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
1 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
2 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
En / kT proportional to e of being in a state of energy En . 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 p1 or the state | 2 with a probability p2 , etc… Obviously, we have
p12 p ... pn 1. (1) n
We then say that we have a statistical mixture of states |1 , | 2 , ... with probabilities p1 , p2 , … 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:
| Cnn | . (2) n 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 CCnn' , 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 (rr ) Cnn ( ), the probability of n finding the particle at a given point r is
2 2 * * PCCC()|()||r r n n ()| r n n '' n ()(). r n r (3) n nn'
3 The above expression involves cross terms of the type CCnn' , which describe interference between the component states n ()r .
However, if the particle is in a statistical mixture of states | n with weight factors pk , the probability of finding the particle at a given point r is
2 Pp(rr ) nn | ( ) | . (4) n
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 pn 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 (5) ** (,(,(,1 1 2 2 1 1 2 2 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.
4
v. Basic properties of the ket and bra vectors (1) | ' ' | * . (6) ** (2) If | 1 | 1 2 | 2 then | 1 1 | 2 2 |.
More generally, for the superposition state vector || Cunn , we have n
* | uCnn | . (7) n 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 ' * (r ) '( r )d 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 | un is a basis set if an arbitrary state vector can be decomposed as
| Cunn | . (8) n The set of basis states is an orthonormal basis set if
uun m nm. (9)
The coefficients Cn in Eq. (8) are called the probability amplitudes of the superposition state in the basis {|un }. More exactly, we say that is the probability amplitude of the basis state in the superposition state .
It follows from the expansion (8) and the orthonormality (9) of the basis states | un that the coefficients Cn of the state vector | can be determined as
Cunn | (10) We say that the state vector is normalized if |1 . If the state vector is normalized, its coefficients satisfy the normalization condition
2 |Cn | 1. (11) n From now we consider only normalized state vectors. Note that the bra vector | associated with the ket vector | can be decomposed in the form
5 * | uCnn | . (12) n
With the use of the orthonormality of the basis states | un , we find
* Cn || u n u n (13) vii. Projection operators We introduce the operator P| u u |, (14) |un n n whose action on an arbitrary vector |V is defined as P| V (| u u |) | V |un n n (15) |un ( u n | V | u n u n | V As seen, the action of P on |V gives a vector parallel to | u . In addition, we have |un n PPP2 |un | u n | u n
(|un u n |)(| u n u n |)
|un ( u n | u n ) u n | (16)
||uunn
P|u . n The operator P|| u u is called the projection operator onto the basis vector | u . |un n n n viii. Completeness of the basis
The completeness of the set of the basis states | un means that PPP ... 1, (17) |un | u12 | u n that is,
|unn u || u1 u 1 || u 2 u 2 |...1. (18) n From now we use only the orthonormal basis. The properties of this basis are summarized below:
uun m nm , (19) |uunn | 1. n
ix. Linear operators A linear operator O associates each ket vector | with a ket vector | ' O | and satisfies the linearity
6 OOO(1 | 1 2 | 2 ) 1 | 1 2 | 2 . (20) Similarly, the linear operator O associates each bra vector | with a bra vector ' | |O and satisfies the linearity
(1 1 | 2 2 |)OOO 1 1 | 2 2 | . (21)
For the superposition state vector || Cunn , we have n O| O Cn | u n C n O | u n C n O | u n , n n n (22) *** |O Cn u n | O C n u n | O C n u n | O . n n n We consider the operator OAB| |, (23) whose action on ket and bra vectors is defined as OABABAB| | | | | | | | , (24) |OABABAB | | | | | | |. We can show that ||'OOABO ||' | |' ||'. (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 O| un u n | O | u m u m | nm
|un u n |(|)| O u m u m |(|)| u n u n O u m u m | (26) nm nm
un|(|)| O u m u n u m | (|)| u n O u m | u n u m |. nm nm Hence, we find
un|(|)(|)| O u m u n O u m u n || O u m O nm . (27)
The matrix Onm u n|| O u m is called the matrix form of the linear operator . Every operator can be represented by a matrix Onm . Every matrix represents an operator . Indeed, it follows from Eqs. (26) and (27) that
7 O Onm| u n u m |. (28) nm We can generalize Eq. (27) to obtain
1|(|)(|)|OOO 2 1 2 1 ||. 2 (29)
xi. Hermitian conjugate Every operator O is associated with an operator O† , whose matrix is
†**†*T ()()()Onm O mn O nm u n || O u m u m ||. O u n (30) Here the upper label T means the matrix transpose and the symbol * means the complex conjugate. The operator is called the Hermitian conjugate of the operator .
