Transcript The semiclassical Rabi problem

The semiclassical Rabi
problem
We have a two level atom,with W1 ,W2 energylevels and 1 , 2 eigenstates
We look for the solution of the Schrödinger
equation as:  (r , t )  c1 1  c2 2
^
The atom has a hamiltonian: H atom
^
The field interaction has a corresponding hamilton operator: H i
Our goal is to look for the solution of a two level atom in a classical electric
^
^
^
field, described, by, H  H atom  H i
If we are talking about a two level atom,the solution without the field, we can
describe by the time dependent Schrödinger equation,
^
d (r , t )
c1   1 
H atom  (r , t )  j
If we assume that  ( r , t )     
dt
c2   2 
^
Now we can apply the time independent schrödinger equation, H atom i  Wi i
W
and we get: W c (t )  j dci (t ) , c (t )  c e j  , where the last c is a constant
i i
i
i
i
dt
Since we have a two level atom, the general solution of the differential equation
is
i
 (r, t )  c1e

jW1

1  c2e

jW2

2
e
__
r
1
2
Transition from state 1 to state 2 of the
electron.d is the dipole moment vector.
This is the two level atom. When there is no radiation there
is no transition between states, thus the interaction between
field and atom is none. That is why we need to introduce
operators. To describe the interactions only when transitions
__
__
occur
We have a classical electromagnetic field: E (t )  E0 Cos (t )
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:
1. operator: 2 1
2. operator: 1 2
The dipole momentum is equal in
__
__
__
length:
d  d  red
21
12
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 electric
field contains the electric field strength measured in V/m, divided also in 3 components.
^
d  d( 2 1  1 2 )
^
^ __
__ __
Here the __two operators decides
weather r e is emitted or absorbed radiation.
H i  d E (t )  d E0 Cos(t )( 1 2  2 1 )
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:
^
( H atom  W1 1 1  W2 2 2 )
 H^ i  W i 


i


^

11 H 1 

^
2H 2 

^
1 H atom 1  W1
^
^
1 H atom 2  2 H atom 1  0
^
2 H atom 2  W2
The operator matrix of the field hamiltonian:
This is true because the property of the interaction
Hamiltonian.
(There is no dipole with 1 state, we wolud
need Infinite energy)
 ^
^
 1H 1
H  ^
 2H1

^
^
1 H i 1  2 Hi 2  0
^
__ __
^
__ __
1 H i 2  d E 0 Cos(t )
2 H i 1  d E 0 Cos(t )
Now we put the atom in an electric field. What happens to  (r , t ) ?
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: c1  c1 (t ) and c2  c2 (t )

jW1


jW2

 c2 (t ) 2 e
Now let us write the new equation:  (r, t )  c1 (t ) 1 e
^
^
^
Our hamiltonian of the whole system is: H  H atom  H i This is the total energy.
^
We apply this on our partial solution,
H  (r , t )  j
d (t )
 (r )
dt
but first we write the upper equation in another form,
jW
 1t 

c1 (t )e    1 

 (r , t ) 
 
jW

 2t  2

c2 (t )e
  
Afterwards we wrote
__
__

0
d 12 E 0 Cos (t )
our hamiltonian matrix: H atom  W1 0 , and H i  

__
__
0 W 
d 12 E 0 Cos (t )


2
0
__
__


^
W
d
E
Cos
(

t
)
12
1
0


Thus…
^
H 
__
 __
 d 12 E0 Cos(t )
^
W2


What about the derivative of the equation, we derivate the matrix form and we
jW
jW

t

t 
j



 W1c1 (t )e

 c1 (t )e
 1 

 (r , t )   
jW
jW  


t

t
j
2
c 2 (t )e   W2c2 (t )e    



1
get:
1
2
2
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:
jW
jW

 1t
 1t 

j
__
__





c

W
c
(
t
)
e
1 (t )e

 1 
1
1


d 12 E0 Cos(t )  c1 (t )e
 1  j 

 __ __ W1
 

jW2
jW2
jW2  




t

t

t
2
2
 d E Cos(t )

j
c 2 (t )e   W c (t )e    
W2
 12 0
 c2 (t )e    
2 2



jW
 1t

The last two terms are left aside, and by multiplying them back we get two
equations:
W2  W1
j __ __

 j0t
Where:


c1 (t )   d 12 E 0 Cos(t )c (t )e
0
2



j __ __
 c 2 (t )   d 12 E 0 Cos(t )c1 (t )e j0t



_
The Rabi frequency is defined as:
We know that:
Rabi 
e jt  e  jt
Cos(t ) 

