The semiclassical Rabi problem

We have a two level atom,with We look for the solution of the Schrödinger equation as:The atom has a hamiltonian: The field interaction has a corresponding hamilton operator: Our goal is to look for the solution of a two level atom in a classical electric

field, described, by, If we are talking about a two level atom,the solution without the field, we can describe by the time dependent Schrödinger equation, If we assume that

Now we can apply the time independent schrödinger equation,and we get:

Since we have a two level atom, the general solution of the differential equation is

seigenstateandlsenergyleveWW 2121 ,, 21),( 21 cctr

atomH^

iH^

iatom HHH^^^

dt

trdjtrH atom

),(),(

^

2

1),(

2

1

c

ctr

iWiH iatom ^

constant a is last the,)(c ,)(

)( i i

Wj

ii

ii cwhereectdt

tdcjtcW

i

21),(21

21jWjW

ecectr

This is the two level atom. When there is no radiation thereis no transition between states, thus the interaction betweenfield and atom is none. That is why we need to introduce operators. To describe the interactions only when transitions occurWe have a classical electromagnetic field:The interaction is only when the electron changes it’s energy level.This is when from state 1 goes to state 2 and

from state 2 goes to state 1.Two transitions can be described with two operators:

)()(__

0

__

tCosEtE

Transition from state 1 to state 2 of the electron.d is the dipole moment vector.

21

121. operator:2. operator:

The dipole momentum is equal in length: derdd

__

12

__

21

__

Basically the dipole momentum vector is a 3 dimensional vector, where each of it’s component contains the momentum in the given direction and the classicall electricfield contains the electric field strength measured in V/m, divided also in 3 components.

)1221)(()(__

0

____^^

tCosEdtEdH i

)2112(^

dd Here the two operators decidesweather is emitted or absorbed radiation. er

__

12

__

r

e

The operator-matrix of an operator can be written as:

The operator matrix of the atom: can be seen:

This is true because of the time independent

Schrödinger equation:

The operator matrix of the field hamiltonian:

This is true because the property of the interactionHamiltonian.

(There is no dipole with 1 state, we wolud need Infinite energy)

2

^

^^

1

^

W22

01221

W11

atom

atomatom

atom

H

HH

H

)(12

)(21

02211

0

____^

0

____^

^^

tCosEdH

tCosEdH

HH

i

i

ii

iWiH i

^

2212

1111

^^

^

1

^

^

HH

HHH

)2211( 21

^

WWH atom

Now we put the atom in an electric field. What happens to ?The state functions remain constant, but the time dependence will change, we can describe this time dependence, by saying that the coefficients are time dependent functions:

Now let us write the new equation:Our hamiltonian of the whole system is: This is the total energy.

We apply this on our partial solution,

but first we write the upper equation in another form,

Afterwards we wrote

our hamiltonian matrix:

Thus…

),( tr

)()( 2211 tccandtcc

21

2)(1)(),( 21

jWjW

etcetctr

iatom HHH

^^^

)()(

),(^

rdt

tdjtrH

2

1

)(

)(),(

2

1

2

1

tjW

tjW

etc

etctr

0)(

)(0,

0

0

0

__

12

__

0

__

12

__^

2

1^

tCosEd

tCosEdHand

W

WH iatom

2

__

012

__

__

012

__

1^

)(

)(

WtCosEd

tCosEdWH

What about the derivative of the equation, we derivate the matrix form and we

get:

By substituting all this information in the time dependent Schrödinger equation,

we get a space invariant set of equations, which will lead us to the Rabi

solution:

The last two terms are left aside, and by multiplying them back we get two equations:

Where:

2

1

)()(

)()(),(

22

11

222

111

tjW

tjW

tjW

tjW

etcWj

etc

etcWj

etctr

2

1

)()(

)()(

2

1

)(

)(

)(

)(22

11

2

1

222

111

2

1

2

__

012

__

__

012

__

1

tjW

tjW

tjW

tjW

tjW

tjW

etcWj

etc

etcWj

etcj

etc

etc

WtCosEd

tCosEdW

tj

tj

etctCosEdj

tc

etctCosEdj

tc

0

0

)()()(

)()()(

10

__

12

__

2

20

__

12

__

1

12

0

WW

The Rabi frequency is defined as:

We know that:

The new equations are of the form:

12

__

0

__^_

0

_

21dE

dE

Rabi

2)()(

2)()(

)()(

12

)()(

21

00

00

tjtj

Rabi

tjtj

Rabi

eetjctc

eetjctc

tjtj eetCos

)(

The Rabi frequency is an interaction frequency betweenfield and atom. It is a mean frequency.

