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[ { "context": "We have\n\n\\[\n\\rho = \\frac{1}{3} (\\rho_0 + \\rho_1 + \\rho_2) = \\frac{1}{2} I, \\quad \\rho^{-\frac{1}{2}} = \\sqrt{2} I,\n\\]\n\nthus the pretty-good-measurement \\(\\{M_0, M_1, M_2\\}\\) is given by\n\n\\[\nM_0 = \\frac{1}{3} \\rho^{-\frac{1}{2}} \\rho_0 \\rho^{-\frac{1}{2}} = \\frac{2}{3} |0\\rangle \\langle 0|, \\quad M_1 = \\frac{1}{3} \\rho^{-\frac{1}{2}} \\rho_1 \\rho^{-\frac{1}{2}} = \\frac{1}{3} I, \\quad M_2 = \\frac{1}{3} \\rho^{-\frac{1}{2}} \\rho_2 \\rho^{-\frac{1}{2}} = \\frac{2}{3} |1\\rangle \\langle 1|,\n\\]\n\nand the success probability using this measurement is\n\n\\[\np_{\\text{good}} = \\frac{1}{3} (\\text{tr}(M_0 \\rho_0) + \\text{tr}(M_1 \\rho_1) + \\text{tr}(M_2 \\rho_2)) = \\frac{5}{9}.\n\\]", "question": "### (a) Suppose Alice sends Bob one of the three states \\( \\rho_0 = |0\\rangle \\langle 0|, \\rho_1 = \\frac{1}{2} I, \\rho_2 = |1\\rangle \\langle 1| \\) with equal probability. Bob wants to figure out which state Alice sent. Compute the success probability achieved by Bob if he uses the pretty-good-measurement." }, { "context": "Let \\(\\sigma^* = \\frac{1}{3} I\\), then it's easy to check that\n\n\\[\n\\sigma^* \\geq \\frac{1}{3} |0\\rangle \\langle 0| = p_0 \\rho_0, \\quad \\sigma^* \\geq \\frac{1}{3} \\times \\frac{1}{2} I = p_1 \\rho_1, \\quad \\sigma^* \\geq \\frac{1}{3} |1\\rangle \\langle 1| = p_2 \\rho_2,\n\\]\n\ntherefore\n\n\\[\nP_{\\text{guess}} = \\inf_{\\sigma \\geq p_i \\rho_i, i=0,1,2} \\text{tr}(\\sigma) \\leq \\text{tr}(\\sigma^*) = \\frac{2}{3}.\n\\]\n\n\\(\\frac{2}{3}\\) is an upper bound of the guessing probability. We will show that this is actually the maximum of the guessing probability. Indeed, consider a POVM \\(\\{M_0^*, M_1^*, M_2^*\\}\\,\n\n\\[\nM_0^* = |0\\rangle \\langle 0|, \\quad M_1^* = 0, \\quad M_2^* = |1\\rangle \\langle 1|.\n\\]\n\nWe can check that this is a legal POVM, and the success probability using this POVM is\n\n\\[\np_{\\text{succ}}^* = \\frac{1}{3} (\\text{tr}(M_0^* \\rho_0) + \\text{tr}(M_1^* \\rho_1) + \\text{tr}(M_2^* \\rho_2)) = \\frac{2}{3}.\n\\]\n\nThus we have\n\n\\[\n\\frac{2}{3} \\geq P_{\\text{guess}} \\geq p_{\\text{succ}}^* = \\frac{2}{3},\n\\]\n\nwhich implies \\(P_{\\text{guess}} = \\frac{2}{3}\\).", "question": "### (b) In Homework 5, problem 1(h), you showed the following formulation of the guessing probability:\n\n\\[\nP_{\\text{guess}}(X | E) = \\inf_{\\sigma: \\sigma \\leq \\sum_i p_i \\rho_i} \\text{Tr} \\sigma,\n\\]\n\nwh" }, { "context": "Already done in (b).", "question": "### (c) Notice that there is a gap between the success probability calculated in parts (a) and (b). Find a measurement whose success probability matches the bound from part (b)." } ]

[ { "context": "Suppose \\( \\rho = p_0 \\rho_0 + p_1 \\rho_1 \\), \\( p_0, p_1 \\geq 0 \\), \\( p_0 + p_1 = 1 \\), then the pretty-good-measurement is given by\n\\[ M_0 = \\rho^{-\frac{1}{2}} p_0 \\rho_0 \\rho^{-\frac{1}{2}}, \\quad M_1 = \\rho^{-\frac{1}{2}} p_1 \\rho_1 \\rho^{-\frac{1}{2}}. \\]\n\nWe only need to show that\n\\[ \\text{tr}(M_0 \\rho_0) \\geq \\text{tr}(M_0 \\rho_1), \\quad \\text{tr}(M_1 \\rho_1) \\geq \\text{tr}(M_1 \\rho_0). \\]\n\nDefine\n\\[ a = \\text{tr}(\\rho^{-\frac{1}{2}} p_0 \\rho_0 \\rho^{-\frac{1}{2}} \\rho), \\quad b = \\text{tr}(\\rho^{-\frac{1}{2}} p_1 \\rho_1 \\rho^{-\frac{1}{2}} \\rho), \\quad c = \\text{tr}(\\rho^{-\frac{1}{2}} p_0 \\rho_0 \\rho^{-\frac{1}{2}} \\rho_1). \\]\n\nIt's easy to check that \\( a, b, c \\geq 0 \\), and we have\n\\[\n\\begin{cases}\np_0 a + p_1 c = \\text{tr}(\\rho^{-\frac{1}{2}} p_0 \\rho_0 \\rho^{-\frac{1}{2}} \\rho) + \\text{tr}(\\rho^{-\frac{1}{2}} p_1 \\rho_1 \\rho^{-\frac{1}{2}} \\rho) = \\text{tr}(\\rho^{-\frac{1}{2}} \\rho \\rho^{-\frac{1}{2}} \\rho) = \\text{tr}(\\rho) = 1, \\\np_0 c + p_1 b = \\text{tr}(\\rho^{-\frac{1}{2}} p_0 \\rho_0 \\rho^{-\frac{1}{2}} \\rho_1) + \\text{tr}(\\rho^{-\frac{1}{2}} p_1 \\rho_1 \\rho^{-\frac{1}{2}} \\rho_1) = \\text{tr}(\\rho^{-\frac{1}{2}} \\rho \\rho^{-\frac{1}{2}} \\rho_1) = \\text{tr}(\\rho_1) = 1,\n\\end{cases}\n\\]\n\\[ \\implies p_0 (a - c) = p_1 (b - c). \\]\n\nAlso using Cauchy-Schwarz inequality we have\n\\[ c^2 = \\left( \\text{tr}(\\rho^{-\frac{1}{2}} p_0 \\rho_0 \\rho^{-\frac{1}{2}} \\rho_1) \\right)^2 = \\left( \\text{tr}(\\rho^{-\frac{1}{4}} p_0 \\rho_0^{\\frac{1}{4}} \\rho^{-\frac{1}{4}} \\rho_1^{\\frac{1}{4}}) \\right)^2 \\leq \\left( \\text{tr}(\\rho^{-\frac{1}{4}} p_0 \\rho_0^{\\frac{1}{4}} \\rho^{-\frac{1}{4}}) \\right) \\left( \\text{tr}(\\rho^{-\frac{1}{4}} \\rho_1^{\\frac{1}{4}} \\rho^{-\frac{1}{4}}) \\right) = \\text{tr}(\\rho^{-\frac{1}{2}} p_0 \\rho_0 \\rho^{-\frac{1}{2}} \\rho) \\text{tr}(\\rho^{-\frac{1}{2}} p_1 \\rho_1 \\rho^{-\frac{1}{2}} \\rho) = ab. \\]\n\nThat is to say, at least one of the following is true: \\( a \\geq c \\); \\( b \\geq c \\). Then using \\( p_0 (a - c) = p_1 (b - c) \\), we must have\n\\[ p_0 (a - c) = p_1 (b - c) \\geq 0. \\]\n\nTherefore\n\\[ \\text{tr}(M_0 \\rho_0) = \\text{tr}(\\rho^{-\frac{1}{2}} p_0 \\rho_0 \\rho^{-\frac{1}{2}} \\rho_0) = p_0 a \\geq p_0 c = \\text{tr}(\\rho^{-\frac{1}{2}} p_0 \\rho_0 \\rho^{-\frac{1}{2}} \\rho_1) = \\text{tr}(M_0 \\rho_1), \\]\n\\[ \\text{tr}(M_1 \\rho_1) = \\text{tr}(\\rho^{-\frac{1}{2}} p_1 \\rho_1 \\rho^{-\frac{1}{2}} \\rho_1) = p_1 b \\geq p_1 c = \\text{tr}(\\rho^{-\frac{1}{2}} p_1 \\rho_1 \\rho^{-\frac{1}{2}} \\rho_0) = \\text{tr}(M_1 \\rho_0). \\]", "question": "### (a) Prove inequality 2 for the case where the ensemble has only two states." }, { "context": "In this case, we have\n\\[ \\rho = \\frac{2}{5} \\rho_0 + \\frac{3}{5} \\rho_1 + \\frac{1}{5} \\rho_2 = \\frac{1}{5} \\begin{pmatrix} 2 & 0 \\\\ 0 & 3 \\end{pmatrix}, \\quad \\rho^{-\frac{1}{2}} = \\sqrt{5} \\begin{pmatrix} \\frac{1}{\\sqrt{2}} & 0 \\\\ 0 & \\frac{1}{\\sqrt{3}} \\end{pmatrix}, \\]\nand the pretty-good-measurement is given by\n\\[ M_0 = \\begin{pmatrix} \\frac{3}{20} & 0 \\\\ 0 & \\frac{2}{9} \\end{pmatrix}, \\quad M_1 = \\begin{pmatrix} \\frac{1}{30} & 0 \\\\ 0 & \\frac{4}{9} \\end{pmatrix}, \\quad M_2 = \\begin{pmatrix} 0 & 0 \\\\ 0 & \\frac{1}{3} \\end{pmatrix}. \\]\n\nWe can see check that\n\\[ \\text{tr}(M_1 \\rho_1) = \\frac{1}{3} \\times \\left( \\frac{1}{3} + \\frac{8}{9} \\right) = \\frac{11}{27}, \\quad \\text{tr}(M_1 \\rho_2) = \\frac{4}{9} = \\frac{12}{27}, \\]\n\n\\[ \\text{tr}(M_1 \\rho_1) < \\text{tr}(M_1 \\rho_2), \\]\n\nwhich violates inequality (2).", "question": "### (b) Let \\( \\rho_0 = \\frac{1}{3} \\begin{pmatrix} 2 & 0 \\\\ 0 & 1 \\end{pmatrix} \\), \\( \\rho_1 = \\frac{1}{3} \\begin{pmatrix} 1 & 0 \\\\ 0 & 2 \\end{pmatrix} \\), and \\( \\rho_2 = \\begin{pmatrix} 0 & 0 \\\\ 0 & 1 \\end{pmatrix} \\). Consider the ensemble \\( \\rho = \\frac{2}{5} \\rho_0 + \\frac{2}{5} \\rho_1 + \\frac{1}{5} \\rho_2 \\). Show that inequality 2 is not satisfied." } ]

[ { "context": "The min entropy for \\( X \\) is defined as\n\n\\[ H_{\\min}(X) = -\\log \\max p_x \\]\n\nSince each of the last \\( n - t \\) bits of \\( X_0 \\) is uniformly random and independent of the other bits, there are \\( 2^{n-t} \\) equally likely outcomes for the distribution \\( X_0 \\), so\n\n\\[ H_{\\min}(X_0) = -\\log \\max \\frac{1}{2^{n-t}} = -\\log \\frac{1}{2^{n-t}} = n - t \\]\n\nFor \\( X_1 \\), there are \\( 2^{n-1} \\) strings of length \\( n - 1 \\), and for each of these, there is exactly one bit we can append to get a string with an even number of 0s. Thus \\( X_1 \\) has \\( 2^{n-1} \\) equally likely outcomes.\n\n\\[ H_{\\min}(X_1) = -\\log \\max \\frac{1}{2^{n-1}} = -\\log \\frac{1}{2^{n-1}} = n - 1 \\]\n\nFor \\( X_2 \\), there are \\( 2^{n/2} \\) possibilities for the first half of the string, and since the first half of the string determines the second half, there are \\( 2^{n/2} \\) equally likely outcomes.\n\n\\[ H_{\\min}(X_2) = -\\log \\max \\frac{1}{2^{n/2}} = -\\log \\frac{1}{2^{n/2}} = n/2 \\]", "question": "### (a) Fix an even integer \\( n \\) and integer \\( t < n \\). Consider the following sources.\n- \\( X_0 \\) is all 1s on the first \\( t \\) bits and uniformly random on the last \\( n - t \\) bits.\n- \\( X_1 \\) is uniformly random over the set of strings with an even number of 0s.\n- \\( X_2 \\) is uniformly random over the set of strings where the first \\( \\frac{n}{2} \\) bits are the same as the last \\( \\frac{n}{2} \\) bits.\n\nCompute the min-entropy \\( H_{\\min}(X_i) \\) for each \\( i \\in \\{0, 1, 2\\} \\)." }, { "context": "\\( f_0(X_0) \\) is not uniformly random, since this is the XOR of t 1s, so this always produces t mod 2.\n\n\\( f_0(X_1) \\) is uniformly random, since \\( t < n \\), so the first t bits of n are uniformly random, so their XOR is uniformly random. (Unless \\( t = 0 \\) in which case the output is not uniform random, it is 0)\n\n\\( f_0(X_2) \\) is uniformly random, since if \\( 1 \\leq t \\leq n/2 \\), it is the XOR of the first t bits of the string and if \\( n/2 \\leq t \\leq n \\), it is the XOR of the last \\( n - t \\) bits of a uniform string. (Unless \\( t = 0 \\) in which case the output is not uniform random, it is 0)\n\n\\( f_1(X_0) \\) is uniformly random, since if \\( 1 \\leq t \\leq n/2 \\), then \\( x_1 x_{1+n/2} \\) is uniform random, and if \\( n/2 \\leq t < n \\), then \\( x_n x_{2n} \\) is uniform random.\n\n\\( f_1(X_1) \\) is uniformly random only if \\( n/2 \\) is odd. If the first half of \\( x \\) has at least a 0 and at least a 1, then there is a 1/2 chance of each outcome, since if we choose the corresponding elements of the right half last, we can get either a result of 0 or 1. (TAs comment: In other\n\nwords, for any possible string on the other \\( n - 4 \\) bits, we can always make the value of \\( f_1(X_1) = 1 \\) by choosing appropriately the bit in the second half that is multiplied with the 1, and fix the number of zeros to be even by choosing appropriately the bit in the second half that is multiplied with 0). If the first half is all 0s, the result will be 0. If the first half is all 1s, then the result is the parity of the second half, which is 1 only if \\( n/2 \\) is odd.\n\n\\( f_1(X_2) \\) is uniformly random, since it is the XOR of a uniform random string of length \\( n/2 \\)\n\n\\( f_2(X_0) \\) is uniformly random, since the last bit is always uniformly random and independent of the previous bits.\n\n\\( f_2(X_1) \\) is not uniformly random, since if there are an even number of 0s, since \\( n \\) is even, there is an even number of 1s, so the XOR is 0.\n\n\\( f_2(X_2) \\) is not uniformly random, since the XOR of the whole string is the XOR of the first and second halves, which have the same parity, so this always results in 0.", "question": "### (b) Consider the following deterministic functions:\n- \\( f_0(x) := \\bigoplus_{i=1}^t x_i \\), the XOR of the first \\( t \\) bits of \\( x \\).\n- \\( f_1(x) := x_L \\cdot x_R = \\bigoplus_{i=1}^{\\frac{n}{2}} x_i x_{i+\\frac{n}{2}} \\), where \\( x = (x_L, x_R) \\) are the left and right halves of \\( x \\).\n- \\( f_2(x) := \\bigoplus_{i=1}^n x_i \\), the XOR of all of the bits of \\( x \\).\n\nFor which pairs \\( (i, j) \\) is \\( f_i(X_j) \\) distributed as a uniformly random bit?\n\n[^1]: [physics.stackexchange.com/questions/245274/probability-distribution-of-a-pretty-good-measurement](https://physics.stackexchange.com/questions/245274/probability-distribution-of-a-pretty-good-measurement)" }, { "context": "Consider the function \\( f \\) defined as follows: We divide the input \\( x \\) into \\( \\left\\lceil \\frac{n}{t+1} \\right\\rceil \\) disjoint segments each containing \\( t+1 \\) bits. There will always be enough bits to do this since\n\n\\[ \\left\\lceil \\frac{n}{t+1} \\right\\rceil \\leq \\frac{n}{t+1} \\]\n\n\\[ (t+1) \\cdot \\left\\lceil \\frac{n}{t+1} \\right\\rceil \\leq (t+1) \\cdot \\frac{n}{t+1} = n \\]\n\nIf there are leftover bits we ignore them. We then define \\( f(x) \\) such that the \\( i \\)-th bit of \\( f(x) \\) is equal to the XOR of the \\( i \\)-th segment (this means the outputs will have the correct number of bits, the same as the number of segments). Moreover, since Eve has only \\( t \\) bits, and the segments are \\( t+1 \\) bits, Eve never has all the bits in a segment, and the \\( i \\)-th bit of the output is therefore independent of Eve. Thus, the whole output is independent of Eve.", "question": "### (c) Alice and Bob share a classical secret \\( X \\in \\{0, 1\\}^n \\) generated uniformly at random. Alice and Bob make an error in their secure communication protocol and as a result, Eve learns \\( t \\) bits of \\( X \\). Give, with proof, a deterministic function \\( f \\) such that \\( f(X) \\) is uniformly random over strings of length \\( \\left\\lfloor \\frac{n}{L+1} \\right\\rfloor \\) and \\( f(X) \\) is independent of Eve." } ]

