SEMINAR:TENSOR PRODUCTSANDTHE PARTIALTRACE Georg Maringer Monday 23rd April, 2018 Definition (dual bases): If fv1, ... , vmg and fw1, ... , wng are bases of V and W then f f1,..., fmg and fg1,..., gng denote the corresponding dual bases. fk(vl) = dk,l, gs(wt) = ds,t (0.1) Inner products using the tensor product The Riesz representation theorem states that for every element in the dual space of a Hilbert space there exists a unique element of the Hilbert space corresponding to the respective element of the dual space. fi(v) = hvi, vi (0.2) For a proof look at any book covering functional analysis. 1 1. Tensor products of Hilbert spaces Definition: Let V =∼ Cm and W =∼ Cn be finite dimensional Hilbert spaces. The tensor product space V ⊗ W is the vector space consisting of all bilinear forms K on V∗ × W∗. Let v 2 V and w 2 W. Then we define v ⊗ w 2 V ⊗ W by v ⊗ w( f , g) = f (v)g(w) 8 f 2 V∗, g 2 W∗ . (1.1) Elements of the form v ⊗ w are called product vectors. Consequence: (mv1 + lv2) ⊗ w = mv1 ⊗ w + lv2 ⊗ w (1.2) and similarly in the second argument. A general element in V ⊗ W is a superposition ∑i,j ai,jvi ⊗ wj with vi 2 V and wj 2 W and ai,j 2 C and ! ∑ ai,jvi ⊗ wj ( f , g) = ∑ ai,j f (vi) f (wj) . (1.3) i,j i,j The inner product on V ⊗ W is defined for product vectors as hv ⊗ w, v0 ⊗ w0i = hv, v0ihw, w0i (1.4) and this definition is linearly (in the second argument) respectively antilinearly (in the first argument) extended to all of V ⊗ W. m n Main message: If fejgj=1 and f fkgk=1 are ONBs of V and W, respectively, then fej ⊗ fkgj=1,...,m (1.5) k=1,...,n is an ONS of V ⊗ W. Let K be a bilinear mapping on V∗ ⊗ W∗. K( f , g) = ∑ K( f (vi) fi, g(wj)gj) fi(v)gj(w) = (1.6) i,j 2 ∑ K( fi, gj) f (vi)g(wj) = ∑ K( fi, gj)[vi ⊗ wj]( f , g) (1.7) i,j i,j Thus, K = ∑i,j K( fi, gj)[vi ⊗ wj] and fvi ⊗ wj : 1 ≤ i ≤ m, 1 ≤ j ≤ ng spans V ⊗ W. Linear independence follows because ∑ bi,jvi ⊗ wj = 0 = (∑ bi,jvi ⊗ wj)( fk, gl) = ∑ bi,j fk(vi)gl(wj) = bk,l (1.8) i,j i,j i,j Considering that [v ⊗ w]i,j = fi(v)gj(w) = hvi, vihwj, wi (1.9) Matrix isomorphism: There exists an isomorphism between V ⊗ W and the matrices Matm×n via the correspondence [K]i,j = K( fi, gj) (1.10) Example: Let v 2 V and w 2 W. The matrix corresponding to v ⊗ w is given by (for bases assume the bases as before (dual and normal one)) [v ⊗ w]i,j = [v ⊗ w]( fi, gj) = fi(v)gj(w) = [v]i[w]j (1.11) rank one matrix iff the corresponding tensor is a product vector in V ⊗ W. What we want to have now is an inner product on V ⊗ W. This is established by the matrix correspondence given before. For matrices a possible inner product is the Hilbert-Schmidt inner product. Let B, C 2 V ⊗ W ∗ hB, Ci = Tr([B] [C]) = ∑ Bi,jCj,i (1.12) i,j 0 0 