density-matrix formalism in quantum opticssop.uohyd.ac.in/~sdg/resources/fam2.pdf · 2018-03-29 ·...

1

DENSITY OPERATOR AND APPLICATIONS IN NONLINEAR

AND QUANTUM OPTICS

Fam Le Kien

Department of Applied Physics and Chemistry,

University of Electro-Communications, Chofu, Tokyo 182-8585, Japan

Lecture notes for the fall semester of 2008

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Contents

I. GENERAL THEORY OF THE DENSITY OPERATOR

1. Concept of a statistical mixture of states

2. Pure states. Introduction of the density operator

3. Statistical mixtures of states

4. Applications of the density operator

II. TWO-LEVEL ATOMS INTERACTING WITH A LIGHT FIELD

1. Atom+field interaction Hamiltonian

2. Optical Bloch equations

3. Absorption spectrum: saturation and power broadening

4. Field propagation equations

5. Susceptibility, refractive index, and absorption coefficient

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I. GENERAL THEORY OF THE DENSITY OPERATOR

1. Concept of a statistial mixture of states

Consider a quantum system. If we have complete information about the system, the

state of the system can be described by a state vector (wave function), denoted by | .

Such a state is called a pure state.

It often happens that we don’t know the exact state vector of the system. In this case,

we say that we have incomplete information about the system. For example, a photon

emitted by a source of natural light can have any polarization state with equal probability.

Similarly, a system in thermal equilibrium at a temperature T has a probability

proportional to /nE kT

e

of being in a state of energy nE .

More generally, the incomplete information usually presents itself in the following

way: the state of this system may be either the state 1| with a probability 1p or the

state 2| with a probability 2p , etc… Obviously, we have

1 2 ... 1.n

n

p p p (1)

We then say that we have a statistical mixture of states 1| , 2| , ... with probabilities

1p , 2p , … A statistical mixture of two or more than two different states is called a

mixed state. In general, a quantum state can be either a pure state or a mixed state.

We emphasize that a system described by a statistical mixture of states must not be

confused with a system whose state | is a linear superposition of states:

| | .n n

n

C (2)

For a linear superposition of states, there exist, in general, interference effects between

these component states. Such effects are due to the cross terms of the type 'n nC C,

obtained when the modulus of the wave function is squared. Meanwhile, for a statistical

mixture of states | n , we can never obtain interference terms between these component

states.

We illustrate the above statements by considering a single particle in the coordinate

space R of the position vector r .

If the particle is in a linear superposition state ( ) ( ),n n

n

C r r the probability of

finding the particle at a given point r is

2 2 * *

' '

'

( ) | ( ) | | ( ) | ( ) ( ).n n n n n n

n nn

P C C C r r r r r (3)

4

The above expression involves cross terms of the type 'n nC C , which describe interference

between the component states ( )n r .

However, if the particle is in a statistical mixture of states | n with weight factors

kp , the probability of finding the particle at a given point r is

2( ) | ( ) | .n n

n

P p r r (4)

The above expression shows no interference between the component states ( )n r .

2. Pure states. Introduction of the density operator

We first examine the simple case where the state vector of the system is perfectly

known, that is, all the probabilities np are zero, except one. The system is then said to be

in a pure state.

a) Description by a state vector (wave function)

i. State space

A pure quantum state of any physical system is characterized by a state vector,

belonging to a linear space S which is called the state space of the system.

ii. Ket vectors

An element, or vector, of the state space S is called a ket vector. It is represented by

the symbol .

iii. Scalar product

With each pair of ket vectors and ' , taken in this order, we associate a

complex number, which is denoted by ( , ' ) and satisfies the various properties

*

1 1 2 2 1 1 2 2

* *

1 1 2 2 1 1 2 2

( , ' ( ' , ,

( , ( , ( ,

( , ( , ( ,

(5)

This complex number is called the scalar or inner product. The linear space with this type

of scalar products is called a Hilbert space.

iv. Bra vectors and dual space

Each ket vector is associated with a bra vector, denoted by the symbol | ,

which associates every ket vector ' with a complex number which is denoted by

' and is equal to ( , ' . The set of bra vectors constitutes a vector space called

the dual space of the state space S.

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v. Basic properties of the ket and bra vectors

*

* *

1 1 2 2 1 1 2 2

(1) | ' ' | .

(2) If | | | then | | | .

(6)

More generally, for the superposition state vector | |n n

n

C u , we have

*| | .n n

n

u C (7)

In the case of a single particle in the coordinate space R of the position vector r , the

ket vector and the bra vector are represented by the complex functions ( ) r and

*( ) r , respectively, and the scalar product ' is defined as *' ( ) '( )d r r r .

It is clear that ' ' .

vi. Orthonormal basis

The state of a pure-state system can be described by a state vector | . The set of

state vectors | nu is a basis set if an arbitrary state vector | can be decomposed as

| | .n n

n

C u (8)

The set of basis states | nu is an orthonormal basis set if

.n m nmu u (9)

The coefficients nC in Eq. (8) are called the probability amplitudes of the superposition

state | in the basis {| }nu . More exactly, we say that nC is the probability amplitude

of the basis state | nu in the superposition state | .

It follows from the expansion (8) and the orthonormality (9) of the basis states | nu

that the coefficients nC of the state vector | can be determined as

|n nC u (10)

We say that the state vector | is normalized if | 1 . If the state vector | is

normalized, its coefficients satisfy the normalization condition

2| | 1.n

n

C (11)

From now we consider only normalized state vectors.

Note that the bra vector | associated with the ket vector | can be decomposed in

the form

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1 1 2 2 1 1 2 2( | | ) | | .O O O (20)

Similarly, the linear operator O associates each bra vector | with a bra vector

' | |O and satisfies the linearity

1 1 2 2 1 1 2 2( | |) | | .O O O (21)

For the superposition state vector | |n n

n

C u , we have

* * *

| | | | ,

| | | | .

n n n n n n

n n n

n n n n n n

n n n

O O C u C O u C O u

O C u O C u O C u O

(22)

We consider the operator

| |,O A B (23)

whose action on ket and bra vectors is defined as

| | | | | | | | ,

| | | | | | | | .

O A B A B A B

O A B A B A B

(24)

We can show that

| | ' | | ' | | ' | | ' .O O A B O (25)

x. Matrix form of a linear operator

With the help of the completeness relation (18), we can represent an arbitrary

operator O in the form

| | | |

| | ( | ) | | ( | ) | |

| ( | ) | | ( | ) | | |.

n n m m

n m

n n m m n n m m

nm nm

n m n m n m n m

nm nm

O u u O u u

u u O u u u u O u u

u O u u u u O u u u

(26)

Hence, we find

| ( | ) ( | ) | | | .n m n m n m nmu O u u O u u O u O (27)

The matrix | |nm n mO u O u is called the matrix form of the linear operator O . Every

operator O can be represented by a matrix nmO .

Every matrix nmO represents an operator O . Indeed, it follows from Eqs. (26) and

(27) that

8

| | .nm n m

nm

O O u u (28)

We can generalize Eq. (27) to obtain

1 2 1 2 1 2| ( | ) ( | ) | | | .O O O (29)

xi. Hermitian conjugate

Every operator O is associated with an operator †O , whose matrix is

† * * † *( ) ( ) ( ) | | | | .T

nm mn nm n m m nO O O u O u u O u (30)

Here the upper label T means the matrix transpose and the symbol * means the complex

conjugate. The operator †O is called the Hermitian conjugate of the operator O .

For | |nm n m

nm

O O u u , we have

† †

*

*

| |

| |

| |.

nm n m

nm

mn n m

nm

nm m n

nm

O O u u

O u u

O u u

(31)

Note that †

†O O .

We can show that

†| |,O O (32)

where † |O is the bra vector associated with the ket vector † †| |O O . Indeed,

since | |nm n m

nm

O O u u , we have

| | | | | |.nm n m nm n m

nm nm

O O u u O u u (33)

On the other hand, since † * | |nm m n

nm

O O u u , we find

† † *

*

*

| | | | |

| |

| | ,

nm m n

nm

nm m n

nm

nm n m

nm

O O O u u

O u u

O u u

(34)

which yields

† | | |.nm n m

nm

O O u u (35)

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Comparison of Eq. (35) with Eq. (33) gives Eq. (32). Thus, |O is a bra vector that is

associated with the ket vector † |O . In the same way, we can say that the ket vector

|O is associated with the bra vector †|O .