[ { "context": "\\[ H(Y|X) = \\sum_x \\Pr[X = x] H(Y|X = x) \\] \\[ = \\sum_x \\Pr[X = x] \\sum_y \\Pr[Y = y|X = x] \\log \\left( \\frac{1}{\\Pr[Y = y|X = x]} \\right) \\] \\[ = \\sum_x \\Pr[X = x] \\sum_y \\Pr[Y = y, X = x] \\log \\left( \\frac{\\Pr[X = x]}{\\Pr[Y = y, X = x]} \\right) \\] \\[ = \\sum_x \\sum_y \\Pr[Y = y, X = x] \\left( \\log \\left( \\frac{1}{\\Pr[Y = y, X = x]} \\right) - \\log \\left( \\frac{1}{\\Pr[X = x]} \\right) \\right) \\] \\[ = \\sum_x \\sum_y \\Pr[Y = y, X = x] \\log \\left( \\frac{1}{\\Pr[Y = y, X = x]} \\right) \\] \\[ - \\sum_x \\sum_y \\Pr[Y = y, X = x] \\log \\left( \\frac{1}{\\Pr[X = x]} \\right) \\] \\[ = H(XY) - \\sum_x \\Pr[X = x] \\log \\left( \\frac{1}{\\Pr[X = x]} \\right) \\] \\[ = H(XY) - H(X). \\]", "question": "### (a) Prove that conditional Shannon entropy satisfies the chain rule: \\[ H(Y \\mid X) = H(XY) - H(X). \\] (4)" }, { "context": "Given that \\(X\\) and \\(Y\\) are independent, we have \\[ H_{\\min}(XY) = -\\log P_{\text{guess}}(XY) \\] \\[ = -\\log (P_{\text{guess}}(X) P_{\text{guess}}(Y)) \\] \\[ = -\\log P_{\text{guess}}(X) - \\log P_{\text{guess}}(Y) \\] \\[ = H(X) + H(Y), \\] and thus \\[ H(Y|X) = H(Y) = H(XY) - H(X). \\]", "question": "### (b) Prove that the conditional min-entropy satisfies the chain rule on \\( X \\) and \\( Y \\) if \\( X \\) and \\( Y \\) are independent." }, { "context": "For \\(i = 1\\), \\[ P_{\text{guess}}(X_1) = \\Pr[X_1 = 0] = \\frac{5}{8}, \\quad P_{\text{guess}}(X_1 Y_1) = \\Pr[X_1 Y_1 = 00] = \\frac{1}{2}; \\] \\[ P_{\text{guess}}(Y_1|X_1) = \\Pr[X = 0] \\Pr[Y_1 = 0|X_1 = 0] + \\Pr[X_1 = 1] \\Pr[Y_1 = 0|X_1 = 1] = \\frac{3}{4}; \\] \\[ H_{\\min}(X_1) = -\\log \\frac{5}{8} = 3 - \\log 5, \\quad H_{\\min}(X_1 Y_1) = -\\log \\frac{1}{2} = 1, \\] \\[ H_{\\min}(Y_1|X_1) = -\\log \\frac{3}{4} = 2 - \\log 3, \\] \\[ H_{\\min}(Y_1|X_1) > H_{\\min}(X_1 Y_1) - H_{\\min}(X_1). \\] For \\(i = 2\\), \\[ P_{\text{guess}}(X_2) = \\Pr[X_2 = 0] = \\frac{5}{8}, \\quad P_{\text{guess}}(X_2 Y_2) = \\Pr[X_2 Y_2 = 00] = \\frac{3}{8}; \\] \\[ P_{\text{guess}}(Y_2|X_2) = \\Pr[X_2 = 0]\\Pr[Y_2 = 0|X_2 = 0] + \\Pr[X = 1]\\Pr[Y_2 = 0|X_2 = 1] = \\frac{11}{16} \\] \\[ H_{\\min}(X_2) = -\\log \\frac{5}{8} = 3 - \\log 5, \\quad H_{\\min}(X_2Y_2) = -\\log \\frac{3}{8} = 3 - \\log 3, \\] \\[ H_{\\min}(Y_2|X_2) = -\\log \\frac{11}{16} = 4 - \\log 11, \\] \\[ H_{\\min}(Y_2|X_2) < H_{\\min}(X_2Y_2) - H_{\\min}(X_2). \\] We have encountered all cases where \\( H_{\\min}(Y|X) =, >, < H_{\\min}(XY) - H_{\\min}(X) \\), thus we may conclude that there is no certain form of the chain rule for conditional min-entropy.", "question": "### (c) For each of the following two distributions, compute \\( H_{\\min}(X_iY_i) \\), \\( H_{\\min}(X_i) \\), and \\( H_{\\min}(Y_i \\mid X_i) \\). Make a conclusion about the form of the general chain rule for conditional min-entropy. \\[ \\begin{aligned} &p(X_1Y_1 = 00) = \\frac{1}{2}, \\quad p(X_1Y_1 = 01) = \\frac{1}{8}, \\quad p(X_1Y_1 = 10) = \\frac{1}{4}, \\quad p(X_1Y_1 = 11) = \\frac{1}{8}, \\\\ &p(X_2Y_2 = 00) = \\frac{3}{8}, \\quad p(X_2Y_2 = 01) = \\frac{1}{4}, \\quad p(X_2Y_2 = 10) = \\frac{5}{16}, \\quad p(X_2Y_2 = 11) = \\frac{1}{16}. \\end{aligned} \\]" } ]

[ { "context": "We wish to show that any observable \\( O \\) is of the form \\[ O = \\alpha X + \\beta Y + \\gamma Z \\] Where the coefficients are real and \\( \\alpha^2 + \\beta^2 + \\gamma^2 = 1 \\). We first reason that since \\( O = O^\\dagger \\), \\( O \\) is of the form \\[ O = \\begin{pmatrix} x & y + zi \\\\ y - zi & w \\end{pmatrix} \\] Where \\( x, y, z, w \\) are real. If we take \\[ y = \\alpha \\] \\[ z = -\\beta \\] \\[ \\frac{x - w}{2} = \\gamma \\] \\[ \\frac{x + w}{2} = \\delta \\] We get \\[ O = \\begin{pmatrix} \\delta + \\gamma & \\alpha - \\beta i \\\\ \\alpha + \\beta i & \\delta - \\gamma \\end{pmatrix} = \\alpha X + \\beta Y + \\gamma Z + \\delta I \\] So any unitary \\( O \\) must be of this form. Since we also know that \\( O^2 = I \\), we have \\[ O^2 = (\\alpha X + \\beta Y + \\gamma Z + \\delta I)(\\alpha X + \\beta Y + \\gamma Z + \\delta I) \\] And since \\( XY = -YX, XZ = -ZX \\) and \\( YZ = -ZY \\), if we expand and cancel, we get \\[ O^2 = \\alpha^2 X^2 + \\beta^2 Y^2 + \\gamma^2 Z^2 + \\delta^2 I^2 + 2\\alpha \\delta X + 2\\beta \\delta Y + 2\\gamma \\delta Z \\] \\[ O^2 = \\alpha^2 I + \\beta^2 I + \\gamma^2 I + \\delta^2 I + 2\\alpha \\delta X + 2\\beta \\delta Y + 2\\gamma \\delta Z \\] \\[ O^2 = (\\alpha^2 + \\beta^2 + \\gamma^2 + \\delta^2)I + 2\\alpha \\delta X + 2\\beta \\delta Y + 2\\gamma \\delta Z \\] And so if this equals I, either δ = 0 or α = β = γ = 0. Since we are assuming nondegeneracy, the former is the case \\[ O^2 = (\\alpha^2 + \\beta^2 + \\gamma^2)I \\] And so \\( \\alpha^2 + \\beta^2 + \\gamma^2 = 1 \\). Thus, any single qubit observable can be represented in this form.", "question": "### (a) Let \\( O \\) be a single-qubit observable such that \\( O \\) is non-degenerate (\\( O \\neq \\pm I \\)). Show that there exists real numbers \\( \\alpha, \\beta, \\gamma \\) such that \\( \\alpha^2 + \\beta^2 + \\gamma^2 = 1 \\) and \\( O = \\alpha X + \\beta Y + \\gamma Z \\), with \\( X, Y, Z \\) the standard Pauli matrices." }, { "context": "Referring to the result of (Due to Mandy Huo) We have \\( |\\Psi\\rangle |E\\rangle \\rightarrow \\alpha |0\\rangle |E_0\\rangle + \\beta |1\\rangle |E_1\\rangle \\) so \\[ |\\Psi\\rangle \\langle \\Psi| \\otimes |E\\rangle \\langle E| = |\\alpha|^2 |0\\rangle \\langle 0| \\otimes |E_0\\rangle \\langle E_0| + \\alpha \\beta^* |0\\rangle \\langle 1| \\otimes |E_0\\rangle \\langle E_1| + \\alpha^* \\beta |1\\rangle \\langle 0| \\otimes |E_1\\rangle \\langle E_0| + |\\beta|^2 |1\\rangle \\langle 1| \\otimes |E_1\\rangle \\langle E_1| \\] Assuming \\( \\langle E_0|E_1 \\rangle \\) is real, we have \\( \\langle E_0|E_1 \\rangle = \\langle E_1|E_0 \\rangle \\). Since \\( |E\\rangle \\), \\( |E_0\\rangle \\), and \\( |E_1\\rangle \\) are normalized, tracing out the environment gives \\[ |\\Psi\\rangle \\langle \\Psi| \\otimes \\text{Tr} (|E\\rangle \\langle E|) = |\\alpha|^2 |0\\rangle \\langle 0| + \\langle E_0|E_1 \\rangle (\\alpha \\beta^* |0\\rangle \\langle 1| + \\alpha^* \\beta |1\\rangle \\langle 0|) + |\\beta|^2 |1\\rangle \\langle 1| \\] Define \\( p = \\frac{1 - \\langle E_0|E_1 \\rangle}{2} \\). We will show later that \\( p \\) is in fact a valid probability. Note that \\( Z|\\Psi\\rangle = \\alpha |0\\rangle - \\beta |1\\rangle \\). Then we have \\[ |\\Psi\\rangle \\langle \\Psi| \\otimes \\text{Tr} (|E\\rangle \\langle E|) = |\\alpha|^2 |0\\rangle \\langle 0| + (1 - 2p)(\\alpha \\beta^* |0\\rangle \\langle 1| + \\alpha^* \\beta |1\\rangle \\langle 0|) + |\\beta|^2 |1\\rangle \\langle 1| \\] \\[ = (1 - p) (|\\alpha|^2 |0\\rangle \\langle 0| + \\alpha \\beta^* |0\\rangle \\langle 1| + \\alpha^* \\beta |1\\rangle \\langle 0| + |\\beta|^2 |1\\rangle \\langle 1|) \\] \\[ + p (|\\alpha|^2 |0\\rangle \\langle 0| - \\alpha \\beta^* |0\\rangle \\langle 1| - \\alpha^* \\beta |1\\rangle \\langle 0| + |\\beta|^2 |1\\rangle \\langle 1|) \\] \\[ = (1 - p)|\\Psi\\rangle \\langle \\Psi| + pZ|\\Psi\\rangle \\langle \\Psi|Z \\] So \\( |\\Psi\\rangle \\langle \\Psi| \\rightarrow (1 - p)|\\Psi\\rangle \\langle \\Psi| + pZ|\\Psi\\rangle \\langle \\Psi|Z \\). Note that \\( |E_i\\rangle \\langle E_i| \\geq 0 \\) since \\( \\langle u|E_i\\rangle \\langle E_i|u\\rangle = |\\langle u|E_i\\rangle|^2 \\geq 0 \\) for any \\( |u\\rangle \\) and \\( \\lambda_{\\max} (|E_i\\rangle \\langle E_i|) \\leq 1 \\) since \\( \\sum_i \\lambda_i (|E_i\\rangle \\langle E_i|) = \\text{Tr} (|E_i\\rangle \\langle E_i|) = 1 \\) and \\( \\lambda_i (|E_i\\rangle \\langle E_i|) \\geq 0 \\). Then by problem 2(b) we have \\[ |\\langle E_0|E_1 \\rangle|^2 = \\text{Tr} (|E_0\\rangle \\langle E_0| |E_1\\rangle \\langle E_1|) \\leq \\lambda_{\\max} (|E_1\\rangle \\langle E_1|) \\text{Tr} (|E_0\\rangle \\langle E_0|) \\leq 1 \\] Then we have \\( |\\langle E_0|E_1 \\rangle| \\leq 1 \\) which implies \\( 0 \\leq p \\leq 1 \\) so \\( p \\) is a valid probability.", "question": "### (b) Let \\( B = A_0 \\otimes B_0 + A_1 \\otimes B_0 + A_0 \\otimes B_1 - A_1 \\otimes B_1 \\). Show that the success probability of the strategy in the CHSH game is \\( p_s = \\frac{1}{2} + \\frac{1}{4} \\langle \\psi | B | \\psi \\rangle \\)." }, { "context": "we found that the probability of success in the CHSH game was \\[ p_s = \\frac{1}{2} + \\frac{1}{8} (\\langle u_0|v_0 \\rangle + \\langle u_0|v_1 \\rangle + \\langle u_1|v_0 \\rangle - \\langle u_1|v_1 \\rangle) \\] Where \\[ |u_x \\rangle = A_x \\otimes I |\\psi \\rangle \\] \\[ |v_y \\rangle = I \\otimes B_y |\\psi \\rangle \\] From these definitions, we see \\[ \\langle u_x|v_y \\rangle = \\langle \\psi | (A_x \\otimes I)(I \\otimes B_y) |\\psi \\rangle = \\langle \\psi | (A_x \\otimes B_y) |\\psi \\rangle \\] And so we can rewrite the result of that problem as \\[ p_s = \\frac{1}{2} + \\frac{1}{8} (\\langle \\psi | (A_0 \\otimes B_0) |\\psi \\rangle + \\langle \\psi | (A_0 \\otimes B_1) |\\psi \\rangle + \\langle \\psi | (A_1 \\otimes B_0) |\\psi \\rangle - \\langle \\psi | (A_1 \\otimes B_1) |\\psi \\rangle) \\] And by linearity \\[ p_s = \\frac{1}{2} + \\frac{1}{8} (\\langle \\psi | B |\\psi \\rangle) \\] Which is the correct identity.", "question": "### (c) Argue that for the purposes of computing the maximum success probability in the CHSH game of players using state \\( |\\psi \\rangle_{AB} \\) as in (5) we may without loss of generality restrict our attention to observables of the form \\( A_x = \\cos(\\alpha_x)X + \\sin(\\alpha_x)Y \\) and \\( B_y = \\cos(\\beta_y)X + \\sin(\\beta_y)Y \\) for some angles \\( \\alpha_x, \\beta_y \\in [0, 2\\pi) \\). [Hint: do a rotation on the Bloch sphere.] Based on the symmetry argument from the previous questions we have reduced our problem to understanding the maximum value that \\( \\langle \\psi | B | \\psi \\rangle \\) can take, when \\( |\\psi \\rangle \\) is as in (5) and \\( B \\) is defined from observables \\( A_x, B_y \\) as in (b). To understand this maximum value we compute the spectral decomposition of \\( B \\)." }, { "context": "We have \\[ B = (A_0 \\otimes B_0) + (A_0 \\otimes B_1) + (A_1 \\otimes B_0) - (A_1 \\otimes B_1) \\] And from the special form of \\( A_x, B_y \\), we have \\[ A_x \\otimes B_y = (\\cos(α_x)X + \\sin(α_x)Y) \\otimes (\\cos(β_y)X + \\sin(β_y)Y) \\] And so we note that \\( ZXZ = -iZ \\) and \\( ZYZ = iZ \\) and we see \\[ (Z \\otimes I)A_x \\otimes B_y(Z \\otimes I) = (-\\cos(α_x)X - \\sin(α_x)Y) \\otimes (\\cos(β_y)X + \\sin(β_y)Y) = -A_x \\otimes B_y \\] And \\[ (I \\otimes Z)A_x \\otimes B_y(I \\otimes Z) = (\\cos(α_x)X + \\sin(α_x)Y) \\otimes (-\\cos(β_y)X - \\sin(β_y)Y) = -A_x \\otimes B_y \\] And so, since \\( B \\) is a linear combination of \\( A_x \\otimes B_y \\), we have by linearity \\[ (Z \\otimes I)B(Z \\otimes I) = (I \\otimes Z)B(I \\otimes Z) = -B \\]", "question": "### (d) Show that \\( (Z \\otimes I)B(Z \\otimes I) = (I \\otimes Z)B(I \\otimes Z) = -B \\). [Hint: use the special form of \\( A_x \\) and \\( B_y \\) you obtained from question (c).]" }, { "context": "", "question": "### (e) Show that \\( B \\) has a basis of eigenvectors of the form \\( |\\phi_{ab} \\rangle = e^{i\\phi_{ab}} |ab \\rangle + \\overline{a} \\overline{b} \\rangle \\), where \\( a, b \\in \\{0, 1\\} \\) and \\( \\overline{a} = 1 - a, \\overline{b} = 1 - b \\). Note that up to local rotations this is the Bell basis." }, { "context": "", "question": "### (f) Write \\( B^2 \\) as a \\( 4 \\times 4 \\) matrix depending on the angles \\( \\alpha_x, \\beta_y \\), and show that \\( \\text{Tr}(B^2) \\leq 16 \\)." }, { "context": "", "question": "### (g) Show that the largest success probability achievable in the CHSH game using \\( |\\psi \\rangle_{AB} \\) is at most \\( \\frac{1}{2} + \\frac{1}{4} \\sqrt{1 + \\sin^2(2\\theta)} \\). [Hint: Decompose \\( |\\psi \\rangle \\) in the eigenbasis of \\( B \\). Use (f) and the symmetries from (d) to bound the success probability via the expression found in (b).]" }, { "context": "", "question": "### (h) Give a strategy for the players which achieves this value, i.e. specify the players' observables." } ]

