Comprehensive Analysis of Geometric Phase for SU(3) Representations
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Comprehensive Analysis of Geometric Phase for SU(3) Representations Abhirup Chatterjee, Dr.Sobhan Kumar Shounda Department of Physics, [Indian Institute Of Technology Hyderabad] Abstract Geometric Phase in Quantum Mechanics is generally formulated entirely in terms of geometric structure of the Complex Hilbert Space. We will exploit this fact in case of mixed states for three level open systems undergoing depolarization using the eight dimensional Poincare sphere in the SU(2) Polarisation picture and non unit vector rays in H3within the limit of pure state approach may be found to be in agreement with the Pancharatnam Phase, Berry Phase and Aharonov-Anandan Phase. 1
Contents 1 Introduction 3 2 Generalized Bloch Sphere representation of three level systems 5 2.1 Bloch Sphere Representation of the Two-Level Quantum System ......... 6 2.1.1 Density Matrix and the Density Operator Formalism ....... 7 2.1.2 Extension for the Three Level Quantum System .......... 10 2.1.3 Parametraisation of the Bloch Sphere for the Three Level system 13 3 Non Unit State Vector Mapping For the Mixed State of Three Level System 14 4 Geometric Phase For Three Level Open Quantum System 17 5 Lindblad Master Equation For Dissipative System 22 5.1 The Solution Of The Master Equation For The Three Level Quantum System ........................................ 23 6 SU(2) Polarization Picture 25 7 SU(2) Depolarization Dynamics for single mode fields 31 7.1 Master equation for the pure dephasing process .................. 31 7.2 Calculation of Geometric Phase ........................... 37 8 Modeling Depolarization with a non resonant randomly distributed atomic bath 38 9 Discussions And Conclusions 40 10 Future Works 43 11 Appendix 43 11.1 Appendix A: Description of Pure and Mixed Quantum States ..... 43 11.2 Appendix B: The Basics of the SU(2) and SU(3) Group ......... 46 11.3 Appendix C: Insights of Lindblad Operators and the Master Equation ...... 49 11.4 Appendix D: Homomorphic Mapping between SU(2) and SO(3) and extension for SU(3) ................................ 50 11.5 Appendix E: Interaction Picture in Quantum Mechanics: ........ 54 2
1 Introduction A Pure Quantum State retains a memory of its evolution in terms of Geometric Phase when it undergoes an evolution in the Parameter Space.The phase factor has its origin which is purely geometric in nature and it can arise even under the most general conditions where the system is undergoing an non-unitary evolution corresponding to the Hamiltonian which does not satisfy the criterion of adiabeticity and cyclicity. Although when geometric phase comes into the picture we talk about Berry’s framework [4] where he assumed the evolution to be cyclic in some parameter space and the Hamiltonian obeys the adiabeticity and cyclicity condition but Pancharatnam’s experimental work [9] on interference gives us the idea about the existence of such phase factor which is known as pancharatnam connection which states that if we consider any three mutually nonorthogonal vectors in a certain hilbert space say |ψ1⟩,|ψ2⟩and |ψ3⟩so that,⟨ψm|ψn⟩ = 0∀m, n = 1,2,3 then if |ψ1⟩is in phase with |ψ2⟩and |ψ2⟩is in phase with |ψ3⟩then |ψ1⟩may not be in phase with |ψ3⟩.The relative phase difference between any two non-orthogonal vectors is defined as follows. Let us consider two vectors |Ψ⟩and |Φ⟩such that they have a non zero value of inner product which is in general complex i.e.⟨Ψ|Φ⟩ = 0 and any complex number can be represented in the polar coordinate so with ⟨Ψ|Φ⟩=reiθ and here θdenotes the relative phase difference between the vectors. Now in Pancharatnams framework the three vectors were the three different electric fields E1, E2, E3such that Ei· Ej= 0,where i=jsay, chosen for the interferrometry experiment and experimentally the Pancharatnams connection was found with the fact that when the two electric fields are in phase with each other it will correspond to the interference maximum and when the relative phase difference between them be πit will correspond to the minimum in the interference pattern i.e. the intensities will be maximum and minimum when they are in phase or out of phase.Apparently there is no quantum mechanics or Schrodinger equation is involved in the Pancharatnams framework as it comes form the result of experimental classical optics but it has a connection in the Three dimensional Poincare sphere representation of the manifold. But after Berry’s discovery of the geometric phase lots of attempts has been done to generalize the idea of the geometric phase and Aharonov and Anandan [1] showed that if we lift up the condition of adiabeticity still we can define the geometric phase and the total phase can be written as a sum of the geometric phase and dynamical phase keeping the condition of cyclical evolution of the quantum system and the geometric phase in their framework is called Aharonov-Anandan Phase. After Aharonov Anandan’s discovery Samuel and Bhandari [25] further relaxed the condition of the cyclicity as the condition of adiabaticity has already been lifted in AharonovAnandans framework and gives the definition of the geometric phase for the non cyclic and non adiabatic evolution of the quantum system.All the approaches of the generalization of the Geometric phase requires the time dependent Schrodinger equation so these are known as the differential approach but Mukunda and simon [23] defined the geometric phase from the kinematic point of view which does not require the schrodinger equation hence the evolution of the quantum system can be non-unitary, thereby proving that the geometric phase is a Ray space quantity and showed that for three arbitrary non-orthogonal state vectors |ϕ1⟩,|ϕ2⟩,|ϕ3⟩ the non-vanishing geometric phase is given by the Bargmann Invariant of order 3 abbreviated by BI(3). Hence, Φgeo =−arg(∆3) = −arg[Tr(ρ1ρ2ρ3)] = −arg[(ϕ1, ϕ2)(ϕ2, ϕ3)(ϕ3, ϕ1)] = 0 with ρ1, ρ2, ρ3are the pure state density matrices corresponding to the states |ϕ1⟩,|ϕ2⟩,|ϕ3⟩ and by definition of the density or projection operators we have ρm=|ϕm⟩⟨ϕm|∀m= 1,2,3 and Tr(ρm) = 1∀m= 1,2,3.So far all the calculations on the geometric phase has been done for the pure states now it can be generalized for the mixed states too.In the present context of the project work we try to find out an expression of the Geometric phase for a three level open quantum system by the SU(3) representations using the fact that the mixed states lies on 3
the interior of the eight dimensional Bloch sphere.Previously progress has been made on the calculation of the geometric phases for two level open quantum systems and there are works on geometric phases for mixed states using the Schmidt’s purification method which lifts the mixed states to pure states. Geometric phase for the three level quantum system of the mixed states can be defined by establishing in connecting the density matrix with the non-unit vector ray in a three dimensional complex Hilbert space. Because the Geometric Phase depends only on the smooth curve on this space,it is formulated entirely in terms of geometric structures.Under the limiting pure state, our approach is in agreement with the Berry’s phase, Pancharatnam phase and Aharonov and Anandan Phase.We find that Berry phase of mixed state correlated to population inversions of three level open quantum system [13]. More precisely,the amplitudes of wave functions are mapped onto given points on the Poincare sphere for a pure state [15]. Application of geometric phases in quantum computation [29] has motivated their studies under more realistic situations [28] [27]. In a real system, it is unavoidable interaction of a quantum system with its surrounding environment. The interaction may lead to an irreversible loss of information on the system so the process limits the ability to maintain pure quantum states in quantum information. Therefore, it is necessary to include the effect of decoherence. Up to now, however, the definition of the geometric phase for the open system is still a controversial issue. It is therefore extremely important to understand all aspects of the geometric phase in open system [8]. It is known that a d-dimensional qudit possesses a much more complex but richer structure than the ordinary qubit. In quantum information science, thus, information processing tasks can not only be implemented using two-dimensional qubits but also are sometimes more efficiently performed using the qudits as carriers of information [11]. It has been found that the qudits are better adapted for certain purposes, such as quantum cryptography. It is known that there have been many proposals tackling the geometric phase of two-level open systems from different generalizations of the parallel transport condition [7]. However, it may be difficult to expand it to the three-level open system. The generalizations especially are not unique so as to give out different results. In addition, a general belief is that the Berry phases are geometric in their nature, i.e., proportional to the solid angle enclosed by the closed curve in parameter space. Therefore, we will express the geometric phase [30] for the three-level open system in terms of geometric structures on a three-dimensional complex Hilbert space. We have calculated the geometric phase for a Three level quantum system incorporating the effect of Depolarization in the SU(2) Polarisation Picture. Polarization of light is a key concept that has deserved a lot of attention over the years. Apart from its fundamental significance, it is also of interest in several active technological fields.One of the interesting phenomena is the Depolarization. When a polarised electromagnetic beam propagates through a certain medium due to the interaction of the beam with its surrounding’s the degree of polarization of the polarised electromagnetic beam decreases and this is known as Depolarization [6]. In classical optics, this depolarization is ascribed, broadly speaking, either to birefringence (as it usually happens, e.g., in optical fibers [14] or to scattering by randomly distributed particles [12]. In both cases the net result is an effective anisotropy that leads to a decorrelation of the phases of the electric field vector. In quantum optics, a sensible approach to deal with this decorrelation is through the notion of decoherence, by which we loosely understand the appearance of irreversible and uncontrollable quantum correlations when a system interacts with its environment [31]. Usually, decoherence is accompanied by dissipation, i.e., a net exchange of energy with the environment. However, we are interested in the case of pure decoherence (also known as dephasing), for which the process of energy dissipation is negligible. The purpose of this project work is to address the SU(2) Polarization picture along with 4
