PHY4610H Final Year Project Mid-Program Report Leung Ming Lam The Department of Physics, The Chinese University of Hong Kong Advisor: Professor Liu Ren Bao Title: Decoherence of a spin in noisy environment Abstract In quantum mechanics, most of the fundamental properties of a particles, such as mass, charge and spin, can be described in different quantum states. In many physical systems, the quantum states and wavefunctions can be affected by the environment. In this project, we are going to study the effect of random background fluctuation towards a quantum system. We use photon echo experiment as an example to illustrate how a collection of two-level systems be affected by the impact noise due to collisions. We calculate the temporal profile of the photon echo signal of a collection of two-level systems with Gaussian random fluctuation. The model will be solved both analytically and numerically by quantum mechanical measure.

I. Background and Introduction In recent years, it is discovered that many physical system can be analyzed in a quantum mechanical measure. We often solve a physical problem quantum mechanically by applying perturbation Hamiltonian in a well-known physical system. However, there are impurities, particle collisions and deflected lattice alignment in real physical system, which leads to random fluctuation affecting the problem. Recently, there is much evidence indicating that there is decoherence in physical system. There exists a mechanism by which quantum systems interact with their environments to exhibit probabilistically additive behavior. For example, the energy state of a two-level system under low temperature can be approximated as a harmonic oscillator if there is no background noise. However, the system will be affected by the impact

noise due to particle collisions. Therefore, the perturbation has to be taken into account during calculating the energy and wavefunction of the system. Here we will illustrate the effect of spin decoherence using a common example, photon echo experiment. We will study how the echo signal is affected by the impact noise due to collisions. When random fluctuation of the environment is considered, a phase shift from the unperturbed system is expected to be found. One of the aims of the project is to calculate the analytical expression of the magnitude echo signal in photon echo experiment, depending on the time parameter T, τ and τc. Each of the time parameters T, τ and τc with specific precision. In this paper, we will briefly go through how to analyze those data, and how to calculate the values and

uncertainties of parameters from the raw data. II. Method Consider N two-level systems randomly distributed in a fixed volume V. Each particles has two states, |0> and |1>, with eigenenergy 0 and 1 respectively. Assume the system interacts with two laser pulse, one with magnitude E1 at time t=0, another with magnitude E2 at time t= τ. Let the two laser pulses exist on an infinitely short time, under room temperature, the system can be characterized as a set of harmonic oscillators with sudden perturbation. The Hamiltonian of each particle j in the system can be expressed as below: H 0 = hσ + σ − Ω r r r r H 1 (t ) = − µ j ⋅ E (t ) cos(Ωt − k ⋅ r j ) H 2 (t ) = hσ + σ − ∆ j (t )

(1)

where H 0 is the unperturbed Hamiltonian of two-level system with harmonic oscillating frequency Ω. H 1 (t ) is the timedependent perturbation due to the laser pulse, and H 2 (t ) is the perturbation due to random fluctuation of the system. We defined σ + and σ − as the upper and lower ⎛0 0⎞ ⎟⎟ and latter operator with σ + = ⎜⎜ ⎝1 0⎠ ⎛0 1⎞ r ⎟⎟ . µ j is the transition operator σ − = ⎜⎜ ⎝ 0 0⎠ for the jth two-level system. ∆ j (t ) is

the modulation of the transition energy of r r the jth site. E (t) and k are the amplitude and wavevector of the applied electric field respectively. For the sudden perturbation occurring at time t=0 and t= τ,

r r r E (t ) = E1δ (t − 0) + E 2δ (t − τ ) r r

r r

= E1e ik1⋅r δ (t − 0) + E 2 e ik2 ⋅r δ (t − τ )

where k1 and k 2 are the wavevectors of the first and second pulse respectively. In order to describe the system, we introduce a density matrix ρ(t) defined as below: ⎛ ρ 00 (t ) ρ 01 (t ) ⎞ ⎟⎟ ⎝ ρ10 (t ) ρ11 (t ) ⎠