For O Onm|| u n u m , we have nm
†† O Onm|| u n u m nm * Omn|| u n u m (31) nm * Onm| u m u n |. nm
† Note that OO† . We can show that |OO † |, (32) where O† | is the bra vector associated with the ket vector ||OO†† . Indeed, since , we have
|O | Onm | u n u m | O nm | u n u m |. (33) nm nm
†* On the other hand, since O Onm|| u m u n , we find nm
††* |O O | Onm | u m u n | | nm * Onm|| u m u n (34) nm * Onm u n| | u m , nm which yields
† O| Onm | u n u m |. (35) nm
8 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 | |O (36) then | O† | . Property 2: If | O | (37) then | |O† . Property 3: |OO | ' *† ' | | . (38) Property 4:
If OOOO 12 n †*††† (39) then OOOO nn11 ,
where is a complex number and O1 , O2 , …, On are arbitrary operators. Property 5: If O f ( A ) (40) then O†*† f ( A ), 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,
†* OOOO or mn nm . (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
{|un } are given by
Anm u n| A | u m u n | A | u m ( u n | A ) | u m . (42) Since A is a Hermitian operator, we have
* AAnm mn. (43)
9 The mean value of A is AA ||
* Cn u n|| A C m u m nm * (44) Cn C m u n|| A u m nm * CCAn m nm. nm 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 | ( t ) , (45) dt 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, HH† . The solution to the Schrödinger equation (45) is given by | ()t exp( iHt /)| (0). (46) The corresponding bra vector is (t ) | (0) | exp( iHt / ). (47) The time-dependent mean value of the observable A is A( t ) ( t ) | A | ( t ) . (48)
Consider the initial state vector | (0) Cunn (0) | . The probability amplitude n
Cn (0) is given by Cunn(0) | (0) For an arbitrary time t , we have
| (t ) Cnn ( t ) | u , (49) n where
Cnn( t ) u | ( t ) (50) Inserting Eq. (49)into Eq. (45) gives d i Cn( t ) | u n C m ( t ) H | u m nmdt
Cm( t ) | u n u n | H | u m (51) mn
Cm( t ) u n | H | u m | u n . mn
10 Hence, we obtain d i Cn( t ) H nm C m ( t ). (52) dt m
Here Hnm u n|| H u m is the matrix of the Hamiltonian.
In terms of the probability amplitudes Ctn (), the mean value of the observable A at the time t is given by
* A() t ()||() t A t Cn ()() t C m t A nm . (53) nm 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 Eaa (upper level) and Ebb (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: a| a b | b 1, (54) a| b b | a 0. The completeness of the basis requires |a a | | b b | 1. (55) In general, any state vector (wave function) of the two-level system can be written in the form
| Cab | a C | b . (56)
Here Ca and Cb are the probability amplitudes of finding the system in states | a and | b , respectively. They satisfy the normalization condition
22 |CCab | | | 1. (57)
11 The bra vector | associated with the ket vector | can be written in the form
** | a | Cab b | C . (58) The mean value of an observable A is AA ||
** a| Ca b | C b A C a | a C b | b (59) 2 2 * * ||||||||Ca aAa C b bAb CCaAb a b || CCbAa b a || 2 2 * * |CACACCACCAa | aa | b | bb abab baba . The total Hamiltonian is
HHH0 I , (60) where H0 and H I are the free (unperturbed) and interaction parts, respectively. The upper state | a and the lower state | b are the eigenstates of the free Hamiltonian H0 , that is, H0 || j Ej j , where j a, b . We can represent H0 in the form
H0 ab| a a | | b b |. (61)
Why does H0 have the above form? The reason is the following: since
|j j | | a a | | b b | 1 and H0 || j Ej j , we can expand H0 as j H(| i i |) H (| j j |) 00ij, |i j | ( i | H | j ) ij, 0 (62) |i j | ( i | j E ) E | i j | i,, jj i j j i, j E| i i | | i i |. iiii
We consider a particular case where the two-level system is free, i.e., H I 0. In this case, the Schrödinger equation is d iH || dt 0 (63)
(ab |a a | | b b |) | . It gives d iCaCb | | ( | aa | | bbCaCb |) | | dt a b a b a b (64) iCa| a iC b | b a C a | a b C b | b . Hence, we find
12 d C i C , dt a a a (65) d C i C . dt b b b The solutions to the above equations are
C( t ) eita C (0), aa (66) itb Cbb( t ) e C (0). Consequently, the explicit time-dependent expression for the state vector (wave function) of the free two-level system is
iab t i t |() t CtaCtbeCa ()| b ()| a (0)| aeC b (0)|. b (67) Note that the probabilities of finding the free two-level system in the upper and lower 22 22 levels are |Caa ( t ) | | C (0) | and |Cbb ( t ) | | C (0) | , respectively. They are independent of time. In the contrary, the interference described by the cross term
**it()ab Ca( t ) C b ( t ) C a (0) C b (0) e oscillates in time. The mean value of an observable A at an arbitrary time t is
* A() t ()||() t A t Cn ()() t C m t A nm nm
* it()nm Cn(0) C m (0) e A nm nm (68) 22 |CACAa (0) | aa | b (0) | bb
**iab t i ab t Ca(0) C b (0) e A ab C b (0) C a (0) e A ba .
Here ab a b is the transition frequency.
If the initial state is | (0) |a , then Ca (0) 1and Cb (0) 0. In this case, the state vector at an arbitrary time is | (t ) eita | a and the mean value of an observable A at an arbitrary time is A( t ) Aaa 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 Cb (0) 1and Ca (0) 0 . In this case, the state vector at an arbitrary time is | (t ) eitb | b and the mean value of an observable A at an arbitrary time is A( t ) Abb 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 Ca (0) 0 and
Cb (0) 0, then At() in general varies in t.
13 Exercises Exercise 1: Consider a two-level system. Suppose that is an operator such that ||ab and |0b . Show that can be written in the form |ba |.
†* † Exercise 2: By definition ()OOkl lk , where O is an arbitrary operator and O is its † Hermitian conjugate. Suppose O|| unm u . Show that O|| umn u . Exercise 3: Suppose OAB|| , where | A and | B are two arbitrary vectors. The action of OAB|| on an arbitrary vector Q is defined as OQABQABQABQ| | | | | | | | . Show that OBA† || .
† T* ††† Exercise 4: By definition OO . Suppose OOO 12. Show that OOO 21.
