Study on Dynamics of N Level System of Atom by Laser Fields

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arXiv:quant-ph/0512126v1 16 Dec 2005

Study on Dynamics of N Level System of Atom by Laser Fields

Kazuyuki FUJII

∗ Department of Mathematical Sciences Yokohama City University

Yokohama, 236–0027 Japan

Abstract

This paper is an extension of Fujii et al (quant–ph/0307066) and in this one we again treat a model of atom with n energy levels interacting with n(n-1)/2 external laser fields, which is a natural extension of usual two level system. Then the rotating wave approximation (RWA) is assumed from the beginning.

To solve the Schr¨ odinger equation we set the consistency condition in our terminology and reduce it to a matrix equation with symmetric matrix Q consisting of coupling constants. However, to calculate exp(−itQ) explicitly is almost impossible.

In the case of three level system we determine it in a perfect manner, so three level system is in our model completed. We also make a comment on some solution of four level one, which is not complete at the present time.

In last, we make a comment on Cavity QED quantum computation based on three energy levels of atoms as a forthcoming target.

∗

E-mail address : [email protected]

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1 Introduction

The purpose of this paper is to develop a useful method applicable to Quantum Computation.

As an easy introduction to it see for example [1], [2], [3].

Quantum Computation is in a usual understanding based on qubits which are based on two level system (two energy levels or fundamental spins of atoms), See [4], [5], [6] as for general theory of two level system.

In a realistic image of Quantum Computer we need at least one hundred atoms. However, we meet a very severe problem called Decoherence which may destroy a superposition of quantum states in the process of unitary evolution of the system. See for example [7] or recent [8] as an introduction. At the present time it is not easy to control Decoherence.

By the way, an atom has in general infinitely many energy levels, while in a qubit method only two energy levels are used, see for example [8] as an introduction of two level approximation.

We should use this possibility to reduce a number of atoms. Since it is not realistic to take all energy levels into consideration at the same time we use n energy levels from the ground state, which is in general called qudit theory. See for example [11], [12], [13], [14] and [15].

In the paper [16] we considered a model of atom with n energy levels interacting with n(n- 1)/2 external laser fields, which is a natural extension of usual two level system. The rotating wave approximation (RWA) is assumed from the beginning.

In the model it is assumed that all coupling constants regarding interactions of atom with different laser fields are equal, which is nothing but an approximation theory. To construct a realistic model we again treat a “full” model of atom with n energy levels interacting with n(n-1)/2 laser fields.

To solve the Schr¨odinger equation we set the consistency condition in our terminology and reduce it to a matrix equation with symmetric real matrix Q consisting of all coupling constants. However, it is almost impossible to calculate exp(−itQ) explicitly.

Therefore we restrict to special cases. In the case of three level system we determine it in a

perfect manner, so three level system is in our model completed. For the four level we make an

attempt, which is not complete at the present time and so leave it as a problem. See also [17].

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In last, we make a comment on Cavity QED quantum computation based on three energy levels of atoms that is our forthcoming target.

2 Review on Two Level System

Let us review a two level system within our necessity and its Rabi oscillation. Let {σ

¹

, σ

2

, σ

3

} be famous Pauli matrices

σ

1

=







0 1 1 0





 , σ

2

=







0 −i i 0





 , σ

3

=







1 0 0 −1





 , (1)

and we set

σ

+

≡ 1

2 (σ

1

+ iσ

2

) =







0 1 0 0





 , σ

₋

≡ 1

2 (σ

1

− iσ

²

) =







0 0 1 0





 .