ρ (t ) = p j | Ψ j >< Ψ j |= ⎜⎜

with initial conditions: ⎛1 0⎞ ⎟⎟; ρ (t = 0) =| 0 >< 0 |= ⎜⎜ ⎝ 0 0⎠

ρ 00 ( 0 ) = 1; ρ 01 ( 0 ) = ρ 10 ( 0 ) = ρ 11 ( 0) = 0

(2) In a photon echo experiment, the sample is irradiated with a pulse at time t = 0, allowed to evolve in the absence of radiation for a period of time τ, and then irradiated with a second pulse. After time τ, the sample will emit an echo signal due to the irradiation, whose temporal profile is peaked at time T, with wavevector r r 2 k 2 − k1 . The amplitude of echo signal is determined by the state transition of the system, which is represented by the density matrix. During the experiment, a detector will be r r set up in the direction 2 k 2 − k1 in order to received the echo signal. Our objective of this project is to obtain the echo signal for the present model by analytical calculation. First, as the received signal we can predict r the form of the echo signal in term of E1 r and E2 . By considering the pulses and echo signal only at suitable direction, r r

Q1st signal ∝ E1 e ik1 ⋅r

r r

2 st signal ∝ E 2 e ik 2 ⋅r

r

r

r

r ∴ Echo signal ∝ ρ 22 1 ∝ E1 E 2 e i ( 2 k 2 − k1 )⋅(3) *

2

Therefore, we predict that the echo signal will be proportional to the magnitude of laser pulses following the above relation. Considering one particular two-level system, we need to solve the density matrix quantum mechanically by applying the Von Neumann equation, i.e. ih∂ t ρ (t ) = [ H I , ρ (t )]

(4) where the brackets denote a commutator, and HI is perturbation Hamiltonian defined by equation (1), with I =1 or 2. To calculate [ H I , ρ (t )] , we need to find the eigenvalues of the each perturbation Hamiltonian. As the system only has two eigenstate, and , we apply the Hamiltonian defined by equation (1) to difference combination of states. Here I skip the mathematical details. For H0 , it follows the result of quantum oscillators, i.e.

n H 0 n = nhΩ 0 H0 1 = 1H0 0 = 0

(5)

By rotating wave approximation, we can drop the high frequency oscillating terms, i.e.

∫e

iΩt

σ + dt = ∫ e −iΩtσ − dt ≈ 0

Therefore, Hamiltonian H1 is given by r r r r H1α = −µ j ⋅ E(t ) cos(Ωt − k ⋅ rj ) r r r r r v r v ⎡d ⋅ E (t )e−iΩtα +ikα ⋅rj σ j + d * ⋅ E * (t )eiΩtα −ikα ⋅rj σ j ⎤ 1 j α α + α α − j ⎥ =− ⎢ r v r r r r r v i t i k r i t i k r − Ω + ⋅ Ω − ⋅ 2 ⎢+ d * ⋅ E (t )e α α j σ j + d ⋅ E * (t )e α α j σ j ⎥ j − α α + ⎦ ⎣ j α α r r r r r r v 1 iΩt −ik ⋅r −iΩt +ik ⋅r j j * v * = − d j ⋅ Eα (tα )e α α j σ + + d j ⋅ Eα (tα )e α α j σ − 2 * = γ α (tα )σ + + γ α (tα )σ − (7)

[

]

where H1α represents the perturbation due to the α th laser pulse, with α =1 or 2. When α=1, tα= 0. With α=2 or 3, tα= τ. Using (7), we can get: r r r* v * iΩt − ik ⋅r * ∴ 0 H 1α 1 = γ α (t ) = d j ⋅ Eα (t α )e α α j r r r v − iΩt + ik ⋅r 1 H 1α 0 = γ α (t ) = d j ⋅ Eα (t α )e α α j

Similarly, for H2 : 0 H 1α 0 = 1 H 1α 1 = 0

n H 2 n = nh∆ j (t ) 0 H2 1 = 1H2 0 = 0

(6)

For H1, we need to express the transition r operator µ j in terms of σ + and σ − : r r r µ j = 0 µ j 1 1 0 + 0 µ j* 1 0 1 r r* = d jσ + + d j σ − Here we define: r r r r d j = 0 µ j 1 d j* = 0 µ j* 1

(8)