14 Solutions Exercise 1: Since ||ab and |0b , we have |a a | | b a | and |bb | 0. When we sum up the last two equations, we obtain (|a a | | b b |) | b a |. On the other hand, the completeness of the basis means that |a a | | b b | 1. Hence, we find |ba |. Using the orthonormality of | a and | b , namely aa| and ab| , we can confirm that the operator ||ba satisfies the relations ||ab and |0b .
Exercise 2:
For O|| unm u , we find
Okl kn lm.
For X|| umn u , we have
X kl km ln. We see that
*† XOOkl lk kl . Hence, we obtain OX† , that is,
† O| umn u |.
Exercise 3: For OAB|| , we have
* Okl a k b l ,
15 where akk u| A and bll u| B . For XBA|| , we have
* Xkl b k a l . We see that
*† XOOkl lk kl . Hence, we obtain OX† , that is, OBA† | |. Exercise 4:
For OOO 12, we have
††****††TTTTT OOOOOOOOOOO()()().1 2 1 2 2 1 2 1 2 1 Generalization:
For OOOO 12 n , where is a complex number and O1 , O2 , …, On are arbitrary operators. Using the result of the above exercise, we can easily show that
†*††† OOOO nn11 .
16 b) Description by a density matrix The relation
* AACCA || n m nm (69) nm shows that the coefficients Cn enter into the mean values through quadratic expressions * of the type CCnm. These quadratic expressions are simply the matrix elements of the operator || , which is the projector onto the ket vector | . Indeed, since
| Cunn | , n (70) * | uCnn | , n we have
Cunn | * (71) Cunn | Hence, we find
* Cn C m u m| | u n . (72) It is therefore natural to introduce the density operator | |. (73)
The density operator is represented in the {|un } basis by a matrix called the density matrix whose elements are
mn uu m|| n
uumn|| (74) * CCnm. The mean value of the observable A is then given by
* AACCAAAA | | n m nm mn nm Tr{ } Tr{ }. (75) nm nm 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, PVV| (| |) | | (77) | ( |V
17 In addition, we have
2 PPP| | | (| |)(| |) | ( | ) | (78) ||
P| .
The above properties indicate that P| || is a projection operator. Since 2 2 PP|| , we find that, in the case of pure states, we have and hence, Tr2 Tr 1 [see Eq. (80)].
Note that the matrix mn is a Hermitian matrix. Indeed, we have
** mnCC m n nm (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
2 Tr nn |C n | 1. (80) nn 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
* AACCAA || n m nm mn nm nm nm (81) Tr{AA } Tr{ }.
iii. Schrödinger equation We derive the time evolution equation for the density operator . As known, the Schrödinger equation is d 1 | H | . (82) dt i
18 The Hermitian conjugate form of the above equation reads d 11 | |HH† | . (83) dt i i Hence, we have dd || dt dt dd | | | | dt dt 11 HH| | | | (84) ii 1 HH i 1 [,].H i Here
[,]OOOOOO1 2 1 2 2 1 (85) is the commutator between the operators O1 and O2 . Thus, the time evolution of the density operator is governed by the equation d iH [ , ]. (86) dt 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, AAA Tr { } Tr { }, (87) d iH [,]. dt 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 , (88) 2 Tr 1. These properties can be used as criteria to find out if a state is a pure state or not.
19 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 ei | describe the same quantum state. In other words, there exists an arbitrary global phase factor for the state vector. However, the state vectors | and ei | correspond to the same density operator || (89) eeii| | . Thus, the use of the density operator eliminates the global phase. Second, the equation AAA Tr { } Tr { } (90) is linear with respect to the density operator . However, the equation AA || (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)
| Cab | a C | b . (92) The density operator of the system is
** | |(| Ca a C b |)( b C a a | C b b |). (93) Taking the matrix elements, we get
2 aa|C a | , CC*, ab a b (94) * ba ab , 2 bb|C b | . The matrix form of the density operator is
2* aa ab ||CCCa a b *2. (95) ba bb CCCb a|| b
* † We see that nm mn , that is, . In addition, we have
20 22 Tr aa bb |CC a | | b | 1. (96) The average value of an observable A is given by AAA Tr{ } Tr{ }
mnA nm (97) nm
aaAAAA aa bb bb ab ba ba ab.
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 P e x e || x 2 2 * * eC(|a | x aa | C b | x bb CCx abab CCx baba ) ** (98) e() Ca C b x ab C b C a x ba * badd x ab x , where e is the electric charge of the electron and
dx ex ab (99) is the dipole matrix element. Here we have used the property xxaa bb 0 , which is a 2 consequence of the fact that x is an odd function while |n (x ) | (with n a, b ) is, in the 2 case of atoms, an even function. The probability density |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
H0 ab| a a | | b b | (100) and the time evolution of the probability amplitudes is governed by the equations d C i C , dt a a a (101) d C i C . dt b b b
* When we use the relation CC , we can easily show that the evolution equations for the density-operator matrix elements of the free two-level system are
21 d 0, dt aa d 0, (102) dt bb d i(). dt ab a b ab We can also derive the above equations directly from the generalized Schrödinger equation (86). The solutions to Eqs. (102) are
aa (t ) constant, bb (t ) constant, (103)
it()ab ab(te ) ab (0). 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
|12 ,| ,...,| k ,..., with probabilities p12, p ,..., pk ,..., where 0p 1, k (104) pk 1. k
The state vectors | k of the components are normalized, that is, kk| 1. However, they are not necessarily orthogonal to each other. The density operator of the statistical mixture of states is defined as
pkkρ , (105) k 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 pk are real, is obviously a Hermitian operator like each of the k :
22 †*† ppk k k k . (107) kk
ii. Probability conservation law The conservation law of probability gives
Tr pkk Tr k (108) pk 1. k
iii. Mean values The expression for the mean value of an observable is
A pkk A k
pAkkTr{ } k (109)
=Tr{Ap kk } k =Tr{A }. 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 d iH | | . (110) dt kk
Therefore, k obeys the evolution equation d iH [ , ]. (111) dt kk Using the linearity of Eqs. (105) and (111), we find the following evolution equation for the density operator :
23 d d d i i pk k p k i k dt dtkk dt
pk[,][,] H k H p k k (112) kk [H , ].