the quantum theory of SU(2) Depolarization and to model the depolarization dynamics by Lindblad type master equation, to solve the master equation for few photonic systems and the calculation of the geometric phase. First we will discuss very fundamental model for the description of the Depolarization along with its drawbacks and the alternative approaches to describe the depolarization. 2 Generalized Bloch Sphere representation of three level systems Because a higher-dimensional Hilbert space may associate with quantum cryptography and entanglement, an increasing interest is to study the geometric phase of the three-level open system. Now,in order to deal with the quantum states (both, the pure and the mixed states) for a three level quantum system we need to represent them in a suitable manner i.e. some sort of mapping is required just like in classical polarisation optics the completely polarised states are mapped to the surface of the Poincare sphere and the partially polarised states are being mapped in the interior of the Poincare sphere. So, in order to understand the representation of the states for the three level Quantum system we first introduce the idea for the representation of the pure and mixed states of a two level quantum system using a three dimensional Bloch Sphere and then extend this idea for the three level system, roughly speaking, the Bloch sphere with radius unity is sometimes called the Poincare Sphere; the terminology is being used in the context of polarisation optics. In the present context we will call it a Bloch sphere. In, the Classical Polarisation optics the idea behind that is to map the polarisation states of light on the surface of the sphere, the polarisation states with degree of polarisation being unity are mapped on the surface of the sphere and the partially polarised states being mapped inside the sphere so for a sphere of radius unity all the normalised completely polarised states can be represented by using the polar angle θ∈[0, π] and azimuthal angle ϕ∈[0,2π] only in a three dimensional Spherical polar Coordinate System. For the description of the partially polarised light we require three variables r, θ, ϕ respectively. Figure 1: Mapping of polarisation states of light on the Poincare sphere. 5
2.1 Bloch Sphere Representation of the Two-Level Quantum System Now, let us do the similar thing as discussed above in the context of Classical Polarisation Optics for the description of the polarised states using Quantum mechanics for a two level quantum system and the most simple system will be a Qubit.A qubit is defined as a two level quantum system having two states represented as,|0⟩and |1⟩which forms the orthonormal basis of the underlying two dimensional Hilbert Space H2and can be represented by 2 ×1 column matrices so that the general state of the system |ψ⟩can be written as a linear combination of |0⟩=1 0and |1⟩=0 1with ⟨m|n⟩=δmn∀m, n = 1,2.So we have, |ψ⟩=α|0⟩+β|1⟩;α, β ∈C(2.1.1) Let us write, α=a1+ib1and β=a2+ib2,where a1, a2, b1, b2∈ ℜ the normalaisation of the state vector |ψ⟩i.e. ⟨ψ|ψ⟩= 1 requires |α|2+|β|2=a2 1+a2 2+a2 3+a2 4= 1.It will be better to use the polar coordinate representation of the complex numbers α, β ∈Cso that, α=r0eiθ0 and β=r1eiθ1we get, |ψ⟩=r0eiθ0|0⟩+r1eiθ1|1⟩ =eiθ0{r0|0⟩+r1ei(θ1−θ0)|1⟩} =eiθ0{r0|0⟩+r1eiϕ |1⟩},[∵(θ1−θ0) = ϕ] (2.1.2) As, the two states differing by a phase factor does not represent two different physical states we can write, |ψ⟩=r0|0⟩+r1eiϕ |1⟩(2.1.3) the normalaisation of the state vector requires |α|2+|β|2=r2 0+r2 1= 1. One of the possible choice of r0and r1that will satisfy the condition is r0= sin θ 2and r1= cos θ 2such that r2 0+r2 1= 1.With this substitutions we get, |ψ⟩= cos θ 2|0⟩+eiϕ sin θ 2,[0 ≤θ≤π, 0≤ϕ≤2π] (2.1.4) in the above equation θand ϕare the usual polar and azimuthal angle for the spherical polar coordinate system.So, we can represent the general state of the qubit by using a sphere of radius unity where all the pure states i.e. |ψ⟩can be mapped to the surface of the 3 dimensional sphere of unit radius. We know that the basis states of the qubit are the pure states and any state which can be written as a linear combination of the basis states will also be a pure state and they will lie on the surface of the 3 dimensional sphere also, known as Bloch Sphere which can be represented by the pair of angular coordinates θ, ϕ. Let, us consider some simple situations. For, θ= 0 and ϕ= 0 we get, |ψ⟩=|0⟩; corresponds to the state |0⟩being mapped to the north pole. Similarly,for θ=πand ϕ= 0 we get, |ψ⟩=|1⟩; corresponds to the state |1⟩being mapped to the diametrically opposite position i.e. south pole and let us also check that for θ=π 2and ϕ= 0 we get, |ψ⟩=1 √2(|0⟩+|1⟩) = |+⟩which corresponds to the homogeneous linear superposition of the basis states being mapped at the equator. But, here we are dealing so far with the pure states which lies on the surface of the Bloch Sphere but it is known that the mixed states for the two level quantum system can be mapped to the interior region of the Bloch sphere. But there is a problem that the mixed states can not be represented as a linear combination of the pure states so the alternative approach is to use the density operator to represent the mixed states.First we introduce the Density Operator and Density Matrix Formalism required to represent the Mixed states of a Quantum system but it can be equally applicable for the pure states too. 6
Figure 2: Mapping of pure states of a qubit on the Bloch sphere. 2.1.1 Density Matrix and the Density Operator Formalism Let us consider a system and large number of virtual copies of the system which constitutes an ensemble.The time dependent schrodinger equation for the kth member of the ensemble is given by, ˆ H|ψ(k)(t)⟩=iℏ∂ ∂t |ψ(k)(t)⟩ |ψk(t)⟩=X n ak n(t)|ϕn⟩(2.1.5) here, |ϕn⟩be the orthonormal basis and can be choosen as the eigenbasis of the hamiltonian satisfying the condition ⟨ϕm|ϕn⟩=δmn∀m, n and Pn|ϕn⟩⟨ϕn|=I. Then the matrix elements of the density operator and the operator itself is defined as follows, ρmn(t) = 1 N N X k=1 ak m(t)ak∗ n(t) ˆρ=1 NX k|ψ(k)(t)⟩⟨ψ(k)(t)|(2.1.6) so, we can define the matrix element of the density operator ˆρin the eigenbasis of the Hamiltonian as, ρmn =⟨ϕm|ˆρ|ϕn⟩and exploiting the completeness relation of the normalised eigenbasis of the hamiltonian we van write, ˆρ=ˆ Iˆρˆ I =X m|ϕm⟩⟨ϕm|ˆρX n|ϕn⟩⟨ϕn| =X mX n⟨ϕm|ˆρ|ϕn⟩|ϕm⟩⟨ϕn| =X mX n ρmn |ϕm⟩⟨ϕn| (2.1.7) where,ρmn =⟨ϕm|ˆρ|ϕn⟩defines the matrix elements of the density operator. The time evolution of the density operator is governed by the Liouville equation of Quantum Statistical Mechanics given by, iℏ∂ ∂t ˆρ(t) = [ˆρ, ˆ H] (2.1.8) 7
in statistical equilibrium ∂ρ ∂t = ˙ρ= 0 ⇒[ˆρ, ˆ H] = 0 so,the density operator and the Hamiltonian has the simultaneous eigenstates i.e.{|ϕn⟩} which allows us to write ρmn =⟨ϕm|ˆρ|ϕn⟩= ρn⟨ϕm|ϕn⟩=ρnδmn where, ρnbe the eigenvalues of the density operator and the density matrix will be diagonal in its own eigenbasis. It is easy to check that the density operator is hermitian i.e.ˆρ†= ˆρSo, finally we have, ˆρ=X n ρn|ϕn⟩⟨ϕn|(2.1.9) The expectation value of any observable ˆ Arepresented by a hermitian operator can be expressed as, ⟨ˆ A⟩=1 N N X k=1 ⟨ψ(k)(t)|ˆ A|ψ(k)(t)⟩ =Tr[ˆρˆ A] Tr[ˆρ] (2.1.10) The above process is called the double averaging process in quantum statistical mechanics i.e. the quantum mechanical expectation value followed by the classical ensemble average.For simplicity if we assume the state vector to be normalized i.e. ⟨ψ(k)(t)|ψ(k)(t)⟩= 1 then Tr[ˆρ] = 1 which leads to the form, ⟨ˆ A⟩=Tr[ˆρˆ A] (2.1.11) The above equation gives the closed mathematical form of the Density operator corresponding to the mixed state and as a special case of this we can obtain the density of state corresponding to a pure state |ϕ⟩say, ˆρ(ϕ) = |ϕ⟩⟨ϕ|=ϕϕ†and if |ϕ⟩be normalized then the pure state density operator will satisfy ˆρ2= ˆρ, ˆρ†= ˆρand Tr[ˆρ] = 1. Now, for a two level quantum system the density matrix will be of dimension 2×2 hermitian matrix with Tr[ˆρ] = 1 and if it corresponds to a pure state then ˆρ2= ˆρ. As we know that the Generators of the SU(2) are the Pauli spin matrices which are hermitian and traceless i.e. σ† i=σi, i =x, y, z and Tr[σi] = 0 denoted by ˆ σ having three components σi, i =x, y, z spans over the two dimensional matrix spaces then any 2 ×2 traceless hermitian matrix can be written as a linear combination of the three pauli matrices. Any 2 ×2 hermitian matrix can be written as a linear combination of the three pauli matrices and one identity matrix I2×2. The density matrix corresponding to the two level quantum system being hermitian can be written as a linear combination of one 2 ×2 Identity matrix i.e. I2×2and the three Pauli Matrices σi. The pauli Matrices are given by, σx=0 1 1 0, σy=0−i i0, σz=1 0 0−1(2.1.12) So, we can write that, ˆρ=Aˆ I+ B·ˆ σ =Aˆ I+Biˆσi (2.1.13) Now, recalling the Lie algebra of the SU(2) we can write the following properties of the pauli matrices σx, σy, σzas follows, [ˆσi,ˆσj] = ˆσiˆσj−ˆσjˆσi= 2iϵijk ˆσk {ˆσi,ˆσj}= ˆσiˆσj+ ˆσjˆσi= 2δij ˆσiˆσj=iϵijk ˆσk+δij∀i, j = 1,2,3 Tr[ˆσiˆσj]=2δij∀i, j = 1,2,3 (2.1.14) 8
Using the above mentioned properties of the pauli matrices we get, ˆρ=Aˆ I+Biˆσi Tr[ˆρ] = ATr[ˆ I] + BiTr[ˆσi]⇒A=1 2(2.1.15) and, ˆρˆσj=Aˆσj+Biˆσiˆσj Tr[ˆρˆσj] = ATr[ˆσj] + BiTr[ˆσiˆσj] Tr[ˆρˆσj] = 2Biδij Tr[ˆρˆσj]=2Bj⇒Bi=1 2Tr[ˆρˆσi] (2.1.16) Putting the values of Aand Biin the equation [2.1.13] we get, ρ=Aˆ I+Biˆσi =1 2ˆ I+1 2Tr(ˆρˆσi)ˆσi =1 2(ˆ I+Tr(ˆρˆσi)ˆσi =1 2ˆ I+n ·ˆ σ (2.1.17) where, we have defined the three dimensional Bloch Vector n ∈ R3having three components enlisted as, ni=Tr[ˆρˆσi],where , i = 1,2,3.If we represent a pure state by the density operator ρthen it can be easily shown that n2 1+n2 2+n2 3= 1. Let us check using the properties of the pauli matrices as follows; ˆρ2= ˆρˆρ =1 2ˆ I+n ·ˆ σ1 2ˆ I+n ·ˆ σ =1 4ˆ I+n ·σ +n ·σ + (n ·σ)2 =1 4ˆ I+niˆσi+niˆσi+ninjˆσiˆσj =1 4ˆ I+ 2niˆσi+ninjˆσiˆσj =1 4ˆ I+ 2niˆσi+hninjδij +iϵijkninjˆσki =1 4ˆ I+ 2niˆσi+hX i nini+i(n×n)kˆσki =1 4ˆ I+ 2niˆσi+X i n2 i (2.1.18) Now, in order to have ˆρ2= ˆρwe must have Pin2 i= 1, the maximum number of independent Bloch Vector components are 2 but the equation [2.1.17] holds for both the pure and Mixed states. using the fact that = ⟨ˆ A⟩=Tr[ˆρˆ A]; for a pure state ˆρ=|ψ⟩⟨ψ|the components of the bloch vector can be written as, ni=Tr[ρσi] = ⟨σi⟩=⟨ψ|σi|ψ⟩;∀i= 1,2,3 (2.1.19) 9
where, ˆρ=|ψ⟩⟨ψ|=ψψ†.So, for a given nonunit vector ray |ψ⟩∈H3the components of the bloch vector are given by, ni=Tr[ˆρˆ λi] = ⟨ˆ λi⟩=⟨ψ|ˆ λi|ψ⟩, i=1,2,..,8 Now, we wnat to find how the components of the bloch vector changes under the unitary transformation of the nonunit vector ray in the hilbert space.Again we bring back the idea for the two level quantum system and try to extend it for the three level. In case of two level quantum system the density matrix being 2 ×2 the bloch vector has the three components which can be written as the expectation value of the pauli spin matrices with respect to the nonunit vector ray in the two dimensional hilbert space. If we consider a SU(2) transformation of the nonunit vector ray |ψ⟩ ∈ H2such that, |ψ⟩′=u|ψ⟩where, u∈SU(2) then the components of the bloch vector changes to n′ isuch that n′ i=Rik(u)nkwhere,Rik being the matrix elements of a 3 ×3 matrix R(u)∈S0(3) and this becomes possible because of the homomorphism between the groups SU(2) and SO(3) i.e. an unitary transformation corresponds to the orthogonal transformation. We have, n′ i=Riknk⇒ n′=R n.(3.4) We know that the components of a vector transforms in the similar way as the coordinates transforms under the passive transformation and as a result the components of the bloch vector will obey the same coordinate transformation rule under the SO(3) group as we know in case of the Euclidean coordinate system. where, the matrix elements Rik are given by, n′ i=⟨ψ′|ˆσi|ψ′⟩=⟨ψ|u†ˆσiu|ψ⟩ n′ i=Riknk=Rik ⟨ψ|ˆσk|ψ⟩=⟨ψ|Rik ˆσk|ψ⟩ Rik ˆσk=u†ˆσiu Rimˆσmˆσk=u†ˆσiuˆσk ⇒RimTr[ˆσmˆσk] = Tr[u†ˆσiuˆσk] = Tr[ˆσiuˆσku†] ⇒2Rimδmk =Tr[ˆσiuˆσku†] ⇒Rik =1 2Tr[ˆσiuˆσku†] (3.5) We will just extend the idea of the homomorphism of the groups SU(2) and SO(3) in case of SU(3) and SO(8) for the three level quantum systems. In the present situation the bloch vector being an eight dimensional vector with eight components if we condider an unitary transformation of the nonunit vector ray |ψ⟩∈H3the components of the bloch vector will, transform under the SO(8) transformation.So, let us define an SU(3) transformation of the nonunit vector ray |ψ⟩′=u|ψ⟩where, u∈SU(3).Then, the componenets of the bloch vector will transform to n′ isuch that, ni=Rik(u)nk⇒ n′=R n.(3.6) where,Rik being the matrix elements of a 8 ×8 matrix R(u)∈SO(8).With SU(3) the situation is more intricate. We have a particular real eight-dimensional adjoint or octet representation of SU(3) given by certain 8×8 real orthogonal unimodular matrices, which is irreducible. All real orthogonal unimodular 8 ×8 matrices taken together form the 28 dimensional group SO(8); the matrices of the octet representation of SU(3) are a ‘very small’ eight-dimensional subset of SO(8), in fact a subgroup. Just like we have calculated the matrix elements of the unimodular orthogonal matrix R∈SO(3) in case of the two level quantum system here we will proceed in the similar way to calculate the transformation matrix R∈SO(8) for the components of the 16
