PROPOSAL OF A UNIVERSAL QUANTUM COPYING MACHINE IN CAVITY QED

• Joanna Gonzalez • Miguel Orszag • Sergio Dagach • Facultad de Física • Pontificia Universidad Católica de Chile Quantum Optics II COZUMEL MEXICO

The No-Cloning Theorem (Wooters and Zurek,Nature299,802(1982)) showed that it is not possible to construct a device that will produce an exact copy of an arbitrary quantum state.

This Theorem is an unexpected quantum effect due to the linearity of Quantum Mechanics, as opposed to Classical Physics, where the copying Process presents no difficulties, and this represents the most significant difference between Classical and Quantum Information.

Thus, an operation like:

Thus, an operation like: 

a

0

b Q x

 

a



b Q

´

x

Is not possible, with: 0 

b a

=INPUT QUBIT =Blank copy

Q Q

´

x x

=initial state of cloner =final state of cloner Because of this Theorem, scientist ignored the subject up to 1996 when Buzek and Hillery (V.Buzek,M.Hillery,Phys.Rev.A,54,1844(1996) proposed the Universal Quantum Copying Machine(UQCM)-that produced two imperfect copies from an original qubit, the quality of which was independent of the input state.

The quality of the copy is measured through the FIDELITY

F

 

copy



ideal



ideal

 

input

UNIVERSAL QUANTUM COPYING MACHINE   BASIS 

B

 2 3  

A

 

B A



F



B

 ,   2 3  

A I



BlankCopy A I



AUXILLIARY

1    2 5 6 1 3 1 3  

A I A

In the present work, we propose a protocol that produces 2 copies from an input state, with Fidelity

F

 5 6 In the context of Cavity QED, in which the information is encoded in the electronic levels of Rb atoms, that interact with two Nb high Q cavities.

SOME PREVIOUS BACKGROUND TO THE PROPOSAL Consider a two level atom that is prepared in a superposition state , using the Microwave pulses in a Ramsey Zone, with frequency 

r

Near the e(excited)-g(ground) transition. It generates superpositions

e g

 cos   

e i



e

 sin 

e i



e

sin  

g

cos 

g where

    Depends on the interaction time ( 

r

 

eg

)

t

Is prop. detuning

On the other hand, the atom-field interaction is described by the Jaynes Cummings Hamiltonian

H

.

    0 , 2 

eg



z

2   ( 

a a

Coupling constant 2 )  0 (  

a

  

a

 ),.

2

The atom-field state evolves like

e

, 0  

e g

,1  

g

  cos(  2 0 cos( 2  0 ,1   sin(  2 0 sin( 2  0 ,1 For example, for 0

t

 2 0

t



e e

, 0 , 0  

g

1 2  

e

, 0 

g

   

e

, 0  1 2  

e

, 0 

g

,1  

Now, consider an external Classical pulse, interacting with the atom

H ext

 

g

       We use the dressed state basis that diagonalizes the J-C Hamiltonian:  , )    sin 

n

cos 

n

 1  1  cos   sin 

n n

cos 2 

n

  , 4  2 (

n

 1)   2

eg

  The Energies of the dressed states are

E

    (

n

 1 2 )   2  2  4  2 (

n

 1 )

In the limit  ,

n

 ,

n

      0 ..

..



e

,1)

g

,1 Consider the external field in resonance with the (+,1)  (-,0) Transition, that is 

g

 

f

(

t

) cos( 

s t

) Where f(t) is some smooth function of time to represent the pulse shape, with(in the dispersive case)  

s



E



n

 1 

E



n

 0   

eg

 3  2  

The above Hamiltonian has been studied by several authors (Domokos et al;Giovannetti et al) and arrive to the conclusion that For a suitable pulse, a C-NOT gate can be achieved, where the photon Number (0 or 1) is the control and the atom the target  ,1 or e,1  ,0 or g,1  ,0 or e,0 -,-1 or g,0   -,0 or g,1  ,1 or e,1    ,0 or e,0 -,-1 or g,0 The mechanism of the above C-NOT gate that forbids, for example the (g,0>-  (e,0> transition is the Stark Effect, caused by one photon in the cavity. In order to resolve these two transitions, we have to make sure that 1  

t

i n t

Where  Is the frequency difference between these two transitions.