Let us consider an atom with two energy levels E

0

and E

1

(E

1

> E

0

) as two level approximation. Its Hamiltonian is in the diagonal form given as

H

0

=







E

0

0 0 E

1





 . (2)

This is rewritten as

H

0

= E

0







1 0 0 1





 + (E

1

− E

⁰

)







0 0 0 1





 = E

0

1

2

+ ∆

2 (1

2

− σ

³

) ,

where ∆ = E

1

−E

⁰

is the energy difference. Since we usually take no interest in constant terms, we can set

H

0

= ∆

2 (1

2

− σ

³

) . (3)

We consider an atom with two energy levels which interacts with external (periodic) field with g cos(ωt + φ). In the following we set ¯ h = 1 for simplicity. The Hamiltonian in the dipole approximation is given by

H = H

0

+ g cos(ωt + φ)σ

1

= ∆

2 (1

2

− σ

³

) + g cos(ωt + φ)σ

1

, (4)

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where ω is the frequency of the external field, g the coupling constant between the external field and the atom. This model is complicated enough to solve, see [9], [10], [18], [19].

In the following we set φ = 0 for simplicity and assume the rotating wave approximation (which neglects the fast oscillating terms), namely

cos(ωt) = 1

2 (e

^iωt

+ e

^−iωt

) = 1

2 e

^iωt

(1 + e

^−2iωt

) ≈ 1 2 e

^iωt

, and

cos(ωt)σ

1

=







0 cos(ωt) cos(ωt) 0





 ≈ 1 2







0 e

^iωt

e

^−iωt

0 



 , therefore the Hamiltonian is given by

H = ∆

2 (1

2

− σ

³

) + g 2

e

^iωt

σ

+

+ e

^−iωt

σ

₋

≡ ∆

2 (1

2

− σ

³

) + g e

^iωt

σ

+

+ e

^−iωt

σ

₋

(5) by the redefinition of g (g/2 −→ g). It is explicitly

H =







0 ge

^iωt

ge

^−iωt

∆





 . (6)

We would like to solve the Schr¨odinger equation i d

dt Ψ = HΨ. (7)

For that purpose let us decompose H in (6) into







0 ge

^iωt

ge

^−iωt

∆





 =







1 e

^−iωt













0 g g ∆













1 e

^iωt





 , (8)

so if we set

Φ =







1 e

^iωt





 Ψ ⇐⇒ Ψ =







1 e

^−iωt





 Φ (9)

then it is not difficult to see

i d dt Φ =







0 g

g ∆ − ω





 Φ, (10)

which is easily solved. For simplicity we set the resonance condition

∆ = ω, (11)

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then the solution of (10) is

Φ(t) = exp



 

 

−igt







0 1 1 0









 

 

Φ(0) =







cos(gt) −i sin(gt)

−i sin(gt) cos(gt)





 Φ(0).

As a result, the solution of the equation (7) is given as

Ψ(t) =







1 e

^−iωt





 Φ(t) =







1 e

^−iωt













cos(gt) −i sin(gt)

−i sin(gt) cos(gt)





 Φ(0) (12)

by (9). If we choose Φ(0) = (1, 0)

^T

as an initial condition, then

Ψ(t) =







cos(gt)

−ie

^−iωt

sin(gt)





 . (13)

This is a well–known model of the Rabi oscillation (or coherent oscillation).

3 General Theory of N Level System

In general, an atom has an infinitely many energy levels, however it is not realistic to consider all of them at the same time. Therefore we take only n energy levels from the ground state into consideration (E

_n−1

> · · · > E

¹

> E

0

). Then from the lesson in the preceding section the energy Hamiltonian can be written as

H

0

=







0 ∆

1

∆

2

·

· ∆

_n−1







, (14)

where ∆

j

= E

j

−E

⁰

for 1 ≤ j ≤ n−1. For this atom we consider the interaction with n(n−1)/2 independent external laser fields corresponding to every energy difference (E

j

− E

ⁱ

for j > i).