[ { "context": "Let \\( \\rho_{AB} = |\\psi\rangle \\langle \\psi|_{AB} \\), then\n\n\\[\n\\rho_A = \\text{Tr}_B(\\rho_{AB}) = \\cos^2(\\theta)|0\\rangle \\langle 0| + \\sin^2(\\theta)|1\\rangle \\langle 1|.\n\\]\n\nAssume that\n\n\\[\nA_x = |u_x^0\\rangle \\langle u_x^0| - |u_x^1\\rangle \\langle u_x^1|, \\quad x \\in \\{0, 1\\},\n\\]\n\nwhere \\( \\{|u_x^0\\rangle, |u_x^1\\rangle\\} \\) is an orthogonal basis. Then for any \\( a \\in \\{0, 1\\}, x \\in \\{0, 1\\} \\), we have\n\n\\[\np_{\\theta}(a|x) = \\text{Tr}(|u_x^a\\rangle \\langle u_x^a| \\rho_A)\n\\]\n\\[\n= \\cos^2(\\theta)|\\langle u_x^a|0\\rangle|^2 + \\sin^2(\\theta)|\\langle u_x^a|1\\rangle|^2\n\\]\n\\[\n\\leq \\cos^2(\\theta)|\\langle u_x^a|0\\rangle|^2 + \\cos^2(\\theta)|\\langle u_x^a|1\\rangle|^2\n\\]\n\\[\n= \\cos^2(\\theta)(|\\langle u_x^a|0\\rangle|^2 + |\\langle u_x^a|1\\rangle|^2)\n\\]\n\\[\n= \\cos^2(\\theta).\n\\]\n\nWe have used the fact that \\( \\sin^2(\\theta) \\leq \\cos^2(\\theta) \\), \\( \\forall \\theta \\in [0, \\frac{\\pi}{4}] \\). Therefore \\( \\max_{a,x} p_{\\theta}(a|x) \\leq \\cos^2(\\theta) \\).", "question": "### (a) Suppose that Alice and Bob play the CHSH game using a two-qubit entangled state \\( |\\psi \rangle_{AB} \\) as in (5). Let \\( p_0(a|x) \\) be the probability that, in this strategy, Alice returns answer \\( a \\in \\{0, 1\\} \\) to question \\( x \\in \\{0, 1\\} \\). Show that \\( \\max_{a,x} p_0(a|x) \\leq \\cos^2(\theta) \\)." }, { "context": "(b) Using the result in problem 5(g), we have\n\n\\[\nI = 8p_s - 4 \\leq 8 \\left( \\frac{1}{2} + \\frac{1}{4} \\sqrt{1 + \\sin^2(2\\theta)} \\right) - 4 = 2 \\sqrt{1 + \\sin^2(2\\theta)}.\n\\]\n\nSince we may also assume that \\( p_s \\geq \\frac{1}{2} \\), i.e. \\( I \\geq 0 \\), then we have\n\n\\[\n\\sin^2(2\\theta) + 1 \\geq \\frac{I^2}{4},\n\\]\n\\[\n\\implies 2 - \\frac{I^2}{4} \\geq 2 - (1 + \\sin^2(2\\theta)) = 1 - \\sin^2(2\\theta) = \\cos^2(2\\theta),\n\\]\n\\[\n\\implies \\sqrt{2 - \\frac{I^2}{4}} \\geq \\cos(2\\theta) = 2 \\cos^2(\\theta) - 1.\n\\]\n\nThen using the result of (a), we have\n\n\\[\n\\max_{a,x} p_{\\theta}(a|x) = \\cos^2(\\theta) \\leq \\frac{1}{2} \\left( 1 + \\sqrt{2 - \\frac{I^2}{4}} \\right),\n\\]\n\ni.e.\n\n\\[\np_{\\theta}(a|x) \\leq \\frac{1}{2} \\left( 1 + \\sqrt{2 - \\frac{I^2}{4}} \\right), \\quad \\forall a, x \\in \\{0, 1\\}.\n\\]", "question": "### (b) Let \\( p_s = \\frac{1}{2} + \\frac{1}{8} I \\) be the players' success probability in CHSH, where \\( I \\in [-4, 4] \\) (\\( I = 2\\sqrt{2} \\) for the optimal quantum strategy). Using (g) from the previous problem, deduce from (a) that\n\\[\n\\forall a, x \\in \\{0,1\\}, \\quad p_{\\theta}(a|x) \\leq \\frac{1}{2} \\left( 1 + \\sqrt{2 - \\frac{I^2}{4}} \\right).\n\\]" }, { "context": "(c) For any two-qubit \\(|\\phi\\rangle\\), consider its Schmidt decomposition\n\n\\[\n|\\phi\\rangle = \\cos(\\theta)|u_0\\rangle|v_0\\rangle + \\sin(\\theta)|u_1\\rangle|v_1\\rangle,\n\\]\n\nwhere \\( \\theta \\in \\left[0, \\frac{\\pi}{4}\\right] \\). We may also assume that\n\n\\[\n|u_0\\rangle = U|0\\rangle, \\quad |u_1\\rangle = U|1\\rangle, \\quad |v_0\\rangle = V|0\\rangle, \\quad |v_1\\rangle = V|1\\rangle,\n\\]\n\nwhere \\( U, V \\) are two unitaries, that is\n\n\\[\n|\\phi\\rangle = \\cos(\\theta)(U|0\\rangle \\otimes V|0\\rangle) + \\sin(\\theta)(U|1\\rangle \\otimes V|1\\rangle) = (U \\otimes V)|\\psi_0\\rangle.\n\\]\n\nNow assume that we use a strategy \\( A_0, A_1, B_0, B_1 \\) to play CHSH game with state \\( |\\phi\\rangle \\), and have a probability distribution \\( \\{p(a, b|x, y), a, b, x, y \\in \\{0, 1\\}\\} \\). Let\n\n\\[\n\\tilde{A}_x = U^T A_x U, \\quad x \\in \\{0, 1\\},\n\\]\n\\[\n\\tilde{B}_y = U^T B_y U, \\quad y \\in \\{0, 1\\}.\n\\]\n\nIt's easy to check that \\( \\tilde{A}_0, \\tilde{A}_1, \\tilde{B}_0, \\tilde{B}_1 \\) are still non-degenerate observables. Then we can check that\n\n\\[\np(a, b|x, y) = \\langle \\phi | (A_x \\otimes B_y) | \\phi \\rangle\n\\]\n\\[\n= \\langle \\psi_0 | (U^T \\otimes V^T) (A_x \\otimes B_y) (U \\otimes V) | \\psi_0 \\rangle\n\\]\n\\[\n= \\langle \\psi_0 | (\\tilde{A}_x \\otimes \\tilde{B}_y) | \\psi_0 \\rangle\n\\]\n\\[\n= \\tilde{p}(a, b|x, y),\n\\]\n\nwhere \\( \\tilde{p}(a, b|x, y), a, b, x, y \\in \\{0, 1\\} \\) is the probability distribution when we use the observables \\( \\tilde{A}_0, \\tilde{A}_1, \\tilde{B}_0, \\tilde{B}_1 \\) to play CHSH game with the state \\( |\\psi_0 \\rangle \\). In particular, we have\n\n\\[\np_s = \\tilde{p}_s = \\frac{1}{2} + \\frac{1}{8} I,\n\\]\n\\[\np(a|x) = \\tilde{p}(a|x) \\leq \\frac{1}{2} \\left( 1 + \\sqrt{2 - \\frac{I^2}{4}} \\right), \\quad \\forall a, x \\in \\{0, 1\\},\n\\]\n\nwhere we have used the result of (b). That is\n\n\\[\np(a|x) \\leq \\frac{1}{2} \\left( 1 + \\sqrt{2 - \\frac{I^2}{4}} \\right) = \\frac{1}{2} \\left( 1 + \\sqrt{2 - 4(2p_s - 1)^2} \\right), \\quad \\forall a, x \\in \\{0, 1\\}.\n\\]\n\nThen for any \\( x \\in \\{0, 1\\} \\),\n\n\\[\nH_{\\min}(A|X = x) = -\\log \\left( \\max_{a \\in \\{0, 1\\}} p(a|x) \\right)\n\\]\n\\[\n\\geq -\\log \\left( \\frac{1}{2} \\left( 1 + \\sqrt{2 - 4(2p_s - 1)^2} \\right) \\right)\n\\]\n\\[\n= 1 - \\log \\left( 1 + \\sqrt{2 - 4(2p_s - 1)^2} \\right).\n\\]", "question": "### (c) Suppose now the players use any single-qubit strategy (not necessarily using \\(|\\psi_{\theta}\\)). Prove a lower bound on the conditional min-entropy \\(H_{\\min}(A|X = x)\\), for any \\(x \\in \\{0,1\\}\\), that is generated in Alice's outputs, as a function of the players' success probability in the CHSH game." }, { "context": "Let\n\n\\[\nA_0 = Z, \\quad A_1 = X, \\quad B_0 = \\cos(t)Z + \\sin(t)X, \\quad B_1 = \\cos(t)Z - \\sin(t)X,\n\\]\n\nwhere \\(\\cos(t) = \\frac{1}{\\sqrt{1 + \\sin^2(2\\theta)}}, \\sin(t) = \\frac{-\\sin(2\\theta)}{\\sqrt{1 + \\sin^2(2\\theta)}}\\). It's easy to check that \\(A_0, A_1, B_0, B_1\\) are non-degenerate observables. Then we have\n\n\\[\nB = A_0 \\otimes B_0 + A_1 \\otimes B_0 + A_0 \\otimes B_1 - A_1 \\otimes B_1\n\\]\n\\[\n= 2 \\cos(t) Z \\otimes Z + 2 \\sin(t) X \\otimes X,\n\\]\n\nand\n\n\\[\n\\langle \\psi_0 | B | \\psi_0 \\rangle = 2 \\cos(t) \\langle \\psi_0 | Z \\otimes Z | \\psi_0 \\rangle + 2 \\sin(t) \\langle \\psi_0 | X \\otimes X | \\psi_0 \\rangle\n\\]\n\\[\n= 2 \\cos(t) + 2 \\sin(t) \\sin(2\\theta)\n\\]\n\\[\n= 2 \\left( \\frac{1}{\\sqrt{1 + \\sin^2(2\\theta)}} \\right) + 2 \\left( \\frac{\\sin^2(2\\theta)}{\\sqrt{1 + \\sin^2(2\\theta)}} \\right)\n\\]\n\\[\n= 2 \\sqrt{1 + \\sin^2(2\\theta)}.\n\\]\n\nRecall that \\(p_s = \\frac{1}{2} + \\frac{1}{8} \\langle \\psi_0 | B | \\psi_0 \\rangle = \\frac{1}{2} + \\frac{1}{8} I\\), thus\n\n\\[\nI = \\langle \\psi_0 | B | \\psi_0 \\rangle = 2 \\sqrt{1 + \\sin^2(2\\theta)},\n\\]\n\\[\n\\frac{1}{2} \\left( 1 + \\sqrt{2 - \\frac{I^2}{4}} \\right) = \\frac{1}{2} \\left( 1 + \\sqrt{1 - \\sin^2(2\\theta)} \\right) = \\frac{1}{2} \\left( 1 + \\cos(2\\theta) \\right) = \\cos^2(\\theta).\n\\]\n\nOn the other hand, since \\(\\forall x \\in \\{0, 1\\}\\),\n\n\\[\np(0|x) - p(1|x) = \\text{Tr}(A_x \\rho_A), \\quad p(0|x) + p(1|x) = 1,\n\\]\n\nwe have\n\n\\[\np(0|x) = \\frac{1}{2} \\left( 1 + \\text{Tr}(A_x \\rho_A) \\right), \\quad p(1|x) = \\frac{1}{2} \\left( 1 - \\text{Tr}(A_x \\rho_A) \\right).\n\\]\n\nThen now we have\n\n\\[\n\\rho_A = \\cos^2(\\theta) |0\\rangle \\langle 0| + \\sin^2(\\theta) |1\\rangle \\langle 1|,\n\\]\n\n\\[\np(0|0) = \\frac{1}{2} \\left( 1 + \\text{Tr}(A_0 \\rho_A) \\right) = \\frac{1}{2} \\left( 1 + \\cos^2(\\theta) - \\sin^2(\\theta) \\right) = \\cos^2(\\theta),\n\\]\n\\[\np(1|0) = \\frac{1}{2} \\left( 1 - \\text{Tr}(A_0 \\rho_A) \\right) = \\frac{1}{2} \\left( 1 - \\cos^2(\\theta) + \\sin^2(\\theta) \\right) = \\sin^2(\\theta),\n\\]\n\\[\np(0|1) = \\frac{1}{2},\n\\]\n\\[\np(1|1) = \\frac{1}{2}.\n\\]\n\nSince \\(\\theta \\in \\left[ 0, \\frac{\\pi}{4} \\right]\\), we have \\(\\cos^2(\\theta) \\geq \\frac{1}{2} \\geq \\sin^2(\\theta)\\), thus\n\n\\[\n\\max_{a,x} p(a|x) = \\cos^2(\\theta) = \\frac{1}{2} \\left( 1 + \\sqrt{2 - \\frac{I^2}{4}} \\right).\n\\]\n\nThe bound is tight.", "question": "### (d) Show that the bound from (b) is tight: for any \\(\\theta \\in [0, \\pi/4]\\) find a strategy for the players using \\(|\\psi_{\theta}\\) such that \\(\\max_{a,x} p_{\theta}(a|x) = \\frac{1}{2} (1 + \\sqrt{2 - I^2/4})\\)." } ]

[ { "context": "- **Attack to protocol 1:**\n\n- Assume that Eve has a quantum machine that can store arbitrary amount of quantum states.\n- After Alice sends out the qubits she prepares via the quantum channel, Eve captures and stores them for the moment.\n- When Alice announces the basis string \\( \\theta \\) via the authenticated channel, Eve learns the basis. So Eve can measure the qubits in the right basis to learn the key \\( x \\) exactly without changing the qubits.\n- Afterwards Eve sends the unchanged qubits to Bob.\n- Since Bob receives the exact qubits that Alice sends at the beginning and measures them in the right basis, the \\( x \\) they share will pass the correctness checking.\n\n- **Protocol improvement:** Bob announces reception when he receives Alice’s qubits. Alice announces the basis string \\( \\theta \\) only after Bob announces reception.\n\n- **Attack to protocol 2:**\n\n- Assume that Eve has a quantum machine that can store and generate arbitrary amount of quantum states.\n- After Alice sends out the n halves of her EPR pairs via the quantum channel, Eve captures and stores them. (They become AE pairs.)\n- Eve generates another n EPR pairs and sends one half of each to Bob to cheat Bob, so that Bob will announce reception. (They become EB pairs.)\n- After Alice and Bob announce their basis strings \\( \\theta \\) and \\( \\hat{\\theta} \\), Eve uses these bases to measure the qubits on her side (of AE pairs and of EB pairs respectively), and learns exactly the raw key \\( x \\) shared with Alice and the raw key \\( \\hat{x} \\) shared with Bob.\n- Eve only keeps the bits \\( x_i \\) and \\( \\hat{x}_i \\) for \\( i \\) such that \\( \\theta_i = \\hat{\\theta}_i \\). Now Eve has the exact key \\( x \\) that Alice will use and the exact key \\( \\hat{x} \\) that Bob will use.\n- Protocol improvement: Alice and Bob carry out an additional correctness checking step before they use the keys \\( x \\) and \\( \\hat{x} \\) as a common key. (Under such attack, the final keys \\( x \\) and \\( \\hat{x} \\), whose expected lengths are \\( \\frac{n}{2} \\), are expected to share only \\( \\frac{n}{4} \\) matching bits.)", "question": "### Let’s imagine that we are playing the role of the eavesdropper Eve. We observe two parties, Alice and Bob, trying to implement certain QKD protocols. Because QKD is hard, Alice and Bob might try to cut corners in the implementation of their protocols. Here are two suggested protocols that Alice and Bob might want to implement. For each of them, either prove security or provide an explicit attack for Eve.\n\n**Protocol 1:** \nAlice and Bob can communicate through a classical authenticated channel, and a quantum (non-authenticated) channel.\n\n- Alice generates bit strings \\( x, \\theta \\in \\{0, 1\\}^n \\) uniformly at random.\n- Alice prepares qubits \\( |x_i\\rangle_{\\theta_i} \\) for \\( i = 1, .., n \\) where \\( |0\\rangle_0 = |0\\rangle \\), \\( |1\\rangle_0 = |1\\rangle \\), \\( |0\\rangle_1 = |+\\rangle \\), \\( |1\\rangle_1 = |-\\rangle \\), and sends them to Bob.\n- Alice announces the basis string \\( \\theta \\).\n- Bob measures the qubits he received according to the bases specified by the string \\( \\theta \\) and recovers \\( x \\)." } ]

[ { "context": "In this case the box measures the first qubit in the standard basis so the first qubit of Alice’s state is \\(|0\\rangle\\). The post-measurement state is \n\\[\\rho = \\frac{1}{2} \\sum_{z=0}^{1} |0z\\rangle \\langle 0z| \\otimes |0z\\rangle \\langle 0z| \\otimes |0z\\rangle \\langle 0z|\\].", "question": "### (a) Alice and Bob put blind faith in their hardware and attempt to implement BB’84. They want to check that their state is an EPR pair, so Alice asks her box to measure in the standard basis. The box returns a measurement outcome of 0. Determine the post-measurement state." }, { "context": "If Bob asks his box to measure in the Hadamard basis, it will measure the second qubit in the standard basis. Thus, Bob will get outcome 0 half the time and outcome 1 half the time. If Bob asks his box to measure in the standard basis, then it will measure the first qubit in the standard basis, so he will get outcome 0.", "question": "### (b) After Alice’s measurement, Bob asks his box to measure in the Hadamard basis. What measurement outcome will Bob receive? Suppose instead Bob had asked his box to measure in the standard basis. What measurement outcome would Bob have received?" }, { "context": "Eve will get 0 since her first qubit is \\(|0\\rangle\\).", "question": "### (c) Suppose Bob did in fact the latter. Suppose that now Eve measures her first qubit in the standard basis. What measurement outcome does she receive?" }, { "context": "Since Alice and Bob have the same state and their boxes work the same way, they must get the same outcome whenever they make the same measurement. Therefore, they will pass with probability 1.", "question": "### (d) Suppose Alice and Bob run the BB’84 protocol, and, as per the usual, they look at all the rounds in which they made the same measurement as each other. They pick a random subset of half of those rounds and test whether they received the same output on all the rounds. What is the probability that they pass the test?" }, { "context": "Since Eve has the same state as Alice and Bob, she can learn \\( x_j \\) by making the same measurement that Alice and Bob’s boxes made: measure the first qubit in the standard basis if \\( \\theta_j = 0 \\) and the second qubit in the standard basis if \\( \\theta_j = 1 \\).", "question": "### (e) Let \\( T' \\) be the set of rounds on which Alice and Bob made the same measurement but didn’t perform a test. Let \\(\\{ \\theta_j \\}_{j \\in T'} \\) be the measurements they made and \\(\\{ x_j \\}_{j \\in T'} \\) and \\(\\{ i_j \\}_{j \\in T'} \\) be the results they received. The \\(\\theta_j\\) have been communicated over the public channel. Eve wishes to learn the \\( x_j \\). Which measurements should she make?" }, { "context": "Since Eve can recover all the \\( x_j, j \\in T' \\), Eve knows the entire key so \\( H_{\\text{min}}(X|E) = 0 \\).", "question": "### (f) Let \\( X \\) be the classical key generated by Alice and Bob. What is \\( H_{\\text{min}}(X \\mid E) \\), where \\( E \\) is Eve’s system?" } ]

