Physics 215 Winter 2018 The Density Matrix The quantu space of states is a Hilbert space H. Any state vector ψ H is a pure state. Since any linear cobination of eleents of H are also an eleent of H, it follows that any linear cobination of pure states is again a pure state. But, in the real world we also encounter ixed states. These are described by the density operator, ρ p ψ ψ, (1) which characterizes an enseble of N states, of which n (on average) are in the state ψ, and p n /N is siply the probability that a state of the enseble is observed to be in the state ψ. By assuption, the state ψ is noralized to unity, i.e. ψ ψ = 1. The probabilities, p ust su to 1, p = 1. (2) The atrix eleents of the density atrix with respect to an orthonoral basis { i } are then given by, ρ ij = i ρ j = p i ψ ψ j. (3) Given a self-adjoint operator Ω that acts on the Hilbert space H, we can define an enseble average, Ω p ψ Ω ψ = Tr(Ωρ). To verify the last step above, recall that the atrix eleents of Ω with respect to an orthonoral basis { i } are given by Ω ij i Ω j. Thus, by definition of the trace, Tr(Ωρ) = ij Ω ij ρ ji = ij i Ω j p j ψ ψ i = p ψ i i Ω j j ψ = ij p ψ Ω ψ = Ω, after using the copleteness relation to su over i and j. Since Ω is self-adjoint, it possesses real eigenvalues. Suppose ω is one of the possible eigenvalues of Ω. Then, ω is the possible outcoe of an experient that easures the observable corresponding to Ω. Given an enseble defined by the density operator ρ, the probability P(ω) of obtaining ω in a easureent is given by P(ω) = Tr(P ω ρ) = p ω ψ 2, (4) 1
where P ω ω ω is the projection operator that projects onto the one-diensional subspace of states spanned by ω. To verify the last step of eq. (4), we use (P ω ) ij = i ω ω j. It then follows that, Tr(P ω ρ) = ij (P ω ) ij ρ ji = ij i ω ω j p j ψ ψ i = p ψ i i ω ω j j ψ = p ω ψ 2. ij The properties of the density operator are listed below. 1. ρ = ρ. Using eq. (3) ρ ji = p j ψ ψ i = p ψ j i ψ = ρ ij. Hence, ρ is heritian. 2. ρ is positive sei-definite. For any state χ, we have χ ρ χ = p χ ψ ψ χ = p χ ψ 2 0, where we have used the fact that the probabilities p are real and non-negative. 3. Trρ = 1. Using the copleteness of the { i } and ψ ψ = 1, it follows that Trρ = ρ ii = p ψ i i ψ = p = 1, i i after eploying eq. (2). 4. For a pure quantu state, ρ 2 = ρ, which iplies that Trρ 2 = 1 in light of property 3 above. For a pure quantu state, we can always find soe eleent of the Hilbert space, ψ, which is noralized to unity, such that ρ = ψ ψ. Then, we trivially obtain ρ 2 = ψ ψ ψ ψ = ψ ψ = ρ. 5. For a ixed quantu state, 0 < Trρ 2 < 1. First we note that in light of property 1 above, ρ 2 = ρρ, which is a non-negative heritian operator. Thus, all the eigenvalues of ρ 2 are non-negative. Denoting the eigenvalues of ρ 2 by λ i, it follows that Trρ 2 = λ i > 0. i 2
Note that it is not possible for all of the eigenvalues of ρ 2 to be zero, since this would iply that ρ = 0. 1 To prove that Trρ 2 < 1, we start fro eq. (3), 2 Trρ 2 = p p l i ψ ψ n n ψ l ψ l i = p p l ψ l ψ ψ ψ l = p p l ψ l ψ 2,,l i,n,l,l (5) after suing over the two coplete sets of states, { i } and { n }. We now ae use of the Schwarz inequality, ψ l ψ 2 ψ l ψ l ψ ψ. Since ψ is noralized to unity, we obtain ψ l ψ 2 1. It then follows that Trρ 2 =,l p p l ψ l ψ 2 1, (6) after aing use of eq. (2). Moreover, the Schwarz inequality is saturated only when ψ and ψ l are proportional for any choice of and l. Since these are noralized states, they can only differ by a ultiplicative phase. But, this case, ψ ψ in eq. (1) would be independent of, which eans one can siply write ρ = ψ ψ, corresponding to a pure state. Thus, for a ixed state, the Schwarz inequality is not saturated, and we conclude that Trρ 2 < 1. Thus, we can conclude that for a ixed state, 0 < Trρ 2 < 1. 