Master Equations Bongkokwei

8/13/2019 Master Equations Bongkokwei

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Derivations

Bong Kok WeiA0072102B

National University of Singapore

[email protected]

1 Master Equations

Beginning with the Hamiltionian in the general form,

H=HS+HR+HSR (1.1)

where HS and HR are Hamiltonians for the System and Reservoir, and HSR represents the inter-action Hamiltonian between the System and its Reservoir.

The Schrodinger equation for | with set to unity reads,

id

dt|= H| (1.2)

and the density operator of the system and reservoir is defined as

(t) =|(t)(t)| (1.3)

and differentiating it with respect to time, we obtain

(t) =|d

dt(t)|+

d

dt|

(t)| (1.4)

Substituting the time-derivative of the state vectors with 1.2, The Schrodinger equation for (t)reads

(t) =iH+iH

=i[H, ] (1.5)

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The density operator,(t) is defined as

(t) =ei(HS+HR)t(t)ei(HS+HR)t (1.6)

Substituting1.2 into the time-derivative of1.6

(t) =i[(HS+HR)(t) (t)(HS+HR)] +ei(HS+HR)t(t)ei(HS+HR)t

=i[(HS+HR)(t) (t)(HS+HR)]i{ei(HS+HR)t[H, ]ei(HS+HR)t}

=i[HS+ HR,]i[HS+HR,]i[HSR ,]

=i[HSR ,] (1.7)

Where

HSR(t) =ei(HS+HR)tHSR(t)ei(HS+HR)t (1.8)

By integrating1.7over time, t. The density operation is in the form

(t) = (0)i

t0

dt[HSR(t),(t)] (1.9)

and substituting1.9 into1.7

(t) =i[HSR(t),(0)] t

0

dt[HSR(t), [HSR(t),(t)]] (1.10)

Assuming that at initial time, t = 0, there is no interaction between the system and reservoir. Wecan factorise the density matrix, (0) into

(0) =0R0 (1.11)

where 0 and R0 are the density operator for the system and reservoir respectively at initial time.

To obtain the density operator for the system, we take the partial trace of over the reservoir

trR((t)) = (t) (1.12)

Therefore,

(t) =trR(i[HSR(t), 0R0])

t0

dttrR[HSR(t), [HSR(t),(t)]] (1.13)

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The first term can be eliminated by assuming that trR(HSRR0) = 0 and replacing (t) with (t)R0

ignoring terms with higher order ofHSR (due to weak coupling between the system and reservoir,Born-approximation), gives

(t) = t

0

dttrR[HSR(t), [HSR(t),(t)R0]] (1.14)

Note thatt in (t)R0 disappeared due to Markoff approximation where information of the currentstate is sufficient to predict the evolution of the density operator.

Equation1.14is the general form of the Master equation, if we want a more specific form, we canwrite HSR explicitly,

HSR(t) =i

si(t)i(t) (1.15)

The master equation is now

(t) =i,j

t0

dttrR[si(t)i(t), [sj(t)j(t

),(t)R0]] (1.16)

(t) =i,j

t0

dt trR(si(t)i(t)sj(t)j(t

)(t)R0)trR(si(t)i(t)(t)R0sj(t)j(t

))

trR(sj(t)j(t

)(t)R0si(t)i(t)) +trR((t)R0sj(t)j(t

)si(t)i(t))

=i,j

t0 dt

si(t)sj(t

)(t)i(t)

j(t

)+ si(t)(t)sj(t

)i(t)

j(t

)+

sj(t))(t)si(t)j(t

)i(t)+ (t)sj(t)si(t)j(t

)i(t)

=i,j

t0

dt {si(t)sj(t)(t)sj(t

))(t)si(t)}i(t)j(t)+

{(t)sj(t)si(t)si(t)(t)sj(t

)}j(t)i(t) (1.17)

The equations below are called correlation functions

trR(i(t)j(t)R0) =i(t)j(t

)

trR(ji(t)(t)R0) =j(t

)i(t)

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2 Master equation for a cavity mode driven by thermal light

We will derive the mater equation for a single mode of the optical cavity. Starting with the Hamil-tionians,

HS= Caa (2.1)

HR =j

jrjrj (2.2)

HSR =j

(jarj+ ja

rj) = (a +a) (2.3)

and the operators in the interaction picture are

s1(t) =eiCa

ataeiCaat =aeiCt (2.4)

s2(t) =eiCa

ataeiCaat =aeiCt (2.5)

The final form of the equation3.6and3.7 can be derived from,

id

dta(t) = [a(t), H] (2.6)

which is the Schrodinger Equation in the interaction picture, known as the SchwingerTomonaga

equation. We can replace the Hamiltonian in equation 2.6,

id

dta(t) =C[a(t), a

a] (2.7)

Making use of the commutator relation where [a, aa] = aand then solving the initial value problem,

id

dta(t) =Ca (2.8)

which leads to

a(t) =aeiCt (2.9)

Using the same technique, we can obtain the rest of the operators in interaction picture.