[ { "context": "The two-dimensional eigenspace to which the post-measurement state will belong will be spanned by the two Bell states which correspond to the eigenvalue \\(-1\\). If we evaluate\n\n\\[\n(X \\otimes X)|\\Psi_{00}\\rangle = (X \\otimes X) \\frac{1}{\\sqrt{2}} (|00\\rangle + |11\\rangle) = \\frac{1}{\\sqrt{2}} (|11\\rangle + |00\\rangle) = |\\Psi_{00}\\rangle\n\\]\n\n\\[\n(X \\otimes X)|\\Psi_{01}\\rangle = (X \\otimes X) \\frac{1}{\\sqrt{2}} (|00\\rangle - |11\\rangle) = \\frac{1}{\\sqrt{2}} (|11\\rangle - |00\\rangle) = -|\\Psi_{01}\\rangle\n\\]\n\n\\[\n(X \\otimes X)|\\Psi_{10}\\rangle = (X \\otimes X) \\frac{1}{\\sqrt{2}} (|01\\rangle + |10\\rangle) = \\frac{1}{\\sqrt{2}} (|10\\rangle + |01\\rangle) = |\\Psi_{10}\\rangle\n\\]\n\n\\[\n(X \\otimes X)|\\Psi_{11}\\rangle = (X \\otimes X) \\frac{1}{\\sqrt{2}} (|01\\rangle - |10\\rangle) = \\frac{1}{\\sqrt{2}} (|10\\rangle - |01\\rangle) = -|\\Psi_{11}\\rangle\n\\]\n\nWe see that if the measurement is \\(-1\\), then the post-measurement state belongs to the eigenspace spanned by \\(|\\Psi_{01}\\rangle, |\\Psi_{11}\\rangle\\).\n\nIf we evaluate the eigenvalues for the Bell states under the \\(Z \\otimes Z\\) operator\n\n\\[\n(Z \\otimes Z)|\\Psi_{00}\\rangle = (Z \\otimes Z) \\frac{1}{\\sqrt{2}} (|00\\rangle + |11\\rangle) = \\frac{1}{\\sqrt{2}} (|00\\rangle + |11\\rangle) = |\\Psi_{00}\\rangle\n\\]\n\n\\[\n(Z \\otimes Z)|\\Psi_{01}\\rangle = (Z \\otimes Z) \\frac{1}{\\sqrt{2}} (|00\\rangle - |11\\rangle) = \\frac{1}{\\sqrt{2}} (|00\\rangle - |11\\rangle) = |\\Psi_{01}\\rangle\n\\]\n\n\\[\n(Z \\otimes Z)|\\Psi_{10}\\rangle = (Z \\otimes Z) \\frac{1}{\\sqrt{2}} (|01\\rangle + |10\\rangle) = \\frac{1}{\\sqrt{2}} (|01\\rangle + |10\\rangle) = -|\\Psi_{10}\\rangle\n\\]\n\n\\[\n(Z \\otimes Z)|\\Psi_{11}\\rangle = (Z \\otimes Z) \\frac{1}{\\sqrt{2}} (|01\\rangle - |10\\rangle) = \\frac{1}{\\sqrt{2}} (-|01\\rangle + |10\\rangle) = -|\\Psi_{11}\\rangle\n\\]\n\nAnd so after the second measurement of 1, the post measurement state is \\(|\\Psi_{00}\\rangle\\).", "question": "### (a) Suppose we measure an arbitrary two-qubit state \\(\\lvert \\phi \\rangle \\) using the observable \\( X \\otimes X \\) and obtain the outcome \\(-1\\). To which two-dimensional eigenspace does the post-measurement state belong? (Specify the subspace using two of the Bell states above.) Next, we measure the observable \\( Z \\otimes Z \\) and obtain outcome 1. What is post-measurement state \\(\\lvert \\phi' \\rangle \\)?" }, { "context": "If we apply \\((Y \\otimes Y)\\) to \\(|\\Psi_{00}\\rangle\\), we get \\(|\\Psi_{00}\\rangle\\), so if the post-measurement state had some overlap with this state, the measured eigenvalue would have to be \\(1\\).", "question": "### (b) Suppose that instead we performed the measurement \\(-Y \\otimes Y = (X \\otimes X)(Z \\otimes Z) \\) directly, and the post-measurement state had nonzero overlap with \\(\\lvert \\phi' \\rangle \\). What measurement outcome would we have obtained?" }, { "context": "In general, when one measures a product of commuting observables, one gets the product of the measurement one would have gotten from each measurement individually. That is,\n\n\\[\n\\langle \\psi|A| \\psi \\rangle \\langle \\psi|B| \\psi \\rangle = \\langle \\psi|AB| \\psi \\rangle\n\\]", "question": "### (c) What do you deduce about the relationship between the outcomes of measuring commuting observables \\( A \\) and \\( B \\) with the outcome of measuring the observable \\( AB \\) directly?" } ]

[ { "context": "(a) They win with probability \\(\\frac{3}{4}\\). They only lose when \\(s = t = 0\\), because then both sides of the equation evaluate to 0.", "question": "### (a) We begin by describing the single-shot game. Eve starts by generating a pair \\((s, t) \\in \\{(0,0), (0,1), (1,0), (1,1)\\}\\) uniformly at random. She gives \\(s\\) to Alice and \\(t\\) to Bob. Alice and Bob generate output bits \\(a, b \\in \\{0,1\\}\\), respectively. They win if \\(a \\vee b \\neq s \\vee t\\). As a warm-up, consider the strategy in which \\(a = s\\) and \\(b = t\\). What is the winning probability? Which inputs cause Alice and Bob to lose?" }, { "context": "There are 9 possible equally likely situations that can occur, because each game has 3 possible inputs. Both Alice and Bob follow the same strategy. If their input is \\((0,0)\\), they output \\((0,0)\\), and otherwise output \\((1,1)\\).\n\nClearly, if they both output \\((0,0)\\), they will fail because all 4 variables in both equations are 0. Additionally, if both of them output \\((1,1)\\), they will fail because both sides of both equations will be 1.\n\nThe only remaining case is, if only one of them outputs \\((0,0)\\) and the other outputs \\((1,1)\\), then we have that either \\(a_0 \\vee s_0 = a_1 \\vee s_1 = 0\\) and \\(b_0 \\vee t_0 = b_1 \\vee t_1 = 1\\) or \\(a_0 \\vee s_0 = a_1 \\vee s_1 = 1\\) and \\(b_0 \\vee t_0 = b_1 \\vee t_1 = 0\\), and either way they succeed in both games, so they win. This can happen 6 different ways, because \\((s_0, s_1) = (0,0)\\), then we can have \\((t_0, t_1) \\in \\{(0,1), (1,0), (1,1)\\}\\), and vice versa if \\((t_0, t_1) = (0,0)\\). Therefore, the probability of success is \\(6 \\times \\frac{1}{9} = \\frac{2}{3}\\).", "question": "### (b) In the two-parallel version \\(G^{(2)}\\) of the game we just described, Eve picks two strings \\((s_0, t_0), (s_1, t_1)\\) from \\(\\{(0,0), (0,1), (1,0), (1,1)\\}\\) independently and uniformly at random. She gives \\((s_0, s_1)\\) to Alice, \\((t_0, t_1)\\) to Bob, and demands outputs \\((a_0, a_1), (b_0, b_1)\\) from Alice and Bob. They win if \\(a_0 \\vee s_0 \\neq b_0 \\vee t_0\\) and \\(a_1 \\vee s_1 \\neq b_1 \\vee t_1\\). Describe a deterministic strategy for Alice and Bob that achieves a winning probability of 2/3." }, { "context": "Alice can just randomly generate values of \\((s_1, t_1)\\) and send Bob \\(t_1\\) and then 'forget' the value. Then, they can play the two-shot game and win with probability \\(w_c\\), which means they must win the one-shot game with probability at least \\(w_c\\), because they need to win both games to win the two-shot version.", "question": "### (c) Suppose Alice and Bob have a valid strategy for the two-parallel game which wins with probability \\(\\omega_c\\). Describe a strategy for them to win the one-shot game with probability at least \\(\\omega_c\\). This proves that the optimal success probability in the one-shot game is an upper bound for the optimal success probability in the two-parallel game." }, { "context": "First, note that \\(A_0\\) and \\(B_0\\) commute because they each act on separate parts of the state that are spacelike separated.\n\nIf \\((s, t) = (0, 0)\\), which happens with probability \\(\\frac{1}{3}\\), Alice and Bob win the game with likelihood \\(P(a = 0)P(b = 1) + P(a = 1)P(b = 0) = P(a \\neq b)\\). We know that \\(\\langle \\Psi | A_0 B_0 | \\Psi \\rangle\\) is the probability of the measured eigenvalues of \\(A_0\\) and \\(B_0\\) being the same minus the probability that they are different, because as we know from the last question the measured eigenvalue of a product of two commuting operators is the same as the product of measuring with each operator. Therefore, this tells us that \\(\\langle \\Psi | A_0 B_0 | \\Psi \\rangle = P(a = b) - P(a \\neq b)\\). However, \\(P(a = b) = 1 - P(a \\neq b)\\), so we have \\(\\langle \\Psi | A_0 B_0 | \\Psi \\rangle = 1 - 2P(a \\neq b)\\), so \\(P(a \\neq b) = \\frac{1}{2} - \\frac{1}{2} \\langle \\Psi | A_0 B_0 | \\Psi \\rangle\\) is the success probability in this case.\n\nIf \\((s, t) = (0, 1)\\), which happens with probability \\(\\frac{1}{3}\\), they succeed exactly when \\(a = 0\\). We know that \\(\\langle \\Psi | A_0 | \\Psi \\rangle = P(a = 0) - P(a = 1)\\), and using the substitution \\(P(a = 1) = 1 - P(a = 0)\\), we get \\(P(a = 0) = \\frac{1}{2} + \\frac{1}{2} \\langle \\Psi | A_0 | \\Psi \\rangle\\).\n\nSimilarly, if \\((s, t) = (1, 0)\\), which happens with probability \\(\\frac{1}{3}\\), they succeed exactly when \\(b = 0\\). We know that \\(\\langle \\Psi | B_0 | \\Psi \\rangle = P(b = 0) - P(b = 1)\\), and using the substitution \\(P(b = 1) = 1 - P(b = 0)\\), we get \\(P(b = 0) = \\frac{1}{2} + \\frac{1}{2} \\langle \\Psi | B_0 | \\Psi \\rangle\\).\n\nTherefore, their success probability will be\n\n\\[\n\\frac{1}{3} \\left( \\frac{1}{2} - \\frac{1}{2} \\langle \\Psi | A_0 B_0 | \\Psi \\rangle \\right) + \\frac{1}{3} \\left( \\frac{1}{2} + \\frac{1}{2} \\langle \\Psi | A_0 | \\Psi \\rangle \\right) + \\frac{1}{3} \\left( \\frac{1}{2} + \\frac{1}{2} \\langle \\Psi | B_0 | \\Psi \\rangle \\right)\n\\]\n\nLetting \\(M = \\frac{1}{3} (A_0 + B_0 - A_0 B_0)\\), this reduces to \\(\\frac{1}{2} + \\frac{1}{2} \\langle \\Psi | M | \\Psi \\rangle\\).", "question": "### (d) Now we will find an upper bound on the success probability of the one-shot game, assuming that Alice and Bob may use shared entanglement in addition to classical resources.\n\nThe most general strategy that Alice and Bob can take is as follows. They each have two \\(\\pm 1\\)-eigenvalue-observables \\(A_0, A_1, B_0, B_1\\). They share a joint state \\(|\\psi\\rangle\\). Alice measures \\(|\\psi\\rangle\\) on \\(A_s\\), Bob measures on \\(B_t\\), and they each output 0 if they measured a 1 and 1 if they measured a -1.\n\nIn general, if \\(X\\) is an observable, then \\(\\langle\\psi| X |\\psi\\rangle\\) is equal to the probability of measuring a 1 minus the probability of measuring -1.\n\nLet \\(M = -\\frac{1}{3}A_0 B_0 + \\frac{1}{3}A_0 + \\frac{1}{3}B_0\\). Prove that the probability that Alice and Bob win the game is \\(\\frac{1}{2} + \\frac{1}{2}\\langle\\psi| M |\\psi\\rangle\\)." }, { "context": "\\[M^2 = \\frac{1}{9} (A_0 + B_0 - A_0 B_0)^2\\]\n\n\\[= \\frac{1}{9} (A_0^2 + B_0^2 + A_0 B_0 + B_0 A_0 - A_0^2 B_0 - B_0 A_0^2 - B_0 A_0 B_0 + A_0 B_0 A_0 B_0)\\]\n\nBecause \\(A_0\\) and \\(B_0\\) are operators, they square to identity\n\n\\[= \\frac{1}{9} (2I + A_0 B_0 + B_0 A_0 - B_0 - B_0 A_0 B_0 + A_0 B_0 A_0 B_0)\\]\n\nAdditionally, \\(A_0\\) and \\(B_0\\) commute, so we can rewrite this as\n\n\\[= \\frac{1}{9} (3I + 2A_0 B_0 - B_0 - A_0)\\]\n\n\\[= \\frac{1}{3} (I - 2M)\\]", "question": "### (e) Prove that \\(M^2 = \\frac{1}{3}I - \\frac{2}{3}M\\)." }, { "context": "By Cayley-Hamilton, \\(M\\) satisfies its characteristic polynomial, and since this is the unique monic polynomial that it satisfies, this must be its characteristic polynomial. Therefore, the eigenvalues of \\(M\\) will satisfy this as well, so \\(3\\lambda^2 + 2\\lambda - 1 = 0\\). This factors to \\((3\\lambda - 1)(\\lambda + 1) = 0\\), so \\(\\lambda_{\\text{max}} = \\frac{1}{3}\\).", "question": "### (f) The answer to the last is the characteristic polynomial of \\(M\\) (indeed, it is the unique monic quadratic satisfied by \\(M\\)). Use it to solve for the largest eigenvalue \\(\\lambda_{\\text{max}}\\) of \\(M\\)." }, { "context": "\\( p_{\\text{win}} \\leq \\frac{1}{2} + \\frac{1}{2} \\langle \\Psi | M | \\Psi \\rangle \\leq \\frac{1}{2} + \\frac{1}{2} \\cdot \\frac{2}{3} = \\frac{2}{3} \\). We also know that this win probability is achievable even without any quantum strategy from the naive strategy in part (a), so this must be the tightest possible bound.", "question": "### (g) Now use the facts that \\(p_{\\text{win}} \\leq \\frac{1}{2} + \\frac{1}{2}\\langle\\psi| M |\\psi\\rangle\\) and \\(\\langle\\psi| M |\\psi\\rangle \\leq \\lambda_{\\text{max}}\\) to find the tightest possible upper bound on \\(p_{\\text{win}}\\)." } ]