Now, with the eigenvalue given in expressions (5), (6) and (8), we can express the Von Neumann equation as below: For HI = H1α , equation (4) becomes: ih∂ t ρ 00 (t ) = 0 H1α 1 ρ10 (t ) − 1 H1α 0 ρ 01 (t ) ih∂ t ρ 01 (t ) = 0 H1α 1 ρ11 (t ) − 0 H1α 1 ρ 00 (t ) ih∂ t ρ10 (t ) = 1 H1α 0 ρ 00 (t ) − 1 H1α 0 ρ11 (t ) ih∂ t ρ11 (t ) = 1 H1α 0 ρ 01 (t ) − 0 H1α 1 ρ10 (t )

Let ρ(t) = ρ(0) + ρ(1) + ρ(2) + ρ(3) + …

by H 1α = γ α (tα )σ + + γ α * (tα )σ − ⎛ ρ 00 ⎞ ⎛ 0 ⎜ ⎟ ⎜ * ⎜ ρ 01 ⎟ ⎜ − γ α = ih∂ t ⎜ ρ ⎟ ⎜ ⎜ 10 ⎟ ⎜ γ α ⎜ρ ⎟ ⎜ 0 ⎝ 11 ⎠ ⎝

−γα

γα*

0

0

0

0

γα

−γα

0 ⎞⎛ ρ 00 ⎞ ⎟⎜ ⎟ γ α * ⎟⎜ ρ 01 ⎟ ⎟ − γ α ⎟⎜⎜ ρ10 ⎟⎟ 0 ⎟⎠⎜⎝ ρ11 ⎟⎠ (9)

*

Where ρ(0), ρ(1), ρ(2), ρ(3) are the 0th, 1st, 2nd and 3rd order term of time t respectively. By the prediction given by expression (3), we know that the echo signal should depend on the third order of time t. i.e. we need to find ρ~ ( 3) 22 1 .

For HI = H 2 , equation (4) becomes:

ih∂ t ρ 00 (t ) = 0

ih∂ t ρ 01 (t ) = −h∆ j ρ 01 (t ) ih∂ t ρ10 (t ) = h∆ j ρ 01 (t ) ih∂ t ρ11 (t ) = 0

(10)

We can combine equation sets (9) and (10), by substitute the density matrix ρ (t ) by new variable ρ~ (t ) : i ∆ dt ρ~10 (t ) = ρ 10 (t )e ∫ j

− i ∆ dt ρ~01 ( 0 ) = ρ 01 ( 0 ) e ∫ =0 j

which implies: (1) * ih∂ ρ~ = γ ρ~ α

00

ih∂ t ρ~11

ρ~01 (t ) = ρ 01 (t )e ∫ (11) ρ~00 (t ) = ρ 00 (t ); ρ~11 (t ) = ρ 11 (t )

(1)

= γ α ρ~01

− γ α ρ~01 = 0 * (0) − γ α ρ~10 = 0

(0)

10 (0)

(0)

Similarly, as ρ~00 (1) = ρ~11 (1) = 0 , ( 2) * (1) * (1) ih∂ t ρ~01 = γ α ρ~11 − γ α ρ~00 = 0 ( 2) (1) (1) ih∂ ρ~ = γ ρ~ − γ ρ~ =0

Then equation set (10) becomes: ih∂ ρ~ (t ) = 0

t

α

10

00

α

11

00

ih∂ t ρ~01 (t ) = 0 ih∂ ρ~ (t ) = 0 t

i ∆ dt Therefore, ρ~10 ( 0 ) = ρ 10 ( 0 ) e ∫ j = 0

t

− i ∆ j dt

t

To solve the equation set (13), we first consider the 0th and 1st order term of ρ(t) using initial conditions given in (2). All the density matrix components of the 0th order term are 0 except for ρ00(0) =1.

(12)

10

ih∂ t ρ~11 (t ) = 0

As a result, we can show that by induction: For all odd number 2n-1, ρ~ ( 2 n −1) = ρ~ ( 2 n −1) = 0 00

And equation set (9) becomes:

10

⎛ ρ~00 ⎞ ⎛ 0 ⎜~ ⎟ ⎜ * ⎜ ρ ⎟ ⎜ − χα ih∂ t ⎜ ~01 ⎟ = ⎜ ρ ⎜ 10 ⎟ ⎜ χ α ⎜ ρ~ ⎟ ⎜ 0 ⎝ 11 ⎠ ⎝