Hz

Wsm

VE

Asmd

Rabi

)dim(

)dim(

)dim(

)dim(

2

__

0

__

12

If we neglect the fast oscillations, which are near 2ω , beacause of the near resonance effect, then we apply the rotating wave approximation (RWA). We can do this because the slow oscillations govern the time evolution (the fast oscillations change very fast the sign of the term)

The new system is :

If we are at perfect resonance (ω0=ω), then te equations are:

The solution is:

Because we defined the probability of finding the system in a state is,

that is why

)2

(22

20)

2(22

10

2

1tRabiSinctRabiCoscc

2)()(

2)()(

)(

12

)(

21

0

0

tj

Rabi

tj

Rabi

etjctc

etjctc

dV*

2 statein electron thefinding ofy probabilit thec

1 statein electron thefinding ofy probabilit thec2

2

2

1

2

1)()(

2

1)()(

12

21

Rabi

Rabi

tjctc

tjctc

conditions initial theare ,2

20

2

10 cc

002

2

0

1

0

0{.}

2

1

2

1

2

1

2

12

1

2

1

2

1

2

1

2

1

2

1

2

2

2

2,

2

2

0

2

20

),(0

2

20

)(

C

jsjsjsjs

jsjsjsjsCC

ssj

jsC

Csj

jsCCC

sj

js

Cj

jCsCtc

j

jtc

RabiRabiRabiRabi

RabiRabiRabiRabi

RabiRabi

Rabi

Rabi

Rabi

Rabi

Rabi

Rabi

Rabi

Laplace

Rabi

Rabi

2

1

2

1

22

1

4

1

2

1

2

1

2

1

4

2

2

2

2

RabiRabiRabiRabi

RabiRabiRabi

jsjsjs

jsjss

s

The solution of the Rabi problem

,2

1)( 0

__

2222

2222__

ceeee

eeeetc tjtjtj

tj

tjtj

tj

tj

RabiRabiRabiRabi

RabiRabiRabiRabi

Applying inverse Laplace transform:

Where we assumed that at t=0 we are in state 1 with probability 1. thus |c10|=1, and |c20|=0.

,)

2()

2(

)2

()2

(

)(

)()(

2010

2010

2

1__

tCosctjSinc

tjSinctCosc

tc

tctc

RabiRabi

RabiRabi

)2

()2

()(

)2

()2

()(

20*

10*

1*

20101

tjSinctCosctc

tjSinctCosctc

RabiRabi

RabiRabi

)2

(22

20)

2(22

10

2

1 tRabiSinctRabiCoscc

If we assume that in time instant 0 we are in state 1 and we are in perfect resonance, then the probabilities are:

In this case the solutions for the probabilities are:

We can see on this picture, that the probabilitiesare inverted in phase, and, they are preserved.If we are at time instant , then

the system is in state 1 with ~ 0.7 and instate 2 with ~ 0.3 probability.

The final conclusion is that the probability change between is:In this diagram at the peak values we are instate 2 with 1 probability, and at min values we are in state 1 with 1 probability.

0c and 1c2

2

2

1

22

2

2

2

22

1

2

1

)2

((t)c0(0)c

)2

(C(t)c1(0)c

tSin

tos

Rabi

Rabi

21 Rabi

2

2

2

1

2

2

2

1 c -1c:property theof because ,c and c

1 2 3 4 5 6

0.2

0.4

0.6

0.8

1

Density Matrices

22

__________________________

)(2

)()(

)(2

)()(

122111

1

**

*

21

21

1

tj

Rabi

tj

Rabi

tj

Rabi

tj

Rabi

ej

ej

tce

tjctc

tce

tjctc

)(2

______________________

)(2

)()(

)(2

)()(

112221

2

*2

*1

*

112

tjRabi

tj

Rabi

tj

Rabi

ej

tce

tjctc

tce

tjctc

These are the elementsof the thensity matrix:

*

22

*

12

*

21

*

11

2221

1211

cccc

cccc

22112211 1

*

2112*

2112

mesaure detuning ,, 0 theiswhere

We can fulfill these constraints even if we introduce other variables , to solve

the equation with Laplace Transformation: ~*211212

~

2222

~

1111

~

,

tje

Derivating these:

~*

21121212

~

2222

~

1111

~

,

tjtj eje

Substituting the derivatives intothe original equations:

)(2

22

11

~

22

~

21

~

21

~

12

~

21

~

11

~

Rabi

RabiRabi

jj

jj

We can write now the system of equations:

22

~21

~12

~11

~

22

~

21

~

12

~

11

~

022

0

20

2

20

2

022

0

RabiRabi

RabiRabi

RabiRabi

RabiRabi

jj

jjj

jjj

jj

This is a standard equation of a system with 4 inputs, 4 outputs, and we assume thatthe system’s initial state is state 1.