For example see the following figure for the case of n = 4 :

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|2i

|1i

|0i E

2

E

1

E

0

ω

01

ω

₁₂

ω

01

ω

₂₃

ω

02

E

3

|3i

ω

₂₃

ω

13

ω

03

Fig.1 Atom with four energy levels and general action by laser fields

Then the interaction term assuming the RWA from the beginning is given as V =







0 g

01

e

^iω⁰¹^t

g

02

e

^iω⁰²^t

· · · · g

_0,n−1

e

^iω⁰^,n−1^t

g

01

e

^−iω⁰¹^t

0 g

12

e

^iω¹²^t

· · · · g

_1,n−1

e

^iω¹^,n−1^t

g

₀₂

e

^−iω⁰²^t

g

₁₂

e

^−iω¹²^t

0 · · · · g

_2,n−1

e

^iω²^,n−1^t

· · · · · · · ·

· · · · · · 0 g

_n−2,n−1

e

^iω^n−2,n−1^t

g

_0,n−1

e

^−iω⁰^,n−1^t

g

_1,n−1

e

^−iω¹^,n−1^t

g

_2,n−1

e

^−iω²^,n−1^t

· · · g

n−2,n−1

e

^−iω^n−2,n−1^t

0 





,

(15) where {g

^αβ

} are coupling constants, so the Hamiltonian is

H = H

₀

+ V. (16)

Here we set

ω

1

= ω

01

, ω

2

= ω

12

, · · · , ω

n−2

= ω

_n−3,n−2

, ω

_n−1

= ω

_n−2,n−1

for simplicity. We would like to solve the Schr¨odinger equation

i d

dt Ψ = HΨ. (17)

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Similarly in (8) let us decompose H : For

U =







1 e

^−iω¹^t

e

^−i(ω¹^+ω²^)t

·

· e

^−i(ω¹^+ω²^+···+ωⁿ⁻²^)t

e

^−i(ω¹^+ω²^+···+ωⁿ⁻¹^)t







(18)

it is easy to see H

U

≡ U

^†

HU =







0 g

01

g

02

e

^iǫ⁰²^t

· · g

_0,n−2

e

^iǫ⁰^,n−2^t

g

_0,n−1

e

^iǫ⁰^,n−1^t

g

01

∆

1

g

12

· · g

_1,n−2

e

^iǫ¹^,n−2^t

g

_1,n−1

e

^iǫ¹^,n−1^t

g

02

e

^−iǫ⁰²^t

g

12

∆

2

· · g

_2,n−2

e

^iǫ²^,n−2^t

g

_2,n−1

e

^iǫ²^,n−1^t

· · · · · · ·

g

_0,n−2

e

^−iǫ⁰^,n−2^t

g

_1,n−2

e

^−iǫ¹^,n−2^t

g

_2,n−2

e

^−iǫ²^,n−2^t

· · ∆

_n−2

g

_n−2,n−1

e

^−iǫ^n−2,n−1^t

g

_0,n−1

e

^−iǫ⁰^,n−1^t

g

_1,n−1

e

^−iǫ¹^,n−1^t

g

_2,n−1

e

^−iǫ²^,n−1^t

· · g

n−2,n−1

e

^−iǫ^n−2,n−1^t

∆

_n−1







,

(19) where

ǫ

ij

= ω

ij

− (ω

ⁱ⁺¹

+ ω

i+2

+ · · · + ω

^j

) for j − i ≥ 2.

By setting

Ψ = U ˜

^†

Ψ ⇐⇒ Ψ = U ˜ Ψ it is not difficult to see

i d

dt Ψ = ˜ ˜ H

U

Ψ, ˜ (20)

where

H ˜

_U

=

(8)