Several useful properties of the Hermitian conjugates:

Property 1:

†

If | |

then | | .

O

O

(36)

Property 2:

†

If | |

then | | .

O

O

(37)

Property 3:

* †| | ' ' | | .O O (38)

Property 4:

1 2

† * † † †

1 1

If

then ,

n

n n

O O O O

O O O O

(39)

where is a complex number and 1O , 2O , …, nO are arbitrary operators.

Property 5:

† * †

If ( )

then ( ),

O f A

O f A

(40)

where f is a real function, A is an arbitrary operator, and is a complex number.

xii. Hermitian operators

An operator O is called Hermitian if it is equal to its Hermitian conjugate †O , that is,

† * or .mn nmO O O O (41)

xiii. Mean values of observables

Every observable is described by a Hermitian operator. Consider an observable

described by a Hermitian operator A . The matrix elements of the operator A in the basis

{| }nu are given by

| | | | ( | ) | .nm n m n m n mA u A u u A u u A u (42)

Since A is a Hermitian operator, we have

* .nm mnA A (43)

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The mean value of A is

*

*

*

| |

| |

| |

.

n n m m

n m

n m n m

nm

n m nm

nm

A A

C u A C u

C C u A u

C C A

(44)

xiv. Schrödinger equation and Schrödinger picture of quantum mechanics

The time evolution of the state vector | ( )t of the quantum system is described by

the Schrödinger equation

| ( ) | ( ) ,d

i t H tdt

(45)

where H is the Hamiltonian operator of the system. The Hamiltonian is the energy, an

observable. Therefore, it must be a Hermitian operator, that is, †H H .

The solution to the Schrödinger equation (45) is given by

| ( ) exp( / ) | (0) .t iHt (46)

The corresponding bra vector is

( ) | (0) | exp( / ).t iHt (47)

The time-dependent mean value of the observable A is

( ) ( ) | | ( ) .A t t A t (48)

Consider the initial state vector | (0) (0) |n n

n

C u . The probability amplitude

(0)nC is given by (0) | (0)n nC u For an arbitrary time t , we have

| ( ) ( ) | ,n n

n

t C t u (49)

where

( ) | ( )n nC t u t (50)

Inserting Eq. (49)into Eq. (45) gives

( ) | ( ) |

( ) | | |

( ) | | | .

n n m m

n m

m n n m

mn

m n m n

mn

di C t u C t H u

dt

C t u u H u

C t u H u u

(51)

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Hence, we obtain

( ) ( ).n nm m

m

di C t H C t

dt (52)

Here | |nm n mH u H u is the matrix of the Hamiltonian.

In terms of the probability amplitudes ( )nC t , the mean value of the observable A at

the time t is given by

*( ) ( ) | | ( ) ( ) ( ) .n m nm

nm

A t t A t C t C t A (53)

xv. Example of a two-level system

Figure 1: A two-level system.

The simplest quantum system is a two-level system, see Fig. 1. Such a system has two

stationary energy levels, denoted by a aE (upper level) and b bE (lower level).

The wave functions of these two levels are denoted by the state vectors | a (upper state)

and | b (lower state). These state vectors are normalized in modulus to unity and are

orthogonal to each other:

| | 1,

| | 0.

a a b b

a b b a

(54)

The completeness of the basis requires

| | | | 1.a a b b (55)

In general, any state vector (wave function) of the two-level system can be written in the

form

| | | .a bC a C b (56)

Here aC and bC are the probability amplitudes of finding the system in states | a and

| b , respectively. They satisfy the normalization condition

2 2| | | | 1.a bC C (57)

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The bra vector | associated with the ket vector | can be written in the form

* *| | | .a ba C b C (58)

The mean value of an observable A is

* *

2 2 * *

2 2 * *

| |

| | | |

| | | | | | | | | | | |

| | | | .

a b a b

a b a b b a

a aa b bb a b ab b a ba

A A

a C b C A C a C b

C a A a C b A b C C a A b C C b A a

C A C A C C A C C A

(59)

The total Hamiltonian is

0 ,IH H H (60)

where 0H and IH are the free (unperturbed) and interaction parts, respectively. The

upper state | a and the lower state | b are the eigenstates of the free Hamiltonian 0H ,

that is, 0 | |jH j E j , where ,j a b . We can represent 0H in the form

0 | | | | .a bH a a b b (61)

Why does 0H have the above form? The reason is the following: since

| | | | | | 1j

j j a a b b and 0 | |jH j E j , we can expand 0H as

0 0,

0,

,, ,

(| |) (| |)

| | ( | | )

| | ( | ) | |

| | | |.

i j

i j

j j i ji j i j

i ii i

H i i H j j

i j i H j

i j i j E E i j

E i i i i

(62)

We consider a particular case where the two-level system is free, i.e., 0IH . In this

case, the Schrödinger equation is

0| |

( | | | |) | .a b

di H

dt

a a b b

(63)

It gives

| | ( | | | |) | |

| | | | .

a b a b a b

a b a a b b

di C a C b a a b b C a C b

dt

iC a iC b C a C b

(64)

Hence, we find

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,

.

a a a

b b b

dC i C

dt

dC i C

dt

(65)

The solutions to the above equations are

( ) (0),

( ) (0).

a

b

i t

a a

i t

b b

C t e C

C t e C

(66)

Consequently, the explicit time-dependent expression for the state vector (wave function)

of the free two-level system is

| ( ) ( ) | ( ) | (0) | (0) | .a bi t i t

a b a bt C t a C t b e C a e C b

(67)

Note that the probabilities of finding the free two-level system in the upper and lower

levels are 2 2| ( ) | | (0) |a aC t C and 2 2| ( ) | | (0) |b bC t C , respectively. They are independent

of time. In the contrary, the interference described by the cross term ( )* *( ) ( ) (0) (0) a bi t

a b a bC t C t C C e

oscillates in time.

The mean value of an observable A at an arbitrary time t is

*

( )*

2 2

* *

( ) ( ) | | ( ) ( ) ( )

(0) (0)

| (0) | | (0) |

(0) (0) (0) (0) .

n m

ab ab

n m nm

nm

i t

n m nm

nm

a aa b bb

i t i t

a b ab b a ba

A t t A t C t C t A

C C e A

C A C A

C C e A C C e A

(68)

Here ab a b is the transition frequency.

If the initial state is | (0) | ,a then (0) 1aC and (0) 0bC . In this case, the state

vector at an arbitrary time is | ( ) |ai tt e a

and the mean value of an observable A at

an arbitrary time is ( ) | | .aaA t A a A a The mean value is independent of time

because the initial state | a is an eigenstate of the system.

If the initial state is | (0) | ,b then (0) 1bC and (0) 0aC . In this case, the state

vector at an arbitrary time is | ( ) |bi tt e b

and the mean value of an observable A at

an arbitrary time is ( ) | | .bbA t A b A b The mean value is independent of time

because the initial state | b is an eigenstate of the system.

If the initial state is a superposition state of | a and | b , with (0) 0aC and

(0) 0,bC then ( )A t in general varies in t.

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Exercises

Exercise 1: Consider a two-level system. Suppose that is an operator such that

| |a b and | 0b . Show that can be written in the form | | .b a

Exercise 2: By definition † *( )kl lkO O , where O is an arbitrary operator and †O is its

Hermitian conjugate. Suppose | |n mO u u . Show that † | |m nO u u .

Exercise 3: Suppose | |O A B , where | A and | B are two arbitrary vectors. The

action of | |O A B on an arbitrary vector Q is defined as

| | | | | | | |O Q A B Q A B Q A B Q . Show that † | |O B A .

Exercise 4: By definition † T*O O . Suppose 1 2O O O . Show that † † †

2 1O O O .

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where |k ka u A and |l lb u B .