1(t) =(t) =

j

jrje

ijt (2.10)

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2(t) =(t) =j

jrjeijt (2.11)

By subsituting the operators in the interaction picture into equation 1.17, the master equation forthe system can be written in

(t) =

t0

dt(aa(t)a(t)a)eiC(t+t)(t)(t)+ ((t)aaa(t)a)eiC(t+t

)(t)(t)+

(aa(t)a(t)a)eiC(tt)(t)(t)+ ((t)aaa(t)a)eiC(tt

)(t)(t)+

(aa(t)a(t)a)eiC(tt)(t)(t)+ ((t)aa a(t)a)eiC(tt

)(t)(t)+

(aa(t)a(t)a)eiC(t+t)(t)(t)+ ((t)aa a(t)a)eiC(t+t

)(t)(t)+

(2.12)

We can see from equation2.12that we can further evaluate the correlation functions

(t)(t)= 0 (2.13)

(t)(t)= 0 (2.14)

We can show that the two correlation function equal to zero by taking the trace of the reservoiroperator

(t)(t)= T r((t)(t)R0) (2.15)

where R0 is the initial density operator of the reservoir.

R0 =j

ejrjrj/kT/(1ej/kT) (2.16)

By inserting identity into the trace, we can obtain an expression which can be easily evaluated.

T r( |nn| (t)(t)R0) =n|R0(t)(t)|n (2.17)where|n= |n1n2n3 . . . nNand N is theN-th dimension harmonic oscillator. From the equationabove, we can spot that the mean will be zero as the creation operators will act on the numberstate twice. Since the number states form an orthonormal basis, equation 2.17will be zero.

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Since the integration is dominated by times that are much shorter than the time scale for theevolution of , we can replace (t) with (t)

(t) =(aa aa) +(aa +aaaa aa) +h.c (2.26)

with

=

t0

d

0

d ei(C)g()|()|2 (2.27)

=

t0

d

0

d ei(C)g()|()|2n(, T) (2.28)

After integration,

= g(C)|(C)|2 +i (2.29)

= g(C)|(C)|2n(C) +i

(2.30)

With

=P

0

dg()|()|2

C(2.31)

=P 0

dg()|()|2

Cn(, T) (2.32)

with P being the Cauchy Principal value. The master equation is,

= i[aa,] +(2aa aaaa) + 2n(aa +aaaa aa) (2.33)

3 Two-state atoms

We consider an atom which is radiatively damped by its interaction with the many modes of theradiation field in thermal equilibrium. This field acts as a reservoir of harmonic oscillators. TheHamiltonian is given by

HS=1

2Az (3.1)

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HR =k,

krk,rk, (3.2)

HSR = k,

(k,rk,++k,rk,) (3.3)

We can immediately spot the correspondence between the Hamiltonians of a two level system andsingle mode cavity.

s1= , s2= + (3.4)

1= =

k,

k,r, 2 = =

k,

k,r (3.5)

In the interaction picture, the operators become,

s1(t) =eiA+te

iA+t =eiAt (3.6)

s2(t) =eiA+t+e

iA+t =+eiAt (3.7)

and

1(t) =(t) =

k,

k,rk,e

ikt (3.8)

2(t) =(t) =k,

k,rk,eikt (3.9)

From equation2.26

(t) =(++) +(++++ +) +h.c (3.10)

With

=

2+i (3.11)

=

2n(A, T) +i

(3.12)

And

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= 2

d3k g(k)|(k, )|2(kC A) (3.13)

= d3k

g(k)|(k, )|2

AkC

(3.14)

=

d3k

g(k)|(k, )|2

AkCn(A, T) (3.15)

(t) =(

2+i)(++) + (

2i)(+ +)+

(

2n+i)(++++ +)+

(

2ni)(+++ ++)

= 2

(n+ 1)(2+++)i( + )[+,]+

2n(2++ +) +i

[+,]

=i1

2(2 + )[z ,] +

2(n+ 1)(+++)+

2n(2++ +) (3.16)

transforming back to the Schrodinger picture, we obtain we obtain the master equation for a two-state atom in thermal equilibrium the master equation for a two-state atom in thermal equilibrium

= i1

2A[z,]+

2(n+ 1)(+++)

+

2n(2++ +) (3.17)

With

A= A+ 2 + (3.18)

To transform back to Schrodinger picture, we begin with the following,

= eiHSteiHSt (3.19)

Taking the time derivative, we obtain,

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= i(HSHS) +eiHSteiHSt

=i[HS, ] +eiHSteiHSt (3.20)

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