Initial condition

In[2]:= PowerExpandFullSimplifyInverseLaplaceTransformInverse

s 2 2 0 2 s 0 2 2 0 s 2

0 2 2 s

.1000

, s, t

Out[2]=2 2 2 2 Cosht2 222 2 , Sinh1

2t2 22 2Cosh1

2t2 2

2 2 Sinh1

2t2 22 232 ,

Sinh12t2 22 2Cosh1

2t2 2

2 2 Sinh1

2t2 22 232 , 2Sinh1

2t2 22

2 2

12 232 Sinh12 t2 22 2Cosh1

2t2 2

2 2 Sinh1

2t2 2

))()((2

1

2

)()())(1(

]2

[]2

[]2

[)](2

[

22222222

322

22222222

322

2222

22

322

2222

22

RabiRabiRabiRabi

Rabi

Rabi

RabiRabiRabiRabiRabiRabi

Rabi

RabiRabi

RabiRabi

Rabi

RabiRabi

RabiRabi

tSinjtCos

tSinjtCos

jtSinh

jtSinh

jtCosh

jtSinh

Solutions:

)2

(

)2

(1

22

2

22

2

22

22

2

22

2

11

Rabi

Rabi

Rabi

Rabi

Rabi

Rabi

tSin

tSin

*

21

222222

2212 ))()((2

1

RabiRabiRabi

Rabi

Rabitj tSinjtCose

1 2 3 4 5 6

-0.4

-0.2

0.2

0.4

0.6

0.8

1

In case of perfect resonance again we know much more:

)2

(

)2

(1

))(2

1

2

22

2

11

*

2112

Rabi

Rabi

Rabi

tSin

tSin

tSinj

1)( 2 Wp

1)( 1 Wp

)1(

)2(

1. We start by the given initial condition,that we are in state 1. then the weight for transitionis 0, because we know for shure that we are in state 1. (1)2. When the electron is halfway on it’s road between states(.5 probability), then the absor-bed energy weight is maximum.(2)3. If we are in state 2 with 1 probability, then it’s the same story as in 2. (3)4. If we are again halfway between states but we go from 2 to 1, then the emmitted energy amount reaches it’s maximum.5. The period closes

!! Remark: This is the perfect resonance case.

22

12

)3(

)4(

)5(

2 and 1 statebetween ns transitioof weight theρ12

Solutions:

)2

/1(

/1

1

)2

/1(

/1

11

22

2

2222

22

2

2211

RabiRabi

Rabi

RabiRabi

Rabi

tSin

tSin

1Rabi

1 2 3 4 5 6

0.2

0.4

0.6

0.8

1

4

2

1

5.0

0

Conclusion:

The larger the detuning is the the larger the probability of remaining in state 1.

We took the Rabi frequency as 1.

2

)(1)

2(

2

)(1)

2(1

))(2

1

2

22

2

11

*

2112

RabiRabi

RabiRabi

Rabi

tCostSin

tCostSin

tSinj

Rabit

2Rabit

After this new form,we can see, that ifwe irradiate, by a pulse of lengththen we put the electron in state 2. if we Irradiate by then we get back in state 1.

2212

~

21

~

22

~

11

~

22

~

21

~

21

~

11

~

22

~

12

~

12

~

2212

~

21

~

11

~

22

)(2

)2

(

)(2

)2

(

22

Ajj

jjA

jjA

Ajj

RabiRabi

Rabi

Rabi

RabiRabi

Relaxation: Dapming is because of thespontaneous emission of Photons. Thus the probability of being in state 1 increases after a given amount of time, and that of being in state 2 decreases. Furthermore the electron’s probability reaches a steady state. (It’s value is constant after a time.)

We introduce two coefficients.A passes the probability fromstate 2 to 1. (spont. emmission) is the coherent interaction – ץ

The reasons of decay (A): Thermal reservoir, electronic discharges (noise), andeverything that affects the whole system – this is the spontaneous emmission coefficient.The reasons of decay (ץ): This represens the the interactions between the coherent states (|1><2|, and |2><1|) due to the interaction between electrons.

~*211212

~

2222

~

1111

~

,

tje

We must not forget, that:

… and do our calculations accordingly.