0 g

01

g

02

e

^iǫ⁰²^t

· · g

_0,n−2

e

^iǫ⁰^,n−2^t

g

_0,n−1

e

^iǫ⁰^,n−1^t

g

₀₁

∆

₁

− ω

1

g

₁₂

· · g

_1,n−2

e

^iǫ¹^,n−2^t

g

_1,n−1

e

^iǫ¹^,n−1^t

g

0,2

e

^−iǫ⁰²^t

g

1,2

∆

2

− (ω

¹

+ ω

2

) · · g

_2,n−2

e

^iǫ²^,n−2^t

g

_2,n−1

e

^iǫ²^,n−1^t

· · · · · · ·

g

_0,n−2

e

^−iǫ⁰^,n−2^t

g

_1,n−2

e

^−iǫ¹^,n−2^t

g

_2,n−2

e

^−iǫ²^,n−2^t

· · ∆

_n−2

− ^P

ⁿ⁻²l=1

ω

l

g

_n−2,n−1

e

^−iǫ^n−2,n−1^t

g

_0,n−1

e

^−iǫ⁰^,n−1^t

g

_1,n−1

e

^−iǫ¹^,n−1^t

g

_2,n−1

e

^−iǫ²^,n−1^t

· · g

n−2,n−1

e

^−iǫ^n−2,n−1^t

∆

_n−1

− ^P

ⁿ⁻¹l=1

ω

l







.

(21) At this stage we take the resonance conditions

∆

₁

= ω

₁

, ∆

₂

= ω

₁

+ ω

₂

, · · · , ∆

n−2

=

n−2

X

l=1

ω

_l

, ∆

_n−1

=

n−1

X

l=1

ω

_l

⇐⇒ ω

j

= E

_j

− E

j−1

(j = 1, 2, · · · , n − 1) (22) to make the situation simpler.

Moreover, we consider a very special case : namely, in (21) we set

ǫ

ij

= 0 ⇐⇒ ω

^ij

= ω

i+1

+ ω

i+2

+ · · · + ω

^j

= E

j

− E

ⁱ

(23) for all j − i ≥ 2 (see for example the figure 1). We call this a consistency condition. Then H ˜

U

becomes a constant symmetric matrix !

H ˜

U

=







0 g

01

g

02

· · g

_0,n−2

g

_0,n−1

g

₀₁

0 g

₁₂

· · g

_1,n−2

g

_1,n−1

g

02

g

12

0 · · g

_2,n−2

g

_2,n−1

· · · · · · ·

g

_0,n−2

g

_1,n−2

g

_2,n−2

· · 0 g

_n−2,n−1

g

_0,n−1

g

_1,n−1

g

_2,n−1

· · g

n−2,n−1

0 





≡ Q. (24)

It is easy to solve the equation (20) with constant Q i d

dt Ψ = Q ˜ ˜ Ψ, (25)

(9)

whose solution is formally given by

Ψ(t) = exp(−itQ) ˜ ˜ Ψ(0). (26)

As a result we have a general solution

Ψ(t) = U exp(−itQ)Ψ(0). (27)

Therefore the problem left is to calculate exp(−itQ) explicitly, which is however a very hard (maybe impossible) task. In the following let us treat the special cases n = 3 and n = 4.

4 Exact Solution in Three Level System

In this section we consider the case of n = 3 and calculate exp(−itQ) in a complete manner

¹

. For simplicity we set ω

01

= ω

1

, ω

02

= ω

3

, ω

12

= ω

2

and g

01

= g

1

, g

02

= g

3

, g

12

= g

2

. See the following figure.

|2i

|1i

|0i E

2

E

1

E

0

ω

₁

ω

2

ω

₁

ω

2

ω

3

ω

3

Fig.2 Atom with three energy levels and general action by laser fields Now the matrix Q in (24) is in this case

Q =







0 g

1

g

3

g

₁

0 g

₂

g

3

g

2

0 





, (28)

1

The result was obtained in collaboration with Shin’ichi Nojiri

(10)

so it is desirable for us to make this diagonal.

First of all we look for eigenvalues of Q. From

0 = |λE − Q| =

λ −g

¹

−g

³

−g

1

λ −g

2

−g

³

−g

²

λ

= λ

³

− g

²₁

+ g

₂²

+ g

₃²

λ − 2g

¹

g

2

g

3

the Cardano formula (see for example [20]) gives three real solutions

λ

1

= α

+

+ α

₋

, λ

2

= σ

²

α

+

+ σα

₋

, λ

3

= σα

+

+ σ

²

α

₋

, (29) where σ = e

^2πi/3

and

α

_±

=







g

1

g

2

g

3

±

s

g

₁²

g

₂²

g

₃²

− (g

₁²

+ g

²₂

+ g

₃²

)

³

27 





1 3

.