The new equations are of the form:
^
_
dim( d12 )  Asm 

__
V 
dim( E0 ) 
  dim(  Rabi )  Hz
m 
dim( )  Ws 2 

__
E0
E 0 d 12
1d 2 



e  j (  0  ) t
c1 (t )   jc2 (t ) Rabi
 
j (  0  ) t
e
 c 2 (t )   jc1 (t ) Rabi

The Rabi frequency is an interaction frequency between
field and atom. It is a mean frequency.
__
__
 e  j (0  )t
2
 e j (0  )t
2
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)

e  j (  0  ) t
c1 (t )   jc2 (t ) Rabi
2
The new system is :  
j (  0  ) t
 c 2 (t )   jc1 (t ) Rabi e

2

c1 (t )   jc2 (t ) Rabi
If we are at perfect resonance (ω0=ω), then te equations are:  
c 2 (t )   jc1 (t ) Rabi



2 2
2
Rabi
c c
Cos (
t)  c
Sin ( Rabi t )
1
10
20
2
2
2
The solution is:
2
2
2
c10 , c20 are the initial conditions
Because we defined the probability of finding the system in a state is,  *dV
c1  theprobability of finding theelectronin state1
2
that is why
c 2  theprobability of finding theelectronin state2
2
1
2
1
2
s
The solution of the Rabi
problem
s2 
1
s2 

0


c(t )  

 j Rabi
2


s



 j Rabi
2


 s
C

 j Rabi
2

j
 Rabi
4
 Rabi 

0

{.}
2  c(t ), Laplace
  sC  C0  



 j Rabi
0
2


 

 j Rabi 
s
2 C  C , C  

 
0

 j Rabi
s
2


j
 Rabi
4
 Rabi 
1
2 

s  s 2   Rabi

2
2
2
2


1
2
1
1



s  j Rabi s  j Rabi
2
2
1

2 j Rabi
2
j


1
1



 Rabi

s  j Rabi
s j
2
2







 Rabi 
2 C 


0

1
 
j Rabi 
2
 C0
s 

1



 Rabi
s
1 s  j 2
C0  C  
1
2

 Rabi

s
 s  j 2
1

j Rabi
2
1

j Rabi
2
1

 Rabi
s j
s
2
1

 Rabi
s j
s
2
1

j Rabi
2
1

j Rabi
2



 C0



Applying inverse Laplace transform:

t
 j Rabi
 j  R2abi t
2
1 e
e
c (t ) 


t
 j Rabi t
2  j R2abi
2
e
e
__
e
e
j
 Rabi
t
2
j
 Rabi
t
2
 __
e
 c 0,
 Rabi t
j

2
e

j
 Rabi t
2
 Rabi
 Rabi 

c
Cos
(
t
)

c
jSin
(
t )
__
20
 c1 (t )   10
2
2
c (t )  

,



c2 (t ) c10 jSin(  Rabi t )  c20Cos(  Rabi t )
2
2


 Rabi


t )  c20 jSin( Rabi t ) 

2
2



 c*1 (t )  c*10Cos( Rabi t )  c* 20 jSin( Rabi t )
2
2

 c1 (t )  c10Cos(

2 2 
2
c c
Cos ( Rabi t )  c
Sin ( Rabi t )
1
10
20
2
2
2
2
Where we assumed that at t=0 we are in state 1 with probability 1. thus
|c10|=1, and |c20|=0.
If we assume that in time instant 0 we are in state 1 and we are in perfect
resonance, then the probabilities are: c1 2  1 and c 2 2  0
2
t 
2
2

c1 (0)  1  c1 (t)   Cos ( Rabi ) 
2 

In this case the solutions for the probabilities are:
2
t 
2
2

c2 (0)  0  c2 (t)   Sin ( Rabi ) 
2 
We can see on this picture, that the probabilities

are inverted in phase, and, they are preserved.
If we are at time instant Rabi , then
1
2
the system is in state 1 with ~ 0.7 and in
state 2 with ~ 0.3 probability.
The final conclusion is that the probability change
1
2
2
2
2
between c1 and c 2 , because of the property : c1  1 - c 2 is:
0.8
In this diagram at the peak values we are in
0.6
state 2 with 1 probability, and at min values we
0.4
are in state 1 with 1 probability.
0.2
1
2
3
4
5
6
Density Matrices
e  jt *
c1 (t )   jc 2 (t ) Rabi
c (t ) 
2
*
e jt
*
c (t )  jc (t ) Rabi
c1 (t )
2
__________ __________ ______


c 2 (t )   jc1 (t ) Rabi
1
1
e  jt
e jt
 11   j 21 Rabi
 j12  Rabi
2
2


11  22  1   11    22
These are the elements
of the thensity matrix:
jt
e
c (t )  jc2* (t ) Rabi
c 2 (t )
2
__________
__________
__