[ { "context": "For any \\( |\\Psi\\rangle_{XE} \\in \\mathcal{H}_X \\otimes \\mathcal{H}_E \\), say\n\n\\[ |\\Psi\\rangle_{XE} = \\sum_{i,j} \\alpha_{ij} |i\\rangle_X |j\\rangle_E, \\]\n\nwe have\n\n\\[ \\langle \\Psi |_{XE} (|x\\rangle \\langle x| \\otimes N_z) |\\Psi\\rangle_{XE} = \\sum_{i,j} \\sum_{k,l} \\langle i|x \\rangle \\langle x|i \\rangle \\langle j| (|x\\rangle \\langle x| \\otimes N_z) |k\\rangle \\langle k|j \\rangle_E \\]\n\n\\[ = \\sum_{i,j} \\sum_{k,l} \\langle i|x \\rangle \\langle x|k \\rangle \\langle j|N_z|l \\rangle_E \\]\n\n\\[ = \\left( \\sum_{i,k} \\langle i|x \\rangle \\langle x|k \\rangle \\right) \\left( \\sum_{j,l} \\langle j|N_z|l \\rangle_E \\]\n\n\\[ = \\left( \\left( \\sum_{i} \\langle i|x \\rangle \\langle x| \\left( \\sum_{i} |i\\rangle \\right) \\right) \\left( \\left( \\sum_{j} \\langle j|E \\right) N_z \\left( \\sum_{j} |j\\rangle_E \\right) \\right) \\right) \\]\n\n\\[ \\geq 0. \\]\n\nThus \\( |x\\rangle \\langle x| \\otimes N_z \\geq 0, \\ \\forall x \\in \\mathcal{X}, \\) and consequently\n\n\\[ Z = \\sum_{x \\in \\mathcal{X}} |x\\rangle \\langle x| \\otimes N_z \\geq 0. \\]\n\nBy definition we have\n\n\\[ \\Phi(Z) = \\sum_{x' \\in \\mathcal{X}} \\left( \\langle x' | \\otimes I_E \\right) Z \\left( | x' \\rangle \\otimes I_E \\right) \\]\n\n\\[ = \\sum_{x' \\in \\mathcal{X}} \\left( \\langle x' | \\otimes I_E \\right) (| x \\rangle \\langle x | \\otimes N_x) \\left( | x' \\rangle \\otimes I_E \\right) \\]\n\n\\[ = \\sum_{x' \\in \\mathcal{X}} \\langle x' | x \\rangle \\langle x | x' \\rangle N_x \\]\n\n\\[ = \\sum_{x \\in \\mathcal{X}} N_x \\]\n\n\\[ = I_E. \\]", "question": "### (a) Suppose \\( \\{N_x\\} \\) is a valid POVM. Show that the matrix \\( Z = \\sum_x |x\\rangle \\langle x| \\otimes N_x \\) satisfies \\( Z \\geq 0 \\), and compute \\( \\Phi(Z) \\) (the result should be a matrix defined on system \\( E \\) only)." }, { "context": "\\[ \\text{tr}(Z \\rho_{X E}) = \\text{tr} \\left( Z \\sum_{x' \\in \\mathcal{X}} (| x \\rangle \\langle x | x' \\rangle \\langle x' |) \\otimes (N_x \\rho_E^x) \\right) \\]\n\n\\[ = \\sum_{x' \\in \\mathcal{X}} \\text{tr} \\left( (| x \\rangle \\langle x' | x' \\rangle \\langle x' |) \\text{tr}(N_x \\rho_E^x) \\right) \\]\n\n\\[ = \\sum_{x \\in \\mathcal{X}} \\text{tr}(N_x \\rho_E^x) \\]", "question": "### (b) For the same matrix \\( Z \\) as in the previous question, compute \\( \\text{Tr}(Z \\rho_{XE}) \\)." }, { "context": "For any \\( | \\phi \\rangle_E \\in \\mathcal{H}_E \\), we have\n\n\\[ \\langle \\phi | E N_x | \\phi \\rangle_E = \\langle \\phi | E \\left( (| x \\rangle \\langle x | \\otimes I_E) Z (| x \\rangle \\langle x | \\otimes I_E) \\right) \\phi \\rangle_E = \\left( \\langle x | \\langle \\phi | E (| x \\rangle \\langle x | \\otimes I_E) | \\phi \\rangle_E \\right) \\geq 0, \\]\n\nthus \\( N_x \\geq 0, \\forall x \\in \\mathcal{X} \\). Also we have\n\n\\[ \\sum_{x \\in \\mathcal{X}} N_x = \\sum_{x \\in \\mathcal{X}} \\left( (| x \\rangle \\langle x | \\otimes I_E) Z (| x \\rangle \\langle x | \\otimes I_E) \\right) = \\Phi(Z) = I_E. \\]\n\nTherefore \\( \\{ N_x \\}_x \\) is a valid POVM over \\( \\mathcal{H}_E \\).", "question": "### (c) Conversely, suppose \\( Z \\geq 0 \\) and \\( \\Phi(Z) = I_E \\). Show that the elements \\( N_x = (\\langle x| \\otimes I_E) Z (\\langle x| \\otimes I_E) \\) form a valid POVM \\( \\{N_x\\} \\) over \\( \\mathcal{H}_E \\) (with outcomes \\( x \\in \\mathcal{X} \\))." }, { "context": "By the previous results, the constraint that \\( \\{ M_x \\}_x \\) is a POVM can be translated into the conditions\n\n\\[ M_x = \\left( (| x \\rangle \\otimes I_E) Z (| x \\rangle \\langle x | \\otimes I_E) \\right), \\quad \\forall x \\in \\mathcal{X}, \\]\n\nfor some \\( Z \\) satisfying\n\n\\[ Z \\geq 0, \\quad \\Phi(Z) = I_E, \\]\n\nwhere\n\n\\[ \\Phi(Z) = \\sum_{x \\in \\mathcal{X}} \\left( (| x \\rangle \\otimes I_E) Z (| x \\rangle \\langle x | \\otimes I_E) \\right). \\]\n\nAnd the objective function can rewrite as\n\n\\[ \\sum_{x \\in \\mathcal{X}} \\text{tr}(M_x \\rho_E^x) = \\text{tr}(Z \\rho_{X E}). \\]\n\nTherefore the primal problem that gives \\( P_{\\text{guess}} \\) is\n\n\\[ P_{\\text{guess}}(X|E) = \\sup_Z \\, \\text{tr}(Z \\rho_{X E}) \\]\n\ns.t. \\( \\Phi(Z) = I_E \\),\n\n\\( Z \\geq 0 \\).\n\nIn the language of HW3 Problem 2, we are using \\( A = \\rho_{X E}, B = I_E \\).", "question": "### (d) Use the previous questions to give a semidefinite program in primal form whose optimum is \\( P_{\\text{guess}}(X|E) \\). That is, specify the map \\( \\Phi \\) and matrices \\( A \\) and \\( B \\) that define the SDP." }, { "context": "Since\n\n\\[ \\Phi(Z) = \\sum_{x \\in \\mathcal{X}} (|x\\rangle \\otimes I_E) Z ( \\langle x| \\otimes I_E) \\quad \\forall Z, \\]\n\nwe have\n\n\\[ \\Phi^*(Y_E) = \\sum_{x \\in \\mathcal{X}} (|x\\rangle \\otimes I_E) Y_E ( \\langle x| \\otimes I_E) \\]\n\n\\[ = \\sum_{x \\in \\mathcal{X}} (|x\\rangle \\langle x| \\otimes Y_E) \\]\n\n\\[ = \\left( \\sum_{x \\in \\mathcal{X}} |x\\rangle \\langle x| \\right) \\otimes Y_E \\]\n\n\\[ = I_X \\otimes Y_E, \\quad \\forall Y_E. \\]", "question": "### (e) Show that the map \\( \\Phi^* \\) associated to \\( \\Phi \\) is such that \\( \\Phi^*(Y) = I_X \\otimes Y \\), for any matrix \\( Y \\) defined over system \\( E \\) (remember the definition of \\( \\Phi^* \\) from \\( \\Phi \\) given in Homework 3, Problem 1)." }, { "context": "Recall that the dual problem is\n\n\\[ \\inf_Y \\, \\text{tr}(B Y) \\]\n\ns.t. \\( \\Phi^*(Y) \\geq A \\),\n\n\\( Y = Y^\\dagger \\).\n\nSince we are using \\( A = \\rho_{X E}, B = I_E \\), thus the dual problem becomes\n\n\\[ \\inf_Y \\, \\text{tr}(Y) \\]\n\ns.t. \\( I_X \\otimes Y \\geq \\rho_{X E} \\),\n\n\\( Y = Y^\\dagger \\).\n\nMoreover, if \\( I_X \\otimes Y \\geq \\rho_{X E} \\), given any \\( |\\phi\\rangle_E \\in \\mathcal{H}_E \\), we have\n\n\\[ \\langle \\phi|_E (Y - \\rho_E^P) |\\phi\\rangle_E = \\langle \\phi| Y |\\phi\\rangle_E - \\langle \\phi| \\rho_E^P |\\phi\\rangle_E \\]\n\n\\[ = (\\langle x| \\otimes \\langle \\phi|_E) (I_X \\otimes Y) (|x\\rangle \\otimes |\\phi\\rangle_E) - (\\langle x| \\otimes \\langle \\phi|_E) \\left( \\sum_{x' \\in \\mathcal{X}} |x'\\rangle \\langle x'| \\otimes \\rho_E^P \\right) (|x\\rangle \\otimes |\\phi\\rangle_E) \\]\n\n\\[ = (\\langle x| \\otimes \\langle \\phi|_E) (I_X \\otimes Y) (|x\\rangle \\otimes |\\phi\\rangle_E) - (\\langle x| \\otimes \\langle \\phi|_E) \\rho_{X E} (|x\\rangle \\otimes |\\phi\\rangle_E) \\]\n\n\\[ = (\\langle x| \\otimes \\langle \\phi|_E) (I_X \\otimes Y - \\rho_{X E}) (|x\\rangle \\otimes |\\phi\\rangle_E) \\]\n\n\\[ \\geq 0, \\]\n\nthus \\( Y \\geq \\rho_E^P \\), \\( \\forall x \\in \\mathcal{X} \\). Conversely, if \\( Y \\geq \\rho_E^P \\), \\( \\forall x \\in \\mathcal{X} \\), then\n\n\\[ I_X \\otimes Y = \\sum_{x \\in \\mathcal{X}} |x\\rangle \\langle x| \\otimes Y \\geq \\sum_{x \\in \\mathcal{X}} |x\\rangle \\langle x| \\otimes \\rho_E^P = \\rho_{X E}. \\]\n\nTherefore we have\n\n\\[ I_X \\otimes Y \\geq \\rho_{XE} \\iff Y \\geq \\rho_x^E, \\ \\forall x \\in \\mathcal{X}, \\]\n\nthe dual problem is more explicitly as\n\n\\[ \\inf_Y \\ \\text{tr}(Y) \\]\n\n\\[ \\text{s.t.} \\ Y \\geq \\rho_x, \\ \\forall x \\in \\mathcal{X}, \\]\n\n\\[ Y = Y^\\dagger. \\]", "question": "### (f) Write the dual program to your SDP explicitly." }, { "context": "Since the primal SDP problem is strictly feasible and the objective function is bounded above by 1, the problem has strong duality. That is, the supremum of the primal problem and the infimum of the dual problem are the same, i.e. we have\n\n\\[ P_{\\text{guess}}(X|E) = \\inf_Y \\ \\text{tr}(Y) \\]\n\n\\[ \\text{s.t.} \\ Y \\geq \\rho_x, \\ \\forall x \\in \\mathcal{X}, \\]\n\n\\[ Y = Y^\\dagger, \\]\n\nwhich is what we want to conclude.", "question": "### (g) Conclude that the guessing probability satisfies\n\n\\[ P_{\\text{guess}}(X|E) = \\inf_{\\sigma} \\text{Tr}(\\sigma), \\]\n\nwhere the infimum is taken over all matrices \\( \\sigma \\) defined on system \\( E \\) such that \\( \\sigma \\geq \\rho_x \\) for all \\( x \\in \\mathcal{X} \\).\n\nIn the final two parts of this problem, we use the previous parts (whose results you may use even if you didn’t prove them) to compute the min-entropy of cq states given as the tensor product of many copies of the same state." }, { "context": "Given that \\( \\sigma \\geq \\rho_x, \\ \\forall x \\in \\mathcal{X}, \\) we have\n\n\\[ P_{\\text{guess}}(X|E) = \\sup_{M_x} \\sum_{x \\in \\mathcal{X}} \\text{tr}(M_x \\rho_x) \\leq \\sup_{M_x} \\sum_{x \\in \\mathcal{X}} \\text{tr}(M_x \\sigma) = \\sup_{M_x} \\text{tr}\\left( \\left( \\sum_{x \\in \\mathcal{X}} M_x \\right) \\sigma \\right) = \\text{tr}(I_E \\sigma) = \\text{tr}(\\sigma). \\]", "question": "### (h) Show that for any \\( \\sigma \\) such that \\( \\sigma \\geq \\rho_x \\) we have \\( P_{\\text{guess}}(X|E) \\leq \\text{Tr}(\\sigma) \\). [Hint: remember Exercise 2 from Homework 3...]" }, { "context": "Suppose that\n\n\\[ \\tau_{X_1 E_1} = \\sum_{x \\in \\mathcal{X}_1} |x\\rangle \\langle x| \\otimes \\tau_x^{E_1}, \\]\n\nthen\n\n\\[ \\rho_{XE} = \\sigma_{X_1 E_1} = \\sum_{x_1, x_2, \\ldots, x_n \\in \\mathcal{X}_1} \\left( |x_1\\rangle \\langle x_1| \\otimes \\tau_{x_1}^{E_1} \\right) \\otimes \\left( |x_2\\rangle \\langle x_2| \\otimes \\tau_{x_2}^{E_2} \\right) \\otimes \\cdots \\otimes \\left( |x_n\\rangle \\langle x_n| \\otimes \\tau_{x_n}^{E_n} \\right) \\]\n\n\\[ = \\sum_{x_1, x_2, \\ldots, x_n \\in \\mathcal{X}_1} \\left( |x_1 x_2 \\ldots x_n \\rangle \\langle x_1 x_2 \\ldots x_n| \\right) \\otimes \\left( \\tau_{x_1}^{E_1} \\otimes \\tau_{x_2}^{E_2} \\otimes \\cdots \\otimes \\tau_{x_n}^{E_n} \\right) \\]\n\n\\[ = \\sum_{x \\in \\mathcal{X}} |x\\rangle \\langle x| \\otimes \\rho_x^E, \\]\n\nwhere\n\n\\[ \\mathcal{X} = \\{ x = (x_1, x_2, \\ldots, x_n) : x_i \\in \\mathcal{X}_1, \\ i = 1, 2, \\ldots, n \\}, \\]\n\n\\[ \\rho_x^E = \\tau_{x_1}^{E_1} \\otimes \\tau_{x_2}^{E_2} \\otimes \\cdots \\otimes \\tau_{x_n}^{E_n}, \\ \\forall x \\in \\mathcal{X}. \\]\n\nUsing the previous results, we have\n\n\\[ P_{\\text{guess}}(X_1|E_1) = \\sup_{\\{M_x\\} \\in \\mathcal{P}_1} \\sum_{x \\in \\mathcal{X}_1} \\text{tr}(M_x \\tau_x^{E_1}) = \\inf_{\\sigma \\in \\Omega_1} \\text{tr}(\\sigma). \\]\n\n\\[ P_{\\text{guess}}(X|E) = \\sup_{\\{M_x\\} \\in P} \\sum_{x \\in \\mathcal{X}} \\text{tr}(M_x \\rho_x^E) = \\inf_{\\sigma \\in \\Omega} \\text{tr}(\\sigma), \\]\n\nwhere\n\n\\[ \\begin{aligned} P_1 &= \\{\\{M_x^1\\} : \\{M_x^1\\} \\text{ is a POVM over } \\mathcal{H}_{E_1}\\}, \\\\ \\Omega_1 &= \\{\\sigma : \\sigma \\geq \\tau_x^{E_1}, \\forall x \\in \\mathcal{X}_1\\}, \\\\ P &= \\{\\{M_x\\} : \\{M_x\\} \\text{ is a POVM over } \\mathcal{H}_E\\}, \\\\ \\Omega &= \\{\\sigma : \\sigma \\geq \\rho_x^E, \\forall x \\in \\mathcal{X}\\}. \\end{aligned} \\]\n\nDefine\n\n\\[ \\begin{aligned} \\tilde{P} &= \\{\\{M_x\\} = \\{M_{x_1} \\otimes M_{x_2} \\otimes \\cdots \\otimes M_{x_n}\\} : \\{M_{x_i}\\} \\in P_1, i = 1, 2, \\ldots, n\\}, \\\\ \\tilde{\\Omega} &= \\{\\sigma = \\sigma_1 \\otimes \\sigma_2 \\otimes \\cdots \\otimes \\sigma_n : \\sigma_i \\in \\Omega_1, i = 1, 2, \\ldots, n\\}, \\end{aligned} \\]\n\nthen it's easy to check that\n\n\\[ \\tilde{P} \\subset P, \\quad \\tilde{\\Omega} \\subset \\Omega, \\]\n\nand thus we have\n\n\\[ \\begin{aligned} P_{\\text{guess}}(X|E) &= \\sup_{\\{M_x\\} \\in P} \\sum_{x \\in \\mathcal{X}} \\text{tr}(M_x \\rho_x^E) \\geq \\sup_{\\{M_x\\} \\in \\tilde{P}} \\sum_{x \\in \\mathcal{X}} \\text{tr}(M_x \\rho_x^E), \\\\ P_{\\text{guess}}(X|E) &= \\inf_{\\sigma \\in \\tilde{\\Omega}} \\text{tr}(\\sigma) \\leq \\inf_{\\sigma \\in \\Omega} \\text{tr}(\\sigma). \\end{aligned} \\]\n\nBut on the other hand, we have\n\n\\[ \\begin{aligned} \\sup_{\\{M_x\\} \\in \\tilde{P}} \\sum_{x \\in \\mathcal{X}} \\text{tr}(M_x \\rho_x^E) &= \\sup_{\\{M_x\\} \\in \\tilde{P}} \\sum_{x \\in \\mathcal{X}} \\text{tr}((M_{x_1} \\otimes M_{x_2} \\otimes \\cdots \\otimes M_{x_n})(\\rho_{x_1}^{E_1} \\otimes \\rho_{x_2}^{E_2} \\otimes \\cdots \\otimes \\rho_{x_n}^{E_n})) \\\\ &= \\sup_{\\{M_x\\} \\in \\tilde{P}} \\sum_{x \\in \\mathcal{X}} \\left( \\prod_{i=1}^n \\text{tr}(M_{x_i}^{E_i} \\rho_{x_i}^{E_i}) \\right) \\\\ &= \\sup_{\\{M_x\\} \\in \\tilde{P}} \\prod_{i=1}^n \\left( \\sum_{x_i \\in \\mathcal{X}_i} \\text{tr}(M_{x_i}^{E_i} \\rho_{x_i}^{E_i}) \\right) \\\\ &= \\prod_{i=1}^n \\left( \\sup_{\\{M_{x_i}\\} \\in P_1} \\sum_{x_i \\in \\mathcal{X}_i} \\text{tr}(M_{x_i}^{E_i} \\rho_{x_i}^{E_i}) \\right) \\\\ &= (P_{\\text{guess}}(X_1|E_1))^n, \\end{aligned} \\]\n\n\\[ \\begin{aligned} \\inf_{\\sigma \\in \\tilde{\\Omega}} \\text{tr}(\\sigma) &= \\inf_{\\sigma \\in \\tilde{\\Omega}} \\text{tr}(\\sigma_1 \\otimes \\sigma_2 \\otimes \\cdots \\otimes \\sigma_n) \\\\ &= \\inf_{\\sigma \\in \\tilde{\\Omega}} \\prod_{i=1}^n \\text{tr}(\\sigma_i) \\\\ &= \\prod_{i=1}^n \\left( \\inf_{\\sigma_i \\in \\Omega_1} \\text{tr}(\\sigma_i) \\right) \\\\ &= \\left( \\inf_{\\sigma \\in \\Omega_1} \\text{tr}(\\sigma) \\right)^n \\\\ &= (P_{\\text{guess}}(X_1|E_1))^n. \\end{aligned} \\]\n\nThus we come to\n\n\\[ \\begin{aligned} P_{\\text{guess}}(X|E) &= \\sup_{\\{M_x\\} \\in \\mathcal{P}} \\sum_{x \\in \\mathcal{X}} \\text{tr}(M_x \\rho_x^E) \\geq \\sup_{\\{M_x\\} \\in \\mathcal{P}} \\sum_{x \\in \\mathcal{X}} \\text{tr}(M_x \\rho_x^E) = (P_{\\text{guess}}(X_1|E_1))^n, \\\\ P_{\\text{guess}}(X|E) &= \\inf_{\\sigma \\in \\mathcal{E}(\\mathcal{H})} \\text{tr}(\\sigma) \\leq \\inf_{\\sigma \\in \\mathcal{E}(\\mathcal{H})} \\text{tr}(\\sigma) = (P_{\\text{guess}}(X_1|E_1))^n, \\end{aligned} \\]\n\n\\[ \\implies P_{\\text{guess}}(X|E) \\geq (P_{\\text{guess}}(X_1|E_1))^n \\geq P_{\\text{guess}}(X|E), \\]\n\n\\[ \\implies P_{\\text{guess}}(X|E) = (P_{\\text{guess}}(X_1|E_1))^n, \\]\n\nand therefore\n\n\\[ H_{\\text{min}}(X|E)_p = -\\log P_{\\text{guess}}(X|E) = -\\log (P_{\\text{guess}}(X_1|E_1))^n = -n \\log P_{\\text{guess}}(X_1|E_1) = n H_{\\text{min}}(X_1|E_1)_r. \\]", "question": "### (i) Suppose that \\( \\rho_{XE} = \\tau_{X_1E_1}^{\\otimes n} \\), where \\( \\tau_{X_1E_1} \\) is a cq-state. That is, \\( \\rho_{XE} \\) is formed of \\( n \\) identical copies of the same state. Show that \\( H_{\\min}(X|E) = n H_{\\min}(X_1|E_1)_{\\tau} \\), where we have used the subscripts \\( \\rho \\) and \\( \\tau \\) to remind ourselves of which states we compute the min-entropy. [Hint: consider the solutions for the primal and dual SDP from the previous problem for a single instance \\( \\sigma_{X_1E_1} \\). Use these solutions to construct matching solutions for the primal and dual SDP associated with the cq state \\( \\rho \\).]" } ]