6. For a unifor distribution over N states, ρ = N 1 I, where I is the N N identity operator. This is a ixed quantu state. An exaple of such a quantu state would be unpolarized light. Using TrI = N, we easily verify the first three properties above. Moreover, Trρ 2 = 1/N which satisfies the fifth property above for any N > 1. Note that by taing N 1, we can obtain an arbitrarily sall value of Trρ 2. Pure states are governed by the tie-dependent Schrodinger equation. A pure state will evolve into a pure state. So, how does one create ixed states? One way to produce a ixed state is to consider a syste that is sensitive only to a subset of the full quantu Hilbert space. For exaple, consider a syste that is ade up of two separate subsystes. We will perfor easureents using operators that are only sensitive to one of the two subsystes. In particular, consider two subsystes called subsyste 1 and subsyste 2. Corresponding to each subsyste is an orthonoral basis { n, 1 } and {, 2 }, respectively. Matheatically, the total Hilbert space is a direct product of two subsyste Hilbert spaces, H = H 1 H 2. 1 In general, if an n n atrix M is diagonalizable, then it possesses n linearly independent eigenvectors. In this case, there exists an invertible atrix S such that D = S 1 MS, where D is a diagonal atrix whose diagonal eleents are the eigenvalues of M. Hence, if all the eigenvalues of M are zero, then it follows that M = 0. Note that this arguent does not apply to atrices that are not diagonalizable, as the faous exaple ( 0 0 1 0 ) attests. Finally, by noting that all heritian atrices are diagonalizable, we can conclude that for any density atrix (which is necessarily nonzero), Trρ > 0. 2 Given that p 0, eq. (2) iplies that at least one of the p appearing in eq. (5) is positive. Hence, it follows that Trρ 2 p 2. This provides another arguent for Trρ > 0. 3
A generic pure state of H is of the for ψ = c n n,1,2, (7) n where ψ ψ = 1 iplies that c n 2 = 1. (8),n The corresponding density operator is ρ = ψ ψ = c n c n n,1,2 n,1,2. (9) n, n, Consider an operator Ω that is only sensitive to subsyste 1. That is, when Ω acts on ψ given by eq. (7), it has no effect on {,2 }. Using the direct product notation, Ω actually eans Ω I, where Ω acts on subsyste 1, and the identity operator I acts on subsyste 2. Then, the enseble average of Ω is given by Ω = Tr1 Tr 2 (ρω) = Tr 1 [ (Tr2 ρ)ω ], where Tr i (i = 1,2) is carried out by suing over the subsyste i part of the total syste. Thus, we can express the enseble average of the operator Ω I as Ω = Tr1 (ˆρΩ), (10) where ˆρ Tr 2 ρ, (11) is the effective density atrix for subsyste 1. Even though there are typically quantu correlations between the two subsystes, observers (by assuption) are only sensitive to subsyste 1. Thus, in coputing an enseble average of a quantu state of subsyste 1, we ust define the enseble average solely in ters of subsyste 1 quantities; that is, as a trace over subsyste 1 as in eq. (10). It is now a siple atter to copute ˆρ = Tr 2 ρ = j j,2 ρ j,2, (12) where ρ is given by eq. (9). Using eq. (12), we then find, ˆρ = j,2 ρ j,2 = c n c n n,1 j,2,2,2 j,2 n,1 j n, n, j = c n c n n,1 n,1 δ j δ j. (13) n, n, j Perforing the sus over j and is now trivial due to the two Knonecer deltas, and we end up with ˆρ = c n c n n,1 n,1. n, n 4