v. Positivity
We see from the definition pkkρ that, for any ket | u , we have k
u| | u pkk u | | u k (113) 2 pukk| | | k and, consequently, uu| | 0. (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, AAA Tr { } Tr { }, (115) d iH [,]. dt vii. Differences between the density operator of a mixed quantum state and that of a pure state The equations 2 , (116) Tr 2 1 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, Tr 2 1. For a mixed state, according to comments (i) and (iii) at the end of this section (see pages 26 and 27), we have Tr 2 1. (118) Thus, we have, in general, for pure and mixed states,
24 Tr 2 1. (119) The above equation can be used as a criterion to find out if a state is a pure state (Tr 2 1) or a mixed state (Tr 2 1).
c) Populations and coherences What is the physical meaning of the matrix elements of the density operator in the basis {|un }?
First, we consider the diagonal element nn . We have
nn uu n|| n
pk u n|| k u n k (120) pk u n|| k k u n k 2 puk| n | k | . k
We express the state | k in terms of the basis {|un } as
()k | k cu n | n . (121) n This gives
()k cun n|. k (122) Then Eq. (120) becomes
(k ) 2 nn pc k| n | . (123) k
(k ) 2 It follows from Eq. (123) that nn is a positive real number. The squared modulus ||cn is the probability of | un in the pure state | k . Therefore, Eq. (123) means that nn is the probability of | un in the state . For this reason, the diagonal matrix element nn is called the population of the state | un . 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 | un .
Now, we consider the non-diagonal element nm . A calculation analogous to the preceding one gives the following expression for nm :
25 nm uu n|| m
pk u n|| k u m k (124) pk u n|| k k u m k (kk ) ( )* pk c n c m . k
(kk ) ( )* The term ccnm is a cross term. It expresses the interference effects between the states
| un and | um 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 | un and | um . 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 {|un } . 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 |. (125) l 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
22 Tr l l (126) l 1. l
When one of the coefficients l is equal to 1, all the other coefficients must be zero. In 2 this case, the state ρ is a pure state and Tr 1 . When all the coefficients l are smaller than 1, the state ρ is a mixed state and Tr 2 1.
(ii) Assume that the Hamiltonian H is time-independent. If | un are eigenvectors of H, then H| u E | u , n n n (127) un| H u n | E n . Hence, we obtain from Eq. (112)
26 d i u || u u H H u dt n m n m
un E n E m| u m (128)
En E m u n | u m that is, d i (). E E (129) dt nm n m nm The solution to the above equation is
i ()E E t nm nm(te ) nm (0). (130)
In particular, we have nn (t ) constant . 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
(kk ) 2 ( ) 2 nn mm p k| c n | p k | c m | kk 2 ()()kk pk|| c n c m (132) k 2 (kk ) ( ) 2 pk c n c m| nm | . k It follows from Eq. (131) that can have coherences only between states whose populations are not zero. It also follows from Eq. (131) that
22 Tr |nm | nm
nn mm nm (133)
nn mm nm 1, see Eq. (119).
27 Exercises Exercise 1: Consider a two-level system with density operator 131 0 |a a | | b b | 4 . 3 440 4 Show that the state of the system is a mixed state.
Exercise 2: Consider a two-level system with density operator
1 3 3 3 1 3 |a a | | b b | | a b | | b a |44 . 4 4 4 4 3 3 44 Show that the state is a pure state.
Exercise 3: Consider two orthonormal vectors | a and | b . Consider a quantum system (ensemble) Q1 that is prepared in the state | a with probability 1/4 and in the state | b with probability 3/4. Suppose we define 13 | 0 |ab | 44
13 |1 |ab | 44
Consider a quantum system (ensemble) Q2 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
Q1 and Q2 correspond to the same density operator.
Exercise 4: Consider two arbitrary vectors | A and | B Show that Tr |ABBA | | .
28 Solutions Exercise 1: The square of the density operator of the system is 191 0 2 |a a | | b b | 16 . 9 16 16 0 16 We find Tr 2 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 1 3 3 3 2 |a a | | b b | | a b | | b a | 4 4 4 4 1 3 44 3 3 44 . Hence we find Tr2 Tr 1. These equalities indicate that the state of the system is a pure state.
Exercise 3:
The density operator of the system Q1 is 13 |a a | | b b |. 1 44
The density operator of the system Q2 is 11 | 0 | |1 |. 2 22 We can easily show that 11 | 0 | |1 | 2 22 1 1 3 1 3 1 1 3 1 3 |a | b a | b | | a | b a | b | 2 4 4 4 4 2 4 4 4 4 13 |a a | | b b | 44
1.