0 0 0 0 hv ⊗ w, v ⊗ w i = ∑ hvi, vihwj, wihvi, v ihwj, w i = hv, v ihw, w i (1.13) i,j 3 Using this definition for the inner product it is easy to see that the system fvi ⊗ wj : 1 ≤ i ≤ m, 1 ≤ j ≤ ng is an orthonormal basis of V ⊗ W. Composite systems Composite quantum systems can be described by the tensor product of the Hilbert spaces which describe the contributing systems. Example: spin up j0i = (1 0)T and spin down j1i = (0 1)T. The rest of the states is the composition of the two. They span C2 = spanfj0i, j1ig. As described earlier the tensor product space C2 ⊗ C2 = spanfj0i ⊗ j0i, j0i ⊗ j1i, j1i ⊗ j0i, j1i ⊗ j1ig. Bell ONB (maximally entangled state) 1 1 1 1 fp (j00i + j11i), p (j00i − j11i), p (j01i + j10i), p (j01i − j10i)g (1.14) 2 2 2 2 check orthogonality: 1 1 (h00j + h11j)(j00i − j11i) = (h00j00i − h00j11i + h11j00i − h11j11i) = 0 2 2 (1.15) check normalization: 1 (h00j + h11j)(j00i + j11i) = 1 (1.16) 2 Review: state: r∗ = r, r ≥ 0, tr(r) = 1. Pure state: r = jYihYj , kYk = 1. Observable A = A∗. Definition: Let r be a state and A = A∗ be an observable. Then expectation value of A in the state r is given by tr(Ar). Remark: If r = jYihYj is a pure state, then tr(Ar) = hYjAjYi. Proof: choose an ONB where jYi is the first basis vector j1i = jYi tr(Ar) = tr(AjYihYj) (1.17) d = ∑hjjAjYihYjji (1.18) j=1 = ... (1.19) 4 Tensor product of matrices Operators on V ⊗ W: For A 2 B(V) and B 2 B(W), we define A ⊗ B 2 B(V ⊗ W) by (A ⊗ B)(v ⊗ w) = Av ⊗ Bw (1.20) where this definition is again linearly extended to all of V ⊗ W. We call an operator of the form A ⊗ B a product operator. Consequence: (A1 ⊗ B1)(A2 ⊗ B2) = A1 A2 ⊗ B1B2 (1.21) Remark: every operator X 2 B(V ⊗ W) can be written as a linear combination of product operators. X = ∑ x(i,j),(k,l)(jeii ⊗ j fji)(hekj ⊗ h flj) (1.22) i,j,k,l = ∑ xi,j,k,ljeiihekj ⊗ j fjih flj . (1.23) i,j,k,l Little examples: Let A and B be invertible matrices. Then (A ⊗ B)−1 = A−1 ⊗ B−1 (1.24) just write down definitions. More interesting is the adjoint: (A ⊗ B)∗ = A∗ ⊗ B∗ (1.25) Proof: Let f and y be arbitrary elements in V ⊗ W. hf, (A ⊗ B)yi = h∑ aijvi ⊗ wj, (A ⊗ B)(∑ bklvk ⊗ wl)i (1.26) i,j k,l ∑ aijbklhvi ⊗ wj, Avk ⊗ Bwli = ∑ aijbklhvi, Avkihwj, Bwli (1.27) i,j,k,l i,j,k,l 5 ∗ ∗ ∗ ∗ = ∑ aijbklhA vi ⊗ B wj, vk ⊗ wli = h(A ⊗ B )(∑ aij(vi ⊗ wj)), ∑ bklvk ⊗ wli i,j,k,l i,j k,l (1.28) Consequently (A ⊗ B)∗ = A∗ ⊗ B∗. Lemma: tr(A ⊗ B) = tr(A)tr(B) . (1.29) Proof: tr(A ⊗ B) = ∑(hejj ⊗ h fkj)(A ⊗ B)(jeji ⊗ j fki) (1.30) j,k = ∑hejjAjeji · h fkjBj fki (1.31) j,k = tr(A)tr(B) . (1.32) Definition: local observable. An observable A 2 B(H1 ⊗ H2) is called local on the first system (i.e., on H1) if A = O1 ⊗ I2 for some O1 2 B(V1) . (1.33) Here I2 denotes the identity on V2. Similarly, local on H2. 