15 1.10 Matrix formalism The right-hand side in the matrix form is then Awb1) 422 (《地1b)(地b2) (地bw) AN (BN 1) (1.94) which after a long multiplication reduces to a single number which is,of course,the single matrix element ( Often it is more convenient to write the above product by utilizing the relation (华bm）=(bml)* A11 A1A13 b1) A21 (b2 1o) (12* 431 A33 (bN 1) (1.95) (vi)The trace ofa matrix A is defined as a sum of its diagonal elements Tr(d)=(anl A lan))=∑Am (1.960 where a's form an orthonormal basis set.An important property of a trace is Tr(AB)=Tr(BA) (1.97 This can be proved by noting that TrB)=∑anl4Bla=∑∑Ald)B) (1.98) where we have used the completeness relation for the basis sets m).Since the matrix elements are numbers and no longer operators,they can be switched.Hence using completeness for the lam)'s we have Tr(AB)=>(aml B lan)(anl A lam)=>(aml BA lam)=Tr(BA),(1.99) which is the desired result.This leads to the generalization involving a product of an arbitrary number of operators that Tr(4BC...)=invarian (1.100) ander a cyclic permutation of the product ABC
15 1.10 Matrix formalism The right-hand side in the matrix form is then ψ |b1 ψ |b2 ... ψ |bN ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ A11 A12 . . A1N A21 A22 .. . . . A33 . . . . .. . AN1 . .. ANN ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ b1 |φ b2 |φ . . bN |φ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ , (1.94) which after a long multiplication reduces to a single number which is, of course, the single matrix element ψ| A |φ. Often it is more convenient to write the above product by utilizing the relation ψ |bm = bm |ψ ∗, b1 |ψ ∗ b2 |ψ ∗ ... bN |ψ ∗ ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ A11 A12 A13 . A1N A21 A22 A23 . . A31 A32 A33 . . . . .. . AN1 . .. ANN ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ b1 |φ b2 |φ . . bN |φ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ . (1.95) (vi) The trace of a matrix A is defined as a sum of its diagonal elements, Tr(A) = n an| A |an = n Ann, (1.96) where |an’s form an orthonormal basis set. An important property of a trace is Tr(AB) = Tr(BA) (1.97) This can be proved by noting that Tr(AB) = n an| AB |an = n m an| A |am am| B |an (1.98) where we have used the completeness relation for the basis sets |am. Since the matrix elements are numbers and no longer operators, they can be switched. Hence using completeness for the |am’s we have Tr(AB) = m n am| B |an an| A |am = m am| BA |am = Tr(BA), (1.99) which is the desired result. This leads to the generalization involving a product of an arbitrary number of operators that Tr(ABC ...) = invariant (1.100) under a cyclic permutation of the product ABC ...
16 Basic formalism 1.11 Eigenstates and diagonalization of matrices Consider the case where we know the matrix elements of A with respect to the basis given by la).That is,we know (la)'s are not necessarily the eigenstates of We call la)'s the old basis.We now wish to obtain the eigenstates and eigenvalues ofA. Let b)'s be the eigenstates of 4,we will call them the new basis,which satisfy (1.101) We proceed in a manner very similar to the previous problem.We multiply both sides of (1.101)by (ml and insert a complete set of states p), ∑(aml4 lapl(apl ba）=bn(am ba) (1.102) The above relation can be written as a matrix relation by taking m=1,2,....N for a fixed value of n 「A11A2 AIN (al lba) (a1b) A2 (a2lbn) (a2 1bm) bn (1.103) ANI ANN (an b) (an Ioa) as stated earlier,the matrix elem ()are known.The relation (1.103)can be written as A11-bn 11 AIN (alb) A22-6 a2 Iba) 0 (1.104) AN ANN (av b) which is valid for each value of n.Thus,effectively,the above relation corresponds to "diagonalization"of the matrix formed by the A A solution of equation(1.104)is possible only if the determinant of the Nx N matrix vanishes for each value of n.Hence the eigenvalues bn are the roots of the determinant equation det[4-]=0 (1.105)
16 Basic formalism 1.11 Eigenstates and diagonalization of matrices Consider the case where we know the matrix elements of A with respect to the basis given by |an. That is, we know am| A |an where |an’s are not necessarily the eigenstates of A. We call |an’s the old basis. We now wish to obtain the eigenstates and eigenvalues of A. Let |bn’s be the eigenstates of A, we will call them the new basis, which satisfy A |bn = bn |bn. (1.101) We proceed in a manner very similar to the previous problem. We multiply both sides of (1.101) by am| and insert a complete set of states ap , p am| A ap ap bn = bn am| bn. (1.102) The above relation can be written as a matrix relation by taking m = 1, 2, ... , N for a fixed value of n ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ A11 A12 . . A1N A21 A22 .. . . . .. . . . .. . AN1 . .. ANN ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ a1 |bn a2 |bn . . aN |bn ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ = bn ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ a1 |bn a2 |bn . . aN |bn ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ (1.103) where, as stated earlier, the matrix elements Amp = am| A ap are known. The relation (1.103) can be written as ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ A11 − bn A12 . . A1N A21 A22 − bn .. . . . .. . . . .. . AN1 . .. ANN − bn ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ a1 |bn a2 |bn . . aN |bn ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ = 0, (1.104) which is valid for each value of n. Thus, effectively, the above relation corresponds to “diagonalization” of the matrix formed by the Amn. A solution of equation (1.104) is possible only if the determinant of the N × N matrix vanishes for each value of n. Hence the eigenvalues bn are the roots of the determinant equation det [A − λI] = 0 (1.105)
17 1.11 Eigenstates and diagonalization of matrices where is a unit matrix.The different values of correspond to the different eigenvalues b.The corresponding elements (m)can then be determined by solving N simultaneous equations in (1.104). 1.11.1 Diagonalization through unitary operators Let us bring in the unitary operators,which will shed further light on determining eigenstates and eigenvalues ofA.Let U be the unitary operator that transforms the old basis to the new basis. bn）=Ulan). (1.106 This transformation preserves the norms of the two basis sets since Uis unitary.Equation (1.101)can be expressed as AU \an）=baU lan) (1.107 We now multiply both sides on the left by then Ut AU lan)=U lan)=bn lan) (1.108) where we have used the unitary property UU=1.Once again we multiply on the left, this time by (a.We find (aml U'AU lan）=bn(am an）=bn8mn (1.109 The right-hand side corresponds to a diagonal matrix.Thus the operator U must be such that UAU is a diagonal matrix and we write AD=U'AU (1.110) Once we find U then we can immediately obtain the eigenstates b)from (1.106)and ing the trace of both sides of(1.10)we obtain Tr(Ap)=Tr(UAU). (1.111) Since Ap isa diagonal operator with matrix elements given by the eigenvalues of A,the trace on the left of the above equation is simply the sum of the eigenvalues.For the right-hand side we note,using the invariance of a trace under cyclic permutation,that Tr(UAU)=Tr(AUU)=Tr(A). (1.112)
17 1.11 Eigenstates and diagonalization of matrices where I is a unit matrix. The different values of λ correspond to the different eigenvalues bn. The corresponding elementsam |bn can then be determined by solving N simultaneous equations in (1.104). 1.11.1 Diagonalization through unitary operators Let us bring in the unitary operators, which will shed further light on determining eigenstates and eigenvalues of A. Let U be the unitary operator that transforms the old basis to the new basis, |bn = U |an. (1.106) This transformation preserves the norms of the two basis sets since U is unitary. Equation (1.101) can be expressed as AU |an = bnU |an. (1.107) We now multiply both sides on the left by U†; then U† AU |an = bnU† U |an = bn |an (1.108) where we have used the unitary property U† U = 1. Once again we multiply on the left, this time by am| . We find am| U† AU |an = bnam |an = bnδmn. (1.109) The right-hand side corresponds to a diagonal matrix. Thus the operator U must be such that U† AU is a diagonal matrix and we write AD = U† AU. (1.110) Once we find U then we can immediately obtain the eigenstates |bn from (1.106) and eigenvalues bn from (1.109). Taking the trace of both sides of (1.10) we obtain Tr(AD) = Tr(U† AU). (1.111) Since AD is a diagonal operator with matrix elements given by the eigenvalues of A, the trace on the left of the above equation is simply the sum of the eigenvalues. For the right-hand side we note, using the invariance of a trace under cyclic permutation, that Tr(U† AU) = Tr(AUU† ) = Tr(A). (1.112)
18 Basic formalism Thus Tr()=sum of the eigenvalues of the operator 4. (1.113) This result holds even though A itself is not diagonal in the basis set lan). In a two-channel system,i.e.,in a system that consists of only two eigenstates,U is relatively easy to express.One writes u-[om。] (1.114) which is manifestly unitary.One then imposes the condition UAU diagonal matrix. (1.115) This is accomplished by taking the off-diagonal elements to be zero.From this relation one can determine the angle 6 and,therefore,the matrix U.The diagonal elements of the matrix(1.116)give the eigenvalues,while the eigenstates b)are obtained in terms of U 1.12 Density operator The expectation value ()taken with respect to a state )of an operator 4 was defined earlier.It was defined with respect to a single state,often referred to as a pure may,however,have a collection of ket la)then one refers to it as a pure ensemble. When all the states in an ensemble are not the same then it is called a mixed ensemble. If one is considering a mixed ensemble where w describes the probability that a state la) is present in the ensemble,describes the probability that a state )is present,and s on,then,instead of the expectation value,a),the relevant quantity is the ensemble average,which is defined as (4)av=>Wa (a Lal a) (1.116) where the sum runs over all the states in the ensemble.Naturally,if1,with all othe w's zero,then only the state )contributes in the sum,in which case we have a pure ensemble and the ensemble average is then the same as the expectation value.We note that a mixed ensemble is also referred to as a statistical mixture. Below we outline some important properties of ()and (