29 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 12 . 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
|uunn | 1, n we find
Tr |A B | unn | A B | u n
B|| unn u A n BA|
30 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 Ze1/H kT . (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, Ze Tr {H/ kT }. (135)
We use the basis from the eigenvectors | un of H. In this basis, we have
1 H / kT 1 En / kT nnZ u n|| e u n Z e (136) and
1/H kT nmZ u n| e | u m 0 for 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 |VVVVVVVAB | A B | A | B | A | B (138) and also their superpositions, where |V A and |V B are arbitrary vectors in the state spaces of A and B, respectively. The vector |V AB has two parts, one belongs to A and the other belongs to B. We have ||VVAB BA , ||VVVVABBA , |VVVVABBA | | | , and |VVVVABBA | | | . 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 |uAB | u A u B | u A | u B | u A | u B (139) where | u A and | uB are the basis states of the subsystems A and B, respectively.
31 To label the individual basis states of the subsystems A and B, we use indices nA and nB , respectively. Then, we can label the basis states of the combined system AB as |uAB | u A u B | u A | u B | u A | u B (140) nABABABAB n n n n n n n 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
AB uu AB AB| AB nABABABAB n, m m n n m m A B AB A B un u n | u m u m ABAB (141) uB| u A AB | u A | u B nBAAB n m m uB| u A AB | u A | u B . nBAAB n m m 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
A AB TrB ( ) B AB B (142) uunn| | , n where TrB is a map of operators known as the partial trace over system B. In the matrix form, we have A AB . (143) nAAABAB m n n, m n nB
AB Note that the partial trace of an operator of the type O| a1 a 2 | | b 1 b 2 | is given by Tr (|a a | | b b |) | a a | Tr (| b b |) BB1 2 1 2 1 2 1 2 (144) |a1 a 2 | b 2 | b 1 .
Here | a1 and | a2 are any two vectors in the state space of A, and | b1 and | b2 are any two vectors in the state space of B. In deriving the above expression, we have used the formula Tr |1 2 | 2 | 1 . 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 n 0,1,2,... . The basis states of the composite system “atom+field” are |a ;0 , |a ;1 , |a ;2 , … , |;an , … , and |b ;0 , |b ;1 , |b ;2 , … , |;bn , … .
32 The initial state of the composite system is |a ;0 . 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 |AB |ab ;0 | ;1 . (145) The density operator of the composite atom+photon system is AB|| AB AB |;0a |;1 b a ;0| ** b ;1| (146) | ||;022a a ;0| | ||;1 b b ;1| **|;0a b ;1| |;1 b a ;0|. 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 Tr AB Tr | AB AB | atom ph ph (147) | |22 |a a | | | | b b |.
Basic properties of the reduced density operator Hermitivity: We can show that
† AA , or, (148)
AA* nn'' n n.
Indeed, the matrix elements of the reduced density operator A are given by
AAAA nn'' uu n|| n A B AB B A un| u m | | u m | u n ' m A B AB A B (149) un u m|| u n' u m m AB nm,' n m. m
AB AB* AB From the Hermitivity of , we have nm, n ' m ' n ' m ', nm . Hence, we find
* A** AB AB nn',',' nm n m nm n m mm (150) AB A n',' m nm n n. m
33 Thus, A is a Hermitian operator.
Normalization: We can show that Tr A 1. (151) Indeed,
A A AB Tr nn nm, nm n n m (152) Tr AB 1. Thus, the probability conservation law is valid for the reduced density operator. Average of an observable: Consider an observable O A that depends only on part A . We can show that the average of this observable is given by the formula
AAAAA OOO TrAA Tr . (153) Indeed, by definition we have
A AB A OO TrAB AB A TrAB Tr O (154) AB A TrAB Tr O AAAA TrAAOO Tr . Another way of proving is the following:
A AB A OO TrAB AB AB A AB unm|| O u nm (155) nm A B AB A A B un u m| O | u n u m . nm Since O A does not depend on part B , it does not change any vectors in the state space of B , that is,
AABAAB O| un u m O (| u n u m AAB O| unm u (156) BAA umn O| u Consequently, we have
34 A A B AB A A B O un u m|| O u n u m nm A B AB B A A un u m| | u m O | u n nm A B AB B A A un u m| | u m } O | u n (157) nm AB A TrAB {{Tr }O } AAAA TrAA {OO } Tr { }. Thus, the average of O A 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 d 1 [H , ]. (158) dt i 0 Here the Hamiltonian of the free atom is given by
H0 ab| 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 d 1 [H , ] 2 . (160) dt i 0 2
Here is the atomic decay rate, and ||ab and ||ba are the atomic transition operators. The decay rate is sometimes called the Einstein A-coefficient and 3 2 3 is given as ab |dc | /3 0 , 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
35 1 [H , ], 00 i (162) 2. decay 2 We find 0 i ab| a a | | a a | | b b | | b b | , (163) |a a | 2 | b a | | a b | | a a | . decay 2 We can easily do some exercises to show that
aa|0 | 0,
bb|0 | 0, (164)
a|0 | b i ab ab , and
aa|decay | aa ,
bb|decay | aa , (165) ab| | . decay 2 ab The equations of motion for the matrix elements of can be obtained from Eq. (161): d , dt aa aa d , (166) dt bb aa d i . dt ab ab ab2 ab
It may be noted that aa bb 0, that is, aa bb constant . 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
t aa(te ) aa (0) , t bb(te ) 1 aa (0) , (167)
iab t t/2 ab(te ) ab (0) .
36 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 .
37 Exercises Exercise 1: Consider an arbitrary operator O and two arbitrary vectors |V and |U . Show that Tr{OVUUOV | |} | | . Exercise 2: Consider an arbitrary operator O and a state vector | . Show that Tr{OO | |} .
Exercise 3: Consider an arbitrary operator O and two basis vectors | un and | um . Show that
Tr{O | un u m |} O mn .
Exercise 4: Consider a density operator and two basis vectors | un and | um . Show that
Tr{ |uun m |} mn .