 21  j Rabi e jt (  22  11 )
2

*
1
2

e jt *
c1 (t ) 
2


*
21
12     12  
*
21
 11

 21
12   c1c1* c1c2* 
 *

22  c2c1 c2c2* 
where,   0  , is the detuningmesaure
We can fulfill these constraints even if we introduce other variables , to solve
~
~
 11  11,  22   22
the equation with Laplace Transformation:

~


~
~
*
21
~
 12  12 e jt  

 11   11 ,  22   22
Derivating these:

~


~
*
21
 12   12 e jt  j12e jt  
Substituting the
derivatives into
the original equations:
 ~
~
~
  11   j  Rabi  21  j  Rabi  12

2
2

~
~
~
 Rabi ~
(  22   11 )
  21   j  21  j
2

We can write now
the system of equations:
 ~  
0
  11  
   
 ~   j Rabi
2
  12   
 ~   j  Rabi
  21  
2
  
0
~  

 22  
j
 Rabi
2
j
j
 Rabi
2
j
0
0
 j
 Rabi
2
j
 Rabi
2

 ~ 

 Rabi   ~ 11 
 
j
2    12 
 Rabi   ~ 
 21
j
2  ~ 
   22 
0


0
This is a standard equation of a system with 4 inputs, 4 outputs, and we assume that
the system’s initial state is state 1.
Initial condition
1
2
2 3 2
Sinh
1
2
t
2
2
2
2
Cosh
1
2
t
2
2
2
2
Sinh
1
2
t
2
2
2
2
2
2
jt 2   Rabi
jt 2   Rabi
jt 2   Rabi
jt 2   Rabi
2
2
2
2
 Rabi Sinh[
](    Rabi )Cosh[
]  Rabi Sinh[
]    Rabi Sinh[
]
2
2
2
2



3
3
2
2
2
2
   Rabi
   Rabi

2
2
2
2
 Rabi  2   Rabi
(1  Cos(t 2   Rabi
))  j Rabi ( 2   Rabi
)Sin(t 2   Rabi
)
2
2 2   Rabi

3
1  Rabi
2
2
2
(   Cos(t 2   Rabi
)  j 2   Rabi
Sin(t 2   Rabi
))
2
2
2    Rabi
Solutions:
12  e  jt
1  Rabi
(  Cos (t 2   2Rabi )  j 2   2Rabi Sin (t 2   2Rabi ))   21*
2
2
2    Rabi
2
2
t



 2Rabi
Rabi
2
11  1  2
Sin
(
)
2
 Rabi  
2
2
2
t



 2Rabi
Rabi
2
 22  2
Sin
(
)
2
 Rabi  
2
In case of perfect resonance again we know much more:
1
12   j Sin (t Rabi ))   21*
2
t
11  1  Sin 2 ( Rabi )
2
t
 22  Sin 2 ( Rabi )
2
 22
p(W2 )  1
1
0.8
0.6
(3)
0.4
0.2
( 2)
1
(1)
ρ12  the weight of transitions betweenstate1 and 2
( 4)
2
3
4
5
6
p(W1 )  1
-0.2
-0.4
1. We start by the given initial condition,

that we are in state 1. then the weight for transition
(5)
is 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 absorbed 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.
12
Solutions:
t Rabi 1  2 /  2Rabi
1
2
11  1 
Sin (
)
2
2
1   /  Rabi
2
Rabi  1
t Rabi 1  2 /  2Rabi
1
2
 22 
Sin (
)
2
2
1   /  Rabi
2
Conclusion:
1
0
The larger the
detuning is the the
larger the probability of
remaining in state 1.
0.8
  0.5
0.6
 1
0.4
We took the
Rabi frequency as 1.
2
4
0.2
1
2
3
4
5
6
1
2
12   j Sin (t Rabi ))   21*
t Rabi 1  Cos (t Rabi )
)
2
2
t
1  Cos (t Rabi )
 22  Sin 2 ( Rabi ) 
2
2
11  1  Sin 2 (
After this new form,we can see, that if
we irradiate, by a pulse of length tRabi  
then we put the electron in state 2. if we
Irradiate by tRabi  2 then we get
back in state 1.
Relaxation:
 ~
~
~