Therefore the normalized eigenvectors {|λ

^j

i | 1 ≤ j ≤ 3} corresponding to {λ

^j

| 1 ≤ j ≤ 3} are given as

|λ

^j

i =







s

λ²_j−g2²

3λ²_j−

(

^g²¹^+g²²^+g³²

)

s

λ²_j−g3²

3λ²_j−

(

^g²¹^+g²²^+g³²

)

s

λ²_j−g1²

3λ²_j−

(

^g²¹^+g²²^+g³²

)







. (30)

This derivation is not so easy, see Appendix. Then the matrix defined by

O = (|λ

¹

i, |λ

²

i, |λ

³

i) (31)

makes Q diagonal like

Q = O







λ

₁

0 0 0 λ

2

0 0 0 λ

3







O

^T

. (32)

Therefore we obtain the desired form

exp(−itQ) = O







e

^−itλ¹

0 0 0 e

^−itλ²

0 0 0 e

^−itλ³







O

^T

≡ (a

^ij

) (33)

(11)

where

a

ij

=

3

X

k=1

e

^−itλ^k

v u u t

λ

²_k

− g

i+1²

3λ

²_k

− (g

²1

+ g

₂²

+ g

₃²

)

v u u t

λ

²_k

− g

j+1²

3λ

²_k

− (g

1²

+ g

²₂

+ g

²₃

) . It is assumed g

4

≡ g

¹

in the right hand side.

Let us calculate a special case g

3

= 0. From (29)

λ

1

= ^q g

₁²

+ g

₂²

, λ

2

= 0, λ

3

= − ^q g

₁²

+ g

₂²

and from (30)

²

| ^q g

₁²

+ g

²₂

i =







g¹

√

2(g²1+g2²)

√1 2 g2

√

2(g²1+g2²)







, |0i =







g²

√

g1²+g²2

0

−g¹

√

g1²+g²2







, |− ^q g

²₁

+ g

₂²

i =







g¹

√

2(g1²+g²2)

−

^√¹₂

g2

√

2(g1²+g²2)







.

Therefore from

O =

| ^q g

₁²

+ g

₂²

i, |0i, | ^q g

²₁

+ g

₂²

i

a straightforward calculation leads us to

exp(−itQ) =







g1²

cos

^(t

√

g²1+g²2)+g2²

g²1+g²2

−i

^g¹

sin

^(t

√

g1²+g²2)

√

g1²+g²2

g1g2

cos

^(t

√

g²1+g2²)−g¹g2

g²1+g²2

−i

^g¹

sin

^(t

√

g²1+g²2)

√

g²1+g2²

cos(t ^q g

₁²

+ g

₂²

) −i

^g²

sin

^(t

√

g²1+g²2)

√

g²1+g2²

g¹g²

cos

(t

√

g²1+g²2)−g¹g²

g²1+g²2

−i

^g²

sin

(t

√

g1²+g²2)

√

g1²+g²2

g²2

cos

(t

√

g²1+g2²)+g²1

g²1+g²2







. (34)

See §3 in [17].

5 Four Level System

In this section we consider the case of n = 4. However, we cannot calculate exp(−itQ) in a complete manner.

For simplicity we also set ω

01

= ω

1

, ω

02

= ω

4

, ω

03

= ω

6

, ω

12

= ω

2

, ω

13

= ω

5

, ω

23

= ω

3

and g

01

= g

1

, g

02

= g

4

, g

03

= g

6

, g

12

= g

2

, g

13

= g

5

, g

23

= g

3

. See the figure 1 once more.