38 Solutions Exercise 1: Using the formula Tr{|ABBA |} | , we can show that Tr{OVUUOV | |} | | . Exercise 2: Use the result of exercise 1 and the definition OO | | .
Exercise 3: Use the result of exercise 1 and the definition Omn u m| O | u n .
Exercise 4: Use the result of exercise 1 and the definition mn uu m| | n .
39 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
HHH0 I , (168) where H0 and H I 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
H0 ab| a a | | b b |. (169)
The projection operators aa ||aa and bb ||bb describe the populations (probabilities) of levels a and b, respectively. Indeed, we have
aa Tr{ |a a |} a | | a aa (170) and
bb Tr{ |b b |} b | | b bb . (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
H0 ab | a a | | b b |
a aa b bb (172)
a aa b bb.
40 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 |aa | (1 ), 2 z (173) 1 |bb | (1 ). 2 z Hence, we find H ab () 0 22z a b (174) ab const, 2 z 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 a 0 /2 and
b 0 /2. Then, the constant is the above expression for the Hamiltonian H0 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 HI d E exE eaax(| | | bbx | | abx | | baxE | ) aa bb ab ba (175) e(| a b | xab | b a | x ba ) E
(|a b | | b a |) dx E .
Here we have used the properties xxaa bb 0 , have introduced the notation dx ex ab , which is called matrix element of the atomic dipole moment, and have assumed for simplicity that xab and consequently d x are real parameters. We present the electric component of the light field in the form
1 i t i t E E00cos t e e E . (176) 2 We introduce the Rabi frequency
dEx 0 /. (177) Then, we obtain
41 H (| a b | | b a |)( ei t e i t ) I 2 (178) ( )(eei t i t ) . 2 Here we have introduced the upward transition operator
ab ||ab (179) and the downward transition operator
ba |ba |. (180) We note that, for a free atom, in the Heisenberg presentation, the transition operators
itab itab ||ab and ||ba oscillate as e and e , respectively. Indeed, for a free
itab atom we have Tr Tr| b a | a || b ab ab (0) e . Therefore, the
i t i t i t i t it()ab it()ab terms e|| a b e and e|| b a e quickly vary as e and e , i t i t i t i t respectively. Meanwhile, e|| a b e and e|| b a e slowly vary as eit()ab and eit()ab , respectively. Let’s come back to the interaction Hamiltonian (178). We neglect the fast rotating terms ||a b eit and ||b a eit . This approximation is called the rotating-wave approximation. The result is H () ei t e i t I 2 (181) (|a b | ei t | b a | e i t ). 2
Appendix: Properties of the Pauli operators , , and z :
The operators ||ab , ||ba , and z |a a | | b b | are called the Pauli operators. The matrix forms of these operators are 01 , 00 00 , (182) 10 10 z . 01
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. We can easily show that
42 220, 2 z 1, 1 (1 z ), 2 1 (1 ), 2 z , z z , and , z z .
The above relations allow us to reduce any function f (,,) z of the Pauli operators to a linear form of the type c c czz . In particular, we find the following commutation relations:
[z , ] 2 ,
[z , ] 2 ,
[,]. z Exercise Consider a free two-level atom, with the Hamiltonian HH0 . Use the 0 2 z 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 di OHO [ , ]. dt
When we use the commutation relations [z , ] 2 , [z , ] 2 , and [,] and the Hamiltonian HH0 , we find z 0 2 z
43 di [ , ] 0, dt z2 0 z z di [,], i dt 2 00z di [,]. i dt 2 00z
The solutions to the above equations are zz(t ) (0) , (t ) (0)exp( i 0 t ) , and
(t ) (0)exp( i 0 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
| Cab | a C | b . (183) The state vector | satisfies the Schrödinger equation d i | H | ( H H ) | . (184) dt 0 I
For the probability amplitudes Ca and Cb , we find the equations di C i C eit C , dt a a a2 b (185) di C i C eit C . dt b b b2 a 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 ab , that is, 0 . We change the fast oscillating amplitudes Ca and Cb to the slowly varying amplitudes
c C eita , aa (186) itb cbb C e . Then, Eqs. (185) yield di cc , dt ab2 (187) di cc . dt ba2
The solutions for ca and cb can be written as
44 tt ca( t ) c a (0)cos ic b (0)sin , 22 (188) tt cb( t ) c b (0)cos ic a (0)sin . 22 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, ca (0) 0 and cb (0) 1. In this case, we find t ca ( t ) i sin , 2 (189) t ctb ( ) cos . 2 Hence, we have
t ita Ca ( t ) i sin e , 2 (190)
t itb Cb ( t ) cos e . 2
The populations of the levels, described by the diagonal matrix elements aa and bb , are found to be
22t aa(t ) | C a ( t ) | sin , 2 (191) 22t bb(t ) | C b ( t ) | cos . 2 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
* ab()()()t C a t C b t
tt it() iesin cos ab (192) 22 ii sin( t ) eit0 sin( t ) eit . 22
45 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, ca (0) 1 and cb (0) 0 . In this case, we find t cta ( ) cos 2 (193) t cb ( t ) i sin . 2 Hence, we have
t ita Ca ( t ) cos e , 2 (194)
t itb Cb ( t ) i sin e . 2
The populations of the levels, described by the diagonal matrix elements aa and bb , are found to be
22t aa(t ) | C a ( t ) | cos , 2 (195) 22t bb(t ) | C b ( t ) | sin . 2 As seen, the populations oscillate in time, with the Rabi frequency Ω. Meanwhile, the coherence (interference) is
* ab()()()t C a t C b t
tt it() iesin cos ab (196) 22 ii sin( t ) eit0 sin( t ) eit . 22