bloch vector. n′ i=⟨ψ′|ˆ λi|ψ′⟩=⟨ψ|u†ˆ λiu|ψ⟩ n′ i=Riknk=Rik ⟨ψ|ˆ λk|ψ⟩=⟨ψ|Rik ˆ λk|ψ⟩ Rik ˆ λk=u†ˆ λiu Rimˆ λmˆ λk=u†ˆ λiuˆ λk ⇒RimTr[ˆ λmˆ λk] = Tr[u†ˆ λiuλk] = Tr[ˆ λiuˆ λku†] ⇒2Rimδmk =Tr[ˆ λiuˆ λku†] ⇒Rik(u) = 1 2Tr[ˆ λiuˆ λku†] (3.7) Similarly one can verify that for two different transformation matrices u, u′∈SU(3) we have R(u)R(u′) = R(uu′).In Short we can summarise the results as, u∈SU(3) →Rik(u) = 1 2Tr[ˆ λiuˆ λku†], R(u)∈SO(8); R(u)R(u′) = R(uu′) (3.8) Now, given any two ’octet vectors’ a,b∈R8, we can form one scalar product and two diffrent octet vectors from them defined as follows: a·b=arbr; (3.9a) a∧b=−b∧a=frstasbt; (3.9b) a∗b=b∗a=√3drstasbt; (3.9c) R(u)a∧R(u)b=R(u)a∧b; (3.9d) R(u)a∗R(u)b=R(u)a∗b(3.9e) 4 Geometric Phase For Three Level Open Quantum System Let us consider an open curve C=|ψ(t)⟩defined in the manifold Bwhich is also known as the state space and the curve is smoothly parametrised by some parameter tlies entirely inside the State space so that at each points on the curve corresponds to some |ψ(t)⟩. Because of the mapping exist between State space and the Ray Space i.e. B → R the point on the curve Cwill be mapped to their corresponding density operators ˆρ(t) = |ψ(t)⟩⟨ψ(t)|=ψ(t)ψ(t)†so that there will be an image curve in the Ray Space Rdefined by, C=nˆρ(t) = |ψ(t)⟩⟨ψ(t)|o on each point of the curve there is a density operator or, projection operator ˆρ(t) = Π |ψ(t)⟩ satisfying the conditions Tr[ˆρ(t)] = 1,ˆρ†(t) = ˆρ(t) and the inverse mapping leads to a lift of the curve Cin the state space Bwhich can be defined as, C → C=|ψ(t)⟩ → ˆρ(t) = Π(|ψ(t)⟩) = |ψ(t)⟩⟨ψ(t)|(4.1) C→ C =n|ψ(t)⟩= Π−1ˆρ(t)o(4.2) Let the curve Cdescribes the evolution of the quantum state |ψ(t)⟩parametrized by a smoothly varying parameter talthough we can consider either an open curve or a closed curve.Let us divide the curve into small segments and the points of subdivisions are t0, t1, t2, ..., tN. The state vectors at that points of subdivision are respectively |ψ(t0)⟩,|ψ(t1)⟩,|ψ(t2)⟩, ..., |ψ(tN)⟩ 17
or, in general |ψi⟩=|ψ(t=ti)⟩=|ψ(ti)⟩, i = 1,2,3, ..., N and the corresponding density operators in the image curve be ρ(ti) = ρ(t=ti) = |ψ(ti)⟩⟨ψ(ti)|and |ψi⟩=|ψ(t=ti)⟩= Π−1ρ(ti)∈ B.Then, each trajectory can be described by a discrete sequence of quantum states n|ψ0⟩,|ψ1⟩,|ψ2⟩, .., |ψN⟩o. Then the equation of the geometric phase for the three level open quantum mixed state is given by the Pancharatnam Formulae.In order to understand the idea let us work out some simple mathematical steps. We have subdivided the entire curve Cinto some small segments and let us call those segments as C0,C1,C2, ..., CNand the corresponding image segments of the curve in the Ray space are C0, C1, C2, ..., CNrespectively where,C0joins |ψ0⟩and |ψ1⟩,C1joins |ψ1⟩and |ψ2⟩and so on. The same goes for the curve segments C0, C1, C2, ..., CNwhich joins the density operators corresponding to the points of subdivions associated with the vectors |ψ0⟩,|ψ1⟩,|ψ2⟩, .., |ψN⟩. Now, the geometric phase along the union of those curve segments in the Ray space will be as follows, γghC0∪C1∪C2∪... ∪CNi=γg[C0] + γg[C1] + γg[C2] + ... +γg[CN]−BN[ψ0, ψ1, ψ2, ..., ψN] (4.3) where, BN= arg Tr[ρ0ρ1ρ2...ρN] = arg n⟨ψ0|ψ1⟩⟨ψ1|ψ2⟩⟨ψ2|ψ3⟩... ⟨ψN|ψ0⟩o = arg n∆N(ψ0, ψ1, ψ2, ...ψN)o (4.4) with, ∆N(ψ0, ψ1, ψ2, ...ψN) be the Bargmann Invariant of order N i.e. BI(N). In the continuum limit i.e. N→ ∞ if the geometric phases vanishes on individual segment curves i.e. γg[Ci] = 0∀i= 1(1)Nthe formulae for the Geometric Phase γgbecomes, γg=−lim N→∞arg Tr[ρ0ρ1ρ2...ρN] = −lim N→∞arg n⟨ψ0|ψ1⟩⟨ψ1|ψ2⟩⟨ψ2|ψ3⟩... ⟨ψN|ψ0⟩o(4.5) Now, as we know that the Geometric Phase can be written as, γg(C) = φtot(C)−φdyn(C) the former being the Total phase that depends on the initial and final points of the curve C ⊂ B and the later being the Dynamical Phase and the striking feature of the geometric phase is that it does not depend on the System Hamiltonian rather it only depends on the curve joining the end points lying in the Ray Space R. Using the fact that the total phase can be written as the sum of the Dynamical phase and the Geometric phase the Pancharatnam’s formulae in the continuum limit leads to, γg=−LtN→∞arg (⟨ψ0|ψ1⟩⟨ψ1|ψ2⟩.... ⟨ψN−1|ψN⟩⟨ψN|ψ0⟩) = arg ⟨ψ(t0)|ψ(t)⟩−Im ˆt t0 dτ ⟨ψ(τ)|d dτ |ψ(τ)⟩ ⟨ψ(τ)|ψ(τ)⟩!(4.6) where, the total phase φtot and the dynamic phase φdyn is given by, φtot[C] = arg ⟨ψ(t0)|ψ(t)⟩(4.7a) φdyn[C] = −Im ˆt t0 dτ ⟨ψ(τ)|d dτ |ψ(τ)⟩ ⟨ψ(τ)|ψ(τ)⟩!(4.7b) From ,the previous ansartz we have |ψ⟩=√r ei(α−γ)sin θcos ϕ ei(β−χ)sin θsin ϕ eiξ cos θ which satisfies the equation ni=√3 2⟨ψ|λi|ψ⟩and now the time dependence of |ψ⟩will be translated through the angular 18
bloch parameters i.e.(θ, ϕ, α, β, γ, χ, ξ). For convenience we define θ(t) = θt, α(t) = αt, β(t) = βt, γ(t) = γt, ϕ(t) = ϕt, χ(t) = χt, ξ(t) = ξtand θ(t0) = θ0, α(t0) = α0, β(t0) = β0, γ(t0) = γ0, ϕ(t0) = ϕ0, χ(t0) = χ0, ξ(t0) = ξ0so that we can write, |ψ(t)⟩=√r ei(αt−γt)sin θtcos ϕt ei(βt−χt)sin θtsin ϕt eiξtcos θt (4.8) Then, the Total phase φtot[C] is calculated as follows, ⟨ψ(t0)|ψ(t)⟩=re−i(α0−γ0)sin θ0cos ϕ0e−i(β0−χ0)sin θ0sin ϕ0e−iη0cos θ0 ei(αt−γt)sin θtcos ϕt ei(βt−χt)sin θtsin ϕt eiξtcos θt =X+iY (4.9) where, Xand Ybeing the real and imaginary part of ⟨ψ(t0)|ψ(t)⟩which are given by, X=rncos (αt−γt−α0+γ0) cos θtcos ϕtcos θ0cos ϕ0+ (4.10) sin (βt−χt−β0+χ0) sin θtsin ϕtsin θ0sin ϕ0+ sin (ηt−η0) cos θ0cos θto(4.11) Y=rnsin (αt−γt−α0+γ0) sin θtcos ϕtsin θ0cos ϕ0+ (4.12) sin (βt−χt−β0+χ0) sin θtsin ϕtsin θ0sin ϕ0+ sin (ηt−η0) cos θ0cos θto(4.13) Then, the Total phase φtot = arg ⟨ψ(t0)|ψ(t)⟩= tan−1(Y X) where, X and Y are the real and imaginary part of ⟨ψ(t0)|ψ(t)⟩and it leads to φtot = tan−1nsin (αt−γt−α0+γ0)A+ sin (βt−χt−β0+χ0)B+ sin (ηt−η0)Co ncos (αt−γt−α0+γ0)A+ cos (βt−χt−β0+χ0)B+ cos (ηt−η0)Co(4.14) where, we have introduced three factors A, B, C given by, A= sin θ(t) cos ϕ(t) sin θ(t0) cos ϕ(t0) = sin θtcos ϕtsin θ0cos ϕ0,B= sin θtsin ϕtsin θ0sin ϕ0= sin θ(t) sin ϕ(t) sin θ(t0) sin ϕ(t0) ,C= cos θ(t) cos θ(t0) = cos θtcos θ0, these terms appear in both the real and imaginary part of the inner product ⟨ψ(t0)|ψ(t)⟩. The expression of the geometric phase given by the equation [4.6] is gauge and reparametrisation invariant, therefore γgis a geometric phase associated with a three level open quantum system. Now, it is interesting to note that under a specific local U(1) Gauge transformation defined as, |ψ′(t)⟩= exp (−iarg ⟨ψ(t0)|ψ(t)⟩)|ψ(t)⟩, the geometric phase takes the exact same form we obtained in the context of Aharonov-Anandan’s Generalization.For simplicity let us consider the phase involves in the gauge transformation be θ(t) = iarg ⟨ψ(t0)|ψ(t)⟩), so that we can write, |ψ′(t)⟩= exp (−iθ(t)) |ψ(t)⟩. This can be easily achieved by considering the following mathematical steps, ⟨ψ′(t)|d|ψ′(t)⟩ ⟨ψ′(t)|ψ′(t)⟩=⟨ψ′(t)|d dt |ψ′(t)⟩ ⟨ψ′(t)|ψ′(t)⟩dt =⟨ψ(t)|d|ψ(t)⟩−idθ(t)⟨ψ(t)|ψ(t)⟩ ⟨ψ(t)|ψ(t)⟩ =⟨ψ(t)|d|ψ(t)⟩ ⟨ψ(t)|ψ(t)⟩−idθ(t) (4.15) 19
Now, let us consider the following integral, ˆt t0 ⟨ψ′(t)|d|ψ′(t)⟩ ⟨ψ′(t)|ψ′(t)⟩=ˆt t0"⟨ψ(t)|d|ψ(t)⟩ ⟨ψ(t)|ψ(t)⟩−idθ(t)# =ˆt t0 ⟨ψ(t)|d|ψ(t)⟩ ⟨ψ(t)|ψ(t)⟩−iˆt t0 dθ(t) =ˆt t0 ⟨ψ(t)|d|ψ(t)⟩ ⟨ψ(t)|ψ(t)⟩−i[θ(t)−θ(t0)] ⇒ −Im ˆt t0 ⟨ψ′(t)|d|ψ′(t)⟩ ⟨ψ′(t)|ψ′(t)⟩=−Im ˆt t0 ⟨ψ(t)|d|ψ(t)⟩ ⟨ψ(t)|ψ(t)⟩+ [θ(t)−θ(t0)] =−Im ˆt t0 ⟨ψ(t)|d|ψ(t)⟩ ⟨ψ(t)|ψ(t)⟩+ arg ⟨ψ(t0)|ψ(t)⟩ (4.16) here, we have used the fact that θ(t0) = arg ⟨ψ(t0)|ψ(t0)⟩= 0 as, ⟨ψ(t0)|ψ(t0)⟩is purely real. Then the Geometric phase can be represented by the formulae, γA g=−Im ˆt t0 ⟨ψ′(t)|d|ψ′(t)⟩ ⟨ψ′(t)|ψ′(t)⟩(4.17) The above formulae is the generalization of the Aharonov-Anandan’s phase for the pure state with the condition that the vector ray is normalized i.e. ⟨ψ′(t)|ψ′(t)⟩=⟨ψ(t)|ψ(t)⟩= 1 so that, γA g=−Im ˆt t0⟨ψ′(t)|d dt|ψ′(t)⟩dt(4.18) In the terminology of the fibre bundle approach |ψ′(t)⟩can be viewed as the lift of the closed curve in the Ray space for a cyclic evolution of the quantum system in the Aharonov-Anandan Framework.Here γA gcan be viewed as the generalized Aharonov Anandan Phase for the mixed state.In other other words the geometric phase can be expressed as the integral of a gauge invariant one-form defined on Bdefined by, K=ℑ⟨ψ(t)|d|ψ(t)⟩ ⟨ψ(t)|ψ(t)⟩(4.19) Now, let is consider an interaction between the physical system and the surrounding, in this case we can’t expect a cyclic evolution of the quantum system and as a result of the system is not undergoing a cyclic evolution the Bloch parameters, the components of the bloch vector and the matrix elemets of the density operator will be decaying function of time with the time evolution of the density operator of the system will be described by the Lindblad type of Master equation.The nonunit state vectors will also involve deacying exponent.When the system is isolated from the environment, it can be considered that the system is undergoing a quasicyclic evolution so that the total phase φtot = arg ⟨ψ(t0)|ψ(t)⟩becomes 2πwhich can be dropped out in quantum computation as it’s a constant term and in general we identify the geometric phase term by dropping the overall phase or, total phase and the geometric phase is found as the negative of the dynamic phase. Thus we obtained the geometric phase under the 20
quasicyclic evolution of the quantum system with |ψ⟩=√r ei(α−γ)sin θcos ϕ ei(β−χ)sin θsin ϕ eiξ cos θ is given by, γg(C) = −ℑ˛C ⟨ψ|d|ψ⟩ ⟨ψ|ψ⟩ =−ℑ˛C"ie−i(α−γ)sin θcos ϕ e−i(β−χ)sin θsin ϕ e−iξ cos θ ei(α−γ)d(α−γ) sin θcos ϕ ei(β−χ)d(β−ξ) sin θsin ϕ eiξdξ cos θ e−i(α−γ)sin θcos ϕ e−i(β−χ)sin θsin ϕ e−iξ cos θ ei(α−γ)sin θcos ϕ ei(β−χ)sin θsin ϕ eiξ cos θ # =−ℑ˛C ind(α−γ) sin2θcos2ϕ+d(β−χ) sin2θsin2ϕ+dξ cos2θo =−ℑ˛C insin2θ[d(α−γ) cos2ϕ+d(β−χ) sin2θ] + cos2θdξo =−˛Cnsin2θ[d(α−γ) cos2ϕ+d(β−χ) sin2θ] + cos2θdξo(4.20) Here, due to the spherical symmetry we have considered the variations of the bloch parameters α, γ, β, χ, ξ only with the other parameters fixed. The equation is known as the Berry phase of mixed state, while Cis a closed circular arc of the eight dimensional bloch sphere for the three level system.It is mentioned before that the geometric phase γgcan be expressed by an integral around the closed curve Cof the one form defined in the manifold Bgiven by K=ℑ⟨ψ(t)|d|ψ(t)⟩ ⟨ψ(t)|ψ(t)⟩which is also known as the Mead-Berry connection one form. We want to show that the expression of the geometric phase is invariant under the gauge transformation. Let us consider a gauge transform defined as, |ψ∗(t)⟩=eiβ(t)|ψ(t)⟩. So. under the transformation if the one form Kchanges to K′i.e. K→k′and along with ⟨ψ(t)|ψ(t)⟩=⟨ψ∗(t)|ψ∗(t)⟩ then we have, K∗=ℑ⟨ψ∗(t)|d|ψ∗(t)⟩ ⟨ψ∗(t)|ψ∗(t)⟩ =ℑh⟨ψ(t)|d|ψ(t)⟩ ⟨ψ(t)|ψ(t)⟩+idβ(t)⟨ψ(t)|ψ(t)⟩ ⟨ψ(t)|ψ(t)⟩i =ℑh⟨ψ(t)|d|ψ(t)⟩ ⟨ψ(t)|ψ(t)⟩i+dβ(t) ⇒K∗=ℑ⟨ψ∗(t)|d|ψ∗(t)⟩ ⟨ψ∗(t)|ψ∗(t)⟩=K+dβ(t) (4.21) We can even consider the effect of the reparametraisation along with the Gauge transformation by replacing tby t′such that,ψ(t) = ψ′(t′), a smooth monotonic parametraisation is defined as,ψ(t) = ψ′(t′), t′=f(t)|df(t) dt ≥0.The parametraisation leads to the transformation C → C′ and also the similar kind of transformation in terms of the image of the curve in the Ray space. Now,it is evident under that transformation the geometric phase γgand the Mead-Berry connection one form remains invariant. So, under the gauge transformation we have K→K∗ 21