− χα

χα *

0

0

0

0

χα

− χα

i ∫ ∆ j dt where χ α = γ α e

11

For all even number 2n, ρ~ ( 2 n ) = ρ~ ( 2 n ) = 0

*

~ 0 ⎞⎛ ρ 00 ⎞ ⎟ ⎜ ⎟ χ α * ⎟⎜ ρ~01 ⎟ ⎟ ~ − χ α ⎟⎜⎜ ρ 10 ⎟⎟ ~ 0 ⎟⎠⎜⎝ ρ 11 ⎟⎠ (13)

01

To find ρ~ ( 3) 22 1 , we can extract the high order density matrix component by tracing the lower order term, using the matrix mentioned in equation (13)

according to the definition in expressions (11). Finally, we can find the expression of ρ~ ( 3) 22 1 , which

Figure 1 The step-by-step calculation of density matrix from low order term to high order term

ρ~10 ( 22 1 )

3rd order E2

ρ~11( 2 1 )

2nd order

Q ρ~11 st

1 order

(2 1 )

ρ E2

(2 1) = − ρ~00

ρ~10 ( 2)

10

*

2

(T ) ∝ E1 E 2 e

r r r − i ( 2 k 2 − k1 )⋅r j

τ

T

∆ j ( t ') dt ' − i ∫ ∆ j ( t ') dt ' τ e ∫0 +i

So we can observed that the echo signal is r r

ρ~00 ( 2 1 ) E1*

22 1

*

2

r − i ( 2 k − k )⋅ r

proportional to E1 E2 and e 2 1 j , consistent with the prediction in the beginning. Besides, it is consist of a term called e iφ (T ,τ ) , where φ is defined as:

E2

ρ~01( 1 ) E1*

0th order

E2

φ (T ,τ ) = ∫0τ ∆ j (t ' )dt ' − ∫τT ∆ j (t ' )dt '

E1*

ρ~00

ρ~00 ( 0 )

(0)

(14)

Each time when a perturbation Hamiltonian is applied, the operation matrix given in equation (13) will * introduce a factor of χα or χ α to the

Notice that the term e iφ (T ,τ ) is correlated to the environment fluctuation ∆j. So, if we know the functional form of ∆j(t), the magnitude of echo signal can be expressed as a function of measuring time T, period between two pulse τ, and the parameters of function ∆(t).

higher order term. Since χα and χ α are r r proportional to Eα and Eα * respectively, when we need to find ρ~ ( 3) 22 1 , we need to

Consider the simplest case with uniform fluctuation, i.e. ∆(t) = ∆ = constant. Then the term e iφ (T ,τ ) becomes:

have a function proportional to E2 2 and E1* , as mentioned as expression (3). Therefore, during the process of computing ρ~ ( 3) 22 1 ,

∆ j ( t ') dt ' − i ∫ ∆ j ( t ') dt ' τ e ∫0

*

r

r

we need to χ 2 twice and χ 1 once. *

Figure 1 shows the detailed procedure of calculation. By symmetry argument such (2 1) (2 1) as ρ~11 , we only need to = − ρ~00 ( 2) ~ ( 1 ) and calculate the value of ρ~ ,ρ 10

ρ~

(2 1) 00

01

one by one, then we can solve

ρ~11( 22 1 ) in terms of those lower order terms. ( 22 1 ) ( 22 1 ) − i ∫ ∆ dt = ρ~10 e By substitute ρ 10 j

+i

τ

T

τ

=e

∫0

T

∫τ

+ i∆ j dt ' − i∆ j dt '

=e

+ i∆ jτ − i∆ j (T −τ )

=e

− i∆ j (T − 2τ )

When T=2 τ , e iφ (T ,τ ) =1, which is the maximum. i.e. The echo signal will reach its maximum at time T=2τ if the lattice fluctuation is independent of time. However, this ideal case seldom happens in reality. Mostly the fluctuation term ∆j(t) would be a random function with certain kind of probability distribution. To

calculating the mean signal magnitude, we iφ (T ,τ ) need to calculate the average of e over all single j, i.e. the expected value e iφ (T ,τ ) over all two-level systems. If the random function ∆j(t) follows a Gaussian distribution, we can approximate the term e iφ (T ,τ ) as



e iφ

e



φ

2

2

(15)