Rabi
Rabi
  11   j
 21  j
 12  A 22
2
2


~
~
 Rabi ~
A
~
  12  (  2  j)  12  j 2 (  22   11 )

~
~
~
 Rabi ~
A
(  22   11 )
  21  (   j)  21  j
2
2

 ~
~
~


Rabi
Rabi
  22  j
 21  j
 12  A 22

2
2
Dapming is because of the
spontaneous 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 from
state 2 to 1. (spont. emmission)
‫ – ץ‬is the coherent interaction
The reasons of decay (A): Thermal reservoir, electronic discharges (noise), and
everything 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.
~
~
We must not forget, that:
 11  11,  22   22
~
~
*
21
 12  12 e jt  
… and do our calculations accordingly.
<-the probability of finding the electron in state 1.
<-the probability of finding the electron in state 2.
In this figure we can
see, that after the
steady state of the
ρ11 and ρ22 probabilities
The coherence of the
is constant too.
These facts have theoretical background too.

~
 Rabi ~
 Rabi ~
(1)  11   j
 21  j
 12  A 22
2
2

~
(2)  12

~
(3)  21

~
(4)  22
~
~
 Rabi ~
A
 (   j)  12  j
(  22   11 )
2
2
~
~
 Rabi ~
A
 (   j)  21  j
(  22   11 )
2
2
11 ,  22 , are almostconstantbecause
 Rabi ~
 Rabi ~
 j
 21  j
 12  A 22
2
2
Let’s assume, that:
A

 j   verysmall A 22 - governsthe
2
2
changeof 11 ,  22 . 12 ,  21 are decreasing rapidly.
1.  Rabi   
A
 j  11 ,  22 are almostconstant,because both of
2
the termsin eq (1),and (4) are verysmall in coefficients.
2. A   
theyare changingslowly. 12 and  21
are decreasingrapidly and aftersteady


state 21 ,  12 are 0. T hus:
 Rabi
(  22  11 )
1
2(  A  j)
2
 Rabi
12  j
(  22  11 )
1
2(  A  j)
2
 21   j
The quantized
electromagnetic field
H 
H
, qi 
We know from quantummechanics the Hamilton equations: p i 
qi
pi

We also know the global definition for impulse:
The Lagrange density function can be defined
as the difference between the kinetic and potential 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):
pi 
L

 qi
Ldensity  kin   pot
1
H density  (E 2  H 2 )
2
A
E
 grad 
t
rot( A)
H

1 A
rot( A) 2
2
Ldensity  [ (  grad )   (
) ]
2
t

def
def
A

pi     (  grad )  E  E   , where q i  A
t

H density 

2
2

rot( A)
2
2
, where  is theimpulse density
If we want to write the electromagnetic field in a unitary cube, we should introduce
two orthogonal vector functions and the wave vector, which make an orhogonal
system.
We introduce two orthogonal unit vectors:
ei1, e j 2 , and the wave vector ki
1
uil  3 / 2 eil e jki r , (  ui1ui*1dV   ui 2ui*2 dV  1)
L
L3
L3
e jki r
  a pi (t )ui (r ) rot(ui1 )    u1   3 / 2  ei1 
L
i
A  b qi (t )ui (r )  j ki  ei1e jki r  jki ui 2  rot(A) jk i A
i
L3 / 2
 a2 2
ki2b 2 2  2
1
2
H   H density dV   
pi (t ) 
qi (t ) ; a   , b 
2

i  2

L3
The sinusoidal and the
orthonormal property:
ki
1  2
This is because

H
pi (t ) 
p,and q functions

2 i 

were arbitrary chosen
so the coeff. can be arbitrary chosen too.

We introduce the
space and impulse
operaors:
2
 1
q (t )    pi2 (t )  i2 qi2 (t )
 2 i

2
i

pl   j and qk  r. T hesetwo operatorsmust satisfy the
Schrodinger equation: qk , pl   j kl
1
H   pi  ji qi pi  ji qi   i
2 i
From here the emission and absorption
operators are:
1

H   i  ai ai  
2

i
ai 
iqi  jpi
 q  jpi
, ai  i i
2i
2i
*   1
1
1


  ( jai 2i  ji qi )
ei e  jki r  ( jai 2i  ji qi )
ei e jki r
2
2
V
V
i
   jei
i
2i   jki r
i   jki r
jki r
(a i e
 aie )  E    jei
(a i e
 aie jki r )
V
2V
i
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 ( t FWHM ) length, because we can measure it. It is at the half of a
gaussian function. (which is the monochromatic light spectrum)
x2

