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Lecture 3: Mixed States and Entangled States 1 the Density Matrix ∑ ∑

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Lecture 3: Mixed States and Entangled States 1 the Density Matrix ∑ ∑ Lecture 3: Mixed states and entangled states Read:Gottfried & Yan, Chapter 1, 2.1, 2.2. 1 The density matrix Quantum mechanical systems can be in states which cannot be described by wavefunctions. Such states are called the mixed states. The states described by wavefunctions are called pure states. Unlike the “statistical description” in classical mechanics which is essentially due to our insufficient knowledge of classical systems, even if we have complete knowledge of quantum systems, they still can be in the mixed state. This subtle point will be explained below. In this case, we need to use the concept of density matrix instead of wavefunctions. A mixed state can be represented as an incoherent summation of orthonormal bases jYii’s as r = ∑rijYiihYij; (1) i where ri is the probability for the system in the state of Yi, and Yi’s are the diagonal basis for r. ri are called eigenvalues of the density matrix r. Please note that Eq. 1 is not the superposition of states jYii. There are no interference between different jYii’s. If we measure an observable F, the distribution of measurement results in the mixed state described by r is the sum of those from individual pure states jYii with the corresponding probability ri. In other words, when performing measurements of F in the mixed states, we have two different steps of average. 1. For each eigenstate jYii or r, we have the quantum mechanical statistical interpretation F¯i = hYijFjYii. Generally speaking, the diagonal bases for r are not the eigenstates for F, in which F = ∑i j jYiihY jjFi j. 2. The second step is the usual statistical average over the probability distribution ri, such that we have the expectation value of F as ¯ F = ∑r jFj = ∑r jhY jjFjY ji; (2) j j which can further expressed in a formal way as F = ∑r jhY jjFjY ji = ∑∑r jhYijY jihY jjFjYii = ∑hYijrFjYii = trfrFg: (3) j i j i Since ri is the probability, it satisfies ri ≥ 0; ∑ri = trr = 1: (4) i 1 If there is one of ri’s equal to 1, say, r1 = 1, and all other probabilities ri = 0(i 6= 1), then r = jY1ihY1j. It means that the system is in the pure state jY1i, thus the pure state can also be represented by the density matrix. The trace of the square of r2 satisfies the inequality of 2 2 trr = ∑ri ≤ 1: (5) i The equality is satisfied if r represents the pure state. In this case, r2 = r, which is just a projection operator. 2 Density matrix in a general representation Now we can use a general representation with orthonormal bases of jji’s to represent the density matrix as r = ∑rijYiihYij = ∑ rijjaihjajYiihYijjbihjbj = ∑jjaihjbjrab; (6) i i;a;b ab where rab = hjajrjjbi = ∑rihjajYiihYijjbi: (7) i 2 The diagonal matrix elements raa = ∑i rijhjajYiij , which remains non-negative and can be interpreted as the probability for the system in the state of jjai. The off-diagonal matrix elements are generally complex valued. Since the trace is invariant under different bases, the expectation value of an observable F remains F = tr[rF] = ∑rabFba = ∑rabhjbjFjjai: (8) ab ab Exercise I leave it as an exercise that you can directly obtain Eq. 8 starting from Eq. 2 by perform- ing the transformation between bases. 3 Entropy Next we define the von Neumann entropy S for a mixed state. Let us first recall how we define classical entropy Scl. If there are a set of classical states with the probability distribution of −1 ri, the classical entropy is defined as Scl = kB ∑ri lnri . Now for the mixed states described by the density matrix, in the diagonal representation jYii, we still define the von Neumann 2 entropy S S = −kB ∑ri lnri: (9) i Formally, we can write a representation independent form as S = −kBtr[rlnr]: (10) One question is how to define a logarithmic of an operator. Since all the eigenvalues of r are positive and smaller than 1, we can define ¥ (1 − r)n lnr = ln[1 − (1 − r)] = − ∑ : (11) n=1 n For the pure state in which only one of ri equals to 1 and all others probabilities are 0, thus Spure = 0. Exercise: We can also prove that Eq. 9 reaches maximal if all the ri are equal to each other. I leave it as an exercise, or, you can refer to Gottfried’s book. 4 Entangled states Generally speaking, for a large system with N-particles, in a pure state Y(q1;::;qN). Let us ask what is the state that particle one is in? The answer