For | |X B A , we have

*.kl k lX b a

We see that

* † .kl lk klX O O

Hence, we obtain †O X , that is,

† | | .O B A

Exercise 4:

For 1 2O O O , we have

† † * * * * † †

1 2 1 2 2 1 2 1 2 1( ) ( ) ( ) .T T T T TO O O O O O O O O O O

Generalization:

For 1 2 nO O O O , where is a complex number and 1O , 2O , …, nO are

arbitrary operators. Using the result of the above exercise, we can easily show that

† * † † †

1 1 .n nO O O O

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b) Description by a density matrix

The relation

*| | n m nm

nm

A A C C A (69)

shows that the coefficients nC enter into the mean values through quadratic expressions

of the type *

n mC C . These quadratic expressions are simply the matrix elements of the

operator | | , which is the projector onto the ket vector | . Indeed, since

*

| | ,

| | ,

n n

n

n n

n

C u

u C

(70)

we have

*

|

|

n n

n n

C u

C u

(71)

Hence, we find

* | | .n m m nC C u u (72)

It is therefore natural to introduce the density operator

| | . (73)

The density operator is represented in the {| }nu basis by a matrix called the density

matrix whose elements are

*

| |

| |

.

mn m n

m n

n m

u u

u u

C C

(74)

The mean value of the observable A is then given by

*| | Tr{ } Tr{ }.n m nm mn nm

nm nm

A A C C A A A A (75)

Mathematically, the operator | | is the projector onto the ket vector | . Indeed,

with the use of the notation

| | |,P (76)

we see that the action of |P on an arbitrary vector |V gives a vector aligned along | ,

namely,

| | (| |) |

| ( |

P V V

V

(77)

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In addition, we have

2

| | |

|

(| |)(| |)

| ( | ) |

| |

.

P P P

P

(78)

The above properties indicate that | | |P is a projection operator. Since

2

| | ,P P we find that, in the case of pure states, we have 2 and hence,

2Tr Tr 1 [see Eq. (80)].

Note that the matrix mn is a Hermitian matrix. Indeed, we have

* *

† .

mn m n nmC C

(79)

The specification of suffices to characterize the quantum state of the system. In

other words, the density operator enables us to obtain all the physical predictions that

can be calculated from | . To show this, we express in terms of the operator the

conservation law of probability, the mean value of an observable, and the time evolution

of a quantum state.

i. Conservation law of probability

We find from the conservation law of probability that

2Tr | | 1.nn n

n n

C (80)

Thus, the sum of the diagonal elements of the density matrix is equal to 1.

ii. Mean value of an observable

For the mean value of an observable, we have the formula

*| |

Tr{ } Tr{ }.

n m nm mn nm

nm nm

A A C C A A

A A

(81)

iii. Schrödinger equation

We derive the time evolution equation for the density operator . As known, the

Schrödinger equation is

1

| | .d

Hdt i

(82)

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The Hermitian conjugate form of the above equation reads

†1 1| | | .

dH H

dt i i

(83)

Hence, we have

| |

| | | |

1 1| | | |

1

1[ , ].

d d

dt dt

d d

dt dt

H Hi i

H Hi

Hi

(84)

Here

1 2 1 2 2 1[ , ]O O O O O O (85)

is the commutator between the operators 1O and 2O . Thus, the time evolution of the

density operator is governed by the equation

[ , ].d

i Hdt (86)

This equation is called the generalized Schrödinger equation.

iv. Summary of the properties of the density operator of a pure state

The properties of the density operator in the case of pure states are

† ,

Tr 1,

Tr { } Tr { },

[ , ].

A A A

di H

dt

(87)

The above properties are general, i.e., they are true also in the case of mixed states.

In the case of pure states, there are two specific properties:

2

2

,

Tr 1.

(88)

These properties can be used as criteria to find out if a state is a pure state or not.

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v. Advantages of the description in terms of the density operator

A pure state can be described by a density operator as well as by a state vector (wave

function). Both descriptions are equivalent. However, the density operator presents a

number of advantages:

First of all, two state vectors | and |ie describe the same quantum

state. In other words, there exists an arbitrary global phase factor for the state

vector. However, the state vectors | and |ie correspond to the same

density operator

| |

| | .i ie e

(89)

Thus, the use of the density operator eliminates the global phase.

Second, the equation

Tr { } Tr { }A A A (90)

is linear with respect to the density operator .

However, the equation

| |A A (91)

is quadratic with respect to | .

vi. Example: the density operator of a pure state of a two-level system

We consider a two-level system in a pure state described by the state vector (wave

function)

| | | .a bC a C b (92)

The density operator of the system is

* *| | ( | | )( | |).a b a bC a C b C a C b (93)

Taking the matrix elements, we get

2

*

*

2

| | ,

,

,

| | .

aa a

ab a b

ba ab

bb b

C

C C

C

(94)

The matrix form of the density operator is

2 *

* 2

| |.

| |

aa ab a a b

ba bb b a b

C C C

C C C

(95)

We see that *

nm mn , that is, † . In addition, we have

21

2 2Tr | | | | 1.aa bb a bC C (96)

The average value of an observable A is given by

Tr{ } Tr{ }

.

mn nm

nm

aa aa bb bb ab ba ba ab

A A A

A

A A A A

(97)

It is obvious that the diagonal matrix elements aa and bb are the probabilities of

being in the upper and lower states, respectively. The off-diagonal matrix elements ab

and ba , where *

ab ba , describe the interference between the upper level | a and the

lower level | b in the state . To illustrate more clearly the physical meaning of the off-

diagonal elements, we use the fact that the atomic polarization (average dipole moment)

of a single two-level atom is given by

2 2 * *

* *

*

| |

(| | | | )

( )

,

a aa b bb a b ab b a ba

a b ab b a ba

ba x ab x

P e x e x

e C x C x C C x C C x

e C C x C C x

d d

(98)

where e is the electric charge of the electron and

x abd ex (99)

is the dipole matrix element. Here we have used the property 0aa bbx x , which is a

consequence of the fact that x is an odd function while 2| ( ) |n x (with ,n a b ) is, in the

case of atoms, an even function. The probability density 2| ( ) |n x is an even function

because so is the Coulomb potential in the atom. Thus, in the case of two-level atoms, the

off-diagonal elements ab and ba determine the atomic polarization.

As known, for a free two-level system, the Hamiltonian is

0 | | | |a bH a a b b (100)

and the time evolution of the probability amplitudes is governed by the equations

,

.

a a a

b b b

dC i C

dt

dC i C

dt

(101)

When we use the relation * ,C C we can easily show that the evolution equations

for the density-operator matrix elements of the free two-level system are

22

0,

0,

( ) .

aa

bb

ab a b ab

d

dt

d

dt

di

dt

(102)

We can also derive the above equations directly from the generalized Schrödinger

equation (86). The solutions to Eqs. (102) are

( )

( ) constant,

( ) constant,

( ) (0).a b

aa

bb

i t

ab ab

t

t

t e

(103)

As seen, when the two-level system is free, the diagonal matrix elements of the density

operator are constant. However, the off-diagonal matrix elements oscillate in time.

3. Statistical mixtures of states

a) Definition of the density operator

We now return to the general case where the system is in a mixture of states

1 2| ,| ,...,| ,...k , with probabilities 1 2, ,..., ,...kp p p , where

0 1,

1.

k

k

k

p

p

(104)

The state vectors | k of the components are normalized, that is, | 1.k k However,

they are not necessarily orthogonal to each other.

The density operator of the statistical mixture of states is defined as

,k k

k

p ρ (105)

where k is the density operator corresponding to the state | k , i.e.,

| | .k k k (106)

b) Properties of the density operator

i. Hermitivity

Since the coefficients kp are real, is obviously a Hermitian operator like each of

the k :

23

† * † .k k k k

k k

p p (107)

ii. Probability conservation law

The conservation law of probability gives

Tr Tr

1.

k k

k

k

k

p

p

(108)

iii. Mean values

The expression for the mean value of an observable is

Tr{ }

=Tr{ }

=Tr{ }.

k k

k

k k

k

k k

k

A p A

p A

A p

A

(109)

The averaging by means of the density operator, expressed by Eq. (109), has a twofold

nature. It comprises both the averaging due to the probabilistic nature of quantum

mechanics (even when the information about the object is complete) and the statistical

averaging necessitated by the incompleteness of our information. For a pure state, only

the first averaging remains. For a mixed state, both types of averaging are always present.

iv. Schrödinger equation for the density operator

We now derive an equation for the time evolution of the density operator. We assume

that the Hamiltonian H of the system is known. The state | k satisfies the Schrödinger

equation

| | .k k

di H

dt (110)

Therefore, k obeys the evolution equation

[ , ].k k

di H

dt (111)

Using the linearity of Eqs. (105) and (111), we find the following evolution equation for

the density operator :

24

[ , ] [ , ]

[ , ].

k k k k

k k

k k k k

k k

d d di i p p i

dt dt dt

p H H p

H

(112)

v. Positivity

We see from the definition k k

k

p ρ that, for any ket | u , we have

2

| | | |

| | |

k k

k

k k

k

u u p u u

p u

(113)

and, consequently,

| | 0.u u (114)

Equation (114) means that the density operator is a positive operator.

vi. Summary of the properties of the density operator of a general quantum state

Thus, for pure states as well as statistical mixtures of states, we have

† ,

Tr 1,

Tr { } Tr { },

[ , ].

