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1 Mixed Quantum State 2 Density Matrix In this lecture, we will discuss the basics of quantum information theory. In particular, we will discuss mixed quantum states, density matrices, von Neumann entropy and the trace distance between mixed quantum states. 1 Mixed Quantum State So far we have dealt with pure quantum states ψ = ∑αx x . | i x | i This is not the most general state we can think of. We can consider a probability distribution of pure states, such as 0 with probability 1/2 and 1 with probability 1/2. Another possibility is the state | i | i + = 1 ( 0 + 1 ) with probability 1/2 | i √2 | i | i = 1 ( 0 1 ) with probability 1/2 ( |−i √2 | i − | i In general, we can think of mixed state as a collection of pure states ψ , each with associated probability | ii p , with the conditions 0 p 1 and ∑ p = 1. One reason we consider such mixed states is because the i ≤ i ≤ i i quantum states are hard to isolate, and hence often entangled to the environment. 2 Density Matrix Now we consider the result of measuring a mixed quantum state. Suppose we have a mixture of quantum states ψ with probability p . Each ψ can be represented by a vector in C 2n , and thus we can associate | ii i | ii the outer product ψ ψ = ψ ψ , which is an 2n 2n matrix | iih i| i i∗ × a a a¯ a a¯ a a¯ 1 1 1 1 2 ··· 1 N a a a¯ a a¯ a a¯ 2 2 2 1 2 2 N . a¯ a¯ a¯N = . ··· . 1 2 ··· . a a a¯ a a¯ a a¯ N N 1 N 2 ··· N N We can now take the average of these matrices, and obtain the density matrix of the mixture p , ψ : { i | ii} ρ ψ ψ = ∑ pi i i . i | ih | We give some examples. Consider the mixed state 0 with probability of 1/2 and 1 with probablity 1/2. Then | i | i 1 1 0 0 0 = 1 0 = , | ih | 0 0 0 and 0 0 0 1 1 = 0 1 = . | ih | 1 0 1 Thus in this case 1 1 1/2 0 ρ = 0 0 + 1 1 = . 2| ih | 2| ih | 0 1/2 CS 294-2, Fall 2004, Lecture 16 1 Now consider another mixed state, this time consisting of + with probability 1/2 and with probability | i |−i 1/2. This time we have 1 1 1 1 + + = (1/2) 1 1 = , | ih | 1 2 1 1 and 1 1 1 1 = (1/2) 1 1 = − . |−ih−| 1 − 2 1 1 − − Thus in this case the offdiagonals cancel, and we get 1 1 1/2 0 ρ = + + + = . 2| ih | 2|−ih−| 0 1/2 Note that the two density matrices we computed are identical, even though the mixed state we started out was different. Hence we see that it is possible for two different mixed states to have the same density matrix. We now show that two mixed states can be distinguished if and only if the density matrix ρ are different: Theorem 16.1: Suppose we measure a mixed state p , ψ in an orthonormal bases β . Then the { j | ji} | k outcome is β with probability β ρ β . i | ki h k| | ki Proof: We denote the probability of measuring β by Pr[k]. Then | ki ψ β 2 Pr[k] = ∑ p j j k j |h | i| β ψ ψ β = ∑ p j k j j k j h | ih | i β ψ ψ β = k ∑ p j j j k * j | ih | + β ρ β = k k . h | | i 2 ψ ρ corollary If we measure the mixed state p j, j in the standard basis, we have Pr[k]= k,k, the diagonal entry of the density matrix ρ. { | i} We list several more properties of the density matrix: 1. trρ = 1. This follows immediately from Corollary 2, since the probabilities Pr[k] must add up to 1. 2. ρ is Hermitian. This follows from the fact that ρ is a sum of Hermitian outer products ((ψψ∗)∗ = ψψ∗). 