and assume that the geometric phase γgchanges to γg∗which implies that, γg∗=−ℑ˛C ⟨ψ∗(t)|d|ψ∗(t)⟩ ⟨ψ∗(t)|ψ∗(t)⟩ =−˛C K∗ =−˛C (K+dβ(t)) =−h˛C K+˛C dβ(t)i =−˛C K=−ℑ˛C ⟨ψ(t)|d|ψ(t)⟩ ⟨ψ(t)|ψ(t)⟩=γg (4.22) So, we have K∗=K+dβ(t) and as a consequence dK∗=dK The closed integral of the total differential β(t) over the path Cvanishes.This establishes the gauge invariance of the Berry phase for a three level open quantum system of the mixed states i.e. γg∗=γg. Thus the Berry phase in the equation (4.20) is gauge invariant and the invariance condition holds for any arbitrary choice of β(t) for a given closed curve C ⊂ B under the quasicyclic evolution. 5 Lindblad Master Equation For Dissipative System As mentioned earlier in order to find out the geometric phase given by the equation (4.20) we require the time dependent bloch sphere parameters which depends on the matrix elements of the density operator which are time dependent, when interaction comes into the picture in case of the open system between the system and the surroundings the time evolution of the density operator is governed by the Lindblad type of Mater Equation. As an example, let us consider a three-level system interacting with environment.When a relevant dynamical time scale of the open quantum system is long compared to the time for the environment to the forgetting quantum information, the evolution of system is effectively local in time (the Markovian approximation) and may be described by the Lindblad’s master equation [21] given by, ∂ρ ∂t =−i ℏ[ˆ H, ρ] + 8 X i=1 Γ† iρΓi−1 2{Γ† iΓi, ρ}(5.1) The first term of the master equation is a usual schr¨odinger term which generates an unitary evolution which appear in the well known Von-Neumann Liouville equation of quantum statistical mechanics and the remaining part of the equation describe all possible transitions that the open system may undergo due to the interaction with the reservior.The operators Γi(i=1,2,3,...,8) are called the Lindblad Operators or the quantum jump operators. It can be readily checked that ˙ρis hermitian and Tr( ˙ρ) = 0, which implies that that he Lindblad Master equation preserves the positivity of the density operator ρ(t) for the open system. In the present work the Lindblad operators are choosen as Γi=pηi(t)λiwhich represents the coupling to the environement where,λi, i = 1,2, ..., 8 are the standard Gell-Mann matrices which are Hermitian and traceless.The Decoherence time is approximately given by 1 Γi(t). Now, for simplicity we will consider a simple situation where the dephasing noise term or, the system and environment coupling term is represented by only a single type of Lindblad Operator Γ = √ηλ3is applied to the three level quantum system in the present context with the Hamiltonian H=1 2ℏΩλ3.The next task is to find the solution to the Master Equation, the matrix element of the Density operator which has been represented by a 3 ×3 matrix with, 22
ρ= ρ11 ρ12 ρ13 ρ21 ρ22 ρ23 ρ31 ρ32 ρ33 and all the matrix elements are time dependent i.e. ρij(t). Now, taking the trace of both sides of equation [5.1] we get, Tr[ ˙ρ(t)] = Tr[Hρ−ρH] + 8 X i=1 nTr[Γ† iρΓi−1 2Γ† iρΓi−1 2ρΓ† iΓi]o = 8 X i=1 nTr[Γ† iρΓi]−1 2Tr[Γ† iρΓi]−1 2Tr[ρΓ† iΓi]o = 8 X i=1 nTr[Γ† iρΓi]−1 2Tr[ρΓ† iΓi]−1 2Tr[ρΓ† iΓi]o = 8 X i=1 nTr[ρΓiΓ† i]−Tr[ρΓ† iΓi]o= 0 (5.2) Where we have used the Cyclical Invariance properties of Trace operation Tr[AB] = Tr[BA], Tr[ABC] = Tr[BCA] = Tr[CAB] and in the last step we have used the Hermiticity condition of the Lindblad Operators i.e. Γ† i= Γi, i = 1,2, ..., 8. As a consequence of the following result Tr[ ˙ρ(t) = 0 we can say that the Lindblad Master Equation preserves the positivity of the Density Operator with ellapsation of time. 5.1 The Solution Of The Master Equation For The Three Level Quantum System We will solve the Lindblad equation with the special case mentioned before where the dephasing noise term or, the system and environment coupling term is represented by only a single type of Lindblad Operator Γ = √ηλ3is applied to the three level quantum system in the present context with the Hamiltonian H=1 2ℏΩλ3. So, in our case the Lindblad Master equation will be, ∂ρ ∂t =−i ℏ[ˆ H, ρ] + Γ†ρΓ−1 2{Γ†Γ, ρ}(5.1.1) And we have the following results, 1 iℏ[H, ρ] = −iΩ 2 0 2ρ12 ρ13 −2ρ21 0−ρ23 −ρ31 ρ32 0 Γ†ρΓ−1 2{Γ†Γ, ρ}=η 0−2ρ12 −ρ13 2 −2ρ21 0−ρ23 2 −ρ31 2−ρ32 20 (5.1.2) Subsituting them back in equation (5.3) we get, d dt ρ11(t)ρ12(t)ρ13(t) ρ21(t)ρ22(t)ρ23(t) ρ31(t)ρ32(t)ρ33(t) = ˙ρ11(t) ˙ρ12(t) ˙ρ13(t) ˙ρ21(t) ˙ρ22(t) ˙ρ23(t) ˙ρ31(t) ˙ρ32(t) ˙ρ33(t) ="−iΩ 2 0 2ρ12(t)ρ13(t) −2ρ21(t) 0 −ρ23(t) −ρ31(t)ρ32(t) 0 +η 0−2ρ12(t)−ρ13(t) 2 −2ρ21(t) 0 −ρ23(t) 2 −ρ31(t) 2−ρ32(t) 20 #(5.1.3) 23
Now, on compairing the both side of the matrix equation [5.1.3] elementwise we obtain the following set of ordinary coupled first order diffetential equations involving the rate of change of the matrix elements ρij(t) as follows, ˙ρ11(t) = ˙ρ22(t) = ˙ρ33(t) = 0 (5.1.3a) ˙ρ12(t) = −Ωρ12(t)−2ηρ12(t) = −ρ12(t)(iΩ+2η) (5.1.3b) ˙ρ21(t) = ρ21(t)(iΩ−2η) (5.1.3c) ˙ρ31(t) = ρ31(t)(iΩ−η 2) (5.1.3d) ˙ρ23(t) = ρ23(t)(iΩ−η 2) (5.1.3e) ˙ρ32(t) = ρ32(t)(−iΩ−η 2) (5.1.3f) ˙ρ13(t) = ρ13(t)(−iΩ−η 2) (5.1.3g) Solving the above equations we get the solution of the Lindblad Master Equation as follows, ρ11(t) = ρ11(0),(5.1.4a) ρ22(t) = ρ22(0),(5.1.4b) ρ33(t) = ρ33(0),(5.1.4c) ρ12(t) = ρ12(0) exp (−iΩ−2η)t, (5.1.4d) ρ21(t) = ρ21(0) exp (iΩ−2η)t, (5.1.4e) ρ13(t) = ρ13(0) exp (−iΩ 2−η 2)t, (5.1.4f) ρ31(t) = ρ31(0) exp (iΩ 2−η 2)t, (5.1.4g) ρ23(t) = ρ23(0) exp (iΩ 2+η 2)t, (5.1.4h) ρ32(t) = ρ32(0) exp (−iΩ 2−η 2)t(5.1.4i) For the unique solution of the Lindblad Master Equation we require the initial condition i.e. ρij(0),(i, j) = 1,2,3..Let, us consider that the initial pure state |ψ(t= 0)⟩=|ψ(0)⟩given by |ψ(0)⟩=δ1|1⟩+δ2|2⟩+δ3|3⟩with δ1, δ2, δ3∈Cwhere,δi(i= 1,2,3) are independent of the evolving time and may be tuned by the external conditions and the canonical Basis vectors |1⟩,|2⟩,|3⟩are given by,|1⟩= 1 0 0 ,|2⟩= 0 1 0 ,|3⟩= 0 0 1 respectively and then the initial state vector will be |ψ(0)⟩= δ1 δ2 δ3 .Then the density matrix at t= 0 will be,ˆρ(0) = |ψ(0)⟩⟨ψ(0)|. Let us denote the matrix elements of the density matrix at initial time be ρij(0) 24
and it will be a 3 ×3 matrix. So, the density matrix at the initial time i.e. t= 0 is given by, ˆρ(0) = |ψ(0)⟩⟨ψ(0)| = ρ11(0) ρ12(0) ρ13(0) ρ21(0) ρ22(0) ρ23(0) ρ31(0) ρ32(0) ρ33(0) = δ1 δ2 δ3 δ∗ 1δ∗ 2δ∗ 3 = |δ1|2δ1δ∗ 2δ1δ∗ 3 δ2δ∗ 1|δ2|2δ2δ∗ 3 δ3δ∗ 1δ∗ 3δ2|δ3|2 (5.1.5) So, from the initial condition we get, ρ11(0) = |δ1|2, ρ22(0) = |δ2|2(5.1.6a) ρ33(0) = |δ3|2, ρ12(0) = δ1δ∗ 2(5.1.6b) ρ21(0) = δ∗ 1δ2, ρ13(0) = δ1δ∗ 3(5.1.6c) ρ31(0) = δ∗ 1δ3, ρ23(0) = δ2δ∗ 3, ρ32(0) = δ3δ∗ 2(5.1.6d) So, the solution of the lindblad’s master equation can be obtained by putting ρij(0); ∀(i, j = 1,2,3 in equations [5.1.4] and it leads to, ρ11(t) = |δ1|2,(5.1.7a) ρ22(t) = |δ2|2,(5.1.7b) ρ33(t) = |δ3|2,(5.1.7c) ρ12(t) = δ1δ∗ 2exp (−iΩ−2η)t, (5.1.7d) ρ21(t) = δ∗ 1δ2exp (iΩ−2η)t, (5.1.7e) ρ13(t) = δ1δ∗ 3exp (−iΩ 2−η 2)t, (5.1.7f) ρ31(t) = δ∗ 1δ3exp (iΩ 2−η 2)t, (5.1.7g) ρ23(t) = δ2δ∗ 3exp (iΩ 2+η 2)t, (5.1.7h) ρ32(t) = δ3δ∗ 2exp (−iΩ 2−η 2)t(5.1.7i) As, we know that the Bloch parameters depend on the matrix elements of the density operator then the time dependent bloch parameters will be obtained by substituting the matrix elements ρij(t); ∀(i, j) = 1,2,3 in the set of equations [2.1.30] we get the time dependent angular bloch parameters (θ, ϕ, ξ, χ, α, β, γ) and then using the formulae for the Geometric phase γggiven by equation [4.20] we will get the expression of the geometric phase. In the equations [5.1.7] the dephasing factors exp (−2ηt) and exp (−ηt 2)parametrize the amount of decoherence.The effect of dephasing are to decrease the size of the nondiagonal elements of the density matrix in a basis determined by the dephasing interaction with the environment. 6 SU(2) Polarization Picture Polarization is a property applying to transverse waves that specifies the geometrical orientation of the oscillations. In a transverse wave, the direction of the oscillation is perpendicular 25
ˆρBthen the partial trace of the global density operator with respect to the reservoir or, the bath states will give the actual density operator for the system i.e. field. Then ˆρfield =TRBˆρT ot Under the weak coupling limit the Born-Markov master equation [10] leads to, ˙ ˆρ(t) = −1 ℏ2ˆ∞ 0 dτTrBhˆ Hint(t),ˆ Hint(t−τ),ˆρ(t)⊗ˆρB (7.1.3) Here,TrBindicates the partial trace with respect to the bath states and ˆρ(t) being the reduced density operator for the field. As, we can see that the master equation does not depend on the specific choice of the bath or, it’s Hamiltonian rather it only depends on the interaction Hamiltonian i.e. ˆ Hint. One common way to couple the bath and the single mode field is by the interaction Hamiltonian of this type: ˆ Hint =ℏX λX s=±κλΓλˆa† s+κ∗ λΓ† λˆas(7.1.4) Here, κλbeing the complex coupling constant describing the interaction strength between the field and the bath, the sum over λcorresponds to the sum over all possible accessible bath modes and {ˆ Γλ,ˆ Γ† λ}corresponds to the annihilation and creation operators corresponding to the bath quanta for each bath modes.A photon can create a bath quanta by loosing energy and vice versa. Now, for simplicity we assume the zero temperature bath in order to neglect the process of stimulated emission and with those underlying assumptions the master equation [7.1.3] with the interaction Hamiltonian of the above form becomes, ˙ ˆρ(t) = X s=± γs 2Lˆasˆρ(t) (7.1.5) where,γsbeing the decoherence rates associated with each mode of polarization and We have the lindblad superoperators [21]Lˆ Csdefined as, Lˆ Cs= 2 ˆ Csˆρˆ C† s−{ˆ C† sˆ Cs,ˆρ}(7.1.6) The decoherence rate can be expressed in terms of the coupling constant κλ.˙ ˆρ(t) is hermitian and this form of master equation is always positive [20] describing the dynamical evolution of an open system ensuring that the state of the system remains always valid at all times. If, we solve this above master equation for a single photon state then, first we need to find the matrix realisation of the operators.Here, the basis can be choosen as the two mode fock basis i.e. |N, k⟩=|k⟩+⊗|N−k⟩−. For, single photon states the dimension of each invariant subspaces will be 2 and the density operator will be a 2 ×2 hermitian matrix in the orthonormal basis: |1,0⟩=|0⟩+⊗|1⟩−=|−⟩ and |1,1⟩=|1⟩+⊗|0⟩−=|+⟩respectively. Let, the density matrix being ˆρ(t) = ρ11(t)ρ12(t) ρ21(t)ρ22(t). Now, the finite dimensional matrix representation of the creation and annihilation operators can be obtained by knowing the action of the operators on the basis {|+⟩,|−⟩}. we have, ˆa+|+⟩=|0⟩+⊗|0⟩−(7.1.7) ˆa−|−⟩ =|o⟩+⊗|0⟩−(7.1.8) ˆa+|−⟩ = 0 = ˆa−|+⟩.(7.1.9) so, the matrix representation for the operator ˆa+and its hermitian adjoint will be, ˆa+(2×2) =⟨+|ˆa+|+⟩ ⟨+|ˆa+|−⟩ ⟨−|ˆa+|+⟩ ⟨−|ˆa+|−⟩=0 0 0 0= ˆa† +(2×2) 32