A A A

di H

dt

(115)

vii. Differences between the density operator of a mixed quantum state and that of a pure state

The equations

2

2

,

Tr 1

(116)

are true for pure states but not true for mixed states. For mixed states, since is no

longer a projector, we have

2 (117)

and, consequently, 2Tr 1. For a mixed state, according to comments (i) and (iii) at

the end of this section (see pages 26 and 27), we have

2Tr 1. (118)

Thus, we have, in general, for pure and mixed states,

25

2Tr 1. (119)

The above equation can be used as a criterion to find out if a state is a pure state 2(Tr 1) or a mixed state ( 2Tr 1 ).

c) Populations and coherences

What is the physical meaning of the matrix elements of the density operator in the

basis {| }nu ?

First, we consider the diagonal element nn . We have

2

| |

| |

| |

| | | .

nn n n

k n k n

k

k n k k n

k

k n k

k

u u

p u u

p u u

p u

(120)

We express the state | k in terms of the basis {| }nu as

( )| | .k

k n n

n

c u (121)

This gives

( ) | .k

n n kc u (122)

Then Eq. (120) becomes

( ) 2| | .k

nn k n

k

p c (123)

It follows from Eq. (123) that nn is a positive real number. The squared modulus ( ) 2| |k

nc

is the probability of | nu in the pure state | k . Therefore, Eq. (123) means that nn is

the probability of | nu in the state . For this reason, the diagonal matrix element nn is

called the population of the state | nu . If the same measurement is carried out N times

under the same initial condition, where N is a large number, nnN systems will be found

in the state | nu .

Now, we consider the non-diagonal element nm . A calculation analogous to the

preceding one gives the following expression for nm :

26

( ) ( )*

| |

| |

| |

.

nm n m

k n k m

k

k n k k m

k

k k

k n m

k

u u

p u u

p u u

p c c

(124)

The term ( ) ( )*k k

n mc c is a cross term. It expresses the interference effects between the states

| nu and | mu that can appear when the state | k is a coherent linear superposition of

these states. According to Eq. (124), nm is the average of these cross terms, taken over

all the possible states of the statistical mixture. If nm is zero, this means that the

statistical average has canceled out any interference effects between | nu and | mu . On

the other hand, if nm is different from zero, a certain coherence subsists between these

states. This is why the non-diagonal elements of are often called coherences.

COMMENTS:

(i) The distinction between populations and coherences obviously depends on the

choice of the basis {| }nu . Since is Hermitian, it is always possible to find an

orthonormal basis {| }n where is diagonal. In this basis, can be written as

| | .l l l

l

(125)

The density operator can thus be considered to describe a statistical mixture of the

orthonormal states | n with the probabilities l . There are no coherences between the

states | n in this mixture. We have

2 2Tr

1.

l

l

l

l

(126)

When one of the coefficients l is equal to 1, all the other coefficients must be zero. In

this case, the state ρ is a pure state and 2Tr 1 . When all the coefficients l are

smaller than 1, the state ρ is a mixed state and 2Tr 1 .

(ii) Assume that the Hamiltonian H is time-independent. If | nu are eigenvectors of

H, then

| | ,

| | .

n n n

n n n

H u E u

u H u E

(127)

Hence, we obtain from Eq. (112)

27

| |

|

|

n m n m

n n m m

n m n m

di u u u H H u

dt

u E E u

E E u u

(128)

that is,

( ) .nm n m nm

di E E

dt (129)

The solution to the above equation is

( )

( ) (0).n m

iE E t

nm nmt e

(130)

In particular, we have ( ) constantnn t .

Thus, the populations are constant, and the coherences oscillate at the Bohr

frequencies of the system.

(iii) We can easily prove that

2| | .nn mm nm (131)

Indeed, according to Eqs. (123) and (124) and to the generalized Cauchy-Scharz-

Buniakowsky inequality, we have

( ) 2 ( ) 2

2

( ) ( )

2

( ) ( ) 2

| | | |

| |

| | .

k k

nn mm k n k m

k k

k k

k n m

k

k k

k n m nm

k

p c p c

p c c

p c c

(132)

It follows from Eq. (131) that can have coherences only between states whose

populations are not zero.

It also follows from Eq. (131) that

2 2Tr | |

1,

nm

nm

nn mm

nm

nn mm

n m

(133)

see Eq. (119).

28

Exercises

Exercise 1: Consider a two-level system with density operator

14

34

01 3| | | | .

04 4a a b b

Show that the state of the system is a mixed state.

Exercise 2: Consider a two-level system with density operator

31

4 4

3 34 4

1 3 3 3| | | | | | | | .

4 4 4 4a a b b a b b a

Show that the state is a pure state.

Exercise 3: Consider two orthonormal vectors | a and | b . Consider a quantum

system (ensemble) 1Q that is prepared in the state | a with probability 1/4 and in the

state | b with probability 3/4.

Suppose we define

1 3| 0 | |

4 4

1 3|1 | |

4 4

a b

a b

Consider a quantum system (ensemble) 2Q that is prepared in the state | 0 with

probability 1/2 and in the state |1 with probability 1/2. Show that the quantum systems

1Q and 2Q correspond to the same density operator.

Exercise 4: Consider two arbitrary vectors | A and | B Show that

Tr | | |A B B A .

29

Solutions

Exercise 1:

The square of the density operator of the system is

1

162

916

01 9| | | | .

016 16a a b b

We find 2Tr 1/16 9 /16 10 /16 5 / 8 1 . This inequality indicates that the state of

the system is a mixed state.

Exercise 2:

The square of the density operator of the system is

2

314 4

3 34 4

1 3 3 3| | | | | | | |

4 4 4 4

.

a a b b a b b a

Hence we find 2Tr Tr 1 . These equalities indicate that the state of the system is a

pure state.

Exercise 3:

The density operator of the system 1Q is

1

1 3| | | | .

4 4a a b b

The density operator of the system 2Q is

2

1 1| 0 | |1 | .

2 2

We can easily show that

2

1

1 1| 0 | |1 |

2 2

1 1 3 1 3 1 1 3 1 3| | | | | | | |

2 4 4 4 4 2 4 4 4 4

1 3| | | |

4 4

.

a b a b a b a b

a a b b

30

Thus, two different realizations may correspond to the same statistical quantum

ensemble.

Since | a and | b are orthogonal to each other, they are the eigenstates of the density

operator 1 2 . In the contrary, | 0 and |1 are not orthogonal to each other.

Therefore, | 0 and |1 are not the eigenstates of the density operator . In general, the

eigenvectors and eigenvalues of a density operator just indicate one of many possible

realizations that may give rise to a specific density operator.

Exercise 4:

With the help of the formula

| | 1,n n

n

u u

we find

Tr | | | |

| |

|

n n

n

n n

n

A B u A B u

B u u A

B A

31

4. Applications of the density operator

a) System in thermal equilibrium

The first example we consider is borrowed from quantum statistical mechanics.

Consider a system in thermal equilibrium at the absolute temperature T. According to

quantum statistical mechanics, the density operator of the system is

1 / .H kTZ e (134)

Here H is the Hamiltonian operator of the system, k is the Boltzmann constant, and Z is a

normalization constant chosen so as to make the trace of equal to 1, that is,

/Tr { }.H kTZ e (135)

We use the basis from the eigenvectors | nu of H. In this basis, we have

/1 / 1| | nE kTH kT

nn n nZ u e u Z e (136)

and

1 /| | 0 for .H kT

nm n mZ u e u n m (137)

At thermal equilibrium, the populations of the stationary states are exponentially

decreasing functions of the energy, and the coherences between stationary states are zero.

b) Description of a part of a physical system

The most important application of the density operator is as a descriptive tool for

subsystems of a composite quantum system. Such a description is provided by the

reduced density operator. The reduced density operator is indispensable in the analysis of

composite quantum systems.