[ { "context": "By definition \\( H_{\\text{min}}(X | E) = -\\log P_{\\text{guess}}(X | E) \\) and \\( H_{\\text{min}}(X) = -\\log \\max_x p_x = -\\log P_{\\text{guess}}(X) \\) so we want to show \\(-\\log P_{\\text{guess}}(X | E) \\ge -\\log P_{\\text{guess}}(X) - \\log |E| = -\\log (P_{\\text{guess}}(X)/|E|)\\). Since \\(-\\log x\\) is monotonically decreasing, we need to show \\( P_{\\text{guess}}(X | E) \\le P_{\\text{guess}}(X)/|E| \\).", "question": "### (a) Write out what we want to show in terms of the guessing probability \\( P_{\\text{guess}}(X|E) \\) using the definition of the min-entropy." }, { "context": "Let \\( A \\ge 0, B \\ge 0 \\). We write the eigendecomposition \\( B = \\sum_i \\lambda_i(B) |u_i\\rangle \\langle u_i| \\). Then using the linearity of trace we have\n\n\\[ \\text{Tr}(AB) = \\text{Tr} \\left( A \\sum_i \\lambda_i(B) |u_i\\rangle \\langle u_i| \\right) = \\sum_i \\lambda_i(B) \\text{Tr}(A |u_i\\rangle \\langle u_i|) \\]\n\n\\[ \\le \\lambda_{\\max}(B) \\sum_i \\text{Tr}(A |u_i\\rangle \\langle u_i|) \\]\n\n\\[ = \\lambda_{\\max}(B) \\text{Tr} \\left( A \\sum_i |u_i\\rangle \\langle u_i| \\right) \\]\n\n\\[ = \\lambda_{\\max}(B) \\text{Tr}(A \\mathbb{I}) \\]\n\n\\[ = \\lambda_{\\max}(B) \\text{Tr}(A) \\]\n\nwhere the inequality step is because \\(\\text{Tr}(A |u_i\\rangle \\langle u_i|) = \\langle u_i| A |u_i\\rangle \\ge 0\\) since \\(A \\ge 0\\).", "question": "### (b) It will be useful to establish the following fact. Suppose given two Hermitian matrices \\( A \\) and \\( B \\), which are positive semidefinite: \\( A \\geq 0 \\) and \\( B \\geq 0 \\). Show that \n\\[ \\text{Tr}(AB) \\leq \\lambda_{\\text{max}}(B) \\text{Tr}(A), \\]\nwhere \\( \\lambda_{\\text{max}}(B) \\) is the largest eigenvalue of \\( B \\)." }, { "context": "Let \\(\\{M_x\\}\\) be a POVM and \\(\\rho_x^E\\) be a quantum state. Then \\(M_x \\ge 0\\) and since \\(\\rho_x^E\\) is a density matrix, we have \\(\\rho_x^E \\ge 0\\) so \\(\\lambda_i(\\rho_x^E) \\ge 0\\) so \\(\\sum_i \\lambda_i(\\rho_x^E) = \\text{Tr}(\\rho_x^E) = 1 \\implies \\lambda_{\\max}(\\rho_x^E) \\le 1\\). Then applying part (b) gives\n\n\\[ \\text{Tr}(M_x \\rho_x^E) \\le \\lambda_{\\max}(\\rho_x^E) \\text{Tr}(M_x) \\le \\text{Tr}(M_x). \\]", "question": "### (c) Use this fact to show that for any POVM \\( \\{M_x\\} \\) and any quantum state \\( \\rho_E^x \\) we have \n\\[ \\text{Tr}(M_x \\rho_E^x) \\leq \\text{Tr}(M_x). \\]" }, { "context": "Since \\( p_x \\geq 0 \\), applying part (c),\n\n\\[ \\sum_x p_x \\text{Tr} \\left( M_x \\rho_E^x \\right) \\leq \\sum_x p_x \\text{Tr} (M_x) \\leq \\left( \\max_x p_x \\right) \\sum_x \\text{Tr} (M_x) \\]\n\n\\[ = \\left( \\max_x p_x \\right) \\text{Tr} \\left( \\sum_x M_x \\right) \\]\n\n\\[ = \\left( \\max_x p_x \\right) \\text{Tr} (\\mathbb{I}_E) \\]\n\n\\[ = \\left( \\max_x p_x \\right) |E|. \\]\n\nThen we have\n\n\\[ P_{\\text{guess}}(X | E) = \\max_{\\{M_x\\}} \\sum_x p_x \\text{Tr} (M_x \\rho_E^x) \\leq \\left( \\max_x p_x \\right) |E|. \\]\n\nSince \\( - \\log x \\) is monotonically decreasing, we have\n\n\\[ H_{\\min}(X | E) = - \\log P_{\\text{guess}}(X | E) \\geq - \\log \\left( \\left( \\max_x p_x \\right) |E| \\right) = - \\log \\left( \\max_x p_x \\right) - \\log |E| \\]\n\n\\[ = H_{\\min}(X) - \\log |E|. \\]", "question": "### (d) Using this trick together with what you know about POVMs, show that \n\\[ H_{\\min}(X|E) \\geq H_{\\min}(X) - \\log |E|. \\]" } ]

[ { "context": "Since \\(\\{A_x^0, A_x^1\\}\\) is a valid POVM, we have\n\n\\[ 0 \\leq A_x^0 \\leq I_{A_x}, \\quad 0 \\leq A_x^1 \\leq I_{A_x}, \\]\n\n\\[ \\implies -I_{A_x} \\leq A_x = A_x^0 - A_x^1 \\leq I_{A_x}, \\]\n\n\\[ \\implies r(A_x) \\leq 1, \\]\n\nwhere \\(r(A_x)\\) denotes the spectral radius of \\(A_x\\). Since \\(A_x^0, A_x^1\\) are Hermitian, \\(A_x\\) is also Hermitian, thus the largest singular value of \\(A_x\\) equals \\(r(A_x)\\). Therefore\n\n\\[ \\|A_x\\| = r(A_x) \\leq 1. \\]\n\nThe same argument also works for \\(B_y\\), i.e. \\(\\|B_y\\| \\leq 1\\).", "question": "### (a) Show that if \\( \\{A_a^0, A_a^1\\} \\) is a valid POVM then \\( \\|A_x\\| \\leq 1 \\) (where as usual \\( \\|\\cdot\\| \\) is the operator norm, the largest singular value). Similarly for \\( B_y \\)." }, { "context": "Suppose that \\(A_x\\) is diagonalized in a basis \\(\\{\\phi_i\\}\\), i.e.\n\n\\[ A_x|\\phi_i\\rangle = \\lambda_i|\\phi_i\\rangle, \\quad i = 0, 1, \\ldots, d_a - 1, \\]\n\n\\[ \\langle \\phi_i|\\phi_j \\rangle = \\delta_{ij}, \\quad i, j = 0, 1, \\ldots, d_a - 1, \\]\n\nwhere \\(d_a\\) is the dimension of \\(\\mathcal{H}_A\\), and \\(\\lambda_i, i = 0, 1, \\ldots, d_a - 1\\), are all eigenvalues of \\(A_x\\). Then\n\n\\[ \\lambda_i^2 \\leq \\|A_x\\|^2 \\leq 1, \\quad i = 0, 1, \\ldots, d_a - 1, \\]\n\nsince \\(A_x\\) is Hermitian. We can always write \\(|\\psi\\rangle_{AB}\\) as\n\n\\[ |\\psi\\rangle_{AB} = \\sum_{i=0}^{d_A-1} \\sum_{j=0}^{d_B-1} \\alpha_{ij} |\\phi_i\\rangle_A |j\\rangle_B. \\]\n\nThen we have\n\n\\[ \\| |u_x\\rangle \\|^2 = \\langle u_x|u_x \\rangle = \\langle \\psi|_{AB}(A_x \\otimes I_B)(A_x \\otimes I_B)|\\psi\\rangle_{AB} = \\sum_{i,j} \\sum_{k,l} \\alpha_{ij} \\alpha_{kl} \\langle \\phi_i|A_x^2|\\phi_k \\rangle_A \\langle j|l \\rangle_B = \\sum_{i,j} \\sum_{k,l} \\alpha_{ij} \\alpha_{kl} \\lambda_k \\lambda_k \\langle \\phi_i|\\phi_k \\rangle_A \\langle j|l \\rangle_B = \\sum_{i,j} |\\alpha_{ij}|^2 \\lambda_i^2 \\leq \\sum_{i,j} |\\alpha_{ij}|^2 = \\| |\\psi\\rangle_{AB} \\|^2 = 1, \\]\n\nthat is \\(\\| |u_x\\rangle \\| \\leq 1\\). Similarly, we can also prove that \\(\\| |v_y\\rangle \\| \\leq 1\\).", "question": "### (b) For \\( x \\in \\{0, 1\\} \\) define \\( |u_x\\rangle = A_x \\otimes I_B |\\psi\\rangle_{AB} \\), and for \\( y \\in \\{0, 1\\} \\) define \\( |v_y\\rangle = I_A \\otimes B_y |\\psi\\rangle_{AB} \\). Give a bound on the Euclidean norms \\( \\|u_x\\|, \\|v_y\\| \\)." }, { "context": "By direct calculation, we have\n\n\\[ \\langle u_x|v_y \\rangle = \\langle \\psi|_{AB}(A_x \\otimes I_B)(I_A \\otimes B_y)|\\psi\\rangle_{AB} = \\langle \\psi|_{AB}(A_x \\otimes B_y)|\\psi\\rangle_{AB} = \\langle \\psi|_{AB}(A_x^0 \\otimes B_y^0)|\\psi\\rangle_{AB} - \\langle \\psi|_{AB}(A_x^1 \\otimes B_y^0)|\\psi\\rangle_{AB} - \\langle \\psi|_{AB}(A_x^0 \\otimes B_y^1)|\\psi\\rangle_{AB} + \\langle \\psi|_{AB}(A_x^1 \\otimes B_y^1)|\\psi\\rangle_{AB}. \\]\n\nThen for \\((x, y) \\neq (1, 1)\\),\n\n\\[ \\langle u_x|v_y \\rangle = 2(\\langle \\psi|_{AB}(A_x^0 \\otimes B_y^0)|\\psi\\rangle_{AB} + 2(\\langle \\psi|_{AB}(A_x^1 \\otimes B_y^1)|\\psi\\rangle_{AB} - \\langle \\psi|_{AB}(A_x^0 \\otimes B_y^1)|\\psi\\rangle_{AB} - \\langle \\psi|_{AB}(A_x^1 \\otimes B_y^0)|\\psi\\rangle_{AB} = 2\\text{Tr}((A_x^0 \\otimes B_y^0)|\\psi\\rangle_{AB}) + 2\\text{Tr}((A_x^1 \\otimes B_y^1)|\\psi\\rangle_{AB}) - \\langle \\psi|_{AB}((A_x^0 + A_x^1) \\otimes (B_y^0 + B_y^1))|\\psi\\rangle_{AB} = 2p(0, 0|x, y) + 2p(1, 1|x, y) - 1, \\]\n\n\\[ p(0, 0|x, y) + p(1, 1|x, y) = \\frac{1}{2}(1 + \\langle u_x|v_y \\rangle). \\]\n\nfor \\((x, y) = (1, 1)\\),\n\n\\[ \\begin{aligned} \\langle u_{x} | v_{y} \\rangle &= \\langle \\psi |_{AB} (A_{0}^{x} \\otimes B_{0}^{y}) | \\psi \\rangle_{AB} + \\langle \\psi |_{AB} (A_{1}^{x} \\otimes B_{0}^{y}) | \\psi \\rangle_{AB} \\\\ &\\quad + \\langle \\psi |_{AB} (A_{0}^{x} \\otimes B_{1}^{y}) | \\psi \\rangle_{AB} + \\langle \\psi |_{AB} (A_{1}^{x} \\otimes B_{1}^{y}) | \\psi \\rangle_{AB} \\\\ &\\quad - 2 \\langle \\psi |_{AB} (A_{1}^{x} \\otimes B_{0}^{y}) | \\psi \\rangle_{AB} - 2 \\langle \\psi |_{AB} (A_{0}^{x} \\otimes B_{1}^{y}) | \\psi \\rangle_{AB} \\\\ &= \\langle \\psi |_{AB} ((A_{0}^{x} + A_{1}^{x}) \\otimes (B_{0}^{y} + B_{1}^{y})) | \\psi \\rangle_{AB} \\\\ &\\quad - 2 \\text{Tr} ((A_{1}^{x} \\otimes B_{0}^{y}) \\rho) \\langle \\psi |_{AB} \\rangle - 2 \\text{Tr} ((A_{0}^{x} \\otimes B_{1}^{y}) \\rho) \\langle \\psi |_{AB} \\rangle \\\\ &= \\langle \\psi |_{AB} (\\mathbb{I}_{A} \\otimes \\mathbb{I}_{B}) | \\psi \\rangle_{AB} - 2p(1, 0 | x, y) - 2p(0, 1 | x, y), \\end{aligned} \\]\n\n\\[ p(1, 0 | x, y) + p(0, 1 | x, y) = \\frac{1}{2} (1 - \\langle u_{x} | v_{y} \\rangle). \\]\n\nFinally we have\n\n\\[ \\begin{aligned} p_{\\text{succ}} &= \\sum_{x, y \\in \\{0, 1\\}} p(x, y) \\sum_{a, b \\in \\{0, 1\\}} p(a, b | x, y) \\\\ &= \\frac{1}{4} \\big( p(0, 0 | 0, 0) + p(1, 1 | 0, 0) + p(0, 0 | 0, 1) + p(1, 1 | 0, 1) \\\\ &\\quad + p(0, 0 | 1, 0) + p(1, 1 | 1, 0) + p(0, 0 | 1, 1) + p(1, 1 | 1, 1) \\big) \\\\ &= \\frac{1}{8} \\big( 1 + (u_{0} | v_{0}) + 1 + (u_{1} | v_{0}) + 1 + (u_{0} | v_{1}) + 1 + 1 - (u_{1} | v_{1}) \\big) \\\\ &= \\frac{1}{2} + \\frac{1}{8} \\big( (u_{0} | v_{0}) + (u_{1} | v_{0}) + (u_{0} | v_{1}) - (u_{1} | v_{1}) \\big). \\end{aligned} \\]", "question": "### (c) Show that the success probability of the quantum strategy in the CHSH game can be expressed as \n\\[ p_{\\text{succ}} = \\frac{1}{2} + \\frac{1}{8} \\left( \\langle u_0 | v_0 \\rangle + \\langle u_1 | v_0 \\rangle + \\langle u_0 | v_1 \\rangle - \\langle u_1 | v_1 \\rangle \\right). \\]" }, { "context": "Let \\(c = \\max \\{ \\| r_{0} \\|, \\| r_{1} \\|, \\| s_{0} \\|, \\| s_{1} \\| \\}\\). Notice that\n\n\\[ \\begin{aligned} \\| r_{0} \\rangle + \\| r_{1} \\rangle - \\| r_{1} \\rangle \\|^{2} &= \\langle (r_{0} | + (r_{1} |) (| r_{0} \\rangle + | r_{1} \\rangle) + (r_{0} | - (r_{1} |) (| r_{0} \\rangle - | r_{1} \\rangle) \\rangle \\\\ &= 2 \\langle r_{0} | r_{0} \\rangle + 2 \\langle r_{0} | r_{0} \\rangle \\\\ &= 2 \\| r_{0} \\|^{2} + 2 \\| r_{1} \\|^{2} \\\\ &\\leq 4c^{2}, \\end{aligned} \\]\n\nthus\n\n\\[ \\| r_{0} \\rangle + \\| r_{1} \\rangle + \\| r_{0} \\rangle - \\| r_{1} \\rangle \\| \\leq \\sqrt{2} \\left( \\| r_{0} \\rangle + \\| r_{1} \\rangle \\|^{2} + \\| r_{0} \\rangle - \\| r_{1} \\rangle \\| \\right)^{\\frac{1}{2}} \\leq 2 \\sqrt{2} c. \\]\n\nThen we have\n\n\\[ \\begin{aligned} | \\langle r_{0} | s_{0} \\rangle + \\langle r_{1} | s_{1} \\rangle - \\langle r_{1} | s_{1} \\rangle | &\\leq | \\langle r_{0} | s_{0} \\rangle + \\langle r_{0} | s_{0} \\rangle | + | \\langle r_{1} | s_{1} \\rangle - \\langle r_{1} | s_{1} \\rangle | \\\\ &= | \\langle (r_{0} | + (r_{1} |) s_{0} \\rangle | + | \\langle (r_{0} | - (r_{1} |) s_{1} \\rangle | \\\\ &\\leq \\| r_{0} \\rangle + \\| r_{1} \\rangle \\| \\| s_{0} \\rangle + \\| r_{0} \\rangle - \\| r_{1} \\rangle \\| \\| s_{1} \\rangle \\| \\\\ &\\leq c \\| r_{0} \\rangle + \\| r_{1} \\rangle \\| + c \\| r_{0} \\rangle - \\| r_{1} \\rangle \\| \\\\ &\\leq 2 \\sqrt{2} c. \\end{aligned} \\]", "question": "### (d) Show that for any vectors \\( |r_0\\rangle, |r_1\\rangle, |s_0\\rangle, |s_1\\rangle \\), the inequality \n\\[ |\\langle r_0 | s_0 \\rangle + \\langle r_1 | s_0 \\rangle + \\langle r_0 | s_1 \\rangle - \\langle r_1 | s_1 \\rangle| \\leq 2\\sqrt{2} \\max \\{ \\|r_0\\|, \\|r_1\\|, \\|s_0\\|, \\|s_1\\| \\} \\]\nholds." }, { "context": "Using the result of (b), we have\n\n\\[ c = \\max\\{\\| |u_0\\rangle \\|, \\| |u_1\\rangle \\|, \\| |s_0\\rangle \\|, \\| |s_1\\rangle \\|\\} \\leq 1. \\]\n\nThen using the result of (d), we have\n\n\\[ \\left| \\langle u_0 | v_0 \\rangle + \\langle u_1 | v_0 \\rangle + \\langle u_0 | v_1 \\rangle - \\langle u_1 | v_1 \\rangle \\right| \\leq 2 \\sqrt{2} c \\leq 2 \\sqrt{2}. \\]\n\nFinally using the result of (c), we have\n\n\\[ p_{\\text{succ}} = \\frac{1}{2} + \\frac{1}{8} \\left( \\langle u_0 | v_0 \\rangle + \\langle u_1 | v_0 \\rangle + \\langle u_0 | v_1 \\rangle - \\langle u_1 | v_1 \\rangle \\right) \\]\n\n\\[ \\leq \\frac{1}{2} + \\frac{1}{8} \\left| \\langle u_0 | v_0 \\rangle + \\langle u_1 | v_0 \\rangle + \\langle u_0 | v_1 \\rangle - \\langle u_1 | v_1 \\rangle \\right| \\]\n\n\\[ \\leq \\frac{1}{2} + \\frac{\\sqrt{2}}{4} \\]\n\n\\[ = \\cos^2 \\frac{\\pi}{8}. \\]", "question": "### (e) Conclude that \\( p_{\\text{succ}} \\leq \\cos^2(\\pi/8) \\)." } ]