Gaussian

which is derived from Wicker’s theorem. Here we omit the mathematical details. iφ 2 To calculate e , we need to find φ which can be expanded as follows:

[

]

φ 2 = ∫0τ ∆ j (t ' )dt ' − ∫τT ∆ j (t ' )dt ' τ

τ

=∫



0

∆ j ' ∆ j " dt 'dt"+ ∫

τ

0

τ

− 2∫



T

T

0 τ

T

∫τ

2

∆ j ' ∆ j " dt 'dt"

∆ j ' ∆ j " dt 'dt"

(16)

So if we can find ∆ j ' ∆ j " , we can obtain e iφ easily. To find an expected value of two variables, we use an integral of those two variables with their probability density over the phase space. Similarly, to find the expected value ∆ j (t ' )∆ j (t" ) , we have: ∆ (t ' )∆(t" ) = ∫ ∆(t ' )∆ (t" ) P[∆ (t ' )]P[∆(t" )]D[∆(t ' )]D[∆ (t" )]

(17) where P[∆ (t ' )] and P[∆ (t" )] stand for the probability density of particular value ∆ at particular time. They both follow the normalization criteria of probability density:

∫ ∆(t ) P[∆(t )]D[∆(t )] = 1 To find P[∆(t )] and ∆ j (t ' )∆ j (t" ) , here we are going to introduce a common model of defects to develop a possible functional form of ∆(t). We assume the fluctuation ∆(t) exists due the particles collision, as ∆(t) has the dimension of angular frequency appended to oscillating frequency Ω. We assume ∆j(t) varies with the energy and the momentum of particle j. Inside a fixed volume V, the particles will collide with one another. Before collision, each of them contain certain amount of kinetic energy and momentum in arbitrary direction. After collision, due to the particle interaction, their kinetic energy and momentum will change, leads to a change in ∆(t) also. Assume there exists a parameter τc as a mean free time between two collision. Then, we can build a model of ∆(t) as follows:

⎧∆(t ) ∆(t + dτ ) = ⎨ ⎩ s

with probability 1-dτ/τc with probability dτ/τc

(18)

where s is some random number following Gaussian distribution:

P( s) = G ( s) ∝ e



s2

2σ 2

In this model, for each infinite tiny time step dτ , the particle may collide with other particles with probability of dp= dτ/τc. Then the new ∆ (t + dτ ) will become a random number not depending on the past value of ∆ (t ) . On the other hand, if there is no collision in the period dτ , the momentum of the particle remains unchanged an so as ∆(t ) , i.e. ∆(t + dτ ) = ∆(t ) , with a probability 1-dp = 1- dτ/τc

Consider time 0 ≤ t ≤ t I , we treat the function ∆(t ) as an infinite discrete sequence xk, where x 0 = ∆ (t = 0) , x N = ∆ (t = t I ) , and N → ∞ but Ndτ = t I . The probability density function can be expressed by: P[∆ (t )] = P[ x 0 : x1 : x 2 : L : x N −1 : x N ]

in expression (15) to find the magnitude of echo signal. From expression (18),

= A e iφ = Ae

N

before, the new ∆(t + dτ ) , =As G ( xmentioned 0 ) Π [(1 − dp )δ ( x k −1 − x k ) + dpG ( x k )] k = 1 i.e. xk , will become a new Gaussian random number with probability of dp. And xk can also be the same as xk-1, with a probability 1-dp. Then we can calculate ∆ (t ' )∆(t" ) by (17)

= Ae



φ

2

2

T τ −T τ ⎡ − − t −σ 2τ c 2 ⎢ −3+ 2 e τ c + 2 e τ c −e τ c ⎢τ c ⎢⎣

where the mean echo signal is a function of three time parameters, measuring time T, period between two pulse τ , and the mean free time of collisionτc.