2

In the calculation we must choose signs of the complex roots in (30) carefully

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Now the matrix Q in (24) is in this case

Q =







0 g

₁

g

₄

g

₆

g

1

0 g

2

g

5

g

4

g

2

0 g

3

g

₆

g

₅

g

₃

0 





, (35)

so we look for the eigenvalues of Q :

0 = |λE − Q| =

λ −g

¹

−g

⁴

−g

⁶

−g

¹

λ −g

²

−g

⁵

−g

⁴

−g

²

λ −g

³

−g

⁶

−g

⁵

−g

³

λ

= λ

⁴

− g

₁²

+ g

²₂

+ g

²₃

+ g

₄²

+ g

₅²

+ g

₆²

λ

²

− 2 (g

¹

g

2

g

4

+ g

1

g

5

g

6

+ g

2

g

3

g

5

+ g

3

g

4

g

6

) λ + g

₁²

g

₃²

+ g

²₂

g

₆²

+ g

²₄

g

₅²

− 2g

¹

g

2

g

3

g

6

− 2g

¹

g

3

g

4

g

5

− 2g

²

g

4

g

5

g

6

.

From this we can formally write down the solutions as

λ

1

= α + β + γ, λ

2

= σ

³

α + σ

²

β + σγ, λ

3

= σ

²

α + β + σ

²

γ, λ

4

= σα + σ

²

β + σ

³

γ with very complicated terms α, β and γ and σ = e

^2πi/4

= i, see [20]. At this stage we cannot determine the eigenvectors {|λ

^j

i | 1 ≤ j ≤ 4} corresponding the eigenvalues.

Therefore we restrict a situation to the relatively simple case g

4

= g

5

= g

6

= 0. See the following figure.

|3i

|2i

|1i

|0i E

3

E

2

E

1

E

0

ω

1

ω

2

ω

3

ω

₁

ω

2

ω

3

(13)

Fig.3 Atom with four energy levels and special action by laser fields

Then Q is

Q =







0 g

1

0 0 g

1

0 g

2

0 0 g

2

0 g

3

0 0 g

3

0 





, (36)

and the characteristic equation becomes

λ

⁴

− g

²₁

+ g

₂²

+ g

₃²

λ

²

+ g

₁²

g

₃²

= 0.

The solutions are easy to be λ

₁

=

√ A + √ B

2 , λ

₂

=

√ A − √ B

2 , λ

₃

= −

√ A − √ B

2 , λ

₄

= −

√ A + √ B 2

with A = g

₂²

+ (g

1

+ g

3

)

²

and B = g

²₂

+ (g

1

− g

³

)

²

. We can construct the corresponding eigenvectors |λ

¹

i, |λ

²

i, |λ

³

i, |λ

⁴

i, which is however not so easy, see [17]. Using

O = (|λ

¹

i, |λ

²

i, |λ

³

i, |λ

⁴

i) (37) Q can be diagonalized like

Q = O







λ

1

0 0 0 0 λ

2

0 0 0 0 λ

3

0 0 0 0 λ

4







O

^T

. (38)

Therefore we can calculate exp(−itQ) explicitly, see §4 in [17] in detail (we don’t repeat it here).

6 Approximate Solution of N Level System

Since we are interested in the whole of exp(−itQ) in (33) we take an approximation to the

equation. Namely, we assume that all coupling constants are equal : g = g

_αβ

for 0 ≤ α, β ≤

(14)

n − 1. Then

Q = g







0 1 1 · · 1 1 1 0 1 · · 1 1 1 1 0 · · 1 1

· · · ·

· · · · 1 1 1 · · 0 1 1 1 1 · · 1 0







≡ gR (39)

and it is not difficult to calculate exp(−itgR) explicitly, [16].

If we define

|1i = (1, 1, · · · , 1, 1)

^T

, then it is easy to see R = |1ih1| − 1

n

, so

exp(−itgR) = e

^igt

exp(−igt|1ih1|).