In[40]:= A 0.1; 10000; 1; 100; o 2000;

G FullSimplifyInverseLaplaceTransformInverse

s 2 2 A 2 s A

2 0 2

2 0 s A2 2

0 2 2 s A

.1000

, s, t;

F TransposeG.1, 0, 0, 0;PlotAbsEvaluateF,t, 0, o;F TransposeG.0, 0, 0, 1;PlotAbsEvaluateF,t, 0, o;

100 200 300 400 500 600

0.999501

0.999501

0.999502

0.999503

0.999503

100 200 300 400 500 600 700

0.0004988

0.0004992

0.0004994

<-the probability of finding the electron in state 1.

<-the probability of finding the electron in state 2.

In this figure we cansee, that after the steady state of theρ11 and ρ22 probabilitiesThe coherence of the is constant too.

These facts have theoretical background too.

Let’s assume, that:

rapidly. decreasing are , ., of change

thegoverns- A smallvery 22

1.

21122211

22

jA

Rabi

ts.coefficienin small very are (4) and(1), eqin termsthe

ofboth because constant,almost are ,2

2. 2211 jA

A

2212

~

21

~

22

~

11

~

22

~

21

~

21

~

11

~

22

~

12

~

12

~

2212

~

21

~

11

~

22)4(

)(2

)2

()3(

)(2

)2

()2(

22)1(

Ajj

jjA

jjA

Ajj

RabiRabi

Rabi

Rabi

RabiRabi

: Thus 0. are , state

steady after andrapidly decreasing are

and slowly. changing arethey

becauseconstant almost are ,,

1221

2112

2211

)()

21

(2

)()

21

(2

112212

112221

jAj

jAj

Rabi

Rabi

The quantized electromagnetic field

We know from quantummechanics the Hamilton equations:

We also know the global definition for impulse:

i

i

ii

ii

q

Lp

p

Hq

q

Hp

,

potkindensityL The Lagrange density function can be definedas the difference between the kinetic and pot-ential energy density functions.In case of electromagnetic fields, we define the Hamilton density function as:

From classical electromagnetic

field theory (where A is the vector potential):

)(2

1 22 HEH density

gradt

AE

)(Arot

H

AwhereEgradt

Ap

Arotgrad

t

AL

defdef

i

density

i

22

q , E )(

]))(

()([2

1

density impulse theis where,2

)(

2

22

Arot

H density

If we want to write the electromagnetic field in a unitary cube, we should introducetwo orthogonal vector functions and the wave vector, which make an orhogonalsystem.

We introduce two orthogonal unit vectors: iji kvectorwavetheandee ,, 21

The sinusoidal and the orthonormal property:

)1(,1

33

*22

*112/3

dVuudVuueeL

uL

ii

L

iirjk

ilili

)()(

)()(

rutqbA

rutpa

ii

i

ii

i

Ajkrot(A)

)(

i212/3

12/311

iirjk

ii

i

rjk

i

ujkeekL

j

eL

euurot

i

i

1

,a ;)(2

)(2

22222

22

3

btq

bktp

adVHH

ii

ii

L

density

iiii

ii

ii tqtptq

ktpH )()(

2

1)()(

2

1 22222

2

This is becausep,and q functionswere arbitrary chosenso the coeff. can be arbitrary chosen too.

We introduce the space and impulse operaors: klj

rj

l

l

p

p

, :equationr Schrodinge

esatisfy thmust operators twoThese . and

k

k

q

q

ii

iii

i jj

2

1

2

1

ii

iiii

aaH

qpqpH

From here the emission and absorption operators are:

i

i

i

i jj

2,

2ii

iii

i

pqa

pqa

i

rjkrjkii

i

rjkrjkii

rjkiii

rjkii

ii

iiii

ii

eeV

jeEeeV

je

eeV

jjeeV

jj

)(2

)(2

1)2(

1)2(

2

1

2

*

iiii

iiii

aaaa

qaqa

Gaussian pulses

In practice one frequency of an electric field cannot be radiated, it is always a

gaussian pulse, which in case it is narrow enough it approximates very good

the discrete frequency. Instead of scatterings we want to use the full width half

maximum ( ) length, because we can measure it. It is at the half of a

gaussian function. (which is the monochromatic light spectrum)

We want the Gaussian function to be:

FWHMt

5.0

FWHMt

2ln8

t

2ln8t ;2ln22t

2t because ,2ln2

2FWHM2

22FWHMFWHM

FWHM

HWHMtx

22

2

2

1 x

e

2ln22ln2

1

Our gaussian function is applied in the power spec-trum description of light:

2

max

2ln420

FWHMt

)t(t

ePP(t)

FWHM

p

t

EP

22ln22

maxThe peak value, atTime instant t0 :

Where Ep is the pulse energyWe get Ep by integrating P(t).