1 Example: sz ⊗ I . sz = j0ih0j − j1ih1j Compute expectation value in jYi = p (j0i ⊗ 2 2 j1i + j1i ⊗ j0i). By the rule for the product states of traces at the beginning tr((sz ⊗ I2)jyihyj) = hyj(s ⊗ I)jyi = (1.34) 1 = (h0j ⊗ h1j + h1j ⊗ h0j)(s ⊗ I)(j0i ⊗ j1i + j1i ⊗ j0i) 2 z (1.35) 1 = (h0j ⊗ h1j + h1j ⊗ h0j)(j0i ⊗ j1i − j1i ⊗ j0i) (1.36) 2 1 = (h0j0ih1j1i − h1j1ih0j0i) = 0 (1.37) 2 6 A product state is defined as jyi = jf1i ⊗ f2i. (1.38) Next example 1 jyi = p (j0i ⊗ j0i + j1i ⊗ j1i) (1.39) 2 The question is now what the state of the first respectively the second system is. Motivation for the partial trace: find mathematical object which reproduces all expecta- tion values of local observables in a given state. Definition Let r12 2 B(H1 ⊗ H2) an arbitrary operator. Then the unique operator r1 2 B(H1) satisfying tr((O1 ⊗ I2)r12) = tr(O1r1) for all O1 2 B(H1) (1.40) is called the partial trace of r12. Existence and uniquess: Riesz. Notation: r1 = tr2r12. Alternative terminology: reduced density operator of r12. Example r12 = jFihFj with jFi = jj1i ⊗ jj2i. Then r1 = jj1ihj1j. tr2r12 = tr2(jf1i ⊗ jf2ihf1j ⊗ hf2j) (1.41) = tr2(jf1ihf1j ⊗ jf2ihf2j) = jf1ihf1jtr(jf2ihf2j) (1.42) = jf1ihf1j (1.43) alternatively, let O1 be arbitrary 2 H1: tr((O1 ⊗ I2)(jf1i ⊗ jf2ihf1j ⊗ hf2j)) = hf1j ⊗ hf2j(O1 ⊗ I2)jf1i ⊗ jf2i (1.44) = hf1j ⊗ hf2)j(O1 ⊗ I2)jf1i ⊗ jf2i (1.45) = hf1j ⊗ hf2jO1jf1i ⊗ jf2i (1.46) = hf1jO1f1ihf2jf2i (1.47) = hf1jO1f1i (1.48) 7 Example: r = jYihYj with jYi = p1 (j0i ⊗ j1i + j1i ⊗ j0i). Then r = I /2. 12 2 1 1 1 tr r = (j0i ⊗ j1i + j1i ⊗ j0i)(h0j ⊗ h1j + h1j ⊗ h0j) (1.49) 2 12 2 1 = (j0ih0j ⊗ j1ih1j + j1ih0j ⊗ j0ih1j + j0ih1j ⊗ j1ih0j + j1ih1j ⊗ j0ih0j) 2 (1.50) Remark: the partial trace defines a map tr2 : B(H1 ⊗ H2) !B(H1) . (1.51) Properties: tr2(A ⊗ B) = A · tr(B) . (1.52) Proof: Let O1 2 H1 be arbitrary. Then tr((O1 ⊗ I2)(A ⊗ B)) = tr (O1 A ⊗ B) (1.53) = tr(O1 A)tr(B) (1.54) = tr(O1(Atr(B))) . (1.55) n Remark: This implies for example, that for any ONB f fjgj=1 on H2, we have m tr2X = ∑(I1 ⊗ h fjj)X(I1 ⊗ j fji) . (1.56) j=1 Proof: write X as a sum of product operators, insert tr(Y) = ∑jh fjjYj fji. Examples: 1 jyi = p (h0j ⊗ h0j + h1j ⊗ h1j) (1.57) 2 8 1 jyihyj = (j00ih00j + j11ih00j + j00ih11j + j11ih11j) (1.58) 2 1 = (j0ih0j ⊗ j0ih0j + j1ih0j ⊗ j1ih0j + j0ih1j ⊗ j0ih1j + j1ih1j ⊗ j1ih1j) 2 (1.59) 0 1 B1 0 0 1C B C B C B C B0 0 0 0C 1 1 B C jyihyj = (j00hi00j + j00ih11j + j11ih00j + j11ih11j) = B C (1.60) 2 2 B C B C B0 0 0 0C B C B C @ A 1 0 0 1 0 1 B1 0C 1 1 1 B C Tr2jyihyj = j0ih0j + j1ih1j = B C (1.61) 2 2 2 @ A 0 1 0 1 B1 0C 1 1 1 B C Tr1jyihyj = j0ih0j + j1ih1j = B C (1.62) 2 2 2 @ A 0 1 fully mixed state (maximal entropy).