Since h1|1i = n,

(|1ih1|)

^k

= n

^k−1

|1ih1|, so that

exp(−igt|1ih1|) = 1

ⁿ

+

X

∞

k=1

(−igt)

^k

k! (|1ih1|)

^k

= 1

n

+

X

∞

k=1

(−igt)

^k

k! n

^k−1

|1ih1|

= 1

n

+ 1 n

(

_∞

X

k=0

(−ingt)

^k

k! − 1

)

|1ih1|

= 1

n

+ exp(−ingt) − 1

n |1ih1|. (40)

Therefore we obtain

exp(−itQ) = exp(−itgR) = e

^igt

(

1

n

+ exp(−ingt) − 1

n |1ih1|

)

. (41)

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7 Discussion

In this paper we developed a general theory of the n level system of atom by assuming the rotating wave approximation and reduce the Schr¨odinger equation to the simple matrix equation with matrix consisting of coupling constants under the consistency condition.

Moreover, in the three level system we solved the equation in a perfect manner and made a comment on some solution for the four level system.

By the way, to assume the rotating wave approximation is a bit weak in the theory, so that we must study the general equation for example in the case of three level one like

i d dt







Ψ

1

Ψ

2

Ψ

3







=







0 g

1

cos(ω

1

t + φ

1

) g

3

cos(ω

3

t + φ

3

) g

1

cos(ω

1

t + φ

1

) ∆

1

g

2

cos(ω

2

t + φ

2

) g

3

cos(ω

3

t + φ

3

) g

2

cos(ω

2

t + φ

2

) ∆

2













Ψ

1

Ψ

2

Ψ

3







. (42)

However, it is almost impossible to solve this at the present time. Let us leave it for young researchers in Quantum Optics or Mathematical Physics as a challenging task.

Here we state our motivation once more. We would like to construct a realistic model of quantum computation with three level system. Its candidate is a quantum computation based on Cavity QED making use of three energy levels of atoms. See the following figure.

⊗ ⊗ · · · ⊗

· · ·

Fig.4 A general setting for a quantum computation based on Cavity QED

Let us add a short explanation to this figure. The dotted line means a single photon

inserted in the cavity and all curves mean external laser fields (which are treated as classical

(16)

ones) subjected to ultracold–atoms trapped linearly in it. See in detail [21] and [22] in which the Cavity QED quantum computation with two level system was “completed”.

This is our forthcoming target !

Acknowledgment.

K. Fujii wishes to thank Akira Asada, Kunio Funahashi and especially Shin’ichi Nojiri for their helpful comments and suggestions.

Appendix Derivation of (30)

For Q in (28) we consider the equations

Qx

j

= λ

j

x

j

for 1 ≤ j ≤ 3.

Namely,







0 g

1

g

3

g

₁

0 g

₂

g

3

g

2

0 











x y z







= λ

j







x y z







⇐⇒



 

 

 

 

g

1

y + g

3

z = λ

j

x g

₁

x + g

₂

z = λ

_j

y g

3

x + g

2

y = λ

j

z

(43)

We solve the above equations with respect to z. The result is

x = λ

j

g

3

+ g

1

g

2

λ

²_j

− g

1²

z, y = λ

j

g

2

+ g

1

g

3

λ

²_j

− g

1²

z, where we have used the characteristic equation

λ

³_j

− (g

²1

+ g

₂²

+ g

₃²

)λ

j

− 2g

¹

g

2

g

3

= 0,

so

x

_j

= z







λjg³+g¹g² λ²_j−g1²

λjg2+g1g3

λ²_j−g1²

1 





. (44)

(17)

Next let us determine the normalization.

hx

^j

|x

^j

i = 1 ⇐⇒ 1 = x

²

+ y

²

+ z

²

= z

²

(λ

j

g

3

+ g

1

g

2

)

²

+ (λ

j

g

2

+ g

1

g

3

)

²

+ (λ

²_j

− g

1²

)

²

(λ

²_j

− g

²1

)

²

(45) From this

(λ

²_j

− g

²1

)

²

+ (λ

j

g

3

+ g

1

g

2

)

²

+ (λ

j

g

2

+ g

1

g

3

)