2
1
 1
We want the Gaussian function to be:  e 2 
2
 x     2 ln 2, because tFWHM  2tHWHM
 tFWHM  2 2 ln 2;
t
 2  FWHM
8 ln 2
tFWHM   2 8 ln 2
0 .5
2
2
(t  t )2 4 ln 2
Our gaussian function is
0

applied in the power spec2
t FWHM
P(t)  Pmax e
trum description of light:
The peak value, at
Time instant t0 :
Pmax
 2 ln 2
t FWHM
2 ln 2
Where Ep is the pulse energy
2 2 ln 2 E p

2 tFWHM We 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:
These integrals are
done on the whole
pulse length.
2
(
t

t
)
P(t )dt
0

length: t pulse  8 ln 2
and by calculating the expected
 P(t )dt
value.
tP (t )dt

Here the beam spectrum has it’s peak value at:
t0 
 P (t )dt
2
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:
x2  y 2
So let’s assume at z=0 and t = 0 the

w02
E( x, y,0)  E0e
electric field is in the form of:
Where w0 is the beam radius. (the time is separable). The intensity is E
2
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.
 p
Where: k0  q 
 
k 
c2
k0r
This way: e
2
 k0  p 2  q 2  k 2
2
 epx qy kz
The electric field is built up from different amplitudes corresponding to different
wave vectors, by the Fourier Integral:
E ( x, y , z ) 

jpx  jqy  jkz
E
(
p
,
q
)
e
dpdq
 A

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. E A ( p, q ) 

 jpx 
x2
w 02

 jqy 
y2
w 02
E0w 02
1
 

 jpx  jqy
E
(
x
,
y
,
0
)
e
dpdq

4 2    
w 02 2 2
 ( p q )
e 4
E0
e
dx  e
dy 
2 
4
4  

If we assume that the wavenumbers in the z direction are the highest then we
 E A ( p, q ) 
can say that in the ( k0  k )( k0  k )  p2  q 2equation, ( k0  k )  2k0
2
2
p

q
this way we can get k. k  k 
0
2 k0
Now we deal with the Fourier integral, by substituting k. So the the derived
equation. E ( x, y , z ) 

E0w 02
4
e
E0w 02 
p2  q 2
w 02 2 2
j k0 z 
z
 ( p q )
2 k0
jpx  jqy
4
e
e
e
dpdq

4  
j k0 z


p2
q2
w 02 2
w 02 2


z

z

p
 q
2
k
2
k
jpy
jqy
0
0
e 4 e e
dp e 4 e e
dq



<- The result
Finally
we get:

E ( x, y , z, t ) 
E0
e
2 jz
1
k0 w 02
r 2 / w 02

2 jz
1 2
kw 0
e
j ( k0 z t )
2
2 jz
jz
k0 
 1

1


k0 w 02
nw 02
First we assumed, that the beam rad
ius was w0, but we know that the electric field is Re{E(x,y,z,t)}
This way the beam radius as a function
of z is in the exponent :
zR 
n
w 02

This is the Rayleigh length
which is also called the
focus depth of a beam.
w (z)  w0
2
3
w(z)
•The diameter is increasing linearly in function
of z, at large distances.
 z 

1  
2 
n

w
0 

2.5
2
1.5
w0
1
0.5

0.5
1
1.5
2
2.5
3
z
w (z) w0

, where w 0 is the minimumbeam width
z
z
z  
R
For small ,   tan( )  lim
and zR is the Rayleigh length
We usually say that the radius of a gaussian beam is where the intensity
reaches it’s 1/e^2 value.
x2  y 2
2

We know that:
E ( z, x, y )  E0
Cos(2  k0 z )
It was easy to rewrite it in polar
coordinates:
 z
1   2
 0 n



2
e
E ( z, r ' , )  E0
w 02
 z
1  2
 0 n




2
Is the electric field
of the beam

Cos(2  k0 z )
2
e
r'
w 02
4
 z
1  2
 0 n




2
 z 
1   2 
 0 n 
 2
 2
1
2
2
2 2 
r
'
E
(
z
,
r
'
,

)
d

dr
'

r
'
E
(
z
,
r
'
,

)
d

dr
'

E
w
1

Cos(2  k0 z ) 

0
0

 
2
2 e 
0 0
w (z) 0
2
 E0 w 02

2
(0.8646)Cos(2 t  k0 z ), thus
86.5% of power is inside the beam radius.