This is the relevant density atrix for subsyste 1. Note that in general it corresponds to a ixed state, since the original state ψ can possess non-trivial quantu correlations between the two subsystes as previously noted. But, that inforation is lost in ˆρ, which only nows about subsyste 1. The only way for ˆρ to correspond to a pure state is if we can decopose c n as follows, where c n = b n d, (14) b n 2 = n d 2 = 1. (15) In this case, it is straightforward to verify that ˆρ 2 = ˆρ, which indicates that the quantu subsyste 1 is a pure state. Indeed, if eq. (14) holds, then we can write eq. (7) as ( )( ) ψ = b n n,1 d,2. n in which case, ψ can be decoposed into a direct product of two pure states, ψ = ψ 1 ψ 2, where ψ 1 = n b n n,1, ψ 2 = d n,1. Thus, ˆρ = ψ 1 ψ 1, which is clearly a pure state. In contrast, if one cannot write c n in the for given by eq. (14), then ˆρ satisfies 0 < Tr 1 ˆρ 2 < 1, and we conclude that subsyste 1 is a ixed quantu state state. More explicitly, Tr 1 ˆρ 2 = c n c n c jc j i,1 n,1 n,1 l,1 l,1 j,1 j,1 i,1 n,,n j,,j = n,,n which we can write as = n,n ( j,,j c n c n c jc j δ inδ ln δ lj δ ij )( ) c n c n c n c n, Tr 1 ˆρ 2 = c n c n 2. (16) n,n Note that Tr 1 ˆρ 2 > 0. We can reinterpret eq. (16) by defining the vectors c n, whose coponents are given by c n. Then, we can rewrite eq. (16) as Tr 1 ˆρ 2 = n,n c n c n 2, (17) 5
where c n c n is the coplex inner product of the vectors c n and c n. We can now eploy the Schwarz inequality, c n c n 2 c n c n c n c n, (18) to eq. (17). We then obtain, ( 2 Tr 1 ˆρ 2 c n c n ). Finally, we note that eq. (8) is equivalent to c n c n = 1. n n Hence, we conclude that 0 < Tr 1 ˆρ 2 1. Note that the inequality is saturated only when the Schwarz inequality [cf. eq. (18)] is saturated, i.e. when the all the vectors c 1, c 2, c 3,... are proportional to the sae vector, which we shall denote by d. That is, the inequality is saturated if and only if c n = b n d, (19) where b n is the proportionality constant relating the vectors c n and d. Recall that the coponents of the vector c n were denoted by c n. Then, if we denote the coponents of the vector d by d, then, eq. (19) is equivalent to eq. (14). The conclusion of this analysis is that 0 < Tr 1 ˆρ 2 < 1, corresponding to a ixed quantu state, unless eq. (14) is satisfied. If c n = b n d, then the inequality is saturated and Tr 1 ˆρ 2 = 1, corresponding to a pure quantu state. We can verify this result explicitly by inserting c n = b n d directly into eq. (16), Tr 1 ˆρ 2 = 2 b n b n d 2 = b n 2 b n 2 = 1, n,n n n after aing use of eq. (15). It should be noted that the condition c n = b n d is very strong. As shown in the Appendix, if the atrix eleents of a atrix C are c n = b n d, then C is a ran-one atrix with zero deterinant. APPENDIX: Properties of a atrix that satisfies C ij = a i b j Consider the n n non-zero atrix C, whose atrix eleents are given by C ij = a i b j. More explicitly, a 1 b 1 a 1 b 2 a 1 b n a 2 b 1 a 2 b 2 a 2 b n C =....... a n b 1 a n b 2 a n b n 6
It is a siple atter to chec that the following n 1 vectors, b 2 b 3 b n b 1 0 0, b 1,,,, (20) 0.. 0. 0 0 b 1 are linearly independent eigenvectors of C, each with an associated zero eigenvalue. Using the fact that TrC is the su of the eigenvalues, it follows that the (potentially) nonzero eigenvalue of C ust be equal to i a ib i, and the corresponding eigenvector is a 1 a 2, (21) 0. a n as is easily verified. If i a ib i 0, then 0 is an (n 1)-fold degenerate eigenvalue of C. That is, C has n 1 zero eigenvalues and one non-zero eigenvalue. Moreover, all n eigenvectors listed above are linearly independent. In contrast, if i a ib i = 0, then C has n zero eigenvalues. In this case the eigenvector shown in eq. (21) can be written as a linear cobination of the eigenvectors listed in eq. (20). In light of these results, one can iediately conclude that C is a ran-1 atrix with zero deterinant. The latter follows fro the fact that C possesses a zero eigenvalue, since detc is a product of its eigenvalues. Moreover, if a i b i 0, then C possesses n linearly independent eigenvalues, which iplies that C is diagonalizable. In the case where ai b i = 0, C possesses n zero eigenvalues, but only n 1 linearly independent eigenvectors. 3 3 The only n n atrix that possesses n zero eigenvalues and n linearly independent eigenvectors is the n n zero atrix [cf. footnote 1]. 7