[ { "context": "If Eve knows both \\(\\theta\\) and \\(U_A\\) then she can guess perfectly in all cases by applying \\(U_E = U_A\\), the element-wise complex conjugation of \\(U_A\\), and measuring in the same basis as Alice. This is because, from Homework 4, we know that \\(U \\otimes I | \\phi^+ \\rangle = I \\otimes U^T | \\phi^+ \\rangle\\), and to undo \\(U^T\\) Eve just applies \\((U^T)^\\dagger = \\overline{U}\\).", "question": "### (a) However, Alice wants to foil Eve so, before measuring, she first applies some unitary \\( U \\) to her qubit, and then measures. Of course Eve, being really smart, gets wind of this so she will know what unitary Alice has used before measuring. So they share the state\n\\[ |\\Phi_U\\rangle = (U_A \\otimes \\mathbb{I}_E) \\frac{1}{\\sqrt{2}} (|00\\rangle + |11\\rangle) \\]\nand Eve knows \\( \\theta \\) and \\( U \\). What would now be Eve's best guessing probability? (Tip: Eve has a quantum computer!)" }, { "context": "Alice should use the following strategy:\n\n1. If \\(\\theta = 0\\) then Alice applies \\(I\\) with probability \\(1/2\\), \\(X\\) with probability \\(1/2\\), or \\(Z\\) with probability zero before measuring.\n2. If \\(\\theta = 1\\) then Alice applies \\(I\\) with probability \\(1/2\\), \\(Z\\) with probability \\(1/2\\), or \\(X\\) with probability zero before measuring.\n\nWith this strategy, when \\(\\theta = 0\\), the shared state is either \\(\\frac{1}{\\sqrt{2}} (|00\\rangle + |11\\rangle)\\) or \\(\\frac{1}{\\sqrt{2}} (|01\\rangle + |10\\rangle)\\), each with probability \\(1/2\\). So if \\(x = 0\\), since Eve does not know which unitary Alice applied, Eve has the state \\(\\frac{1}{2} |0\\rangle (|0\\rangle + \\frac{1}{2} |1\\rangle)\\), and if \\(x = 1\\), Eve has the state \\(\\frac{1}{2} |1\\rangle (|1\\rangle + \\frac{1}{2} |0\\rangle)\\). Then Eve cannot distinguish between the two outcomes so at best she can guess randomly.\n\nSimilarly, when \\(\\theta = 1\\) the shared state is either \\(\\frac{1}{\\sqrt{2}} (|00\\rangle + |11\\rangle)\\) = \\(\\frac{1}{\\sqrt{2}} (|+ +\\rangle + |-\rangle)\\) or \\(\\frac{1}{\\sqrt{2}} (|+ -\\rangle + |-\rangle)\\), each with probability \\(1/2\\). So if \\(x = 0\\), Eve has the state \\(\\frac{1}{2} |+\\rangle (|+\\rangle + \\frac{1}{2} |-\rangle)\\), and if \\(x = 1\\), Eve has the state \\(\\frac{1}{2} |-\\rangle (|-\rangle + \\frac{1}{2} |+\\rangle)\\). Hence in both cases Eve's best strategy is to guess randomly so she will guess correctly with probability \\(1/2\\). Since the worst strategy Eve can use is a random guess, this strategy makes Eve's guessing probability the lowest possible.", "question": "### (b) Consider now a scenario in which Eve doesn’t know the local unitary that Alice applies. Suppose that Alice can choose her unitary from a set of three unitaries (and assume that she always applies one unitary from this set before measuring). You should assume that, once Alice decides on a (probabilistic) strategy, this is fixed. Provide a strategy for Alice (including her choice of the set of three unitaries) that makes Eve’s guessing probability the lowest possible." }, { "context": "Alice should, for all \\(\\theta\\), first apply either \\(I\\) or \\( Y = XZ \\) each with probability \\(1/2\\). We will show that this achieves the same probability as in part (b).\n\nWith this strategy, when \\( \\theta = 0 \\), the shared state is either \\( \\frac{1}{\\sqrt{2}} (|00\\rangle + |11\\rangle) \\) or \\( \\frac{1}{\\sqrt{2}} (|01\\rangle + |10\\rangle) \\), each with probability 1/2. Then if \\( x = 0 \\), since Eve does not know which unitary Alice used, Eve has the state \\( \\frac{1}{2} |0\\rangle (\\frac{1}{2} |0\\rangle + \\frac{1}{2} |1\\rangle) \\), and if \\( x = 1 \\), Eve has the state \\( \\frac{1}{2} |1\\rangle (\\frac{1}{2} |1\\rangle + \\frac{1}{2} |0\\rangle) \\). Then Eve cannot distinguish between the two so at best she can guess randomly.\n\nSimilarly, when \\( \\theta = 1 \\) the shared state is either \\( \\frac{1}{\\sqrt{2}} (|00\\rangle + |11\\rangle) = \\frac{1}{\\sqrt{2}} (| + + \\rangle + | - - \\rangle) \\) or \\( \\frac{1}{\\sqrt{2}} (| + - \\rangle + | - + \\rangle) \\), each with probability 1/2. So if \\( x = 0 \\), Eve has the state \\( \\frac{1}{2} |+\\rangle (\\frac{1}{2} |+\\rangle + \\frac{1}{2} |-\\rangle) \\), and if \\( x = 1 \\), Eve has the state \\( \\frac{1}{2} |-\\rangle (\\frac{1}{2} |-\\rangle + \\frac{1}{2} |+\\rangle) \\). Hence in both cases Eve’s best strategy is to guess randomly so she will guess correctly with probability 1/2.", "question": "### (c) Suppose we restrict Alice’s set of possible unitaries to contain only two. Can she still make Eve’s guessing probability as low as in part (b)?" } ]

[ { "context": "We have \\( |\\Psi\\rangle |E\\rangle \\rightarrow \\alpha |0\\rangle |E_0\\rangle + \\beta |1\\rangle |E_1\\rangle \\) so\n\n\\[ |\\Psi\\rangle \\langle \\Psi| \\otimes |E\\rangle \\langle E| = |\\alpha|^2 |0\\rangle \\langle 0| \\otimes |E_0\\rangle \\langle E_0| + \\alpha \\beta^* |0\\rangle \\langle 1| \\otimes |E_0\\rangle \\langle E_1| + \\alpha^* \\beta |1\\rangle \\langle 0| \\otimes |E_1\\rangle \\langle E_0| + |\\beta|^2 |1\\rangle \\langle 1| \\otimes |E_1\\rangle \\langle E_1| \\]\n\nAssuming \\( \\langle E_0|E_1 \\rangle \\) is real, we have \\( \\langle E_0|E_1 \\rangle = \\langle E_1|E_0 \\rangle \\). Since \\( |E\\rangle \\), \\( |E_0\\rangle \\), and \\( |E_1\\rangle \\) are normalized, tracing out the environment gives\n\n\\[ |\\Psi\\rangle \\langle \\Psi| \\otimes \\text{Tr} (|E\\rangle \\langle E|) = |\\alpha|^2 |0\\rangle \\langle 0| + \\langle E_0|E_1 \\rangle (\\alpha \\beta^* |0\\rangle \\langle 1| + \\alpha^* \\beta |1\\rangle \\langle 0|) + |\\beta|^2 |1\\rangle \\langle 1| \\]\n\nDefine \\( p = \\frac{1 - \\langle E_0|E_1 \\rangle}{2} \\). We will show later that \\( p \\) is in fact a valid probability. Note that \\( Z|\\Psi\\rangle = \\alpha |0\\rangle - \\beta |1\\rangle \\). Then we have\n\n\\[ |\\Psi\\rangle \\langle \\Psi| \\otimes \\text{Tr} (|E\\rangle \\langle E|) = |\\alpha|^2 |0\\rangle \\langle 0| + (1 - 2p)(\\alpha \\beta^* |0\\rangle \\langle 1| + \\alpha^* \\beta |1\\rangle \\langle 0|) + |\\beta|^2 |1\\rangle \\langle 1| \\]\n\\[ = (1 - p) (|\\alpha|^2 |0\\rangle \\langle 0| + \\alpha \\beta^* |0\\rangle \\langle 1| + \\alpha^* \\beta |1\\rangle \\langle 0| + |\\beta|^2 |1\\rangle \\langle 1|) \\]\n\\[ + p (|\\alpha|^2 |0\\rangle \\langle 0| - \\alpha \\beta^* |0\\rangle \\langle 1| - \\alpha^* \\beta |1\\rangle \\langle 0| + |\\beta|^2 |1\\rangle \\langle 1|) \\]\n\\[ = (1 - p)|\\Psi\\rangle \\langle \\Psi| + pZ|\\Psi\\rangle \\langle \\Psi|Z \\]\n\nSo \\( |\\Psi\\rangle \\langle \\Psi| \\rightarrow (1 - p)|\\Psi\\rangle \\langle \\Psi| + pZ|\\Psi\\rangle \\langle \\Psi|Z \\). Note that \\( |E_i\\rangle \\langle E_i| \\geq 0 \\) since \\( \\langle u|E_i\\rangle \\langle E_i|u\\rangle = |\\langle u|E_i\\rangle|^2 \\geq 0 \\) for any \\( |u\\rangle \\) and \\( \\lambda_{\\max} (|E_i\\rangle \\langle E_i|) \\leq 1 \\) since \\( \\sum_i \\lambda_i (|E_i\\rangle \\langle E_i|) = \\text{Tr} (|E_i\\rangle \\langle E_i|) = 1 \\) and \\( \\lambda_i (|E_i\\rangle \\langle E_i|) \\geq 0 \\). Then by problem 2(b) we have\n\n\\[ |\\langle E_0|E_1 \\rangle|^2 = \\text{Tr} (|E_0\\rangle \\langle E_0| |E_1\\rangle \\langle E_1|) \\leq \\lambda_{\\max} (|E_1\\rangle \\langle E_1|) \\text{Tr} (|E_0\\rangle \\langle E_0|) \\leq 1 \\]\n\nThen we have \\( |\\langle E_0|E_1 \\rangle| \\leq 1 \\) which implies \\( 0 \\leq p \\leq 1 \\) so \\( p \\) is a valid probability.", "question": "### Suppose that this state undergoes “decoherence”, as described by the CPTP map\n\n\\[ |0\\rangle |E\\rangle \\mapsto |0\\rangle |E_0\\rangle , \\]\n\\[ |1\\rangle |E\\rangle \\mapsto |1\\rangle |E_1\\rangle , \\]\n\nwhere the states \\( |E\\rangle \\), \\( |E_0\\rangle \\) and \\( |E_1\\rangle \\) are normalized but not necessarily orthogonal. Show that the density matrix of the qubit evolves as\n\n\\[ |\\Psi\\rangle \\langle \\Psi| \\mapsto (1 - p) |\\Psi\\rangle \\langle \\Psi| + pZ |\\Psi\\rangle \\langle \\Psi| Z. \\]\n\nAssuming \\( \\langle E_0 | E_1 \\rangle \\) is real, express \\( p \\) in terms of \\( \\langle E_0 | E_1 \\rangle \\). This means that with the probability \\( 1 - p \\) the qubit is not affected by the environment and with probability \\( p \\) the qubit undergoes a phase-flip error." } ]

[ { "context": "(Due to Daniel Gu)\nLet \\( X \\) be the random variable which is the number of bits that Alice uses in total, and \\( X_i \\) be the number of bits that Alice uses at step \\( i \\) in the protocol. Then \\( X = \\sum_{i=1}^{n} X_i \\) and so by linearity of expectation\n\n\\[\n\\mathbb{E}[X] = \\mathbb{E}\\left[\\sum_{i=1}^{n} X_i\\right] = \\sum_{i=1}^{n} \\mathbb{E}[X_i] = \\frac{n}{2}\n\\]", "question": "### (a) How many bits of key will Alice use on average with the new protocol?" }, { "context": "(Due to Daniel Gu)\nThe scheme is certainly not correct: since Alice doesn’t send her random choices (random bits) to Bob but only the ciphertext \\( c \\), Bob has no idea which bits got XOR’d with the key and which got XOR’d with a random bit. No deterministic algorithm can decrypt the ciphertext accurately, since once we fix the ciphertext, key, and \\( \\text{DEC}(k, c) \\), we can choose our random bits such that our message does not match our decryption algorithm’s answer.\n\nHowever, the scheme is secure. The probability that the \\( i \\)-th bit of the message is 0 (over the random choices made by Alice and a uniformly random key distribution) given that the \\( i \\)-th bit of the ciphertext is \\( b \\) is 1/2, since with probability 1/2 we XOR \\( b \\) with the \\( i \\)-th bit of the key, which without knowledge of the key is equally likely to be 0 or 1, so it has a 1/2 chance of being 0 and producing 0, and with probability 1/2 we XOR it with a uniformly random bit, which also has a 1/2 chance of being \\( b \\). So given the ciphertext, the distribution of possible messages is the uniform distribution over all \\( n \\) bit messages, so the scheme is secure.", "question": "### (b) Is this protocol correct? Is it secure? Provide a proof of security or an attack scheme." } ]

[ { "context": "(Due to Alex Meiburg)\n\\[\n\\frac{1}{2} |0\\rangle \\langle 0| + \\frac{1}{2} |1\\rangle \\langle 1| = \\begin{bmatrix} \\frac{1}{2} & 0 \\\\ 0 & \\frac{1}{2} \\end{bmatrix}\n\\]", "question": "### (a) Give the density matrix \\( \\rho_0 \\) describing Bob’s state." }, { "context": "(Due to Alex Meiburg)\nNote that \\( \\rho_0 = \\frac{1}{2} I \\). The probability for any given state is given by\n\n\\[\n\\langle \\psi | \\rho_0 | \\psi \\rangle = \\langle \\psi | \\frac{1}{2} I | \\psi \\rangle = \\frac{1}{2} \\langle \\psi | \\psi \\rangle = \\frac{1}{2}\n\\]\n\n\\[\n\\langle \\psi | \\rho_0 | \\psi \\rangle = \\frac{1}{2}\n\\]\nSo that each measurement of a state gives a 50% probability of that occurring.", "question": "### (b) Suppose Bob measures \\( \\rho_0 \\) in the standard basis. What is the probability that the measurement results in \\( |0\\rangle \\)? \\( |1\\rangle \\)? What if Bob measures in the Hadamard basis?" }, { "context": "(Due to Alex Meiburg)\n\\[\n\\rho_0 = |+\\rangle \\langle +| = \\begin{bmatrix} \\frac{1}{2} & \\frac{1}{2} \\\\ \\frac{1}{2} & \\frac{1}{2} \\end{bmatrix}\n\\]\n\n\\[\n\\langle 0|\\rho_0|0\\rangle = \\frac{1}{2}, \\quad \\langle 1|\\rho_0|1\\rangle = \\frac{1}{2}\n\\]\n\n\\[\n\\langle +|\\rho_0|+\\rangle = \\langle +|+\\rangle \\langle +|+\\rangle = 1, \\quad \\langle -|\\rho_0|-\\rangle = \\langle -|+\\rangle \\langle +|-\\rangle = 0\n\\]\nSo that in the standard basis it is completely random, while in the Hadamard basis it is guaranteed \\(|+\\rangle\\).", "question": "### (c) Now say that the machine on Alice’s side is not noisy but simply misaligned: it consistently prepares qubits in the state \\( |+\\rangle \\). Again, what is the distribution of outcomes if Bob measures in the standard basis? In the Hadamard basis?" } ]

[ { "context": "(Due to Anish Thilagar)\nThis protocol is correct. Bob will receive the state \\(H^k |\\psi\\rangle\\). He can then apply \\(H^k\\) again, to get the qubit \\(H^{2k} |\\psi\\rangle = |\\psi\\rangle\\) because \\(H^{2k} = (H^2)^k = I^k = I\\). Therefore, he can correctly extract the message from Alice.", "question": "### (a) Is this protocol a correct encryption scheme?" }, { "context": "(Due to Anish Thilagar)\nThis protocol is not secure. Take the state \\(|\\psi\\rangle = \\frac{1}{\\sqrt{2}} (|0\\rangle + |+\\rangle)\\). Under the action of \\(H\\), this is an eigenvector with eigenvalue 1, so it will remain unchanged. Therefore, the ciphertext \\(c\\) will be equal to the message \\(m = |\\psi\\rangle\\), so \\(p(|\\psi\\rangle |c) = 1 \\neq p(|\\psi\\rangle) < 1\\). Therefore, this protocol is not secure.", "question": "### (b) Is this protocol a secure encryption scheme? Provide either a proof of security or an attack." } ]