This pulse width is Gaussian, because, when the pulse begins, it’s power is minimal, and increases exponentially. At peak value the device that makes the pulse is shut down, thus the pulse’s power decayes exponentially in time.

In general if we don’t know the shape of the pulse we can transform it in a Gaussian, by calculating the sqare scatterig and transforming it into the fwhm and expected value in time:

length: and by calculating the expected

value. Here the beam spectrum has it’s peak value at:

dttP

dttPttt pulse

)(

)()(2ln8

2

02

dttP

dtttPt

)(

)(0

These integrals aredone on the wholepulse length.

The caracteristics of Gaussian beams:

A laser beam, can be caracterized by it’s wave vector, which tells us in

which direction the wave goes to at the given coordinate. Because it’s

intensity usually in one direction is higher, that is why we assume that it’s

intensity in the x, y direction is decaying exponentially.

We „cut” the beam in mind and realise that in the core the intensity is higher:

So let’s assume at z=0 and t = 0 the

electric field is in the form of:

Where w0 is the beam radius. (the time is separable). The intensity is

The beam’s electric field goes in three directions, so

We can expand it. This way the wavenumber is

preserved in the length of wave vectors.

Where:

20

22

0)0,,( w

yx

eEyxE

222202

2

kqpkc

k

q

p

k0

kzqypxrk ee 0This way:

2E

dpdqeyxEqpE jqyjpxA

)0,,(4

1),( 2

The electric field is built up from different amplitudes corresponding to different

wave vectors, by the Fourier Integral:

The amplitude depends only from p and q, because going in the z direction the amplitude is approximatly always the same. (the waves going in x direction have high p, because the weight of the x direction is p)

By inverse Fourier integral of E(x,y,0), we get that at a given wavevector of x

and y what is the electirc field amplitude.

If we assume that the wavenumbers in the z direction are the highest then we

can say that in the equation,

this way we can get k.

)(4

200

20

2220

20

2

20

2

44),(

qpw

w

yjqy

w

xjpx

A ewE

dyedxeE

qpE

2200 ))(( qpkkkk 00 2)( kkk

dpdqeqpEzyxE jkzjqyjpxA

),(),,(

0

22

0 2k

qpkk

Now we deal with the Fourier integral, by substituting k. So the the derived

equation.

dqeeedpeeeewE

dpdqeeewE

zyxE

zk

q

jqyqwz

kp

jpypw

zkj

zkqp

zkjjqyjpxqp

w

0

2

220

0

2

220

0

0

22

02220

2424200

2)(4

200

4

4),,(

)(2

1

/

200

0 020

20

2

21

),,,( tzkjkw

jzwr

ee

wk

jzE

tzyxE

In[23]:=w2

4 Integrate14 p2 w2 p

2z2 k

px,p, , , AssumptionsRew2 2Imz

k

Integrate14 q2 w2q2 z2 k

q y,q, , , Assumptions Rew2 2Imz

k

Out[23]= k x2

k w22z k y2

k w22z w2

w2 2zk

<- The result

Finallywe get:

20

200

0 12

12

nw

jz

wk

jzk

First we assumed, that the beam radius was w0, but we know that the elec-tric field is Re{E(x,y,z,t)}

This way the beam radius as a function of z is in the exponent :

2

20

10)(

wn

zwzw

2

0wnzR This is the Rayleigh length

which is also called the focus depth of a beam.

)(zw

z0.5 1 1.5 2 2.5 3

0.5

1

1.5

2

2.5

3

•The diameter is increasing linearly in functionof z, at large distances.

length Rayleigh the is z and

widthbeam minimum w where,)(

)tan( ,

R

00lim theis

z

w

z

zwsmallFor

Rz

0w

We usually say that the radius of a gaussian beam is where the intensity

reaches it’s 1/e^2 value.

We know that:

2

20

20

22

1

2

2

20

00

1

)2(),,(

n

zw

yx

e

n

z

zkCosEyxzE Is the electric field

of the beam

2

20

20

1

4'

2

20

00

1

)2(),',(

n

zw

r

e

n

z

zkCosErzE

It was easy to rewrite it in polar coordinates:

thuszktCoswE

zkCose

wEdrdrzErdrdrzErzw

),2()8646.0(2

)2(1

12

'),',(''),',('

020

20

0220

20

)(

2

0

2

0

2

0

2

86.5% of power is inside the beam radius.