similarly,the matrix for the ˆa−and its hermitian adjoint will be the same i.e. ˆa−(2×2) = 0 0 0 0= ˆa† −(2×2). In, the similar way we can find the matrix representation of the individual number operators i.e. ˆa† +ˆa+ij and ˆa† −ˆa−ij which leads to, ˆa† +ˆa+ij =⟨+|ˆa† +ˆa+|+⟩ ⟨+|ˆa† +ˆa+|−⟩ ⟨−|ˆa† +ˆa+|+⟩ ⟨−|ˆa† +ˆa+|−⟩=1 0 0 0 and similarly we can write,ˆa† −ˆa−ij =0 0 0 1. With the definition given by [7.1.6] simple matrix multiplication leads to, Lˆasˆρ= 2ˆasˆρˆa† s−{ˆa† sˆas,ˆρ}=−2ρ11 ρ12 ρ21 0, s = + and, Lˆasˆρ= 2ˆasˆρˆa† s−{ˆa† sˆas,ˆρ}=−0ρ12 ρ21 2ρ22, s =− Then with γ+and γ−denoting the decoherence rates for the individual modes of polarization the master equation [7.1.5] becomes, ˙ρ11(t) ˙ρ12(t) ˙ρ21(t) ˙ρ22(t)=γ+ 2−2ρ11(t)−ρ12(t) −ρ21(t) 0 +γ− 20−ρ12(t) −ρ21(t)−2ρ22(t)(7.1.10) The above matrix equation leads to simple ordinary differential equations in terms of the time derivative of the matrix elements of the density operator given by, ˙ρ11(t) = −γ+ρ11(t) (7.1.11) ˙ρ22(t) = −γ−ρ22(t) (7.1.12) ˙ρ12(t) = −γ+ 2ρ12(t)−γ− 2ρ12(t) (7.1.13) ˙ρ21(t) = −γ+ 2ρ21(t)−γ− 2ρ21(t) (7.1.14) The solution to the set of ordinary differential equations leads to, ρ11(t) = ρ11(0) exp {−γ+t}(7.1.15) ρ22(t) = ρ22(0) exp {−γ−t}(7.1.16) ρ12(t) = ρ12(0) exp {−(γ++γ−)t/2}(7.1.17) ρ21(t) = ρ21(0) exp {−(γ++γ−)t/2}(7.1.18) But,the above master equation can not describe the depolarization process correctly as along with the interaction Hamiltonian of the form given by equation describes two standard independent Lindblad decaying processes for individual polarization states with the decoherence parameters γ±as if we are studying the decay process for the individual polarisation modes.The master equation can be written as the sum of two lindbladian describing the decay of individual polarisation modes i.e. s=±. Since the coupling transfers energy between the system and the bath the result will be a damping of the beam intensity. The density matrix is completely known along with the given initial conditions and is given by, ˆρ(t) = ρ11(0) exp {−γ+t}ρ12(0) exp {−(γ++γ−)t/2} ρ21(0) exp {−(γ++γ−)t/2}ρ22(0) exp {−γ−t}(7.1.19) 33
In fact from the solutions of the master equation we can see that the matrix elements ρij(t) decays with time and irrespective of the given initial conditions i.e. ρij(0), the stationary states is always be ˆρ(t) = |0,0⟩⟨0,0|at t→ ∞ with |0,0⟩=|0⟩+⊗|0⟩−. It is the density operator of the vacuum state for both circular polarization i.e. s=±. So, obviously this cannot describe the depolarization process.We expect the density matrix to be a diagonal matrix as opposed to the present situation where the density matrix corresponds to the Vacuum state for every polarization modes. As we have anticipated earlier that we are interested to study the pure dephasing dynamics so that the amount of energy exchange between the system and the bath is negligible.In mathematical terms we can say that the Interaction Hamiltonian must commute with the system Hamiltonian i.e. ˆ Hfield,ˆ Hint= 0 which implies no exchange of energy and only the phase changes. One way to model the pure dephasing dynamics is by choosing the interaction Hamiltonian of the following form [10] so that it commutes with the field Hamiltonian. ˆ Hint =ℏX λX s=±κλΓλˆa† sˆas+κ∗ λΓ† λˆa† sˆas(7.1.20) With this choice of the interaction term the Lindblad Master equation [7.1.3] becomes, ˙ ˆρ(t) = X s=± γs 2Lˆa† sˆasˆρ(t) (7.1.21) The above interaction Hamiltonian can be viewed as a scattering process, in which a bath quanta is absorbed or,emitted but the number of photons in each polarization is preserved.We will solve the master equation for the single and two photon states. As we have already worked out the matrix representation of the Number operators with single photon states for each polarization states i.e. ˆa† sˆasfor, s=±along with the fact that (ˆa† sˆas)2= ˆa† sˆas, with the definition [7.1.6] simple matrix multiplication leads to, Lˆa† sˆasˆρ= 2ˆa† sˆasˆρˆa† sˆas−{ˆa† sˆas,ˆρ}=0−ρ12(t) −ρ21(t) 0 , s = + Lˆa† sˆasˆρ= 2ˆa† sˆasˆρˆa† sˆas−{ˆa† sˆas,ˆρ}=0−ρ12(t) −ρ21(t) 0 , s =− Substituting them back into the master equation [7.1.21] leads to, ˙ρ11(t) ˙ρ12(t) ˙ρ21(t) ˙ρ22(t)=γ+ 20−ρ12(t) −ρ21(t) 0 +γ− 20−ρ12(t) −ρ21(t) 0 (7.1.22) Compairing the two matrices element-wise we get the following ordinary differential equations, ˙ρ11(t) = ˙ρ22(t) = 0 (7.1.23) ˙ρ12(t) = −1 2γ++γ−ρ12(t) (7.1.24) ˙ρ21(t) = −1 2γ++γ−ρ21(t) (7.1.25) From the above equations it is confirmed that the positivity of the density operator is preserved as Tr˙ ˆρ(t)= 0 and ˙ ˆρ(t) is hermitian. One can write the 2 ×2 density matrix as a linear 34
combination of one 2 ×2 identity matrix and three Pauli matrices i.e. σi, i=1,2,3. So, we may write ˆρ=1 2ˆ I+s · ˆσwhich leads to, ˆρ(t) = 1 21 + sz(t)sx(t)−isy(t) sx(t) + isy(t) 1 −sz(t),˙ ˆρ(t) = 1 2˙sz(t) ˙sx(t)−i˙sy(t) ˙sx(t) + i˙sy(t)−˙sz(t)(7.1.26) which corresponds to the following equations, ˙sz(t) = 0 (7.1.27) ˙sx(t)−i˙sy(t) = −1 2γ++γ−sx(t)−isy(t)(7.1.28) ˙sx(t) = −1 2γ++γ−sx(t) (7.1.29) ˙sy(t) = −1 2γ++γ−sy(t) (7.1.30) (7.1.31) The solution pertains to, sz(t) = sz(0) (7.1.32) sx(t) = sx(0) exp {−(γ++γ−)t/2}(7.1.33) sy(t) = sy(0) exp {−(γ++γ−)t/2}(7.1.34) The solution of the Lindblad master equation leads to the fact that the population of the system does not change with time as the diagonal matrix elements does not change over time i.e. ρ11(t) = ρ11(0) and ρ22(t) = ρ22(0). And the off diagonal elements of the density matrix which successfully describe the effect of decoherence in the system as a result of the interaction with the surroundings as the off diagonal matrix elements describe the quantum coherence and the solutions suggests that the correlation is exponentially decaying. The model preserves the invariant subspaces and can also explain that the stationery state is a diagonal state i.e. the off diagonal matrix elements of the density matrix will vanish at t→ ∞ so that the stationery state becomes, ˆρ(t) = ρ11(0) 0 0ρ22(0), t → ∞ So, the stationery state which is described by the density operator is a diagonal state. Similarly, the solution of the Lindblad Master equation has been carried out for the two photon states.In the case of two photon system we have N= 2 so, the invariant subspaces will be 3 dimensional and as a result the density matrix will be a 3 ×3 hermitian matrix. Again we choose the two mode fock basis i.e. |N, k⟩=|k⟩+⊗|N−k⟩−with N= 2 and k= 0,1,2, being the eigenstates of ˆ S0and ˆ S3for the matrix representation of the necessary operators. Let, the density matrix be, ˆρ= ρ11 ρ12 ρ13 ρ21 ρ22 ρ23 ρ31 ρ32 ρ33 . The basis states are, |2,2⟩=|2⟩+⊗ |0⟩−=|1⟩,|2,0⟩=|0⟩+⊗ |2⟩−=|2⟩and |2,1⟩= |1⟩+⊗|1⟩−=|3⟩.The matrix representations for the number operators ˆa† +ˆa+and ˆa† −ˆa−in the basis set {|1⟩,|2⟩,|3⟩are as follows, ˆa† +ˆa+ij = ⟨2,2|ˆa† +ˆa+|2,2⟩ ⟨2,2|ˆa† +ˆa+|2,0⟩ ⟨2,2|ˆa† +ˆa+|2,1⟩ ⟨2,0|ˆa† +ˆa+|2,2⟩ ⟨2,0|ˆa† +ˆa+|2,0⟩ ⟨2,0|ˆa† +ˆa+|2,1⟩ ⟨2,1|ˆa† +ˆa+|2,2⟩ ⟨2,1|ˆa† +ˆa+|2,0⟩ ⟨2,1|ˆa† +ˆa+|2,1⟩ = 2 0 0 0 0 0 0 0 1 35
similarly the matrix representation of ˆa† −ˆa−is given by, ˆa† −ˆa−ij = ⟨2,2|ˆa† −ˆa−|2,2⟩ ⟨2,2|ˆa† −ˆa−|2,0⟩ ⟨2,2|ˆa† −ˆa−|2,1⟩ ⟨2,0|ˆa† −ˆa−|2,2⟩ ⟨2,0|ˆa† −ˆa−|2,0⟩ ⟨2,0|ˆa† −ˆa−|2,1⟩ ⟨2,1|ˆa† −ˆa−|2,2⟩ ⟨2,1|ˆa† −ˆa−|2,0⟩ ⟨2,1|ˆa† −ˆa−|2,1⟩ = 0 0 0 0 2 0 0 0 1 The stokes operator ˆ S3will be ˆ S3=ˆa† +ˆa+−ˆa† −ˆa−= 2 100 0−1 0 000 Now simple ,matrix multiplication leads to, Lˆa† sˆasˆρ= 2ˆa† sˆasˆρˆa† sˆas−{ˆa† sˆasˆa† sˆas,ˆρ}= 0−4ρ12(t)−ρ13(t) −4ρ21(t) 0 −ρ23(t) −ρ31(t)−ρ32(t) 0 , s = + and, Lˆa† sˆasˆρ= 2ˆa† sˆasˆρˆa† sˆas−{ˆa† sˆasˆa† sˆas,ˆρ}= 0−4ρ12(t)−ρ13(t) −4ρ21(t) 0 −ρ23(t) −ρ31(t)−ρ32(t) 0 , s =− Along with this the Master equation [7.1.21] becomes, ˙ ˆρ(t) = ˙ρ11(t) ˙ρ12(t) ˙ρ13(t) ˙ρ21(t) ˙ρ22(t) ˙ρ23(t) ˙ρ31(t) ˙ρ32(t) ˙ρ33(t) =γ+ 2 0−4ρ12(t)−ρ13(t) −4ρ21(t) 0 −ρ23(t) −ρ31(t)−ρ32(t) 0 +γ− 2 0−4ρ12(t)−ρ13(t) −4ρ21(t) 0 −ρ23(t) −ρ31(t)−ρ32(t) 0 (7.1.35) Compairing the matrices element-wise we get the set of first order differential equations as follows, ˙ρ11(t) = 0,(7.1.36) ˙ρ22(t) = 0,(7.1.37) ˙ρ33(t) = 0,(7.1.38) ˙ρ12(t) = −2γ++γ−ρ12(t),(7.1.39) ˙ρ21(t) = −2γ++γ−ρ21(t),(7.1.40) ˙ρ13(t) = −1 2γ++γ−ρ13(t),(7.1.41) ˙ρ31(t) = −1 2γ++γ−ρ31(t),(7.1.42) ˙ρ23(t) = −1 2γ++γ−ρ23(t),(7.1.43) ˙ρ32(t) = −1 2γ++γ−ρ32(t) (7.1.44) 36
The Solution to the above equations leads to, ρ11(t) = ρ11(0),(7.1.45) ρ22(t) = ρ22(0),(7.1.46) ρ33(t) = ρ33(0),(7.1.47) ρ12(t) = ρ12(0) exp {−2(γ++γ−)t},(7.1.48) ρ21(t) = ρ21(0) exp {−2(γ++γ−)t},(7.1.49) ρ13(t) = ρ13(0) exp {−(γ++γ−)t/2},(7.1.50) ρ31(t) = ρ31(0) exp {−(γ++γ−)t/2},(7.1.51) ρ23(t) = ρ23(0) exp {−(γ++γ−)t/2},(7.1.52) ρ32(t) = ρ32(0) exp {−(γ++γ−)t/2}(7.1.53) Let us calculate the degree of polarization Pas a function of time for the single photon state. By definition of degree of polarization we have, P(t) = √⟨ˆ S1⟩2+⟨ˆ S2⟩2+⟨ˆ S3⟩2 ⟨ˆ S0⟩with, ⟨ˆ Si(t)⟩=Trˆρ(t)ˆ Si. Now for the single photon state the Degree of polarization has been calculated as follows, In case of two mode polarization for the single photon state, the matrix representation of ˆ Si for, i=1,2,3 will resemble the forms of the Pauli matrices i.e. σiwhich means, ˆ S1=0 1 1 0,ˆ S2=0−i i0,ˆ S3=1 0 0−1 Then for single photon state from [7.1.26] and [7.1.32] we have, ⟨ˆ S1⟩=Tr[ˆρ(t)ˆ S1]=(ρ12(t) + ρ21(t)) = sx(t) = sx(0) exp {−γ′t/2} ⟨ˆ S2⟩=Tr[ˆρ(t)ˆ S2] = i(ρ12(t)−ρ21(t)) = sy(t) = sy(0) exp {−γ′t/2} ⟨ˆ S3⟩=Tr[ˆρ(t)ˆ S3] = (ρ11(t)−ρ22(t)) = sz(0) ⟨ˆ S0⟩=Tr[ˆρ(t)ˆ S3] = (ρ11(t) + ρ22(t)) = Tr[ˆρ(t)] = Tr[ˆρ(0)] = 1. Where, we have defined the total decoherence rate: γ′=Ps=±γs. By using [6.22] the degree of polarization will be, P(t) = qsx(0)2exp {−γ′t}+sy(0)2exp {−γ′t}+sz(0)2(7.1.54) As,we can see that the degree of polarization decreases with time so,it can explain the depolarization process. 7.2 Calculation of Geometric Phase Once the matrix elements of the density operator are found [7.1.45] then the Geometric phase can be calculated easily. We have obtained the expression of the Geometric phase for a three level quantum system undergoing an quasicyclic evolution given by, γg(C) = −ℑ˛C ⟨ψ|d|ψ⟩ ⟨ψ|ψ⟩ =−˛Cnsin2θ[d(α−γ) cos2ϕ+d(β−χ) sin2θ] + cos2θdξo(7.1.55) The Geometric phase is expressed in terms of the Bloch parameters related to the matrix elements of the density operator by the equations [2.1.30] used to parameterise the 8 dimensional 37
Bloch sphere. Now as the Matrix elements are time dependent the Bloch parameters will also be time dependent [13] so, by expressing the matrix elements in terms of the Bloch parameters and by using the formulae one can determine the Geometric Phase for the system undergoing Depolarization.But here in the present situation the Bloch parameters are time independent although the matrix elements are time dependent because of the choice of Parametraisation used in the present context. 8 Modeling Depolarization with a non resonant randomly distributed atomic bath In the classical picture we describe the phenomena of depolarization resulting in an effective anisotropy of the material medium due to the decorrelation of the phases associated with the Electric filed vector. Now, we will consider a more realistic model where we will assume that a plane polarized electromagnetic wave is propagating through a material medium represented by a randomized distribution of two level atoms along with a dispersive coupling as the amount of energy exchange between the field and atoms is very negligible. Let us consider the frequency of the electromagnetic wave be ωand the atomic transition frequency between the two levels of the ath atom be ωa.In the natural unit system (ℏ= 1) the Hamiltonian of the super system [17][18] (system coupled with the atomic reservoir) can be written as the sum of the field Hamiltonian,the bath Hamiltonian and the interaction Hamiltonian describing the coupling between the system and the reservoir.So that, ˆ HTot =ˆ Hfield +ˆ Hatomic +ˆ Hint (8.1) where, ˆ Hfield =ωX λ=± ˆa† λˆaλ,(8.2) ˆ Hatomic =1 2X a ωaˆσz a,(8.3) ˆ Hint =X aX λ=±gaλˆσ− aˆa† λ+g∗ aλˆσ+ aˆaλ(8.4) Where, the field operator resembles the Hamiltonian of the two dimensional isotropic harmonic oscillator and the Hamiltonian of a single two level atom is expressed in terms of standard Pauli operators.The Interaction Hamiltonian ( ˆ Hint) has been written in Dipole and Rotating Wave approximations [5].The sum over agoes for all the atoms in the atomic reservoir. The term gaλ represents the coupling of the ath atom with the field mode λ=±which is in general a complex quantity.As the atoms are randomly distributed it is reasonable to use random phase approximation [3] for which gaλ =|ga|eiλϕa/2for λ=±which means that the coupling constants carries random phases. It is pointed out that the random phase approximation works well in the long wavelength limit. It is well known that the atoms decay irreversibly.This is usually due to the interaction of the atom with the thermo-electromagnetic environment if we assume a thermal distribution of the randomly distributed atoms. Then in the Interaction picture of Quantum Mechanics the time evolution of the Density operator for the super-system will be given by the Master equation of this type: ˙ ˆρ(t) = −iˆ HTot,ˆρTot+X a γa 2{¯na+ 1L[ ˆσa−]ˆρTot + ¯naL[ ˆσa+]ˆρTot}(8.5) 38