is that the state of particle one cannot be described by a wavefunction, but in a mixed state. Let us consider a system composed of two subsystems with coordinates q1 and q2 for each subsystem, respectively. Suppose the entire system is in a pure state jY12i, and we use the coordinate representation in which the state is represented by the wavefunction Y(q1;q2), and thus r12 = jY12ihY12j. In the coordinate representation, ´ ´ ´ ´ ∗ ´ ´ hq1q2jr12jq1q2i = hq1q2jY12ihY12jq1q2i = Y(q1q2)Y (q1;q2): (12) However, the subsystem 1 is not in a pure state. It lives in the mixed state described by the density matrix r1. It is defined as r1 = tr2 r12; (13) where tr2 is a partial trace over the degree of freedom of particle 2. In the coordinate repre- sentation, Z Z ´ ´ ∗ ´ hq1jr1jq1i = dq2hq1q2jr12jq1q2i = dq2Y(q1q2)Y (q1;q2): (14) 3 Let us consider an operator A1 that only depends on q1 of the subsystem 1, i.e., ´ ´ ´ ´ hq1q2jA1jq1q2i = hq1jA1jq1id(q2 − q2): (15) Thus for the state of jY12i, we have the expectation value of A1 as Z ´ ´ ´ ´ ´ ´ hY12jA1jY12i = dq1dq2dq1dq2 hY12jq1q2ihq1q2jA1jq1q2ihq1q2jY12i Z ´ ∗ ´ ´ = dq1dq1dq2 Y (q1;q2)hq1jA1jq1iY(q1;q2) Z ´ ´ ´ = dq1dq1hq1jA1jq1ihq1jr1jq1i = tr1(r1A1): (16) Only if jY12i can be factorize into a product of jY1i⊗jY2i where jY1i and jY2i are pure states for the subsystem 1 and 2, respectively, then r1 = jY1ihY1j describing a pure state. In this case, there is no entanglement between subsystems 1 and 2. Otherwise, we say that subsystems 1 and 2 are entangled, and r1 describes a mixed state for the subsystem 1. In general, according to the Schmidt decomposition, for any jY12i, it can always be decomposed as n jY12i = ∑ cijY1;ii ⊗ jY2;ii; (17) i=1 where jY1;ii(i = 1;n) are orthogonal with each other for the subsystem 1, and so do jY2;ii(i = 1;n) for the subsystem 2. Then we have r1 = ∑rijY1;iihY1;ij; (18) i 2 2 4 with ri = jcij . If there are more than one terms of ci 6= 0, tr[r1] = ∑i jcij < 1, thus the subsystem 1 is in a mixed state. Here I presen the proof of the theorem of Schmidt decomposition. Theorem: Consider two Hilbert spaces H1 and H2 for systems 1 and 2 with dimensions n1 and n2, respectively. Without loss of generality, we assume n1 ≥ n2. The Hilbert space for the combined system is simply H1 ⊗ H2. For any state of the combined system jY12i in H1 ⊗ H2, we can always find a set of orthonormal states jY1;ii(i = 1;:::n2) in H1, and n2 jY2;ii(i = 1;:::;n2) in H2, such that jY12i = ∑i=1 cijY1;ii ⊗ jY2;ii, where ci is non-negative. The set of jY1;ii and jY2;ii are uniquely determined by the state jY12i. Let us consider two general sets of orthonormal bases jaii(i = 1;:::n1) for space H1, and jbii(i = 1;:::n2) for space H2, respectively. jY12i can always be expanded as n1 n2 jY12i = ∑ ∑ ci1i2 jai1 i ⊗ jbi2 i: (19) i1=1 i2=1 4 The coefficient ci1;i2 can be viewed as a rectangular n1 × n2 matrix Mi1;i2 = ci1;i2 . According to the theorem in linear algebra of singular value decomposition, there always exist an n1 × n1 unitary matrix U, and n2 × n2 unitary matrix V such that S M = U V T ; (20) 0 where S is a n2 × n2 a positive semidefinite diagonal matrix, and 0 refers to a (n1 − n2) × n2 matrix with all elements 0. We use ci(i = 1;:::;n2) to denote the diagonal element of S, u1;:::;un2 for the first n2 columns of U, and v1;:::;vn2 for the columns of V. Then we have n2 T M = ∑i=1 ciuivi , which means that n2 jY12i = ∑ cijuii ⊗ jvii (21) i=1 where n1 n1 juii = ∑ Ui; jja ji; jvii = ∑ Vi; jjb ji: (22) j=1 j=1 5 Two-particle interference Let us consider an entangled two-particle state 2 2 Y(q1;q2) = c1u1(q1)v1(q2) + c2u2(q1)v2(q2); jc1j + jc2j = 1: (23) Here we do not assume that u1;2 are orthogonal, and neither do v1;2. The two-body probability distribution is 2 2 2 2 2 2 r(q1;q2) = jc1j ju1(q1)j jv1(q2)j + jc2j ju2(q1)j jv2(q2)j + I2(q1;q2); n ∗ ∗ ∗ o I2(q1;q2) = 2Re c1c2u1(q1)u2(q1)v1(q2)v2(q2) : (24) The cross term I2(q1;q2) survives even if particles 1 and 2 are faraway from each other, i.e., u1;2 do not have spatial overlap with v1;2. Nevertheless if v1;2(q2) are orthogonal to each other, in the reduced one-body density r(q1), the two-particle interference term I2 vanishes after the integral of q2. If v1;2(q2) are not orthogonal, then in the reduced one-body density 2 2 ∗ r(q1) = jc1jju1(q1)j + jc2jju2(q2)j + 2Re[u1(q1)u2(q1)V]; (25) where Z ∗ V = dq2 v1(q2)v2(q2): (26) 5 Thus particle 1 will show interference if the two states of particle 2 are not completely or- thogonal.
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