Suppose we have a composition of physical systems A and B. The state space of the

combined system AB is the direct product of the state spaces of the subsystems A and B.

By definition, the state space of AB contains the vectors of the type

| | | | | |AB A B A B A BV V V V V V V (138)

and also their superpositions, where | AV and | BV are arbitrary vectors in the state

spaces of A and B, respectively. The vector | ABV has two parts, one belongs to A and

the other belongs to B. We have | |AB BAV V , | |A B B AV V V V ,

| | | |A B B AV V V V , and | | | |A B B AV V V V . The dimension of the state space

of AB is the sum of the dimensions of the state spaces of A and B.

The basis states of the combined system AB are

| | | | | |AB A B A B A Bu u u u u u u (139)

where | Au and | Bu are the basis states of the subsystems A and B, respectively.

32

To label the individual basis states of the subsystems A and B, we use indices An and

Bn , respectively. Then, we can label the basis states of the combined system AB as

| | | | | |A B A B A B A B

AB A B A B A B

n n n n n n n nu u u u u u u (140)

We describe the state of the composite system AB by a density operator AB . The

matrix form of the density operator AB is given by

, |

|

| | |

| | | .

A B A B A B A B

A B A B

B A A B

B A A B

AB AB AB AB

n n m m n n m m

A B AB A B

n n m m

B A AB A B

n n m m

B A AB A B

n n m m

u u

u u u u

u u u u

u u u u

(141)

We assume that only part A is observed. Then information about part B is lost. This is

the so-called incomplete detection. Therefore, a statistical average over part B is

necessary. The state of part A is described by the reduced density operator

Tr ( )

| | ,

A AB

B

B AB B

n n

n

u u

(142)

where TrB is a map of operators known as the partial trace over system B. In the matrix

form, we have

, .A A A B A B

B

A AB

n m n n m n

n

(143)

Note that the partial trace of an operator of the type 1 2 1 2| | | |ABO a a b b is given

by

1 2 1 2 1 2 1 2

1 2 2 1

Tr (| | | |) | | Tr (| |)

| | | .

B Ba a b b a a b b

a a b b

(144)

Here 1| a and 2| a are any two vectors in the state space of A, and 1| b and 2| b are

any two vectors in the state space of B. In deriving the above expression, we have used

the formula 1 2 2 1Tr | | | .

If the composite system is in a pure state, the incomplete detection process may cause

a part of the system to be in a statistical mixture. As an example, consider a two-level

atom initially prepared in the excited state. Assume that the atom interacts with a single-

mode radiation field that is initially prepared in the vacuum state.

The excited and ground states of the atom are indicated by | a and | b , respectively.

The n-photon state of the field is denoted by | n , where 0,1,2,...n . The basis states of

the composite system “atom+field” are | ;0a , | ;1a , | ;2a , … , | ;a n , … , and | ;0b ,

| ;1b , | ;2b , … , | ;b n , … .

33

The initial state of the composite system is | ;0a . After a short time the atom has a

probability to make a transition to the ground state by spontaneous emission of a photon.

The state vector of the composite system at an arbitrary time t is then given by

| | ;0 | ;1 .AB a b (145)

The density operator of the composite atom+photon system is

* *

2 2

* *

| |

| ;0 | ;1 ;0 | ;1 |

| | | ;0 ;0 | | | | ;1 ;1 |

| ;0 ;1 | | ;1 ;0 | .

AB AB AB

a b a b

a a b b

a b b a

(146)

If we observe the state of the atom but not the emitted photon, then the atom will be

found in either the excited | a or the ground state | b ; however, it will no longer be in a

pure state. The new state can be described by the reduced density operator

atom ph ph

2 2

Tr Tr | |

| | | | | | | | .

AB AB AB

a a b b

(147)

Basic properties of the reduced density operator

Hermitivity: We can show that

†

*

' '

,

or,

.

A A

A A

nn n n

(148)

Indeed, the matrix elements of the reduced density operator A are given by

' '

'

'

, '

| |

| | | |

| |

.

A A A A

nn n n

A B AB B A

n m m n

m

A B AB A B

n m n m

m

AB

nm n m

m

u u

u u u u

u u u u

(149)

From the Hermitivity of AB , we have *

, ' ' ' ',

AB AB

nm n m n m nm . Hence, we find

** *

' , ' , '

' , ' .

A AB AB

nn nm n m nm n m

m m

AB A

n m nm n n

m

(150)

34

Thus, A is a Hermitian operator.

Normalization: We can show that

Tr 1.A (151)

Indeed,

,Tr

Tr 1.

A A AB

nn nm nm

n n m

AB

(152)

Thus, the probability conservation law is valid for the reduced density operator.

Average of an observable: Consider an observable AO that depends only on part A .

We can show that the average of this observable is given by the formula

Tr Tr .A A A A A

A AO O O (153)

Indeed, by definition we have

Tr

Tr Tr

Tr Tr

Tr Tr .

A AB A

AB

AB A

A B

AB A

A B

A A A A

A A

O O

O

O

O O

(154)

Another way of proving is the following:

Tr

| |

| | .

A AB A

AB

AB AB A AB

nm nm

nm

A B AB A A B

n m n m

nm

O O

u O u

u u O u u

(155)

Since AO does not depend on part B , it does not change any vectors in the state space of

B , that is,

| (|

|

|

A A B A A B

n m n m

A A B

n m

B A A

m n

O u u O u u

O u u

u O u

(156)

Consequently, we have

35

| |

| | |

| | } |

Tr {{Tr } }

Tr { } Tr { }.

A A B AB A A B

n m n m

nm

A B AB B A A

n m m n

nm

A B AB B A A

n m m n

n m

AB A

A B

A A A A

A A

O u u O u u

u u u O u

u u u O u

O

O O

(157)

Thus, the average of AO can be determined from the reduced density operator A by

using the same formula as for an arbitrary quantum system. This property makes the

reduced density operator A useful for describing the state of part A.

Invalidity of the Schrödinger equation: In general, the Schrödinger equation is not

valid for the reduced density operator. An example is given below.

c) Spontaneous emission

Consider a two-level atom. If the atom is completely free, that is, if there is no

interaction and no spontaneous emission, the evolution of the density operator of the

atom is governed by the generalized Schrödinger equation

0

1[ , ].

dH

dt i

(158)

Here the Hamiltonian of the free atom is given by

0 | | | | .a bH a a b b (159)

When spontaneous emission is included, the evolution of the density operator of

the atom cannot be described by the Schrödinger equation. According to the Weisskopf-

Wigner theory, the reduced density operator of the atom is governed by the master

equation

0

1[ , ] 2 .

2

dH

dt i

(160)

Here is the atomic decay rate, and | |a b and | |b a are the atomic

transition operators. The decay rate is sometimes called the Einstein A-coefficient and

is given as 3 2 3

0| | /3ab d c

, where | | | |d e a r b

is the dipole moment of the

atom and ab a b is the atomic transition frequency. Equation (160) can be

represented in the form

0 decay , (161)

where

36

0 0

decay

1[ , ],

2 .2

Hi

(162)

We find

0

decay

| | | | | | | | ,

| | 2 | | | | | | .2

a bi a a a a b b b b

a a b a a b a a

(163)

We can easily do some exercises to show that

0

0

0

| | 0,

| | 0,

| | ,ab ab

a a

b b

a b i

(164)

and

decay

decay

decay

| | ,

| | ,

| | .2

aa

aa

ab

a a

b b

a b

(165)

The equations of motion for the matrix elements of can be obtained from Eq. (161):

,

,

.2

aa aa

bb aa

ab ab ab ab

d

dt

d

dt

di

dt

(166)

It may be noted that 0aa bb , that is, constantaa bb . This result is valid

only in the framework of the two-level atomic model. In the case where the atom has

three or more levels, the atom can decay from level a to levels other than b. In this case,

the sum of the populations of levels a and b is not conserved.

The solution of Eqs. (166) is found to be

/2

( ) (0) ,

( ) 1 (0) ,

( ) (0) .ab

t

aa aa

t

bb aa

i t t

ab ab

t e

t e

t e

(167)

37

Thus, the upper-state population aa and the coherence ab decay exponentially with

the rates and / 2 , respectively. In other words, the decay rate of ab is half the decay

rate of aa .

