Physics 4022 Notes on Density Matrices

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1 Physics 40 Notes on Density Matrices Definition: For a system in a definite normalized state ψ > the density matrix ρ is ρ = ψ >< ψ 1) From Eq 1 it is obvious that in the basis defined by ψ > and other states orthogonal to ψ >, ρ is a matrix in which every entry is zero except for one diagonal element which is 1. Thus Trρ = 1. Suppose that the system consists of two parts a part A which is described by the set of states i A > and a part B which is described by the states µ B >, so that a basis for the full Hilbert space is given by the direct product of the states i A > and µ B > for example, a system consisting of two spins, A and B). Any state φ > may be expressed as φ >= i A µ B φ iµ i A > µ B > ) In this basis the density matrix may be written ρ = ψ iµ ψj,ν µ B > i A >< j A < ν B 3) i A j A µ B ν B Consider an operator O A which couples only to the A states. In the basis given in Eq we have O A = i A j A O ij i A >< j A 1 B 4) where the 1 B means that the operator acts as the unit matrix i.e. does nothing to) the B states. Thus the expectation value of the operator in the state ψ is < ψ O ψ >= φ jν < j A < ν B O ij j A >< i A 1 B φ iµ i A > µ B > i ja ν B i A j i A A µ B = ) φ iµ φ jµ < j A O ij j A >< i A i A > 5) µ B i ja j A i A j A Define now the reduced density matrix ρ A = T r B ρ = µ B < µ B ρ µ B > = µ B < µ B = i A j A ρ i,j i A j A µ B ν B ψ iµ ψ j,ν µ B > i A >< j A < ν B µ B > A i A >< j A 6) 1

2 with ρ ij A = µ B ψ iµ ψ j,µ 7) From this definition it is obvious that the reduced density matrix is hermitian. Further, T r A ρ A = i A ρ ii A = i A µ B ψ iµ ψ i,µ = 1 8) where the last equality follows from the normalization of ψ. Because ρ A is hermitian, it be diagonalized by a unitary transformation. In the basis in which it is diagonal ρ ij A = δ ij ψ iµ ψi,µ µ B so the eigenvalues are obviously non-negative. Finally, we see using Eqs 5 and 7 that for any operator O A acting only on the A subspace, < ψ O A ψ >= O ij ρ ji = T r A O A ρ A 9) i,j Thus, the reduced density matrix gives you all of the information you can get about a system of which you can only measure one part. Note: sometimes in books you will see a slightly different definition, in which the density matrix is is defined in terms of a system which is not in a specific state. To my mind, this raises epistemological problems which are best avoided. I think it is least confusing to say that a system the universe, if you like) is described by a definite wave function i.e density matrix with only one nonzero eigenvalue), and the part of it we can measure is described by a reduced density matrix. ****Definition: The part of a system which we can observe is said to be in a pure state if it is described by a reduced density matrix with exactly one non-zero eigenvalue. If more than one eigenvalue is non-zero we say the system is entangled with some other state.***** Remark: the diagonal elements ρ n,n A of the reduced density matrix give the probabilities that a measurement will find subsystem A in state n. Example: two state sub-system. If the part, A, which we can observe, is a two state system, then the most general density matrix may be written explicitly as ρ A = λ ) σ 10) This matrix obviously has unit trace. The condition that all all eigenvalues are real and non-negative is that λ is a real vector with magnitude less than or equal to unity. We observe that the expectation value of the a component of the spin in the state described by this density matrix is < σ a >= T rρ A σ a = λ a 11)

3 recall T rσ α = 0; T rσ α σ β = δ αβ ). Thus this density matrix describes a partially polarized spin with moment directed in the λ direction. If we choose spin quantization axis parallel to λ then the reduced density matrix is diagonal, with eigenvalues ρ, = 1 + λ)/ and ρ, = 1 λ)/. The probability that the spin is up in this basis is 1 + λ)/. Time dependence From Schroedinger s equation ι ψ = Hψ we obtain ι ρ = [H, ρ] 1) What we really need to know is the time dependence of the reduced density matrix ρ A. This cannot be determined in general without solving the full Schroedinger equation of the whole system. However, progress may be made in certain physically relevant special cases. a) Hamiltonian acts only on A subspace: First, assume that H acts nontrivially only on the A-subspace is, e.g. a magnetic field which couples only to the A spins) and thus acts as the identity operator on the B space. In this case H = H AA 1 B 13) and by taking the trace of Eq 1 over the B states we find for the projected density matrix ι ρ A = [H AA, ρ A ] 14) Exercise: If the A subspace is a single S=1/ spin then from Eq 10 we see that the time dependence of ρ A is parametrized by the time dependence of λ. If H AA = 1 gµ B B σ, obtain the equation giving the time dependence of λ. Answer: d λ /dt = ω B B λ with ωb = gµ B B/ the polarization precesses in an applied field exactly as a classical magnetic moment would. b) Two spins: depolarization Consider a universe composed of two S=1/ spins, described by the Hamiltonian H spin = J S 1 S 15) and suppose the wave function at t = 0 is a superposition of the singlet and m = 0 triplet states Then we know that ψt) >= ψt = 0) >= cos θ >) + sin θ > + >) 16) cos θeijt/4 > >) + sin θe i3jt/4 > + >) 17) Exercise: show that the reduced density matrix describing the behavior of spin 1 is of the form of Eq 10 with λ t) = ẑ sin θ cos Jt 3 18)

