Question #16009

How can I multiply the initial density matrix by the time evolution (U) to get the density matrix at any time ?

Expert's answer

How can I multiply the initial density matrix by the time evolution (U) to get the density matrix at any time?

Answer

Let's discuss the time evolution of mixed states. In the case of a bipartite pure state governed by the usual axioms of quantum theory, let us suppose that the Hamiltonian on HAHBH_A \otimes H_B has the form


HAB=HA1B+1AHB.H_{AB} = H_A \otimes 1_B + 1_A \otimes H_B.


Under this assumption, there is no coupling between the two subsystems A and B, so that each evolves independently. The time evolution operator for the combined system


UAB(t)=UA(t)UB(t),U_{AB}(t) = U_A(t) \otimes U_B(t),


decomposes into separate unitary time evolution operators acting on each system.

In the Schrodinger picture of dynamics, then, an initial pure state ψ(0)AB|\psi(0)\rangle_{AB} of the bipartite system given by eq. (2.57) evolves to


ψ(0)AB=i,μaiμi(t)Aμ(t)B,|\psi(0)\rangle_{AB} = \sum_{i,\mu} a i \mu \, |i(t)\rangle_A \otimes |\mu(t)\rangle_B,


where


i(t)A=UA(t)i(0)A,|i(t)\rangle_A = U_A(t) \, |i(0)\rangle_A,μ(t)B=UB(t)μ(0)B,|\mu(t)\rangle_B = U_B(t) \, |\mu(0)\rangle_B,


define new orthonormal basis for HAH_A and HBH_B (since UA(t)U_A(t) and UB(t)U_B(t) are unitary). Taking the partial trace as before, we find


ρA(t)=i,j,μai,μajνi(t)AAj(t)=UA(t)ρA(0)UA(t).\rho_A(t) = \sum_{i,j,\mu} a_{i,\mu} \, a_{j\nu}^* \, |i(t)\rangle_{AA} \, \langle j(t)| = U_A(t) \rho_A(0) U_A(t)^\dagger.


Thus UA(t)U_A(t), acting by conjugation, determines the time evolution of the density matrix.

In particular, in the basis in which ρA(0)\rho_A(0) is diagonal, we have


ρA(t)=apaUA(t)ψ(0)AAψa(0)UA(t).\rho_A(t) = \sum_a p_a U_A(t) \, |\psi(0)\rangle_{AA} \, \langle \psi_a(0) | U_A(t).


Eq. (2.80) tells us that the evolution of ρA\rho_A is perfectly consistent with the ensemble interpretation. Each state in the ensemble evolves forward in time governed by UA(t)U_A(t). If the state ψa(0)|\psi_a(0)\rangle occurs with probability pap_a at time 0, then ψa(0)|\psi_a(0)\rangle occurs with probability pap_a at the subsequent time t.

On the other hand, it should be clear that eq. (2.80) applies only under the assumption that systems A and B are not coupled by the Hamiltonian. Later, we will investigate how the density matrix evolves under more general conditions.

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