∆ (t ' )∆ (t" ) = ∫ ∆(t ' )∆ (t" ) P[∆(t ' )]P[∆ (t" )]D[∆ (t ' )]D[∆ (t" )] N

= ∫ ∫ ∫ ...∫ G ( x 0 )G ( x 0 ' ) Π x k x k ' P( x k ) P( x k ' )dx1 ...dx N dx1 '...dx N ' k =1 N

= ∫ ∫ ∫ ...∫ G ( x 0 )G ( x 0 ' ) Π x k [(1 − dp )δ ( x k −1 − x k ) + dpG ( x k )]dx1 ...dx N k =1

= G ( x 0 )G ( x 0 ' ) = σ 2e

−τ

τc

After getting ∆ (t ' )∆(t" ) , we can find

φ 2 easily by equation (16):

[

]

φ 2 = ∫0τ ∆ j (t ' )dt ' − ∫τT∆ j (t ' )dt ' =∫

τ τ



0 0

∆ j ' ∆ j " dt 'dt"+ ∫

T

τ

− 2∫

τ T



0 τ

T

∫τ

2

∆ j ' ∆ j " dt 'dt"

∆ j ' ∆ j " dt 'dt"

T τ −T τ − − ⎤ ⎡t τc τc τc = 2σ τ c ⎢ − 3 + 2e + 2e − e ⎥ ⎥⎦ ⎢⎣τ c 2

2

Since G(x) following Gaussian distribution, we can use the approximation

⎤ ⎥ ⎥ ⎥⎦

For τ=1 and τc =1, the measured signal against time T is shown as Fig.3.

III. Results and Discussion

From the analytical calculation using quantum mechanical methods, we find the magnitude of echo signal Y as below:

Y (T , τ , τ c ) = Ae

T τ −T τ ⎡ − − t −σ 2τ c 2 ⎢ − 3+ 2 e τ c + 2 e τ c − e τ c ⎢τ c ⎣⎢

⎤ ⎥ ⎥ ⎦⎥

(19) The magnitude of echo signal Y depends on three time parameters, measuring time T, period between two pulse τ , and the mean free time of collisionτc.

Figure 3 The graph of received echo signal Y against time T (τ=1 and τc =1) Echo signal Y 0.8

0.7

0.6

0.5

When the noise term Δ is a constant, then φ = ∆ (T − 2τ ) and the magnitude of

1.5

echo signal = Re{ e iφ }

2.0

2.5

3.0

time T

We can see that the maxima of the function located at about 1.5. Compared with the system with uniform fluctuation, the signal is received 0.5τ earlier than the maximum signal when Δ is a constant. This time difference is caused by the phase shift φ , which depends on the timedependence of noise term Δ.

= Cos [ ∆ (T − 2τ )] Which is a cosine function with period 2π/∆ and maxima located at T=2τ When τ =1, the graph of echo signal for uniform fluctuation is shown as Fig.2. Figure 2 The graph of received echo signal Y against time T for constantΔ Echo signal Y

Keeping τ =1 but varies τ c =0.1, the measured signal is shown as Fig.4.

1.0

Figure 4 The graph of received echo signal Y against time T (τ=1 and τc =0.1)

0.9

Echo signal Y

0.8

0.92

0.7

0.90 0.88

1.0

1.5

2.0

2.5

3.0

time T

Consider the case of non-uniform fluctuation. Using expression (19), by fixingτand τc , we can plot a graph of received echo signal against time.

0.86 0.84 0.82 0.80 1.5

2.0

2.5

3.0

time T

In Fig. 4, the maximum signal located at about T= 1.1. Compared with the case shown in fig.3, the function is sheared towards the left, which means the maximum echo signal will come faster with a shorter τc. Actually, the parameter τ c reflects the randomness of the noise termΔ. From the definition ofΔ(t) given by expression (18), we can see that with a longer τ c , the impact noise function will fluctuate more rapidly. If τ c → ∞ , the function Δwill almost never change with time and remain the same as the initial condition Δ(0). i.e. it is refer to the case of uniform fluctuation shown in Fig.2. Comparing Fig.2, 3 and 4, one can show that the maxima of echo signal will be shifted towards the left whenτc. becomes smaller. When τ c → 0 , the echo signal will reach its maximum exactly as time T= τ. Consider the case with τ=2 and τc =1, the measured signal is shown as Fig.5. Figure 5 The graph of received echo signal Y against time T Echo signal Y