Here,ˆσ± abeing the spin raising and lowering operators defined in terms of the Pauli matrices such that, ˆσ± a= ˆσx a±ˆσy a. Along with the definition of the Lindblad Superoperators given by, Lˆ C= 2 ˆ Cˆρˆ C†−{ˆ C†ˆ C, ˆρ}(8.6) Here, γabeing the decay constant of the ath atom due to its coupling with the thermal environment containing ¯naexcitations, here we have different Bosonic reservoirs associated with individual atoms in order to avoid any collective effect. For further simplicity we will consider the high temperature limit for which ¯na≫1 [19], so that the effect of the spontaneous emission can be disregarded compared to the process of stimulated emission.Emission into the reservoir and the absorption from the reservoir then becomes identical and finally they balance each other in the stationery state. The rate of absorption or emission entirely depends on the initial population of the atomic states. Lets define the off resonant detuning parameter ∆a=ωa−ω and in case of far off resonant regime we have |ga|≪|∆a|and in this limit the elimination of the atomic variables using the adiabatic approximation and the averaging over the random phases [18] leads to the Master equation [8.5] reduces to the following form, ˙ ˆρ(t) = −iωˆ S0,ˆρ(t)+ 2γLˆ S0ˆρ(t) + γLˆ S+ˆρ(t) + γLˆ S−ˆρ(t) (8.7) Where,γis the decoherence rate and ˆρ(t)=TratˆρT otis the reduced density operator describing the evolution of the field mode and ˆ S±=ˆ S1±iˆ S2. γ=X a |ga|4 γa∆2¯na (8.8) The term L[ˆ S±] in the Master equation [8.7] describes the depolarization rates in each invariant subspaces.The first term on the R.H.S of the Master equation describes the unitary or the free evolution of the density operator known as the Louivillian where as, the second term describes the non-unitary evolution of ˆρ(t), known as the Lindbladian. We will solve the Master Equation for the single photon state.For the single photon state we have N= 1 so, the dimensionality of the invariant subspace will be 2 so, the entire Hilbert space will split into invariant subspaces of dimension 2. So, we have to solve the master equation in the 2 dimensional subspace and the density operator will be a 2 ×2 matrix in the basis |N, k⟩as been done earlier. For, the single photon state the Stokes operators and the raising lowering operators are given by, ˆ S1=0 1 1 0,ˆ S2=0−i i0,ˆ S3=1 0 0−1 ˆ S0=1 0 0 1,ˆ S+=0 2 0 0,ˆ S−=0 0 2 0 Simple Matrix multiplication with ˆρ(t) = ρ11(t)ρ12(t) ρ21(t)ρ22(t)leads to, ˆ S0,ˆρ= 0 Lˆ S0]ˆρ= 2 ˆ S0ˆρˆ S0−{ˆ S0ˆ S0,ˆρ}= 0 Lˆ S+]ˆρ= 2 ˆ S+ˆρˆ S−−{ˆ S−ˆ S+,ˆρ}=8ρ22(t)−4ρ12(t) −4ρ21(t)−8ρ22(t) Lˆ S−]ˆρ= 2 ˆ S−ˆρˆ S+−{ˆ S+ˆ S−,ˆρ}=−8ρ11(t)−4ρ12(t) −4ρ21(t)−8ρ11(t) 39
And the Master equation given by [8.7] becomes, ˙ρ11(t) ˙ρ12(t) ˙ρ21(t) ˙ρ22(t)=γ8ρ22(t)−8ρ11(t)−8ρ12(t) −8ρ21(t)−8ρ22(t)+8ρ11(t)(8.9) The master equation preserves the positivity of the density operator i.e. Tr[ ˙ ˆρ(t)] = 0 and the positive definiteness of ˙ ˆρ(t) is guaranteed by the hermiticity of ˙ ˆρ(t) i.e. ˙ ˆρ†(t) = ˙ ˆρ(t) Now, with ˆρ(t) = 1 21 + sz(t)sx(t)−isy(t) sx(t) + isy(t) 1 −sz(t)the Master equation leads to the set of ordinary differential equations given by, ˙sx(t) = −8γsx(t),(8.10) ˙sy(t) = −8γsy(t),(8.11) ˙sz(t) = −16γsz(t) (8.12) And the solution to the equations with γ′= 8γgives, sx(t) = sx(0)e−γ′t,(8.13) sy(t) = sy(0)e−γ′t,(8.14) sz(t) = sz(0)e−2γ′t(8.15) Then for single photon state we have, ⟨ˆ S1⟩=Tr[ˆρ(t)ˆ S1]=(ρ12(t) + ρ21(t)) = sx(t) = sx(0) exp {−γ′t} ⟨ˆ S2⟩=Tr[ˆρ(t)ˆ S2] = i(ρ12(t)−ρ21(t)) = sy(t) = sy(0) exp {−γ′t} ⟨ˆ S3⟩=Tr[ˆρ(t)ˆ S3] = (ρ11(t)−ρ22(t)) = sz(t) = sz(0) exp {−2γ′t} ⟨ˆ S0⟩=Tr[ˆρ(t)ˆ S3]=(ρ11(t) + ρ22(t)) = Tr[ˆρ(t)] = Tr[ˆρ(0)] = 1. By the definition given by [6.22] the degree of polarization will be, P(t) = exp {−γ′t}qsx(0)2+sy(0)2+sz(0)2exp {−2γ′t}(8.16) So, The degree of polarization for this single mode field for the single photon state will evolve accordingly.This degree tends then to zero with a typical time scale γ−1,which is exceedingly large. So, this model can successfully describe the phenomena of the Depolarization. 9 Discussions And Conclusions From the calculations of the geometric phase for the three level open system of mixed states for the quasicyclic evolution with quasicyclicity T=2π Ωit is evident that the becausse of the intercation between the system and the surroundings the system is no longer undergoing a cyclic evolution, where the exponent decay factors are included in the matrix elements of the density operator of the system, the components of the Bloch vector and the nonunit state vectors.It can be easily seen by putting the decaying matrix elements i.e. ρij(t) on equation [2.1.25] that the Bloch radius also decays with time and it corresponds to the fact that the physical state of the system is having a transition from the pure state to a mixed state. The mixed degree is independent of the decay rate. The approach towards the generalisation of the geometric phase for the mixed state requires the extensive usage of the kinematic approach due to Mukunda and Simon altough it was done 40
for the pure states we can use the similar idea here. Interesting to note that the expression of the geometric phase is found to be gauge invariant and independent of the choice of the gauge parameter; for both the global and local gauge transformation and also the reparametraisation invariance. The off diagonal matrix elements of the density operator ρmn;m=nin the eigenasis of the hamiltonian which measures the amount of overlap between the mth and nth energy level respectively decays as a function of time and it is obvious from the sign of the phase depends on the population determined by the bloch vector components w2(t) and w3(t) which involves the diagonal matrix elements of the density operator which remains constant irrespective of time as we have found above ,only depends on the value of δ1, δ2, δ3,w1(t) = ρ11(t)−ρ22(t)−ρ33(t) and w2(t) = ρ11(t)−ρ22(t). So, depending on the sign of the bloch vector components associated with the measure of the population inversion the sign of the phase will change accordingly. The Berry phases of the mixed states given by equation [4.20] doesn’t depend on all bloch parameters, it is independent of the radial bloch parameter rand as a result it holds for any mixed state that is being mapped inside the sphere. Berry phase depends on the decay rate, if the decay rate increases the berry phase decreases but the absolute value of the Berry phase increases which is evident from the relation between the Bloch radius and the Berry phase. In the context of the Lindblad Master equation we have assumed the dephasing noise can be represented by a single type of Lindbald operator or quantum jump operator which is a comparatively easy situation compared to the case where the master equation involves all 8 lindblad operator, although this will give us a more general situation when Γi=pηi(t)λirepresents the coupling to the environment. It is interesting to note that when the population inversions have differing signs i.e. w1(t)<0 and w2(t)>0, the Berry phase also changes as a function of the decaying parameter.The nonvanishing value of the Geometric phase establishes the fact that the quantum system under the consideration retains a memory of its evolution in terms of the Berry phase of the Mixed state.Thus, our definition of the geometric phase for the three level system of mixed state may have a hidden rich physics. As we have mentioned previously that the parametrisation used in the present context is not the only possible choice we can use any other parametraisation which satisfies the equation of the bloch sphere.We will get the similar kind of results for the geometric phase under those parametraisation and this must happen as the geometric phase does not depend on how we have parametrized the bloch sphere, it will be the same irrespective of the choice of it. In order to derive the expression of the geometric phase we have used the Pancharatnams formulae and we have obtained the general expression of the geometric phase more, specifically the Berry Phase for the mixed state which we have expressed as the closed path integral over the Mead-Berry connection one form defined over the manifold Bwhich we may say an extension of Berry’s framework including the conditions of adiabeticity and cyclicity. And along with that under a specific Gauge transformation the Pancharatnam’s formulae for the geometric phase reduces to the well known formulae of Aharonov-Anandan phase we obtain in case of pure state with the condition of cyclicity. It is interesting to note the universal feature of the geometric which as expected only depend on the choice of the closed curve Cfor the quasicyclic evolution of the system lying in the manifold B. Previously the kinematic approach has been used to calculate the geometric phase using the purification approach, here the geometric phase formulae obtained before for the pure states has been generalized. In the present context we can easily identify the expression of the geometric phase by dropping the total phase term which resembles the closed mathematical form of the geometric phase obtained in the Berry’s and Aharonov-Anandan’s framework under the limiting of pure quan41
with all the gnerators of the Su(2) group and the operator is called the Casimir Operator. The Casimir Operator of the SU(2) group is defined as follows, σ2=σ2 1+σ2 2+σ2 3=I2×2(11.2.8) So, the Casimir Operator will commute with all the generators of the SU(2) Group and as a result we have, [σ2, σi]=0,(i= 1,2,3).For, the conventional choice in quantum mechanics we choose σ2, σ3as the mutually commuting operators and any two mutually commuting operators have simultaneous eigenstates.The algebra of the spin-1 2particle is identical to the SU(2) Lie algebra.Furthermore, we can define the raising and lowering operator from the pauli matrices as, σ+=σ1+iσ2and σ−=σ1−iσ2and [σ2, σ±] = 0. The SU(3) Group Just like we have defined the SU(2) Group we can define the group of all 3 ×3 unitary matrices with determinant 1.It is a continuous group and a bigger symmetry group compared to SU(2). The number of real continuous parameters required to represent the group element being equal to the number of generators of that group. In general the group SU(n) has (n2−1) number of generators. The SU(3) group has 8 generators.Let U3×3∈SU(3) be an unitary matrix with determinant +1 then in general the matrix Ucan be represented as follows, U= exp A· λ 2!= exp 8 X i=1 Ai λi 2!(11.2.9) So, the generators of this group are identified as Ti=λi 2,(i= 1,2, .., 8) are called the Colour matrices which are traceless and hermitian with λi,(i= 1,2, .., 8) are the usual 3×3 Gell-Mann matrices given by, λ1= 0 1 0 1 0 0 0 0 0 ;λ2= 0−i0 i0 0 0 0 0 ;λ3= 100 0−1 0 000 ;λ4= 0 0 1 0 0 0 1 0 0 λ5= 0 0 −i 0 0 0 i0 0 ;λ6= 0 0 0 0 0 1 0 1 0 ;λ7= 0 0 0 0 0 −i 0i0 ;λ8=1 √3 1 0 0 0 1 0 0 0 −2 Some important properties of λi’s ; the generators of SU(3) are listed below, Tr[λi] = 0 [λr, λs] = 2ifrstλt f123 = 1, f458 =f678 =√3 2, f147 =f246 =f257 =f345 =f516 =f637 =1 2; {λr, λs}=4 3δrs + 2drstλt d118 =d228 =d338 =−d888 =1 √3, d448 =d558 =d668 =d778 =−1 2√3, d146 =d157 =−d247 =d256 =d344 =d355 =−d366 =−d377 =1 2; λrλs=2 3δrs +ifrstλt+drstλt, Tr[λrλs]=2δrs (11.2.10) Their commutators, anti-commutators and products involves two different three-index symbols or invariant tensors respectively fijk, drst, δrs.Only the independent components of the completely anti-symmetric f’s and the completely symmetric d’s have been listed above so that, 48