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 ca (0) 1/ 2 and cb (0) 1/ 2 . In this case, we obtain the solution
46 1 tt cab( t ) c ( t ) cos i sin , (197) 2 22 which leads to
1 tt ita Ca ( t ) cos i sin e , 2 22 (198)
1 tt itb Cb ( t ) cos i sin e . 2 22 Hence, we find that the populations of the levels are constant in time:
aa(tt ) bb ( ) 1/ 2. (199) We also find that the coherence is oscillating in time, with a constant envelope: 1 1 1 ()()().t C t C* t ei()ab t e i 0 t eit (200) ab a b 2 2 2 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 d 1 [H , ]. (201) dt i To obtain the evolution equations for the matrix elements of , we can use Eq. (201) directly. For this purpose, we write d i( O O† ), (202) dt where 1 OH . (203) In the matrix form, we have d i( O O†* ) i ( O O ). (204) dt
Let’s calculate the matrix elements O . When we use the explicit expression for the
Hamiltonian HHH0 I , we find
47 11 OHHH () 0 I (205) |a a | | b b | (| a b | ei t | b a | e i t ). ab2 We can easily show that Oe it , aa a aa2 ba Oe it , bb b bb2 ab (206) Oe it , ab a ab2 bb Oe it . ba b ba2 aa When we insert the above matrix elements into Eq. (204), we obtain di (),eei t i t dt aa2 ba ab di (),eei t i t (207) dt bb2 ba ab di ie it (). dt ab ab ab2 aa bb Note that we can also derive Eqs. (207) with the use of Eqs. (185) and the definitions
2 aa|C a | , |C |2 , bb b (208) * ab CC a b , * ba ab.
We introduce the notation ab for the detuning of the field. It is convenient to use new variables
it ab e ab , eit , ba ba (209) aa aa ,
bb bb. In terms of these variables, we have
48 di (), dt aa2 ba ab di (), (210) dt bb2 ba ab di i (). dt ab ab2 aa bb 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 ppab1/ 2 . In this case, the matrix elements of the initial density of the atom are given by (0) (0) 1/ 2, aa bb (211) ab(0) ba (0) 0. Consequently, we have (0) (0) 1/ 2, aa bb (212) ab(0) ba (0) 0. The Bloch equations show that the above values are true for any time t, that is, (tt ) ( ) 1/ 2, aa bb (213) ab(tt ) ba ( ) 0. 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
49 di C i C eit C , dt a a a2 b
di*** it Ca i a C a e C b , dt 2 di C i C eit C , dt b b b2 a di C*** i C eit C . dt b b b2 a When we use the above equations and the relations
2 aa|C a | , |C |2 , bb b * ab CC a b , * ba ab , 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 d 1 [,],H (214) dt i where the last term describes the decay and is given by [ 2 ]. (215) 2 We can rewrite Eq. (214) as
()(), coherent decay (216) where 1 ()[,] H (217) coherent i and ( ) [ 2 ]. (218) decay 2 In the matrix form, we have
()(). coherent decay (219)
50 The matrix elements () coherent are the matrix elements of the operator 1 ()[,] H . They are given by Eqs. (207). The matrix elements () are coherent i decay the matrix elements of the operator ( ) [ 2 ]. decay 2 They are given by the expressions on the right-hand side of Eqs. (166). Indeed, we can easily show that (), aadecay aa aa (), (220) bbdecay bb aa (). abdecay ab2 ab To obtain the evolution equations for the matrix elements of , we add to Eqs. (207) the decay terms (220). This procedure leads to di (eei t i t ), dt aa aa2 ba ab di (eei t i t ), (221) dt bb aa2 ba ab
diit ab ie ab ab ( aa bb ). dt 22
We introduce the notation ab for the detuning of the field. It is convenient to use new variables
it ab e ab , eit , ba ba (222) aa aa ,
bb bb. Using these variables, we obtain di (), dt aa aa2 ba ab di (), (223) dt bb aa2 ba ab di abi ab (). aa bb dt 22 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.
51
3. Absorption spectrum: saturation and power broadening We introduce the population difference
w aa bb (224) and the optical coherence
c ab. (225) Then, we can rewrite Eqs. (223) as d w ( w 1) i (* ), dt cc (226) di cciw . dt 22 We consider the adiabatic (steady-state) regime where dw d c 0. (227) dt dt In this regime, Eqs. (226) yield
* wi (),cc i (228) iwc . 22 The solutions to the above equations are 1 w (229) 1 s and i . (230) c 2( / 2 is )(1 ) Here s is the saturation parameter and is given by 22/2 s s 0 , (231) 2|/2 i |2 2 2 /41(2/) 2 where s0 is the on-resonance saturation parameter and is defined as 22 s . (232) 0 2
52 For low saturation, s 1, the population is mostly in the ground state ( w 1). For high saturation, s 1, the population is almost equally distributed between the ground and excited states ( w 0 , i.e. aa, bb 1/ 2 ).
The parameter s0 can be written in another form I s0 . (233) I s Here
2 I c00 E /2 (234) is the intensity of the laser beam and
22 c0 Is 2 (235) 22d x is the so-called saturation intensity. According to the Weisskopf-Wigner theory, we have 3 2 3 3 2 3 0|d | /3 0 c 0 dx / 0 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
33 Is hc //, 00 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
1ss0 / 2 aa (1 w ) 2 . (239) 2 2(1ss ) 1 0 (2 / )
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
s0 /2 scatt aa 2 . (240) 1s0 (2 / )
Note that, at high intensity, scatt saturates to /2. Equation (240) can be rewritten as
s0 /2 scatt 2 , (241) 1s0 1 (2 / ') where
53 ' 1 s0 . (242)
The dependence of the scattering rate scatt on the detuning is shown in Fig. 3 for several values of the saturation parameter s0 . 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 s0 , 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 s0 .