Compared with the case shown in fig.3, if the system irradiated with the second laser pulse at a later timeτ=2 , the echo signal reaches its maximum at T=2.66. So the maximum echo signal is received at 0.33 τ after the irradiation with the second laser pulse. In the caseτ=1, the maximum echo signal is received at 0.5τ after the irradiation with the second laser pulse. Therefore, with a longer time τbetween the first and second laser pulse, the echo signal will be also shifted towards the left. The reason behind is that when τ increases, the ratio of τc /τ decreases. So the effect of increasing τby twice is the same as decreasing τ c by half, as the system can be scale up with time without affecting the result. Therefore it is reasonable to see the maxima found at about T=2.66 in figure 5. Numerical Approach

To verify the result given by the expression (19), we can also used numerical method to solve the system and get the echo signal. One of the common methods to use is Monta Carlo Simulation. A program is written base on the above expressions using language Fortran 90. It generates simulated data points of Δ (t) following the definition (18). Figure 6 shows some examples of random function Δ (t) generated by the program. All the random value at each time step follows Gaussian distribution.

0.40

0.38

0.36

0.34

2.2

2.4

2.6

2.8

3.0 time T

Figure 6 The simulated impact noiseΔ(t), following Gaussian distribution (Fig 6a: τc =0.1, 0< t <10)

φ (T ,τ ) = ∫0τ ∆ j (t ' )dt ' − ∫τT ∆ j (t ' )dt ' By given value of parameters T, τ, τc, we can get φ easily, thus the echo signal can also be found as eiφ .

Δ(t)

Figure 7 shows some numerical plot of echo signal Y generated by the program. Figure 7 The simulated impact noiseΔ(t), following Gaussian distribution Time t (Fig 6b: τc =1, 0< t <10)

(Fig 7a: τ=1, τc =1, 1< T < 3) Y(T)

Δ(t)

Time T Time t (Fig 6c:τc =10, 0< t <10)

(Fig 7b: τ=1, τc =0.1, 1< T < 3)

Y(T)

Δ(t)

Time t From the figure above, one can show that for shorter τ c , the randomness of the functionΔ(t) increases. This is consistent with our discussion in analytical result analysis on previous page. With the function Δ(t), the program will construct a numerical integral to calculate φ by expression (14):

Time T Comparing figure 7a with figure 3 and comparing figure 7b with figure 4, we can see that the analytical and numerical results are consistent.

IV. Conclusions

To study quantum decoherence, we introduce photon echo experiment as an example to illustrate a method to calculate to dependence of background noise to a quantum system. Based on the Hamiltonian of the system and perturbation, as well as the initial conditions described by formula (1) and (2) above, we can use the Von Neumann equation given in equation (4), to solve the time-dependent density matrix up to third order term quantum mechanically. Next, we construct a model of impact noise, following Gaussian fluctuation. Then the echo signal impulse can be found as an analytical function. It is found that the echo signal not only depends on the measured time and period between two laser pulses, but also the parameter of impact noise. By Monta Carlo simulation, we find the numerical data generated by the program are consistent with what the theory predicts. Using the numerical data simulated by the program, we can estimate what accuracy we expect to get in the experiment with the real experimental data. The photon echo signal experiment can act as one of the example helping us to know more about the characteristic of fluctuation noise. It is necessary for studying quantum picture in noisy environment. No matter what kind of particles and properties we are considering, we can also use perturbation theory to find out how the background noise affecting the system. Precise measurements of background noise leads to the further study in quantum decoherence, it help us to develop new

theory quantum mechanical in higher precision. V. Acknowledgments

I gratefully acknowledge the support of Physics Department, the Chinese University of Hong Kong. I thank my advisor, Professor Liu Ren Bao, for defining the project and offering continuing advice. VI. References

[1] Roger F. Loring and Shaul Mukamel, Unified Theory of Photon Echoes: The Passage From Inhomogeneous to Homogeneous Line Broadening, in Chemical Physics Letter Vol 114, No. 4 (1985), p. 426-429 [2] R. Kubo, A Strochastic Theory of Line Shape, in Stochasic processes in Chemical Physics, QP1 A242 Vol.15 (2006), P.101126

PHY4610H Final Year Project Mid-Program Report ...

impact noise due to collisions. When random fluctuation of the environment is considered, a phase shift from the unperturbed system is expected to be found. One of the aims of the project is to calculate the analytical expression of the magnitude echo signal in photon echo experiment, depending on the time parameter T, τ ...

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