4 We see from this simple example that coupling to environmental degrees of freedom can change the magnitude of polarization of a system. c) Two spins: dephasing Consider first a single S = 1/ spin, in the state In this state ψ 1 >= cosθ > +e iφ sinθ > ) 19) < σ z > = cosθ 0) < σ x > = sinθcosφ 1) Observe that < σ x > depends on the relative phase between the two parts of the wave function. Now suppose that there is a second spin in the universe, which is entangled with the first spin by the Hamiltonian H spin = J S z 1S z ) ) and suppose that at time t = 0 the system is in the state ψ > = cosθ 1 > +e iφ1 sinθ 1 > ) cosθ > +e iφ sinθ > ) 3) = cosθ 1 cosθ > +e iφ1 φ) sinθ 1 sinθ > +cosθ 1 sinθ e iφ > +sinθ 1 cosθ e iφ1 > 4) Exercise: show that the reduced density matrix describing the behavior of spin 1 at time t = 0 is of the form of Eq 10 with λ = xsinθcosφ + ŷsinθsinφ + ẑcosθ 5) Exercise: show that if the two spin system is governed by Eq then the expectation value of σ z is time independent but < σ x > oscillates in time with frequency Jt/. We see from this simple example that coupling to other degrees of freedom can change the phase relations between different components of a state: in other words can dephase a state. Note that dephasing and depolarization are basis dependent statements. The language assumes that there exists a preferred basis for the reduced density matrix for example, in the hydrogen atom problem the basis of hydrogenic states 1S, P,...). Then the depolarization relaxation) processes are those which cause transitions between these states, changing the diagonal components of the density matrix and driving the system towards some sort of equilibrium with the environment while dephasing processes change the phase relations characteristic of the off-diagonal elements of the density matrix, and typically drive the off diagonal components to zero. Now, let us generalize the calculation to the case of a system coupled to many other degrees of freedom. I will state the results in the language of spins, but the conclusions are general. 4

5 As a first step, let us consider a more general form of our two-spin Hamiltonian: H spin, = J z S1S z z + J S + 1 S + ) S 1 S+ 6) Observe that the J z term in H spin, commutes with S1 z so the only part of H spin, which does anything to ψ > is the exchange term J S + 1 S + ) S 1 S+ which has one term which flips the system spin 1) from down to up and the environment spin ) from up to down, or conversely in other words, which allows the system to exchange z-direction spin polarization with the environment and leading to relaxation of the polarization. On the other hand, the J z term affects the energy of the different eigenstates of S z, allowing the system to exchange energy with its environment, leading to dephasing. Now, consider a system which we model here by a spin, operators S 0 ) weakly coupled to an environment which we model here by a set of many, many other spins, operators S n with n = 1,...). We assume that the system constitutes a negligible perturbation on the environment. Consider the two parts of H spin, Eq 6) separately. Depolarization: J ) The depolarization or relaxation processes come from the J terms, and in the spin language are described by the Hamiltonian H depol = n J n S + 0 S n + S 0 S+ n ) This Hamiltonian causes transitions between the different states of the system here, the up and down states of spin 0). This is a problem we have seen before. If the states n form a continuum and the coupling J is weak we may apply Fermi s Golden Rule, learning that H depol gives rise to transitions from to at a rate R down and transitions from to at a rate R up. These rates are determined by the J n and the properties of the environment compare our results for the hydrogen atom in a radiation field). The result is rate equations ρ, ρ, = R down ρ, + R up ρ, 7) = R up ρ, + R up ρ, 8) Eqs 8 lead to relaxation to a steady state described by ρ, ρ, = R down R up 9) I hope you see that this reasoning generalizes immediately to a system with n > states. Summary: relaxation depolarization) is a coupling to the environment which changes the diagonal elements of the density matrix, leading in usual cases) 5