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 n scatt , where n is the number density of the atoms. Thus, we have
dI s0/ 2 n ( / 2)( I / I s ) nnscatt 22 dz1 s00 (2 / ) 1 s (2 / ) (243) n I I.
54 Here is the scattering cross section and n is the absorption coefficient. These coefficients are given by 1 2 . (244) n2 Is 1 s0 (2 / )
22 c0 hc When we use the expressions Is 23, we can rewrite Eq. (244) as 22d x 0
2 21d x 2 n c1 s (2 / ) 00 (245) 2 0 1 2 . 2 1s0 (2 / )
For low intensity, s0 1, we have
2 21d x 2 nc 1 (2 / ) 0 (246) 2 1 0 . 2 1 (2 / )2 In this regime, the absorption coefficient is independent of the field intensity I. Therefore, the solution for the field intensity is
z I(). z I0 e (247)
At exact resonance ( 0 ), the cross section (89) reduces to 2 0 . (248) n 2
In the case where II s , the absorption coefficient tends to zero. This does not mean the vanishing of the absorption. Indeed, in the limit
II s (249) we have In / 2. (250) Hence, Eq. (243) yields dI In ( / 2). (251) dz The solution to the above equation is I I(0) n ( / 2) z . (252)
55 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 I s , 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 21 2 2 2 2EPP 0 2 2 2 , (253) z c t t0 c t where P is the polarization density. We expand the field and the polarization in the forms 1 E E ei()*() t kz E e i t kz , 2 00 (254) 1 P P ei()*() t kz P e i t kz . 2 00 Here and kc / are the frequency and wave number of the field, respectively.
We introduce new variables z, (255) t z/. c In terms of these variables, we have
1 ii* E E00 e E e , 2 (256) 1 ii* P P00 e P e . 2 We also have , zc (257) , t and, consequently, 2 22 2 2 , z2 2 c c 2 2 (258) 22 . t22 On substituting Eqs. (256) and (258) into Eq. (253), we find
56 22 i 22 i eE2 0 c.c. cc (259) 2 i 1 2 e22 2 i P0 c.c. 0c The above equation will be satisfied if
2 2 2 2 2i 1 2 2 E00 2 2 2. i P (260) c c 0 c
We assume that the envelopes E0 and P0 vary slowly in space and time. When we keep only the lowest nonvanishing order on each side of Eq. (260), we obtain i EP00 . (261) 2 0c Since z , we can rewrite Eq. (261) as i ik EPP0 0 0, (262) zc2200 keeping in mind that the partial derivative with respect to z in Eq. (262) is taken under the condition t z/ c constant . 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
PE0 0 0, (263) where is the susceptibility. We write 'i '', (264) where ' and '' are the real and imaginary parts of . Then, Eq. (262) becomes i E( ' i '') E ( i ' '') E . (265) z022 c 0 c 0 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 ' n 1, (266) ref 2 while the absorption coefficient is defined by
57 ''k ''. (267) c
In terms of nref and , we can rewrite Eq. (265) as E0 ik( n ref 1) E 0 . (268) z 2
At low intensity, we can ignore the dependences of nref and on the field intensity. In this case, the solution of Eq. (268) for the field envelope E0 is
ik( nref 1) z z/2 E0 Ae e . (269) When we use the presentation A|| A ei and the formula, 1 E E ei()*() t kz E e i t kz , (270) 2 00 we obtain
z/2 E| A | e cos( t knref z ). (271)
The expression (271) explains why nref is called the refractive index and why is called the absorption coefficient.
We now calculate , nref , and . As shown in Sec. I.2.e, the polarization of a single two-level atom is Pdx( ab ba ). Here, we have assumed for simplicity that d x is a real parameter. For a medium with the atomic density n, the polarization density of the medium is given by
ii* P ndx( ab ba ) nd x ( c e c e ). (272) Consequently, the envelope of the polarization density is
P0 2. ndxc (273) Inserting Eq. (230) into Eq. (273) yields i P 2 nd 0 x ( 2is )(1 ) (274) 2nd2 E i x 0 . ( 2is )(1 ) Hence, we have
PE0 0 0, (275) where the susceptibility is given by
58 P 0 00E 21nd 2 i x (276) 0 ( 2is )(1 ) 2 22ndx i i 22. 001 s (2 / ) Then, we find
2 4nd x ' 22 (277) 001 s (2 / ) and
2 21nd x '' 2 . (278) 001 s (2 / ) With the help of Eqs. (266) and (267), we obtain
2 2nd x nref 1 22 001 s (2 / ) (279) 3 n0 / 1 22 4 1s0 (2 / ) and
2 21nd x 2 00cs1 (2 / ) (280) 2 n0 1 2 . 2 1s0 (2 / )
3 2 3 Here we have used the expression 00dcx / and the approximation 0 .
Equation (280) is in agreement with Eq. (245). Note that nref and depend on the field intensity via the saturation parameter s0 .
59
Figure 4: Real and imaginary parts of the susceptibility as functions of the normalized detuning
/ of the field for the saturation parameter s0 0.1.
According to Eqs. (266) and (267), the real part ' and the imaginary part '' of the susceptibility determine the refractive index nref 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.
60