18 Basic formalism Thus Tr(A) = sum of the eigenvalues of the operator A. (1.113) This result holds even though A itself is not diagonal in the basis set |an. In a two-channel system, i.e., in a system that consists of only two eigenstates, U is relatively easy to express. One writes U = cos θ sin θ − sin θ cos θ , (1.114) which is manifestly unitary. One then imposes the condition U† AU = diagonal matrix. (1.115) This is accomplished by taking the off-diagonal elements to be zero. From this relation one can determine the angle θ and, therefore, the matrix U. The diagonal elements of the matrix (1.116) give the eigenvalues, while the eigenstates |bn are obtained in terms of U and |an through (1.107). We will return to this formalism in Chapter 13 when we focus on two-channel phenomena. 1.12 Density operator The expectation value α |A| α, taken with respect to a state |α, of an operator A was defined earlier. It was defined with respect to a single state, often referred to as a pure quantum state. Instead of a pure quantum state one may, however, have a collection of states, called an ensemble of states. If each state in this ensemble is described by the same ket |α then one refers to it as a pure ensemble. When all the states in an ensemble are not the same then it is called a mixed ensemble. If one is considering a mixed ensemble where wα describes the probability that a state |α is present in the ensemble, wβ describes the probability that a state |β is present, and so on, then, instead of the expectation value, α |A| α, the relevant quantity is the ensemble average, which is defined as Aav = α wα α |A| α (1.116) where the sum runs over all the states in the ensemble. Naturally, if wα = 1, with all other wi’s zero, then only the state |α contributes in the sum, in which case we have a pure ensemble and the ensemble average is then the same as the expectation value. We note that a mixed ensemble is also referred to as a statistical mixture. Below we outline some important properties of α |A| α and Aav
19 1.12 Density operator Inserting a complete set of states b)with n =1,2,...in the expression for the expectation value )we can write (a la)=>(alba)(bn 141 bm)(bmla) (1.117 =∑∑b4bn}bmla)abl (1.118) where we have made an interchange of some of the matrix elements,which is allowed because they are numbers and no longer operators.We now introduce a"projection operator" defined by Pa la)(al. (1.119) It has the property PalB)=la)(alB)=SaBla) (1.120) where we have taken the states la),IB)...as orthonormal.Thus P projects out the state la)when operating on any state.Furthermore, P2=la)(ala)(al=la)(al=Pa, (1.121) Pt ((a)(la))=la)(al=Pa. (1.122) The completeness theorem gives ∑P。=∑aal=1 (1.123) From(1.119)and (1.120)we can write l)in terms of Pa by noting that (bmla)(alba)(bm IPal bn) (1.124) where b)'s are eigenstates of an operator.Therefore. (alAa)=∑∑(bn4lbm)(mPalbn)) (1.125) =∑(bnl APa IDn） (1.126 Thus, (a Ala)=Tr(APa) (1.127 where"Tr"indicates trace.If we take A=1 (1.128)
19 1.12 Density operator Inserting a complete set of states |bn with n = 1, 2, ... in the expression for the expectation value α |A| α we can write α |A| α = n m α|bn bn |A| bm bm|α (1.117) = n m bn |A| bm bm|α α|bn (1.118) where we have made an interchange of some of the matrix elements, which is allowed because they are numbers and no longer operators.We now introduce a “projection operator” defined by Pα = |αα|. (1.119) It has the property Pα|β=|αα|β = δαβ|α (1.120) where we have taken the states |α, |β... as orthonormal. Thus Pα projects out the state |α when operating on any state. Furthermore, P2 α = |αα|αα|=|αα| = Pα, (1.121) P† α = (α|) (|α) = |αα| = Pα. (1.122) The completeness theorem gives α Pα = α |αα| = 1. (1.123) From (1.119) and (1.120) we can write α |A| α in terms of Pα by noting that bm|α α|bn = bm |Pα| bn (1.124) where |bn’s are eigenstates of an operator. Therefore, α |A| α = n m bn |A| bm bm |Pα| bn (1.125) = n bn| APα |bn (1.126) where we have used the completeness relation m |bmbm| = 1 for the eigenstates |bm. Thus, α |A| α = Tr(APα) (1.127) where “Tr” indicates trace. If we take A = 1 (1.128)