38

Exercises

Exercise 1: Consider an arbitrary operator O and two arbitrary vectors |V and |U .

Show that

Tr{ | |} | | .O V U U O V

Exercise 2: Consider an arbitrary operator O and a state vector | . Show that

Tr{ | |} .O O

Exercise 3: Consider an arbitrary operator O and two basis vectors | nu and | mu .

Show that

Tr{ | |} .n m mnO u u O

Exercise 4: Consider a density operator and two basis vectors | nu and | mu .

Show that

Tr{ | |} .n m mnu u

39

Solutions

Exercise 1: Using the formula Tr{| |} |A B B A , we can show that

Tr{ | |} | | .O V U U O V

Exercise 2: Use the result of exercise 1 and the definition | | .O O

Exercise 3: Use the result of exercise 1 and the definition | | .mn m nO u O u

Exercise 4: Use the result of exercise 1 and the definition | | .mn m nu u

40

3. TWO-LEVEL ATOMS INTERACTING WITH A LIGHT FIELD

1. Atom+field interaction Hamiltonian

Figure 2: A two-level atom interacting with a light field.

We consider a two-level atom interacting with a light field, see Fig. 2. The total

Hamiltonian of the atom in the field is given by

0 ,IH H H (168)

where 0H and IH are the free (unperturbed) and interaction parts.

The free part of the Hamiltonian is the operator for the energy of the atom when it is

free and is given by

0 | | | | .a bH a a b b (169)

The projection operators | |aa a a and | |bb b b describe the populations

(probabilities) of levels a and b, respectively. Indeed, we have

Tr{ | |} | |aa aaa a a a (170)

and

Tr{ | |} | | .bb bbb b b b (171)

Here, the diagonal matrix elements aa and bb of the density operator describe the

populations of the atom in levels a and b, respectively. Note that the energy of the atom

when it is free is given by

0 | | | |

.

a b

a aa b bb

a aa b bb

H a a b b

(172)

41

We introduce the Pauli operator | | | |z a a b b , which describes the inversion of

the populations of the atom. Note that z aa bb . In terms of the population

inversion operator z , we have

1| | (1 ),

2

1| | (1 ).

2

z

z

a a

b b

(173)

Hence, we find

0 ( )

2 2

const,2

abz a b

abz

H

(174)

where ab a b is the transition frequency of the atom. We can choose the middle

point between a and b as the origin for the zero energy. This leads to 0 / 2a and

0 / 2b . Then, the constant is the above expression for the Hamiltonian 0H is zero.

More general, a constant in the Hamiltonian can always be neglected because it

commutes with the density operator and the observables operators and consequently does

not affect their time evolution.

We now describe the atom-field interaction. The interaction part of the Hamiltonian is

the operator for the interaction energy between the atom and the field. We assume that

the field is linearly polarized along the x direction. We use the dipole approximation for

the interaction. Then, the interaction part of the Hamiltonian is given by

(| | | | | | | | )

(| | | | )

(| | | |) .

I

aa bb ab ba

ab ba

x

H d E exE

e a a x b b x a b x b a x E

e a b x b a x E

a b b a d E

(175)

Here we have used the properties 0aa bbx x , have introduced the notation x abd ex ,

which is called matrix element of the atomic dipole moment, and have assumed for

simplicity that abx and consequently xd are real parameters.

We present the electric component of the light field in the form

0 0

1cos .

2

i t i tE E t e e E (176)

We introduce the Rabi frequency

0 / .xd E (177)

Then, we obtain

42

(| | | |)( )2

( )( ) .2

i t i t

I

i t i t

H a b b a e e

e e

(178)

Here we have introduced the upward transition operator

| |ab a b (179)

and the downward transition operator

| | .ba b a (180)

We note that, for a free atom, in the Heisenberg presentation, the transition operators

| |a b and | |b a oscillate as abi te

and abi te

, respectively. Indeed, for a free

atom we have Tr Tr | | | | (0) abi t

ab abb a a b e

. Therefore, the

terms | |i t i te a b e and | |i t i te b a e

quickly vary as ( )abi t

e

and ( )abi t

e

,

respectively. Meanwhile, | |i t i te a b e

and | |i t i te b a e slowly vary as

( )abi te

and

( )abi te

, respectively.

Let’s come back to the interaction Hamiltonian (178). We neglect the fast rotating

terms | | i ta b e and | | i tb a e . This approximation is called the rotating-wave

approximation. The result is

( )2

(| | | | ).2

i t i t

I

i t i t

H e e

a b e b a e

(181)

Appendix: Properties of the Pauli operators , , and z :

The operators | |a b , | |b a , and | | | |z a a b b are called the Pauli

operators. The matrix forms of these operators are

0 1,

0 0

0 0,

1 0

1 0.

0 1z

(182)

The operators x and ( )y i together with z are the original Pauli

operators invented by W. Pauli to describe the spin 1/2 of an electron in the first half of

the last century.


43

2 2

2

0,

1,z

1(1 ),

2

1(1 ),

2

z

z

,

,

z

z

and

,

.

z

z

The above relations allow us to reduce any function ( , , )zf of the Pauli operators

to a linear form of the type z zc c c .

In particular, we find the following commutation relations:

[ , ] 2 ,

[ , ] 2 ,

[ , ] .

z

z

z

Exercise

Consider a free two-level atom, with the Hamiltonian 00

2zH H

. Use the

commutation relations for the Pauli operators to derive the Heisenberg equations for these

operators.

Solution

The Heisenberg equation for the observable O of a quantum system with the

Hamiltonian H reads

[ , ].d i

O H Odt

When we use the commutation relations [ , ] 2z , [ , ] 2z , and

[ , ] z and the Hamiltonian 00

2zH H

, we find

44

0

0 0

0 0

[ , ] 0,2

[ , ] ,2

[ , ] .2

z z z

z

z

d i

dt

d ii

dt

d ii

dt

The solutions to the above equations are ( ) (0)z zt , 0( ) (0)exp( )t i t , and

0( ) (0)exp( )t i t .

2. Optical Bloch equations

a) Evolution equations for the state vector of a pure state of a two-level atom

We first consider the case where the state of the atom is a pure state, described by a

state vector

| | | .a bC a C b (183)

The state vector | satisfies the Schrödinger equation

0| | ( ) | .I

di H H H

dt (184)

For the probability amplitudes aC and bC , we find the equations

,2

.2

i t

a a a b

i t

b b b a

d iC i C e C

dt

d iC i C e C

dt

(185)

The case of exact resonance:

Let consider the case of resonance, where the frequency ω of the field coincides with

the atomic transition frequency 0 a b , that is, 0 . We change the fast

oscillating amplitudes aC and bC to the slowly varying amplitudes

,

.

a

b

i t

a a

i t

b b

c C e

c C e

(186)

Then, Eqs. (185) yield

,2

.2

a b

b a

d ic c

dt

d ic c

dt

(187)

The solutions for ac and bc can be written as

45

( ) (0)cos (0)sin ,2 2

( ) (0)cos (0)sin .2 2

a a b

b b a

t tc t c ic

t tc t c ic

(188)

This solution is general with respect to the initial condition.

Example 1:

We consider the situation where the atom is initially in the ground state b, that is,

(0) 0ac and (0) 1bc . In this case, we find

( ) sin ,2

( ) cos .2

a

b

tc t i

tc t

(189)

Hence, we have

( ) sin ,2

( ) cos .2

a

b

i t

a

i t

b

tC t i e

tC t e

(190)

The populations of the levels, described by the diagonal matrix elements aa and bb ,

are found to be

2 2

2 2

( ) | ( ) | sin ,2

( ) | ( ) | cos .2

aa a

bb b

tt C t

tt C t

(191)

As seen, the populations oscillate in time, with the frequency Ω. Such oscillations are

caused by the driving field. They are different from the optical oscillations, which occur

with the frequency . The oscillations of the populations, caused by the driving field, are

called the Rabi oscillations, and Ω is called the Rabi frequency. The Rabi oscillations

mean that the atom jumps forth and back between the ground and excited states.