dijk =djik which is completely symmetric and fijk =−fjik which is completely antisymmetric and the kronecker delta δrs is completely symmetric unlike the case of the pauli matrices whose commutation and anti-commutation relations only involve a completely antisymmetric levicivita symbol i.e. ϵijk or, an invariant and completely antisymmetric tensor of rank 3.As we can see that the generators of the SU(3) group i.e. the Gell Mann matrices are mutually noncommuting, so again we will define the Casimir Operator for the SU(3) group. Interestingly, unlike the situation of the SU(2) Group here, we can define two different casimir operators denoted by C1and C2such that they commutes with all the generators of the SU(3) group i.e. the Gell-Mann matrices so that,[C1, λi] = [C2, λi] = 0,(i= 1,2, .., 8). The Casimir Operators of the SU(3) group are defined as follows, C1(λi) = 8 X i=1 λ2 i(11.2.11) C2(λi) = 8 X i=1 8 X j=1 8 X k=1 dijkλiλjλk(11.2.12) As, we have done in case of SU(2) here we can choose the mutually commuting operators as λ3, λ8and C1and can form the simultaneous eigenbasis from it.In the present context the SU(3) group has been used extensively. 11.3 Appendix C: Insights of Lindblad Operators and the Master Equation The Lindblad type of Master equation required to study the time evolution of the density operator corresponding to the quantum system in presence of interaction with the surroundings is given by, ∂ρ ∂t =−i ℏ[ˆ H, ρ] + γX iΓ† iρΓi−1 2{Γ† iΓi, ρ}(11.3.1) The symbol used in the above equation have their usual significance. The right hand side of the equation consists of Lindblad operators or,quantum jump operators.The Lindblad superoperator models the environmental conditions that make up the open quantum system such as dephasing and relaxation. The operators λiare also known as the collapse operator and it is important for deciding what the Lindblad superoperator describes. This operator is through which the environment couples to the system. Different collapse operators describe different aspects of the environment.γis an important constant that usually describes dephasing rate, rephasing rate, relaxation rate, etc. It is basically a corresponding rate for the coulping of the environment to the system. It also is important to the master equation. Note that whenever this constant is equal to zero, then we get the quantum Liovillian equation for a closed system without any environmental effects.One more thing we would like to add is that one can add as many Lindblad superoperators to the anticommutator to describe for different environmental conditions. One can describe for rephasing, another for dephasing, etc. It all depends on the environment of the quantum system.In summary, the Lindblad superoperator models an environmental coupling to the system. Without it, we get a model for a closed system with no environmental effects. That’s why the second term in the right hand side is important. A superoperator is like an operator that acts on other linear operators same for the lindblad operatorts too. Required to note that,If we consider only the first term on the right hand side of Lindblad Master equation [11.3.1] we obtain the Liouville-von Neumann equation. This term is the Liouvillian and describes the unitary evolution of the density operator. The second term on the 49
right hand side of the equation is the Lindbladian and it emerges when we take the partial trace - a non-unitary operation - of the degrees of freedom of reservior.The Lindbladian [2] describes the non-unitary evolution of the density operator. By the interaction form adopted here the physical meaning of the Lindblad operators can be understood: they represent the system S contribution to the System-Bath interaction remembering once more that the Lindblad equation was derived from the Liouville-von Neumann one by tracing the bath degrees of freedom. If the Lindblad operators Γiare Hermitian (observables), the Lindblad equation can be used to treat the measurement process. A simple application for a two level quantum system in this sense is the system Hamiltonian ˆ HS∝ˆσzwhere ˆσzbe the zcomponent of the 2 ×2 pauli matrices. when we want to measure one specific component of the spin (Γ ∝σα, α =x, y, z without any summation).If the Lindblad operators are non-hermitain then the master equation can be used to treat dissipation, decay and decoherence.For this type of scenario let us choose the system Hamiltonian same as in the previous case i.e. ˆ HS∝ˆσzwhere ˆσzbe the zcomponent of the 2 ×2 Pauli matrices with the Lindblad operators are taken as, Γ ∝ˆσ−,ˆσ−=ˆσx−iˆσy 2, where γbe the spontaneous emission rate [24]. 11.4 Appendix D: Homomorphic Mapping between SU(2) and SO(3) and extension for SU(3) We see that although the SU(2) group is different than the SO(3) group, the Lie algebras are homomorphic to each other so there exists a homomorphic mapping between the groups which seems to be disconnected from each other.The idea is that any unitary transformation initiated by some 2 ×2 unitary matrix corresponds to the SO(3) transformation of a vector in the 3 dimensional Euclidean Space which is equivalent to a rotation. Let us consider the following situation where, a spinor |ψ⟩which can be represented by a 2 ×1 coloum matrix (a two componnet vector) is undergoing an SU(2) transformation defined as, |ψ′⟩=H|ψ⟩,where H∈SU(2) (11.4.1) We, can write |ψ⟩=η1 η2and under the transformation the inner product or, the length of the vector remains invariant i.e. ⟨ψ′|ψ′⟩=⟨ψ|ψ⟩, now lets find how a matrix transforms under the unitary transformation. Let us consider the transformation of the outer product;which itself is a 2 ×2 matrix given as |ψ⟩⟨ψ|under the unitary transformation.So the outer product transforms to |ψ′⟩⟨ψ′|redefining |ψ⟩⟨ψ|by Mand |ψ′⟩⟨ψ′|by M′we have, |ψ′⟩⟨ψ′|=M′=H|ψ⟩⟨ψ|H†=HMH†=HMH−1(11.4.2) So, we get the transformation property of the the matrix Munder the unitary transformation of the spinor |ψ⟩. The matrix M is a traceless hermitain matrix. Here, H is given as, H=a b −b∗a∗(11.4.3) As,we know that any traceless 2 ×2 hermitian matrix can be written as a linear combination of the pauli matrices i.e. the generators of the SU(2) group. Then we can write for the matrix M, M=r · ˆσ(11.4.4) with,r being the position vector of a point drawn from the origin of the euclidean coordinate system in 3 dimension.Putting the expressions of the pauli matrices given in equation [11.2.6] 50
in equation [11.4.4] we can write, M=xσ1+yσ2+zσ3=x0 1 1 0+y0−i i0+z1 0 0−1=z x −iy x+iy −z(11.4.5) under the transformation the matrix Mchanges to M′such that, M′= r′·ˆ σ and similarly we can write, M′=x′σ1+y′σ2+z′σ3=x′0 1 1 0+y′0−i i0+z′1 0 0−1=z′x′−iy′ x′+iy′−z′ (11.4.6) From the transformation property of the traceless hermitian matrix Mwe get,det(M′) = det(HMH−1) = det(M)det(H)det(H−1) = det(M).This gives,. x′2+y′2+z′2=x2+y2+z2(11.4.7) So, the length of the position vector remains invariant under the transformation.We, know that the length of a vector remains invariant under the orthogonal transformation which is possible only under the SO(3) transformation. In fact we can show by using the condition M′=HMH†along with equation [??] that, r′= Rr ⇒x′ i=Rijxjwhere, R∈SO(3).So,every SU(2) transformation of a 2 component vector will correspond to a SO(3) transformation of a 3 component vector in the 3 dimensional Euclidean space which is equivalent to the rotation of a vector in Eucledian space. More specifically we can write the unitary matrix Has, H= exp (iσ ·ˆnθ 2) (11.4.8) with,ˆσbe the pauli matrices and θbe the angle of rotation for the vector |ψ⟩in the two dimemsional spinor space and ˆnbe the unit vector along the direction of rotation in the spinor space.For example if the direction of rotation of rotation is along the z direction then ˆn= ˆzthen the unitary matrix Hwill be, H= exp (iσzθ 2) and so on.The corresponding SO(3) transformation matrix Rcan be written as, R= exp (i J·ˆnθ) (11.4.9) so,the SU(2) transformation of the vector |ψ⟩in the spinor space or, the rotation of the vector |ψ⟩in the spinor space with respect to the z direction corresponds to the rotation of the 3 dimensional vector here, in our case is the position vector of a point(although it can be in general true for any vector) in the three dimensional Euclidean space.Then, H= exp (iσz θ 2) = eiθ 20 0e−iθ 2!←→ Rz(θ) = cos θsin θ0 −sin θcos θ0 0 0 1 (11.4.10) It means, that under the unitary transformation defined by,|ψ′⟩= exp (iσzθ 2)|ψ⟩will correspond to the rotation with respect to the z direction in the Euclidean space i.e. x′ y′ z′ = cos θsin θ0 −sin θcos θ0 0 0 1 x y z And, similarly we can write the same for the rotation in the spinor 51
space with respect to the X and Y direction along with the rotation about z direction given by, H= exp (iσz θ 2) = eiθ 20 0e−iθ 2!←→ Rz(θ) = exp (iJzθ) = cos θsin θ0 −sin θcos θ0 0 0 1 (11.4.11) H= exp (iσy θ 2) = cos θ 2sin θ 2 −sin θ 2cos θ 2←→ Ry(θ) = exp (iJyθ) = cos θ0 sin θ 010 sin θ0 cos θ (11.4.12) H= exp (iσx θ 2) = cos θ 2isin θ 2 isin θ 2cos θ 2←→ Rx(θ) = exp (iJxθ) = 1 0 0 0 cos θsin θ 0−sin θcos θ (11.4.13) In general we can write, H∈SU(2) = exp (iσ ·ˆnθ 2)←→ R∈SO(3) = exp (i J·ˆnθ). Where,J be the generators of rotation of SO(3) group defined by, Jk(θ) = 1 i dRk(θ) dθ θ=0; where k=x, y, z (11.4.14) Simple calculation shows that the generators of the group can be written using the equation (7.4.14) as, Jz=1 i dRz(θ) dθ θ=0 = 0−i0 i0 0 0 0 0 Jx=1 i dRx(θ) dθ θ=0 = 0 0 0 0 0 −i 0i0 , Jy=1 i dRy(θ) dθ θ=0 = 0 0 i 0 0 0 −i0 0 (11.4.15) Where, the generators of the So(3) group Ji,where (i=x, y, z) satisfies the cyclic commutation relation given by, [Jk, Jm] = iϵkmnJn.(11.4.16) The lie algebra of the generators of SO(3) being closed and the commutation relations obeyed by the generators Jiis identical to the Lie algebra of SU(2) Group which establishes the homomorphic mapping between the two groups SO(3) and SU(2). An SU(2) transformation on |ψ⟩=η1 η2≡SO(3) transformation on x y z . Extension for SU(3): The similar idea can be extended in the context of the SU(3) group which is a bigger group than SU(2). The idea is to establish a homomorphic mapping between the groups SU(3) and SO(8) Groups. Just like we have considered the unitary transformation of the 2 component vector |ψ⟩here, we will consider the unitary transformation of a three compomnent nonunit vector ray |Ψ⟩= η1 η2 η3 in the hilbert space H3defined as follows, |Ψ′⟩=U|Ψ⟩; where U∈SU(3) (11.4.17) 52
