E (t)) = E0 ¢ÃJ0(M) ¢ cos!t+

+1Xn=1

Jn(M)Re¡ei!t+in¼

2 cos(n­t)¢!

31

With

Re¡ei!t+in ¼

2 cos(n­t)¢

= Re

µei!t+in ¼

21

2

¡ein­t + e¡in­t

¢¶=

1

2Re¡ei(!+n­)t+n¼

2 + ei(!¡n­)t+n ¼2

¢=

1

2cos³(! + n­) t+ n

¼

2

´+

1

2cos³(! ¡ n­) t+ n

¼

2

´on obtains

E0¢ÃJ0(M) ¢ cos!t+ 1

2

1Xn=1

Jn(M)³cos³(! + n­) t+ n

¼

2

´+ cos

³(! ¡ n­) t+ n

¼

2

´´!The …rst term is the carrier. It has the intensity J2

0 (M). The sum contains pairs ofside bands of equal intensity J2

n(M). Their frequency is shifted relative to the carrierby §n­. The n-th side bands oscillate in phase but shifted relative to the carrier byn ¢ 90±.

² generating an error signal, quantitative analysis

We look at the beat signal between the carrier and the left side band. The phase ofthe carrier is shifted due to the resonator by '. The beat signal is proportional to

cos(!t+ ') ¢ cos(! ¡ ­ + ¼=2)t = ¡ cos(!t+ ') ¢ sin(! ¡ ­)t

= ¡12sin (2!t+ '¡ t­)| {z }

vanishes by averaging over one optical cycle

+1

2sin ('+ t­)

' 1

2sin(­t+ '):

The beat signal with the right side band is obtained by replacing ­!¡­:¡ cos(!t+ ') ¢ sin(! +­)t

' 1

2sin(¡­t+ ')

32

Adding both signals results in

1

2(sin(¡­t+ ') + sin(­t+ ')) = cos(­t) ¢ sin'

The electric signal from the diode is mixed (i.e. multiplied with cos (­t)) and averagesby the low pass …lter:

U ' 1

T

TZ0

cos(­t) ¢ cos(­t) ¢ sin'

=1

2sin' ' 1

2'

Close to the resonance ' is small and the sin can be approximated by its argument.The signal is proportional to the phase shift of the carried which is antisymmetric withdetuning and may thus serve as an error signal.

² servo loops

A actual time depending signal s (t) is compared to a reference value S and then fedinto a an electronic servo controller. Its output acts on the system such that the actualsignal equals the reference value. The actual value of the system is measured and givesa new signal T (s (t)¡ S). In general, the transfer function T can be an operator thatacts on the function s (t) and describes the properties of the servo and the system.

In the steady state one requires that

T ¢ (s(t)¡ S) = s(t):

² proportional controller

The minimum requirement is a negative feed back. The transfer-oparator thus invertsand ampli…es the signal, i.e. multiplies it with a factor ¡G:

T ¢ s (t) = ¡G ¢ s (t)

33

The steady state condition now reads:

¡G ¢ (s(t)¡ S) = s(t)

or

s(t) =G

1 +G¢ S:

The actual value s equals the reference value S only for in…nite gain:

limG!1

G

1 +G= 1:

² Integral controller

The transfer operator is now an integrator. It transforms then signal according to:

T ¢ s (t) = ¡GtZ0

s (t0) dt0

with a gain factor G .

The steady state condition yields

s(t) = T (s(t)¡ S) = ¡G ¢tZ0

(s(t0)¡ S)dt0

or_s = ¡G ¢ (s¡ S):

If the signal is constant in time, _s = 0, it matches the reference value s = S also for…nite gain G.

² noise

If we add noise R to the signal one obtains

s(t) = ¡GtZ0

(s(t0)¡ S)dt0 +R

and_s = ¡G(s(t) ¡ S) + _R

In steady state one obtains

s =_R

G+ S:

34

We look at noise with a given frequency ! :

R(t) = R(!) ¢ ei!t_R = i! ¢R(!) ¢ ei!t;

inserting into the steady state solution

s(!) = i!

GR(!) + S;

i.e. small frequencies are more suppressed than larger frequencies. (1/f-noise).

² complex transfer-function

We now expand the signal in the base of harmonic oscillations ei!t

s (t) =

Zs (!) ei!td!:

We also require that the transfer-operator T is orthogonal in this base such that itobeys an eigenequation:

T ¢ ei!t = T (!) ei!t:

The complex function T (!) is called the ”spectrum” of the operator. One can nowdescribe the action of the operator T on the function s (t) by an integral:

T ¢ s(t) = T ¢Z

s (!) ei!td!

=

Zs (!)T ¢ ei!td!

=

Zs (!)T (!) ei!td!

A linear servo-loop is completely described by its spectrum, i.e. its transfer-function.

² question to think about

What is the spectrum of an integrator, a proportional controller, or a di¤erentiator?Plot them in a log-log-plot. Determine the slopes. In an ideal servo loop perturbationsat all frequencies are compensated with the maximum gain What would be the bestcontroller?

² stability

A servo controller cannot follow arbitrarily fast perturbations and allways reacts onlywith some time delay. Each controller thus can be characterized by its critical frequency!c where the phase delay exceeds 90± : ' (!c) = 90±. The sign of the real part of thetransfer function changes from negative to positive. A periodic perturbation with afrequency above !c cannot be compensated any more but will be ampli…ed. This canbe avoided if the gain is smaller than 1 for ! > !c: jT (!)j < 1. The ”unity gain”-frequency must be smaller than the critical frequency.

35

² Nyquist-criterium

One can show from complex integral analysis that oscillations can be avoided if theslope at unity gain is smaller than 6 dB

Oct.

The transfer function T (!) can be represented by a curve in the complex plain para-meterize by the frequency !. The servo loop is stable if the the point ¡1 lies outsidethis curve.

36

² Question to think about

A race car is kept on track by the driver. What is the controller and what is thesystem? What is the actual signal and what is the reference value? Which part of theloop determins the critical frequency? Start thinking about the problem by assuminga race track that resembles a sin curve. Think about strategies to improve the servoloop.

37