Meanwhile, the coherence (interference) is described by the off-diagonal matrix

element

0

*

( )

( ) ( ) ( )

sin cos2 2

sin( ) sin( ) .2 2

a b

ab a b

i t

i t i t

t C t C t

t ti e

i it e t e

(192)

46

Note that the coherence ab oscillates with the optical frequency 0 , and the slowly

varying envelope of the coherence oscillates with the Rabi frequency Ω. We usually have

. Thus, the Rabi oscillations are usually very slow compared to the optical

oscillations.

Example 2:

We consider the situation where the atom is initially in the excited state a, that is,

(0) 1ac and (0) 0bc . In this case, we find

( ) cos2

( ) sin .2

a

b

tc t

tc t i

(193)

Hence, we have

( ) cos ,2

( ) sin .2

a

b

i t

a

i t

b

tC t e

tC t i e

(194)

The populations of the levels, described by the diagonal matrix elements aa and bb ,

are found to be

2 2

2 2

( ) | ( ) | cos ,2

( ) | ( ) | sin .2

aa a

bb b

tt C t

tt C t

(195)

As seen, the populations oscillate in time, with the Rabi frequency Ω. Meanwhile, the

coherence (interference) is

0

*

( )

( ) ( ) ( )

sin cos2 2

sin( ) sin( ) .2 2

a b

ab a b

i t

i t i t

t C t C t

t ti e

i it e t e

(196)

We also observe that the coherence ab oscillates with the optical frequency 0 , and

the slowly varying envelope of the coherence oscillates with the Rabi frequency Ω.

Example 3:

We consider the situation where the atom is initially in a linear superposition of the

excited state a and the ground state b, with the probability amplitudes (0) 1/ 2ac and

(0) 1/ 2bc . In this case, we obtain the solution

47

1

( ) ( ) cos sin ,2 22

a b

t tc t c t i

(197)

which leads to

1( ) cos sin ,

2 22

1( ) cos sin .

2 22

a

b

i t

a

i t

b

t tC t i e

t tC t i e

(198)

Hence, we find that the populations of the levels are constant in time:

( ) ( ) 1/ 2.aa bbt t (199)

We also find that the coherence is oscillating in time, with a constant envelope:

0( )* 1 1 1( ) ( ) ( ) .

2 2 2a bi t i t i t

ab a bt C t C t e e e (200)

b) Evolution equations for the density operator of a general quantum state of

a two-level atom with no decay

We now consider the case where the state of the atom is an arbitrary state, described

by a density operator ρ. The density operator is governed by the Schrödinger equation

1

[ , ].d

Hdt i

(201)

To obtain the evolution equations for the matrix elements of , we can use Eq. (201)

directly. For this purpose, we write

†( ),d

i O Odt (202)

where

1

.O H

(203)

In the matrix form, we have

† *( ) ( ).d

i O O i O Odt

(204)

Let’s calculate the matrix elements O . When we use the explicit expression for the

Hamiltonian 0 IH H H , we find

48

0

1 1( )

| | | | (| | | | ).2

I

i t i t

a b

O H H H

a a b b a b e b a e

(205)


,2

,2

,2

.2

i t

aa a aa ba

i t

bb b bb ab

i t

ab a ab bb

i t

ba b ba aa

O e

O e

O e

O e

(206)

When we insert the above matrix elements into Eq. (204), we obtain

( ),2

( ),2

( ).2

i t i t

aa ba ab

i t i t

bb ba ab

i t

ab ab ab aa bb

d ie e

dt

d ie e

dt

d ii e

dt

(207)

Note that we can also derive Eqs. (207) with the use of Eqs. (185) and the definitions

2

2

*

*

| | ,

| | ,

,

.

aa a

bb b

ab a b

ba ab

C

C

C C

(208)

We introduce the notation ab for the detuning of the field. It is convenient to

use new variables

,

,

,

.

i t

ab ab

i t

ba ba

aa aa

bb bb

e

e

(209)

In terms of these variables, we have

49

( ),2

( ),2

( ).2

aa ba ab

bb ba ab

ab ab aa bb

d i

dt

d i

dt

d ii

dt

(210)

The above equations are called the optical Bloch equations for two-level atoms without

decay.

Example:

We consider the situation where the atom is initially in a statistical mixture of the

excited state a and the ground state b, with the weight factors 1/ 2a bp p . In this case,

the matrix elements of the initial density of the atom are given by

(0) (0) 1 / 2,

(0) (0) 0.

aa bb

ab ba

(211)

Consequently, we have

(0) (0) 1 / 2,

(0) (0) 0.

aa bb

ab ba

(212)

The Bloch equations show that the above values are true for any time t, that is,

( ) ( ) 1 / 2,

( ) ( ) 0.

aa bb

ab ba

t t

t t

(213)

Note that the above solutions and the solutions for example 3 in the previous subsection

give the same populations but different coherences.

Exercise

Derive Eqs. (207) from Eqs. (185). Such a derivation is valid only for the case of a

pure state. However, the result is general: it coincides with that for an arbitrary state.

Solution

Equations (185) and their complex conjugates can be written as

50

* * *

* * *

,2

,2

,2

.2

i t

a a a b

i t

a a a b

i t

b b b a

i t

b b b a

d iC i C e C

dt

d iC i C e C

dt

d iC i C e C

dt

d iC i C e C

dt

When we use the above equations and the relations

2

2

*

*

| | ,

| | ,

,

,

aa a

bb b

ab a b

ba ab

C

C

C C

we can easily derive Eqs. (207).

c) Evolution equations for the density operator of a system with decay

We now include the spontaneous emission of the atom into our treatment. In this case,

the state of the atom is described by a reduced density operator , which is governed by

the master equation

1

[ , ] ,d

Hdt i

(214)

where the last term describes the decay and is given by

[ 2 ].2

(215)

We can rewrite Eq. (214) as

coherent decay( ) ( ) , (216)

where

coherent

1( ) [ , ]H

i

(217)

and

decay( ) [ 2 ].2

(218)

In the matrix form, we have

coherent decay( ) ( ) . (219)

51

The matrix elements coherent( ) are the matrix elements of the operator

coherent

1( ) [ , ]H

i

. They are given by Eqs. (207). The matrix elements decay( ) are

the matrix elements of the operator decay( ) [ 2 ].2

They are given by the expressions on the right-hand side of Eqs. (166). Indeed, we can

easily show that

decay

decay

decay

( ) ,

( ) ,

( ) .2

aa aa aa

bb bb aa

ab ab ab

(220)

To obtain the evolution equations for the matrix elements of , we add to Eqs. (207) the

decay terms (220). This procedure leads to

( ),2

( ),2

( ).2 2

i t i t

aa aa ba ab

i t i t

bb aa ba ab

i t

ab ab ab aa bb

d ie e

dt

d ie e

dt

d ii e

dt

(221)

We introduce the notation ab for the detuning of the field. It is convenient to use

new variables

,

,

,

.

i t

ab ab

i t

ba ba

aa aa

bb bb

e

e

(222)

Using these variables, we obtain

( ),2

( ),2

( ).2 2

aa aa ba ab

bb aa ba ab

ab ab aa bb

d i

dt

d i

dt

d ii

dt

(223)

The above equations are called the optical Bloch equations for two-level atoms with

decay.

The above optical Bloch equations play an important role in optics. In particular, they

are used to calculate the absorption spectrum of the atoms.

52

3. Absorption spectrum: saturation and power broadening

We introduce the population difference

aa bbw (224)

and the optical coherence

.c ab (225)

Then, we can rewrite Eqs. (223) as

*( 1) ( ),

.2 2

c c

c c

dw w i

dt

d ii w

dt

(226)

We consider the adiabatic (steady-state) regime where

0.cdw d

dt dt

(227)

In this regime, Eqs. (226) yield

*( ) ,

.2 2

c c

c

w i

ii w

(228)

The solutions to the above equations are

1

1w

s

(229)

and

.2( / 2 )(1 )

c

i

i s

(230)

Here s is the saturation parameter and is given by

2 2

0

2 2 2 2

/ 2,

2 | / 2 | / 4 1 (2 / )

ss

i

(231)

where 0s is the on-resonance saturation parameter and is defined as

2

0 2

2.s

(232)

53

For low saturation, 1,s the population is mostly in the ground state ( 1w ). For

high saturation, 1,s the population is almost equally distributed between the ground

and excited states ( 0w , i.e. , 1/ 2aa bb ).