6 to a steady state in which the system is distributed among the different states accessible to in, in a manner determined by the environment. Dephasing: J z ) In the simple two-state model the dephasing or energy exchange processes come from the J z terms and are described by the Hamiltonian H dephase = n J n z S z 0S z n + S z 0S z n) 30) This Hamiltonian says that the energy of a given state of the system depends on the environment; in other words, it lets the system exchange energy with the environment. Observe that H dephase commutes with S1; z thus it cannot change the z-direction polarization. More generally dephasing processes commute with the diagonal elements of the density matrix. As you know, the phase of a quantum state evolves according to e iet/, thus changes in energies change phase relations between different states. Dephasing is a rather subtle process, with typically) a simple conclusion: dephasing randomizes the phase relations between different states, thereby driving the off-diagonal components of the density matrix to zero; in other words, forcing the system into an incoherent superposition of eigenstates. Let us see how this works for the simple H dephase above. This Hamiltonian is simple enough that we can write down the exact time evolution: if at time t = 0 we have ) ψ µ t) >= cosθ > Π n e ij n z t/ cosθ n n > +e ij n z t/ e iφn sinθ n n > + e iφ sinθ > ) e ij n z t/ cosθ n n > +e ij n z t/ e iφn sinθ n n > 31) n This ugly looking expression can be simply understood: if our system spin is up, then environment spin n feels an effective magnetic field J n, while if our system spin is down, then environment spin n feels a magnetic field of the same magnitude but opposite sign. We can now construct the reduced density matrix as before, obtaining with ρt) = cos θ >< + sin θ >< + Et)sinθe iφ >< + E t)sinθe iφ >< 3) Et) = 1 n cos θ n e ijnt/ + sin θ n e ijnt/ ) 33) The key point is that as the environment becomes large number of terms in the product becomes large) then lim t Et) 0. You can demonstrate this easily numerically. I will now give two analytical arguments; one sloppy and one precise, but for a special case. 6

7 i) Sloppy argument Et) = n 1 + cos θ n 1)sin J n t/ ) ) 34) At long times, the sin Jt term oscillates very fast, so replace it by its time average, 1/. Also for a typical environment state,all of the θ n are different, so replace cos θ n by its average, which is 1/. The result Et) n=1...n 1 1 ) = 4 ) N 3 N 0 35) 4 ii) Simpler case: suppose that all of the J n are equal to J and all of the θ n = ψ. Then Et) = cos ψe ijt/ + sin ψe ijt/ ) N 36) Using the binomial expansion gives Et) = = L=0...N L=0...N N! N L)!L! cos ψ) N L sin ψ) L e in L)Jt/ 37) [ ) ] N! Exp ln + N L)lncos ψ) + Llnsin ψ) e in L)Jt/ N L)!)L! This equation may be easily numerically evaluated for any reasonable) N. We may perform an analytical evaluation for large N, using the approximation N! πnexp [NlnN N], replacing the sum over L by an integral and defining L = xn. Exercise: show that on these assumptions Eq 38 becomes with 1 Et) = einjt/ πn 0 dx x1 x) Exp [ N Φx) i Jt x )] 38) Φx) = 1 x)ln1 x) + xlnx + lntan ψ)) + lncos ψ) 39) As N becomes large the integral becomes dominated by the x for which Φ is minimized. Call this value x. We then have Φx) = Φx ) + 1 Φ x x ) ) If the terms represented by the ellipsis are ignored and we set x = x in the prefactor then we may evaluate the integral in Eq 38, obtaining Et) = e in1 x )Jt/ e 4Nsin ψ Jt ) 41) We therefore see that apart from oscillations, the magnitude of E decays quadratically in time, and falls to essentially zero in a time of order /J N. as 7

8 we saw previously, in order to obtain a reasonable continuum limit one typically must scale the interaction by the square root of the number of degrees of freedom J 1/ N. it is interesting to compare the decoherence time t decoh N/J N with the golden rule decay time t depol N/J D) which involves the density of states D. We have t depol t decoh N JD 4) The density of states is like an inverse energy; in many cases the corresponding energy is large, so JD << and decoherence happens very fast compared to depolarization. Summary: exchange of energy with an environment leads to decoherence, which drives typically rapidly) the off diagonal components of the density matrix to zero. 8