The exchange

g

,1



e

,1

IS POSSIBLE

C-NOT GATE

N=0 ATOMIC STATE IS NOT CHANGED N=1 ATOMIC STATE IS EXCHANGED

 

CONTROL TARGET

UQCM

PROPOSED PROTOCOL

ATOM 1 ( Ramsey Field

g g

1 1  2 3

g

1  1 3

e

1  

ar

cos 1 3 A1 interacts with the cavity Ca(initially in )through a Rotation, so

e

,

o



g

, 1 0

a



g

, 1  

e

, 0 2 3

g

1  1 3

e

1 ) 0

a

 ( 2 3 0

a

 1 3 1

a

)

g

1 State swapping.The excitation of atom 1 is transferred to the cavity a

ATOM 2 IT CONTAINS THE INFORMATION TO BE CLONED  2  

e

2  

g

2 This state can be prepared in the same fashion as the atom 1, for example with a Ramsey Field.

Then we apply a Classical pulse, as described before,  with 0 photons (  C-not 

e

2 2 3  

g

2 )( 

e

2   2 3

g

2 0

a

0 

a

1 3 1

a

)  1 3 

g

2  

e

2 1

a

A3 and A4 are the atoms carrying the two copies(IDENTICAL)

     FINAL STATE 2 3

e

3

e

    2 3

g

3

where e

,

g

  4

g A I

4 , 

A

  1 3  3,4

A

   1 3  3,4

A I

  

B A A I

  ( 2 3  

g g

1 1

g g g

1  2 0 0

a a

1 3 0 1

e b b

1 ) 3,4  1 2 (

e

3

g



g

3

g e

3

g

4 ) 4

DISCUSSION

Experimental numbers(Haroche et al)  2   100

KHz

,  2   50

KHz

,  2   50

KHz

An interaction time of  int  50 

s

Marginally satisfies the earlier requirement.

With the flight time of 100 Take about 700 

s



s

The whole scheme should Which is reasonable in a cavity with a Relaxation time of 16ms.They achieved a resolution required to Distinguish between 1 or 0 photons

DISCUSSION OF THE C-NOT GATE The complete Hamiltonian in the Interaction Picture is:

H INT

 

f

   exp(  exp( 

t

) 2

t

 )   exp( 2

s

 exp(

i

 1

t

) 

i



t

 )

a

    exp( 

i



t

)

a

  exp( 

i

 1

t

)     

with

  

eg

 1  

eg

   

s

Since the external pulse is resonant with the (+,1)  (-,0) Transition,this imposes a condition on  1

 1   2    1   8 (  ) 2  1   4 (  ) 2  2    Also, we notice that we have introduced exponential factors In both terms of the Hamiltonian just to mimic the passage Time and duration of the pulse, referred to as  and 

s

Respectively.

We have done this in order to solve Schrodingers equation With continuous functions.

Assuming  (

t

) 

n N

  0 

a n

(

t

)

n

,

e



b n

(

t

)

n

,

g

 We have to solve the following set of differential equations

i db n dt

  exp ( 

t

) 2 exp( 

i



t

)

n a n

 1 (

t

) 

f

exp(  ( 

t s

) 2 ) exp( 

i

 1

t

)

a n

(

t

)

i da n dt

   exp ( 

t

) 2 exp(

i



t

)

f

exp(  ( 

t s

) 2 ) exp(

i

 1

t

)

b n

(

t

)

n

 1

b n

 1 (

t

)

  27 

s



s

 2   2   1 2   13 .

5 

s

 50

KHZ

 137

KHZ

 46 .

25

KHZ f

2   81 .

1

KHZ

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299

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