The parameter 0s can be written in another form

0 .

s

Is

I (233)

Here

2

0 0 / 2I c E (234)

is the intensity of the laser beam and

2 2

0

22 2s

x

cI

d

(235)

is the so-called saturation intensity. According to the Weisskopf-Wigner theory, we have

3 2 3 3 2 3

0 0 0 0| | /3 / .xd c d c

(236)

Here we have introduced for convenience a new notation 0 ab a b . With the

help of Eq. (236), we can show that

3 3

0 0/ / ,sI hc hc (237)

where

1/ (238)

is the lifetime of the excited state and 0 is the atomic resonant wavelength.

The population of the excited state is given by

0

2

0

1 / 2(1 ) .

2 2(1 ) 1 (2 / )aa

s sw

s s

(239)

Since the population in the excited state decays at a rate , the total scattering rate scatt

of light from the laser field is given by

0scatt 2

0

/ 2.

1 (2 / )aa

s

s

(240)

Note that, at high intensity, scatt saturates to / 2 .

Equation (240) can be rewritten as

0scatt 2

0

/ 2,

1 1 (2 / ')

s

s

(241)

where

54

0' 1 .s (242)

The dependence of the scattering rate scatt on the detuning is shown in Fig. 3 for

several values of the saturation parameter 0s . This dependence describes the absorption

spectrum. The width of the spectral profile is characterized by ' . Note that the width

' increases with increasing intensity of the field. This phenomenon is called the power

broadening of the spectral profile.

The power broadening is a direct result of the fact that, for large 0s , the absorption

continues to increase with increasing intensity in the wings, whereas, in the center, half of

the atoms are already in the excited state. The absorption in the center is saturated,

whereas in the wings it is not.

Figure 3: Scattering rate scatt as a function of the detuning for several values of the saturation

parameter 0s .

The scattering results in intensity loss when the beam travels through a sample of

atoms. The amount of scattered power per unit of volume is scattn , where n is the

number density of the atoms. Thus, we have

0

scatt 2 2

0 0

/ 2 ( / 2)( / )

1 (2 / ) 1 (2 / )

.

sdI s n I In n

dz s s

n I I

(243)

55

Here is the scattering cross section and n is the absorption coefficient. These

coefficients are given by

2

0

1.

2 1 (2 / )sn I s

(244)

When we use the expressions 2 2

0

2 3

02 2s

x

c hcI

d

, we can rewrite Eq. (244) as

2

2

0 0

2

0

2

0

2 1

1 (2 / )

1.

2 1 (2 / )

xd

n c s

s

(245)

For low intensity, 0 1s , we have

2

2

0

2

0

2

2 1

1 (2 / )

1.

2 1 (2 / )

xd

n c

(246)

In this regime, the absorption coefficient is independent of the field intensity I.

Therefore, the solution for the field intensity is

0( ) .zI z I e (247)

At exact resonance ( 0 ), the cross section (89) reduces to

2

0 .2n

(248)

In the case where sI I , the absorption coefficient tends to zero. This does not

mean the vanishing of the absorption. Indeed, in the limit

sI I (249)

we have

/ 2.I n (250)

Hence, Eq. (243) yields

( / 2).dI

I ndz

(251)

The solution to the above equation is

(0) ( / 2) .I I n z (252)

56

According to the above equation, the field intensity will linearly reduce with increasing

propagation length. However, after the field intensity I becomes small as compared to

the saturation intensity sI , the field intensity I will exponentially decrease with

increasing propagation length z.

4. Field propagation equation

We study the propagation of the field along the z direction. The one-dimensional

wave equation reads

2 2 2 2

02 2 2 2 2 2

0

1,E P P

z c t t c t

(253)

where P is the polarization density. We expand the field and the polarization in the forms

( ) * ( )

0 0

( ) * ( )

0 0

1,

2

1.

2

i t kz i t kz

i t kz i t kz

E E e E e

P P e P e

(254)

Here and /k c are the frequency and wave number of the field, respectively.

We introduce new variables

,

/ .

z

t z c

(255)

In terms of these variables, we have

*

0 0

*

0 0

1,

2

1.

2

i i

i i

E E e E e

P P e P e

(256)

We also have

,

,

z c

t

(257)

and, consequently,

2 2 2 2

2 2 2 2

2 2

2 2

2,

.

z c c

t

(258)

On substituting Eqs. (256) and (258) into Eq. (253), we find

57

2 2

02

22

02 2

0

2 2c.c.

12 c.c.

i

i

ie E

c c

e i Pc

(259)

The above equation will be satisfied if

2 2 2

2

0 02 2 2

0

2 2 12 .

iE i P

c c c

(260)

We assume that the envelopes 0E and 0P vary slowly in space and time. When we

keep only the lowest nonvanishing order on each side of Eq. (260), we obtain

0 0

0

.2

iE P

c

(261)

Since z , we can rewrite Eq. (261) as

0 0 0

0 0

,2 2

i ikE P P

z c

(262)

keeping in mind that the partial derivative with respect to z in Eq. (262) is taken under the

condition / constantt z c . Equation (262) is called the propagation equation for

the field in the slowly varying envelope approximation.

5. Susceptibility, refractive index, and absorption coefficient

We have

0 0 0,P E (263)

where is the susceptibility. We write

' '',i (264)

where ' and '' are the real and imaginary parts of . Then, Eq. (262) becomes

0 0 0( ' '') ( ' '') .2 2

iE i E i E

z c c

(265)

The real part ' determines the shift of the wave number of the field, while the

imaginary part '' determines absorption. The refractive index of the medium is defined

by

ref

'1 ,

2n

(266)

while the absorption coefficient is defined by

58

'' ''.kc

(267)

In terms of refn and , we can rewrite Eq. (265) as

0 ref 0( 1) .

2E ik n E

z

(268)

At low intensity, we can ignore the dependences of refn and on the field intensity. In

this case, the solution of Eq. (268) for the field envelope 0E is

ref( 1) /2

0 .ik n z zE Ae e (269)

When we use the presentation | | iA A e and the formula,

( ) * ( )

0 0

1,

2

i t kz i t kzE E e E e (270)

we obtain

/2

ref| | cos( ).zE A e t kn z (271)

The expression (271) explains why refn is called the refractive index and why is called

the absorption coefficient.

We now calculate , refn , and . As shown in Sec. I.2.e, the polarization of a single

two-level atom is ( ).x ab baP d Here, we have assumed for simplicity that xd is a

real parameter. For a medium with the atomic density n, the polarization density of the

medium is given by

*( ) ( ).i i

x ab ba x c cP nd nd e e (272)

Consequently, the envelope of the polarization density is

0 2 .x cP nd (273)

Inserting Eq. (230) into Eq. (273) yields

0

2

0

2( 2 )(1 )

2.

( 2 )(1 )

x

x

iP nd

i s

nd Ei

i s

(274)

Hence, we have

0 0 0,P E (275)

where the susceptibility is given by

59

0

0 0

2

0

2

2 2

0 0

2 1

( 2 )(1 )

2 2.

1 (2 / )

x

x

P

E

ndi

i s

nd ii

s

(276)

Then, we find

2

2 2

0 0

4'

1 (2 / )

xnd

s

(277)

and

2

2

0 0

2 1'' .

1 (2 / )

xnd

s

(278)

With the help of Eqs. (266) and (267), we obtain

2

ref 2 2

0 0

3

0

2 2

0

21

1 (2 / )

/1

4 1 (2 / )

xndn

s

n

s

(279)

and

2

2

0 0

2

0

2

0

2 1

1 (2 / )

1.

2 1 (2 / )

xn d

c s

n

s

(280)

Here we have used the expression 3 2 3

0 0/xd c and the approximation 0 .

Equation (280) is in agreement with Eq. (245). Note that refn and depend on the field

intensity via the saturation parameter 0s .

60

Figure 4: Real and imaginary parts of the susceptibility as functions of the normalized detuning

/ of the field for the saturation parameter 0 0.1s .

According to Eqs. (266) and (267), the real part ' and the imaginary part '' of the

susceptibility determine the refractive index refn and the absorption coefficient . We

illustrate in Fig. 4 the frequency dependence of ' and '' . The figure shows that ' is

large in the region / 1. However, in this region, '' is also large. Thus, in a

medium of two-level atoms, a high refractive index is accompanied by large dispersion.

density-matrix formalism in quantum opticssop.uohyd.ac.in/~sdg/resources/fam2.pdf · 2018-03-29 ·...

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