²

=λ

⁴_j

+ (−2g

1²

+ g

₂²

+ g

²₃

)λ

²_j

+ 4g

1

g

2

g

3

λ

j

+ g

₁²

(g

₁²

+ g

₂²

+ g

₃²

)

=λ

⁴_j

− 3g

²1

λ

²_j

+ (g

²₁

+ g

₂²

+ g

₃²

)λ

²_j

+ 4g

1

g

2

g

3

λ

j

+ g

₁²

(g

₁²

+ g

²₂

+ g

₃²

). (46) Now we want to remove the term 4g

1

g

2

g

3

λ

j

. By using the characteristic equation 2g

1

g

2

g

3

= λ

³_j

− (g

1²

+ g

₂²

+ g

₃²

)λ

j

, we have

The right hand side of (46) = 3λ

⁴_j

− 3g

²1

λ

²_j

− (g

1²

+ g

²₂

+ g

²₃

)λ

²_j

+ g

₁²

(g

₁²

+ g

₂²

+ g

₃²

)

= 3λ

²_j

(λ

²_j

− g

1²

) − (g

1²

+ g

²₂

+ g

₃²

)(λ

²_j

− g

1²

)

= (λ

²_j

− g

1²

)(3λ

²_j

− g

1²

− g

²2

− g

3²

). (47)

From (45)

z =

q λ

²_j

− g

1²

q 3λ

²_j

− g

²1

− g

2²

− g

3²

, so we have

x

j

= z







λjg³+g¹g² λ²_j−g²1

λjg2+g1g3

λ²_j−g²1

1 





= 1

q 3λ

²_j

− g

²1

− g

2²

− g

3²







λjg3+g1g2

√

λ²_j−g1²

λjg2+g1g3

√

_λ2 j−g1²

q λ

²_j

− g

²1







(48)

from (44). At this stage it is easy to conjecture λ

_j

g

₃

+ g

₁

g

₂

q λ

²_j

− g

1²

= ^q λ

²_j

− g

2²

, λ

_j

g

₂

+ g

₁

g

₃

q λ

²_j

− g

1²

= ^q λ

²_j

− g

3²

.

In fact,

λ

j

g

3

+ g

1

g

2

q λ

²_j

− g

1²

=

v u u t

(λ

_j

g

₃

+ g

₁

g

₂

)

²

λ

²_j

− g

²1

=

v u u t

g

₃²

λ

²_j

+ 2g

1

g

2

g

3

λ

j

+ g

₁²

g

₂²

λ

²_j

− g

1²

=

v u u t

(λ

²_j

− g

1²

)(λ

²_j

− g

2²

) λ

²_j

− g

1²

= ^q λ

²_j

− g

2²

,

(18)

where we have used the equation

g

²₃

λ

²_j

+ 2g

1

g

2

g

3

λ

j

+ g

²₁

g

²₂

= g

²₃

λ

²_j

+ λ

⁴_j

− (g

²1

+ g

²₂

+ g

₃²

)λ

²_j

+ g

₁²

g

₂²

= λ

⁴_j

− (g

²1

+ g

₂²

)λ

²_j

+ g

₁²

g

²₂

= (λ

²_j

− g

²1

)(λ

²_j

− g

²2

).

Similarly, we obtain

λ

_j

g

₂

+ g

₁

g

₃

q λ

²_j

− g

1²

= ^q λ

²_j

− g

3²

. From (48) we finally obtain

x

_j

= 1

q 3λ

²_j

− g

1²

− g

2²

− g

3²







q λ

²_j

− g

²2

q λ

²_j

− g

²3

q λ

²_j

− g

²1







=







r

λ²_j−g2²

3λ²_j−g²1−g2²−g²3

r

λ²_j−g3²

3λ²_j−g²1−g2²−g²3

r

λ²_j−g1²

3λ²_j−g²1−g2²−g²3







≡ |λ

j

i.

A comment is in order. √

t means the complex square root defined by √

t = e

^{i arg t/2}

^q |t| , so in our case we must choose its sign (for example, √

4 = 2 or − 2) carefully.

(19)

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