3. Eigenvalues of ρ are non-negative. First of all, eigenvalues of a Hermitian matrix is real. Suppose that λ and e are corresponding eigenvalue and eigenvector. Then if we measure in the eigenbasis, we have | i Pr[e]= e ρ e = λ e e = λ. h | | i h | i Since the probability must be non-negative, we see that λ 0. ≥ CS 294-2, Fall 2004, Lecture 16 2 Now suppose we have two mixed states, with density matrices A and B such that A = B. We can ask, what is a good measurement to distinguish the two states? We can diagonalize the difference6 A B to Λ − get A B = E E∗, where E is the matrix of orthogonal eigenvectors. Then if ei is an eigenvector with − λ λ eigenvalue i, then i is the difference in the probability of measuring ei: Pr [i] Pr [i]= λ . A − B i We can define the distance between two probability distributions (with respect to a basis E) as D D = ∑(Pr [i] Pr [i]). A − B E A − B If E is the eigenbasis, then D D λ A B E = ∑ i = tr A B = A B tr, − i | | | − | k − k which is called the trace distance between A and B. claim Measuring with respect to the eigenbasis E (of the matrix A B) is optimal in the sense that it maximizes the distance D D between the two probability distributions.− | A − B|E Before we prove this claim, we introduce the following definition and lemma without proof. definition Let N N ∑ ∑ ai i=1 and bi i=1 be two non-increasing sequences such that i ai = i bi. Then the sequence ai is said to{ majorize} {b } if for all k, { } { i} k k ∑ ai ∑ bi. i=1 ≥ i=1 lemma[Schur] Eigenvalues of any Hermitian matrix majorizes the diagonal entries (if both are sorted in nonincreasing order). Now we can prove claim 2. λ λ λ proof Since we can reorder the eigenvectors, we can assume 1 2 n. Note that tr(A B)= 0, ∑ λ λ ≥ ≥···≥ − so we must have i i = 0. We can split the i’s into two groups: positive ones and negative ones, we must have 1 1 ∑ = A B tr ∑ = A B tr. λ 2k − k λ −2k − k i>0 i<0 Thus k λ 1 max ∑ i = A B tr. k i=1 2k − k Now consider measuring in another basis. Then the matrix A B is represented as H = F(A B)F , and let − − ∗ µ µ µn be the diagonal entries of H. Similar argument shows that 1 ≥ 2 ≥···≥ k n D D µ 1 µ A B F max ∑ i = ∑ i = | − | . k i=1 2 i=1 | | 2 λ µ But by Schur’s lemma the i’s majorizes i’s, so we must have D D D D = A B . | A − B|F ≤ | A − B|E k − ktr CS 294-2, Fall 2004, Lecture 16 3 3 Von Neumann Entropy Consider the following two mixtures and their density matrices: cθ c2θ cθsθ cos θ 0 + sin θ 1 w.p. 1/2 = 1 cθ sθ = 1 | i | i 2 sθ 2 cθsθ s2θ cos2 θ 0 = cθ c2θ cθsθ 0 sin2 θ cos θ 0 sin θ 1 w.p. 1/2 = 1 cθ sθ = 1 − | i− | i 2 sθ − 2 cθsθ s2θ − − 1 1 0 0 w.p.cos2 θ = cos2 1 0 = cos2 | i 0 0 0 cos2 θ 0 = 0 0 0 0 sin2 θ 1 w.p. sin2 θ = sin2 θ 0 1 = sin2 | i 1 0 1 Thus, since the mixtures have identical density matrices, they are indistinguishable. Let H(X) be the Shannon Entropy of a random variable X which can take on states p1 ... pn. 1 H( pi )= ∑log = 1 { } − i pi In the quantum world, we define an analogous quantity, S(ρ), the Von Neumann entropy of a quantum ensemble with density matrix ρ. 1 1 S(ρ)= H cos2 θ,sin2 θ = cos2 θρ + sin2 θρ { } cos2 θ sin2 θ ψ ψ ρ ρ For the set of quantum states 1 ... n with probabilities p1 ...n and density matrix , we can diagonalize , π th letting i be the i element along the diagonal. This yields p , ψ λ e i | ii≡ i| ii We can also express the Von Neumann entropy in terms of the diagonal elements, as S(ρ)= H( π ) { i} CS 294-2, Fall 2004, Lecture 16 4.
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