Now if we consider the transformation of the projection operator |Ψ⟩⟨Ψ|represented by a 3×3 matrix say,Munder the unitary transformation defined by equation [11.4.17] then we may write similarly, |Ψ′⟩⟨Ψ′|=M′=U|Ψ⟩⟨Ψ|U†=UMU†=UMU−1(11.4.18) Keeping analogy with the SU(2) situation we can say that any 3×3 traceless Hermitian Matrix can also be written as the linear combination of the generators of SU(3) i.e. the Gell Mann matrices which are traceless hermitian. So, we can write for the matrix M, M= A· λ= 8 X i=1 Aiλi(11.4.19) Where, Abe any arbitrary octet vector in the eight dimensional Euclidean space i.e. A∈ R8.From the transformation property of the traceless hermitian matrix Mwe get,det(M′) = det(UMU−1) = det(M)det(U)det(U−1) = det(M) along with M′= A′· λ=P8 i=1 A′ iλiThis gives, 8 X i=1 A′ i 2= 8 X i=1 A2 i(11.4.20) We get the condition for the invariance of the length of the octet vector Aunder the unitary transformation, the smimilar result has been obtained in the case of a 3 dimensional vector as a result of SU(2) transformation.So, we can say that, An SU(3) transformation on |Ψ⟩= η1 η2 η3 ≡SO(8) transformation on A1 A2 A3 A4 A5 A6 A7 A8 . On summarising the above results we can conclude that, |Ψ′⟩=U|Ψ⟩ ←→ A′=R A. Where U∈SU(3) and R∈SO(8).(11.4.21) where,R be the 8 ×8 transformation matrix which, gives the correspondence of the SU(3) transformation of a three component vector |Ψ⟩with the rotation of a vector Ain the eight dimensional Euclidean Space which is equivalent to a SO(8) transformation of an octet vector Awith the components of the vector Atransforms obeying the equation, A′ i=RijAj(11.4.22) With,Rij be the matrix elements of the transformation matrix R8×8∈SO(8). But the homomorphism for the SU(3) case is more intricate. In general we can write R= exp (iLiϕi) = exp (i L· ϕ).Where, we can identify Lias the generators of the SO(8) Group and ϕias the parameters of the Group.But there are 8 2= 28 generators for the SO(8) Group and as a result 28 parameters. All real orthogonal unimodular 8×8 matrices taken together form the 28 dimensional group SO(8); the matrices of the octet representation of SU(3) are a ‘very small’ eight-dimensional subset of SO(8), in fact a subgroup. 53
11.5 Appendix E: Interaction Picture in Quantum Mechanics: Let, us briefly introduce the idea of the interaction picture in order to understand the idea of the Lindblad Type of master equation describing the time evolution of the Density operator of a quantum system interacting with the environment which is obtained as a special case of the Markovian Master Equation along with the Local time approximation also known as the Born Markovian Approximation. We have some ideas over the Schr ¨ Odinger picture and the Heisenburg Picture and about the equivalence of the pictures. The central idea of this formalism’s are to be noted. •In Schrodinger picture the observables i.e. the operators are time independent but the state vector evolves in time.The time evolution of the state vector id governed by the time dependent Schrodinger equation given by, iℏd dt |ψ(s)(t)⟩=ˆ H(s)(t)|ψ(s)(t)⟩(11.5.1) where, |ψ(s)(t)⟩be the state vector in the schrodinger picture and the hamiltonian being in general time dependent. •In the Heisenburg Picture the Operators are time dependent and the State Vectors are freezed i.e. independent of time.The time evolution of the operator say, ˆ A(t) is governed by the Heisenburg’s equation of motion given by, d dt ˆ A(t) = 1 iℏ[ˆ A(t),ˆ H(H)(t)] + ∂ˆ A(t) ∂t (11.5.2) •The intercation Picture where the hamiltonian can be splitted up as the sum of the free hamiltonian and the interaction hamiltonian,both the state vector and the operator are time dependent. As, in the interaction picture both the state vector and the operators are time dependent we need two separate equations governing the time evolution of the State vector and that for the operator as well. Here, for simplicity we will assume the hamiltonian of the system to be independent of time in the schrodinger picture and under this condition in the Heisenburg Picture the operators are defined by, ˆ AH(t) = eiˆ H(s)tˆ Ase−iˆ H(s)t(11.5.3) here, it is convenient to use the natural unit system i.e.ℏ= 1. ˆ Asbe the operator in the schrodinger picture which is independent of time. The hamiltonian in the schrodinger picture is written as the sum of the Free hamiltonian and the interaction hamiltonian.So, we write, ˆ H=ˆ H(S) 0+ˆ H(S) int (11.5.4) Here, ˆ H(S) 0and ˆ H(S) int be the Free hamiltonian and the interaction hamiltonian in the Schrodinger picture. Now in the schrodinger picture with a time independent hamiltonian we have, |ψ(s)(t)⟩=e−iˆ H(s)t|ψ(s)(0)⟩(11.5.5) In the interaction picture the state vector is denoted by |ψ(t)⟩IP and it is defined by, |ψ(t)⟩IP =eiˆ H(s) 0t|ψ(s)(t)⟩=eiˆ H(s) 0te−iˆ H(s)t|ψ(s)(0)⟩(11.5.6) 54
we can in general write, |ψ(0)⟩IP =|ψ(s)(0)⟩=|ψ(H)(0)⟩. So we can write, |ψ(t)⟩IP =eiˆ H(s) 0t|ψ(s)(t)⟩=eiˆ H(s) 0te−iˆ H(s)t|ψ(0)⟩IP (11.5.7) using the equation (7.5.4) we can write, |ψ(t)⟩IP =e−iˆ H(s) intt|ψ(0)⟩IP Now, the operator in the Interaction picture is defined as, ˆ O(IP )(t) = eiˆ H(s) 0tˆ O(s)e−iˆ H(s) 0t =eiˆ H(s) 0te−iˆ H(s)tˆ O(H)(t)eiˆ H(s)te−iˆ H(s) 0t(11.5.8) where, we have used ˆ O(s)=e−iˆ H(s)tˆ O(H)(t)eiˆ H(s)t, the relation between the operators in the schrodinger and hisenburg picture. As we can see from equation (7.5.8) the operator in the interaction picture is time dependent. Here, also we can write that, ˆ O(IP )(0) = ˆ O(s)=ˆ O(H)(0) So, far we have defined, |ψ(t)⟩IP =eiˆ H(s) 0t|ψ(s)(t)⟩=eiˆ H(s) 0te−iˆ H(s)t|ψ(0)⟩IP and ˆ O(IP )(t) = eiˆ H(s) 0tˆ O(s)e−iˆ H(s) 0t(11.5.9) The equation governing the time evolution of the state vector in the interaction picture is derived as follows, i∂ ∂t |ψ(t)⟩IP =i∂ ∂theiˆ H(s) 0te−iˆ H(s)t|ψ(0)⟩IP i =−ˆ H(s) 0eiˆ H(s) 0te−iˆ H(s)t|ψ⟩H+eiˆ H(s) 0tˆ H(s)e−iˆ H(s)t|ψ⟩H =−ˆ H(s) 0|ψ(t)⟩IP +eiˆ H(s) 0tˆ H(s)e−iˆ H(s) 0teiˆ H(s) 0te−iˆ H(s)t|ψ⟩H =−ˆ H(IP ) 0|ψ(t)⟩IP +eiˆ H(s) 0tˆ H(s)e−iˆ H(s) 0t|ψ(t)⟩IP =−ˆ H(IP ) 0|ψ(t)⟩IP +ˆ H(IP )|ψ(t)⟩IP =ˆ H(IP ) int |ψ(t)⟩IP (11.5.10) So,the equation that describes the time evolution of the state vector is given by, i∂ ∂t |ψ(t)⟩IP =ˆ H(IP ) int |ψ(t)⟩IP (11.5.11) Here, we have used, ˆ H(IP ) int (t) = eiˆ H(s) 0tˆ H(s) inte−iˆ H(s) 0twhich gives the Interaction hamiltonian in the interaction picture related to the Interaction hamiltonian in the schrodinger picture by an unitary transformation. Similarly we can derive the equation governing the time evolution of the operator in the interaction picture which is time dependent. ∂ ∂t ˆ O(IP )(t) = ∂ ∂theiˆ H(s) 0tˆ O(s)e−iˆ H(s) 0ti =iˆ H(s) 0eiˆ H(s) 0tˆ O(s)e−iˆ H(s) 0t−ieiˆ H(s) 0tˆ O(s)e−iˆ H(s) 0tˆ O(s) =iˆ H(s) 0ˆ O(IP )(t)−iˆ O(IP )(t)ˆ H(s) 0 =1 i[ˆ O(IP )(t),ˆ H(IP ) 0(t)] (11.5.12) Here, we have used that ˆ H(IP ) 0(t) = eiˆ H(s) 0tˆ H(s) 0e−iˆ H(s) 0t=ˆ H(s) 0eiˆ H(s) 0te−iˆ H(s) 0t=ˆ H(s) 0. Upon summarising the results obtained for the time evolution of the state vector and the operator in 55
the Interaction picture are as follows, i∂ ∂t |ψ(t)⟩IP =ˆ H(IP ) int (t)|ψ(t)⟩IP ∂ ∂t ˆ O(IP )(t) = 1 i[ˆ O(IP )(t),ˆ H(IP ) 0(t)] (11.5.13a) (11.5.13b) Here, the free hamiltonian in the interaction picture i.e. ˆ H(IP ) 0is time dependent. In our present context the Operator is the global density operator of the super-system i.e.(system+surrounding) and the density operator of the system is obtained upon taking the partial trace of the Global density operator with respect to the Bath states (here, the surrounding can be taken as a Bath). The time evolution of the Global Density operator corresponding to the supersystem will be described by the equation just by replacing ˆ Oby the Global density operator ˆρSB(t) in the equation [11.5.13].We get, ∂ ∂t ˆρ(IP) SB (t) = 1 i[ˆρ(IP ) SB (t),ˆ H(IP ) 0(t)] (11.5.14) With, ˆρ(IP ) SB (t) is defined by, ˆρ(IP ) SB (t) = eiˆ H(s) 0tˆρ(s) SBe−iˆ H(s) 0t(11.5.15) Here, we have used the above equation as the starting point to derive the Lindblad Master Equation using the Born-Markovian Approximation.The density operator corresponding to the system is obtained upon taking the partial trace of the Global density operator ˆρ(IP) SB (t) with respect to the Bath states defined as follows, ˆρIP s(t) = TrB[ˆρ(IP ) SB (t)] (11.5.16) As, a result of the Born-Markov Approximation the time evolution of the system’s Density operator ˆρs(t) is governed by the Lindblad Master Equation given below, ∂ˆρs(t) ∂t =−i ℏ[ˆ H, ˆρs] + 8 X i=1 Γ† iˆρsΓi−1 2{Γ† iΓi,ˆρs}(11.5.17) The derivation of the Master Equation has been skipped as it requires the detailed knowledge of generators related to Quantum Dynamical Semigroups [21]. Alternatively, we can define the state vector and the operator in the interaction picture as follows, |ψ(t)⟩IP =eiˆ H(s) intt|ψ(s)(t)⟩=eiˆ H(s) intte−iˆ H(s)t|ψ(0)⟩IP ˆ O(IP )(t) = eiˆ H(s) inttˆ O(s)e−iˆ H(s) intt (11.5.18a) (11.5.18b) Similar calculations leads to the equations governing the time evolution of the state vector and the operator in the interaction picture given by, i∂ ∂t |ψ(t)⟩IP =ˆ H(IP ) 0(t)|ψ(t)⟩IP ∂ ∂t ˆ O(IP )(t) = 1 i[ˆ O(IP )(t),ˆ H(IP ) int (t)] (11.5.19a) (11.5.19b) Here, the standard definition of the state vector and the Operators in the Interaction picture given by the set of equations [11.5.13] has been used, but ultimately it’s a choice which definition will be convenient to use for the purpose of calculation. 56
Remarks regarding the equivalence of the three pictures: Although the three pictures i.e. Schrodinger Picture, Heisenburg Picture and The Interaction Picture the quantum mechanical expectation values at three pictures are identical. Let us check the equivalence of three pictures by evaluating the expectation value of the operator in the interaction picture as follows, .⟨ˆ A(t)⟩IP =⟨ψIP (t)|ˆ AIP (t)|ψIP (t)⟩ =⟨ψ(s)(t)|e−iˆ H(s) 0teiˆ H(s) 0tˆ A(s)e−iˆ H(s) 0teiˆ H(s) 0t|ψ(s)(t)⟩ =⟨ψ(s)(t)|ˆ A(s)|ψ(s)(t)⟩=⟨ˆ A⟩s =⟨ψH(0)|ˆ AH(t)|ψ(H)(0)⟩=⟨ˆ A(t)⟩H (11.5.20) here, we have used that, ˆ A(s)=e−iˆ H(s)tˆ A(H)(t)eiˆ H(s)tin the above equation.Then we can conclude from the above calculation that, ⟨ˆ A(t)⟩IP =⟨ˆ A(t)⟩H=⟨ˆ A⟩s(11.5.21) Which establishes the equivalence of three different pictures as the expectation values of the operator ˆ Ain all the three pictures are identical and what we do measure experimentally is the expectation value of some observable. References [1] Yakir Aharonov and J Anandan. Phase change during a cyclic quantum evolution. Physical Review Letters, 58(16):1593, 1987. [2] Carlos Alexandre Brasil, Felipe Fernandes Fanchini, and Reginaldo de Jesus Napolitano. A simple derivation of the lindblad equation. arXiv e-prints, pages arXiv–1110, 2011. [3] Neil W Ashcroft and N David Mermin. Solid state physics (saunders college, philadelphia, 1976). Appendix N, 166:87, 2010. [4] Michael Victor Berry. Quantal phase factors accompanying adiabatic changes. Proceedings of the Royal Society of London. A. Mathematical and Physical Sciences, 392(1802):45–57, 1984. [5] Max Born and Emil Wolf. Principles of optics: electromagnetic theory of propagation, interference and diffraction of light. Elsevier, 2013. [6] Christian Brosseau. Fundamentals of polarized light: a statistical optics approach. WileyInterscience, 1998. [7] G Dattoli, Roberto Mignani, and A Torre. Geometrical phase in the cyclic evolution of non-hermitian systems. Journal of Physics A: Mathematical and General, 23(24):5795, 1990. [8] Stefan Filipp, Juergen Klepp, Yuji Hasegawa, Christian Plonka-Spehr, Ute Schmidt, Peter Geltenbort, and Helmut Rauch. Experimental demonstration of the stability of berry’s phase for a spin-1/2 particle. Physical review letters, 102(3):030404, 2009. [9] Grant R Fowles. Introduction to modern optics. Courier Corporation, 1989. 57