[ { "context": "(Due to Mandy Huo)\n If Alice measures in the standard basis then given that the state is \\(|0\\rangle\\) she will always get \\(|0\\rangle\\) so she will never misidentify it and given that the state is \\(|+\\rangle\\) she will get \\(|0\\rangle\\) half the time so she will misidentify it probability 1/2.", "question": "### (a) Suppose Alice measures in the basis \\( \\{|0\\rangle , |1\\rangle \\} \\). She identifies the state as \\( |0\\rangle \\) if she gets the outcome \\( |0\\rangle \\) and as \\( |+\\rangle \\) if she gets the outcome \\( |1\\rangle \\). What is her probability of misidentifying the state given that it is \\( |0\\rangle \\)? What is her probability of misidentifying the state given that it is \\( |+\\rangle \\)?" }, { "context": "(Due to Mandy Huo)\n If Alice measures in the Hadamard basis then given that the state is \\(|+\\rangle\\) she will always get \\(|+\\rangle\\) so she will never misidentify it. Given that the state is \\(|0\\rangle\\) she will get \\(|+\\rangle\\) half the time so she will misidentify it probability 1/2.", "question": "### (b) Suppose instead Alice measures in the basis \\( \\{|+\\rangle, |-\\rangle\\} \\). She identifies the state as \\( |+\\rangle \\) if she gets the outcome \\( |+\\rangle \\) and as \\( |0\\rangle \\) if she gets the outcome \\( |-\\rangle \\). Again, what are her probabilities of misidentifying the state in each case?" }, { "context": "(Due to Mandy Huo)\nAssuming both states are equally likely a priori, Alice can do better overall if she measures in the basis \\(\\{|b_1\\rangle, |b_2\\rangle\\}\\) where \\(|b_1\\rangle = \\sin \\frac{3\\pi}{8} |0\\rangle - \\cos \\frac{3\\pi}{8} |1\\rangle\\) and \\(|b_2\\rangle = \\cos \\frac{3\\pi}{8} |0\\rangle + \\sin \\frac{3\\pi}{8} |1\\rangle\\), and identifies \\(|0\\rangle\\) when she gets the outcome \\(|b_1\\rangle\\) and \\(|+\\rangle\\) when she gets the outcome \\(|b_2\\rangle\\). Then the total probability of misidentifying is\n\n\\[\n\\frac{1}{2}(|\\langle b|0\\rangle|^2 + |\\langle b|1\\rangle|^2) = \\frac{1}{2} \\cos^2 \\frac{3\\pi}{8} + \\frac{11}{22} \\left( \\sin^2 \\frac{3\\pi}{8} + \\cos^2 \\frac{3\\pi}{8} - 2 \\sin \\frac{3\\pi}{8} \\cos \\frac{3\\pi}{8} \\right)\n\\]\n\n\\[\n= \\frac{1}{2} \\cos^2 \\frac{3\\pi}{8} + \\frac{11}{22} \\left( 1 - \\sin \\frac{3\\pi}{4} \\right)\n\\]\n\n\\[\n= \\frac{1}{2} \\cos^2 \\frac{3\\pi}{8} + \\frac{11}{22} \\left( 1 + \\cos \\frac{3\\pi}{4} \\right)\n\\]\n\n\\[\n= \\cos^2 \\frac{3\\pi}{8} \\approx 0.15\n\\]\nwhich is less than \\(\\frac{11}{22} = \\frac{1}{4}\\) in parts (a) and (b).", "question": "### (c) Is it possible for Alice to do better than this with any projective measurement? Assume \\( |0\\rangle \\) and \\( |+\\rangle \\) are equally likely a priori.\nNow suppose we allow Alice to perform a general measurement. In particular consider the following POVM with three elements:\n\n\\[\nE_1 = \\frac{\\sqrt{2}}{1 + \\sqrt{2}} |1\\rangle \\langle 1|\n\\]\n\n\\[\nE_2 = \\frac{\\sqrt{2}}{1 + \\sqrt{2}} \\frac{(|0\\rangle - |1\\rangle)(\\langle 0| - \\langle 1|)}{2}\n\\]\n\n\\[\nE_3 = I - E_1 - E_2\n\\]\n\nAlice identifies the state as \\(|+\\rangle\\) if she gets outcome 1, as \\(|0\\rangle\\) if she gets outcome 2, and makes no identification if she gets outcome 3." }, { "context": "(Due to Mandy Huo)\nIf the state is \\(|+\\rangle\\) then Alice will get outcomes 2 and 3 with probabilities\n\n\\[\n\\text{tr}(E_2|+\\rangle\\langle+|) = \\text{tr} \\left\\{ \\left( \\frac{\\sqrt{2}}{1+\\sqrt{2}} |-|\\rangle\\langle-| \\right) |+\\rangle\\langle+| \\right\\} = 0,\n\\]\n\n\\[\n\\text{tr}(E_3|+\\rangle\\langle+|) = 1 - \\text{tr}(E_1|+\\rangle\\langle+|) - \\text{tr}(E_2|+\\rangle\\langle+|) = 1 - \\frac{1}{\\sqrt{2}(1+\\sqrt{2})} = \\frac{1}{\\sqrt{2}}.\n\\]\nSo given that the state is \\(|+\\rangle\\), Alice will never misidentify the state and will fail to make an identification with probability \\(1/\\sqrt{2}\\).\n\nIf the state is \\(|0\\rangle\\) then Alice will get outcomes 1 and 3 with probabilities\n\n\\[\n\\text{tr}(E_1|0\\rangle\\langle0|) = 0,\n\\]\n\n\\[\n\\text{tr}(E_3|0\\rangle\\langle0|) = 1 - \\text{tr}(E_1|0\\rangle\\langle0|) - \\text{tr}(E_2|0\\rangle\\langle0|) = 1 - \\frac{\\sqrt{2}}{2(1+\\sqrt{2})} = \\frac{1}{\\sqrt{2}}.\n\\]\nSo given that the state is \\(|+\\rangle\\), Alice will never misidentify the state and will fail to make an identification with probability \\(1/\\sqrt{2}\\).", "question": "### (d) What is her probability of mis-identifying the state? What is her probability of failing to make an identification?" }, { "context": "(Due to Mandy Huo)\nThere is no POVM that increases the chances of making a correct identification without increasing the chance of making an incorrect identification.\n\nFirst we will show that any POVM such that the probability of mis-identification is zero must have the form \\(E_1 = \\alpha |1\\rangle\\langle1|\\) and \\(E_2 = \\beta |-|\\rangle\\langle-|, \\alpha, \\beta > 0\\). Since we must have \\(\\text{tr}(E_1|0\\rangle\\langle0|) = \\langle0|E_1|0\\rangle = 0\\) for zero chance of mis-identification in the \\(|0\\rangle\\) case, we have that either \\(|0\\rangle\\) is in the nullspace of \\(E_1\\) or \\(E_1\\) projects \\(|0\\rangle\\) onto \\(|1\\rangle\\). In the second case, we would have \\(E_1 = \\alpha |1\\rangle\\langle0|\\), which is not Hermitian and thus not positive so \\(E_1\\) must map \\(|0\\rangle\\) to the zero vector. Then \\(E_1\\) has rank 1 so it has the form \\(E_1 = \\alpha |b\\rangle\\langle1|\\). Then since \\(E_1\\) must be positive (and thus Hermitian) we have \\(E_1 = \\alpha |1\\rangle\\langle1|, \\alpha > 0\\) (note if \\(\\alpha = 0\\) then Alice will always fail to make an identification in the \\(|+\\rangle\\) case.) Similarly, \\(\\text{tr}(E_2|+\\rangle\\langle+|) = \\langle+|E_2|+\\rangle = 0\\) implies that \\(|+\\rangle\\) is either in the nullspace of \\(E_2\\) or is projected onto \\(|-\\rangle\\), but the second case results in \\( E_2 \\) not positive semidefinite so we must have \\( E_2 = \\beta |-|\\rangle \\langle -| \\), \\( \\beta > 0 \\).\n\nThen \\( E_3 = I - E_1 - E_2 \\) as before so that \\( \\sum_i E_i = I \\). What is left to check is whether \\( E_3 \\) is positive semidefinite. Since \\( E_3 \\) is given by\n\n\\[\n\\left| \\begin{array}{cc}\n(1 - \\lambda) - \\beta/2 & \\beta/2 \\\\\n\\beta/2 & (1 - \\lambda) - a - \\beta/2 \\\\\n\\end{array} \\right| = \\lambda^2 + (\\alpha + \\beta - 2)\\lambda - \\left( \\alpha + \\beta - \\frac{\\alpha \\beta}{2} - 1 \\right) = 0\n\\]\nso the eigenvalues are\n\n\\[\n\\lambda = \\frac{-(\\alpha + \\beta - 2) \\pm \\sqrt{(\\alpha + \\beta - 2)^2 + 4 \\left( \\alpha + \\beta - \\frac{\\alpha \\beta}{2} - 1 \\right)}}{2} = \\frac{-(\\alpha + \\beta - 2) \\pm \\sqrt{\\alpha^2 + \\beta^2}}{2}.\n\\]\nSince we want a POVM that fails to make an identification with smaller probability, we need \\( \\text{tr}[E_3|0\\rangle \\langle 0|] = 1 - \\frac{\\beta}{2} < \\frac{1}{\\sqrt{2}} \\) and \\( \\text{tr}[E_3|+\\rangle \\langle +|] = 1 - \\frac{\\alpha}{2} < \\frac{1}{\\sqrt{2}} \\), that is,\n\n\\[\n\\alpha > 2 \\left( \\frac{\\sqrt{2} - 1}{\\sqrt{2}} \\right), \\quad \\beta > 2 \\left( \\frac{\\sqrt{2} - 1}{\\sqrt{2}} \\right).\n\\]\nThen we have\n\n\\[\n-(\\alpha + \\beta - 2) < 2 - 4 \\frac{\\sqrt{2} - 1}{\\sqrt{2}} = -2 + 4 \\frac{\\sqrt{2} - 2}{\\sqrt{2}} = 2(\\sqrt{2} - 2) = 2 \\left( \\sqrt{2} - 1 \\right)\n\\]\n\n\\[\n\\sqrt{\\alpha^2 + \\beta^2} > \\sqrt{2 \\left( 2 \\frac{\\sqrt{2} - 1}{\\sqrt{2}} \\right)^2} = 2(\\sqrt{2} - 2) = 2 \\left( \\sqrt{2} - 1 \\right) > 0\n\\]\nso \\( \\sqrt{\\alpha^2 + \\beta^2} > -(\\alpha + \\beta - 2) \\) and so \\( E_3 \\) will have one negative eigenvalue and thus is not positive. Hence there is no POVM that gives Alice a better chance of making a correct identification without increasing the chance of making an incorrect identification.", "question": "### (e) Is there any POVM that gives Alice a better chance of making a correct identification without increasing the chance of making an incorrect identification?" } ]

[ { "context": "(Due to Mandy Huo)\n(i) Since \\( \\text{Tr}(|i\\rangle \\langle j|) = \\langle j|i \\rangle \\) is 0 for \\( i \\neq j \\) and 1 for \\( i = j \\), we have \\( \\text{Tr}_3[(GHZ_3)(GHZ_3)] = \\frac{1}{2}(|00\\rangle \\langle 00| + |11\\rangle \\langle 11|) \\), and so\n\n\\[\n\\text{Tr}[(|GHZ_2\\rangle \\langle GHZ_2| \\text{Tr}_3[(|GHZ_3\\rangle \\langle GHZ_3|)]) = \\frac{1}{4} \\text{Tr}[(|00\\rangle + |11\\rangle)(\\langle 00| + \\langle 11|)(|00\\rangle \\langle 00| + |11\\rangle \\langle 11|)]\n\\]\n\n\\[\n= \\frac{1}{4} \\text{Tr}[(|00\\rangle + |11\\rangle)(\\langle 00| + \\langle 11|)]\n\\]\n\n\\[\n= \\frac{1}{4} \\text{Tr}[(|00\\rangle + |11\\rangle)(\\langle 00| + \\langle 11|)]\n\\]\n\n\\[\n= \\frac{1}{4} \\text{Tr}[(|00\\rangle + |11\\rangle)(\\langle 00| + \\langle 11|)]\n\\]\n\n\\[\n= \\frac{1}{2}\n\\]\n\n=================\n\n(ii) Note that \\( \\text{Tr}_3(|W_3\\rangle \\langle W_3|) = \\frac{1}{3}(|10\\rangle \\langle 10| + |01\\rangle \\langle 01| + |00\\rangle \\langle 00| + |10\\rangle \\langle 01| + |01\\rangle \\langle 10|) \\) so we have\n\n\\[\n\\text{Tr}(|W_2\\rangle \\langle W_2| \\text{Tr}_3(|W_3\\rangle \\langle W_3|)) = \\frac{1}{6} \\text{Tr}[(|10\\rangle + |01\\rangle)(2\\langle 10| + 2\\langle 01|)] = \\frac{2}{3}\n\\]", "question": "### (a) Calculate the overlaps\n(i) \\(\\text{Tr}(|GHZ_2\\rangle \\langle GHZ_2| \\text{Tr}_3 |GHZ_3\\rangle \\langle GHZ_3|)\\) and\n(ii) \\(\\text{Tr}(|W_2\\rangle \\langle W_2| \\text{Tr}_3 |W_3\\rangle \\langle W_3|)\\).\n\nNow we generalize to the \\(N\\)-qubit case. As you might expect, \\(|GHZ_N\\rangle = \\frac{1}{\\sqrt{2}}(|0\\rangle^{\\otimes N} + |1\\rangle^{\\otimes N})\\) and \\(|W_N\\rangle\\) is an equal superposition of all \\(N\\)-bit strings with exactly one 1 and \\(N - 1\\) 0's." }, { "context": "(Due to Mandy Huo)\n(i) We have \\( \\text{Tr}_3(|GHZ_N\\rangle \\langle GHZ_N|) = \\frac{1}{2}(|0\\rangle \\langle 0|^{\\otimes N-1} + |1\\rangle \\langle 1|^{\\otimes N-1}) \\) so\n\n\\[\n(GHZ_N - 1| \\text{Tr}_N(|GHZ_N\\rangle \\langle GHZ_N|) = \\frac{1}{2}(GHZ_N - 1|\n\\]\n\nand thus\n\n\\[\n\\text{Tr}(|GHZ_{N-1}\\rangle \\langle GHZ_{N-1}| \\text{Tr}_3(|GHZ_N\\rangle \\langle GHZ_N|)) = \\frac{1}{4} \\text{Tr}[(|0\\rangle \\langle 0|^{\\otimes N-1} + |1\\rangle \\langle 1|^{\\otimes N-1})(|0\\rangle \\langle 0|^{\\otimes N-1} + (|1\\rangle \\langle 1|^{\\otimes N-1}) = \\frac{1}{2}\n\\]\n\n(ii) We have \\( W_{N-1} \\text{Tr}_N(|W_N\\rangle \\langle W_N|) = \\frac{N-1}{N}(W_{N-1}) \\) so\n\n\\[\n\\text{Tr}(|W_{N-1}\\rangle \\langle W_{N-1}| \\text{Tr}_3(|W_N\\rangle \\langle W_N|)) = \\frac{1}{N} \\text{Tr}[(|10\\ldots 0\\rangle + |010\\ldots 0\\rangle + \\ldots + |0\\ldots 01\\rangle)(\\langle 10\\ldots 0| + \\langle 010\\ldots 0| + \\ldots + \\langle 0\\ldots 01|)] = \\frac{N-1}{N}\n\\]\n\nSince \\( \\frac{N-1}{N} > \\frac{1}{2} \\) for \\( N > 2 \\) the overlap between the \\( N \\)-qubit \\( W \\) states is greater than between the \\( N \\)-qubit \\( GHZ \\) states so we can conclude that the \\( W \\) states are more 'robust' to tracing out a qubit.", "question": "### (b) Calculate the following overlaps as functions of \\(N\\).\n(i) \\(\\text{Tr}(|GHZ_{N-1}\\rangle \\langle GHZ_{N-1}| \\text{Tr}_N |GHZ_N\\rangle \\langle GHZ_N|)\\) and\n(ii) \\(\\text{Tr}(|W_{N-1}\\rangle \\langle W_{N-1}| \\text{Tr}_N |W_N\\rangle \\langle W_N|)\\).\n\nConclude that W states are “more robust” against loss of a single qubit than GHZ states." } ]

[ { "context": "(Due to De Huang)\n(a) (i) We can see \\(\\rho\\) and \\(T_1(\\rho)\\) as matrices in \\(\\mathbb{C}^{2 \\times 2}\\) and \\(\\mathbb{C}^{4 \\times 4}\\). Then we have\n\n\\[T_1(\\rho) = \\rho \\otimes \\frac{I}{2}\\]\n\n\\[= \\frac{1}{2} \\begin{pmatrix} \\rho_{11} & 0 & \\rho_{12} & 0 \\\\ 0 & \\rho_{11} & 0 & \\rho_{12} \\\\ \\rho_{21} & 0 & \\rho_{22} & 0 \\\\ 0 & \\rho_{21} & 0 & \\rho_{22} \\end{pmatrix} \\]\n\n\\[= \\frac{1}{2} \\begin{pmatrix} 1 & 0 & 0 & 0 \\\\ 0 & 0 & 1 & 0 \\\\ 0 & 1 & 0 & 0 \\\\ 0 & 0 & 0 & 1 \\end{pmatrix} \\begin{pmatrix} \\rho_{11} & \\rho_{12} \\\\ \\rho_{21} & \\rho_{22} \\end{pmatrix} \\begin{pmatrix} 1 & 0 & 0 & 0 \\\\ 0 & 0 & 1 & 0 \\\\ 0 & 1 & 0 & 0 \\\\ 0 & 0 & 0 & 1 \\end{pmatrix} \\]\n\n\\[+ \\frac{1}{2} \\begin{pmatrix} 0 & 0 & 0 & 0 \\\\ 0 & 1 & 0 & 0 \\\\ 0 & 0 & 0 & 0 \\\\ 0 & 0 & 0 & 1 \\end{pmatrix} \\begin{pmatrix} \\rho_{11} & \\rho_{12} \\\\ \\rho_{21} & \\rho_{22} \\end{pmatrix} \\begin{pmatrix} 0 & 1 & 0 & 0 \\\\ 0 & 0 & 0 & 1 \\end{pmatrix} \\]\n\n\\[= A_1 \\rho A_1^{\\dagger} + A_2 \\rho A_2^{\\dagger},\\]\n\nwhere\n\n\\[A_1 = \\frac{1}{\\sqrt{2}} \\begin{pmatrix} 1 & 0 & 0 & 0 \\\\ 0 & 0 & 1 & 0 \\\\ 0 & 0 & 0 & 1 \\end{pmatrix}, \\quad A_2 = \\frac{1}{\\sqrt{2}} \\begin{pmatrix} 0 & 0 & 0 & 0 \\\\ 0 & 1 & 0 & 0 \\\\ 0 & 0 & 0 & 1 \\end{pmatrix}.\\]\n\nIt's easy to check that\n\n\\[A_1^{\\dagger} A_1 + A_2^{\\dagger} A_2 = I.\\]\n\nTherefore \\(T_1\\) is CPTP. Indeed we can check that for any single qubit \\(|\\psi\\rangle\\),\n\n\\[A_1 |\\psi\\rangle = \\frac{1}{\\sqrt{2}} |\\psi\\rangle \\otimes |0\\rangle, \\quad A_2 |\\psi\\rangle = \\frac{1}{\\sqrt{2}} |\\psi\\rangle \\otimes |1\\rangle,\\]\n\n\\[A_1^{\\dagger} (|\\psi\\rangle \\otimes |0\\rangle) = \\frac{1}{\\sqrt{2}} |\\psi\\rangle, \\quad A_2^{\\dagger} (|\\psi\\rangle \\otimes |1\\rangle) = \\frac{1}{\\sqrt{2}} |\\psi\\rangle,\\]\n\ntherefore\n\n\\[A_1 |\\psi\\rangle \\langle \\psi| A_1^{\\dagger} + A_2 |\\psi\\rangle \\langle \\psi| A_2^{\\dagger} = \\frac{1}{2} |\\psi\\rangle \\langle \\psi| \\otimes |0\\rangle \\langle 0| + \\frac{1}{2} |\\psi\\rangle \\langle \\psi| \\otimes |1\\rangle \\langle 1|\\]\n\n\\[= |\\psi\\rangle \\langle \\psi| \\otimes \\frac{1}{2} (|0\\rangle \\langle 0| + |1\\rangle \\langle 1|)\\]\n\n\\[= |\\psi\\rangle \\langle \\psi| \\otimes \\frac{I}{2}\\]\n\n\\[= T_1 (|\\psi\\rangle \\langle \\psi|),\\]\n\nand\n\n\\[(A_1^{\\dagger} A_1 + A_2^{\\dagger} A_2) |\\psi\\rangle = \\frac{1}{\\sqrt{2}} A_1^{\\dagger} (|\\psi\\rangle \\otimes |0\\rangle) + \\frac{1}{\\sqrt{2}} A_1^{\\dagger} (|\\psi\\rangle \\otimes |1\\rangle) = |\\psi\\rangle,\\]\n\nwhich again verifies our proof of CPTP. The cloned qubit has density matrix \\(\\frac{1}{2}I\\), which actually carries no information. No matter what basis we use to measure the cloned qubit, we always get fair probability \\(\\frac{1}{2}\\) on both results. In the meanwhile, the first qubit is still in state \\(\\lvert \\psi \\rangle\\).\n\n(ii) Since \\(T_1(\\lvert \\psi \\rangle \\langle \\psi \\rvert) \\geq 0\\), we have\n\n\\[\\lvert \\langle \\psi \\lvert T_1(\\lvert \\psi \\rangle \\langle \\psi \\rvert) \\lvert \\psi \\rangle \\lvert = \\langle \\psi \\lvert T_1(\\lvert \\psi \\rangle \\langle \\psi \\rvert) \\lvert \\psi \\rangle\\]\n\n\\[= \\langle \\psi \\lvert (\\lvert \\psi \\rangle \\langle \\psi \\rvert \\frac{1}{2}I) \\lvert \\psi \\rangle\\]\n\n\\[= \\langle \\psi \\lvert \\psi \\rangle \\langle \\psi \\lvert \\psi \\rangle \\times \\langle \\psi \\lvert \\frac{1}{2}I \\lvert \\psi \\rangle\\]\n\n\\[= \\frac{1}{2}.\\]", "question": "### (a) Consider the map which takes as input a pure single-qubit state \\(\\rho = |\\psi\\rangle \\langle \\psi|\\), and returns \\(T_1(\\rho) = \\rho \\otimes \\frac{1}{2} I\\), where \\(\\frac{1}{2} I\\) is the maximally mixed state of a single qubit.\n(i) Show that this map is a valid quantum operation: it is CPTP. Give an interpretation of this map in terms of making a random guess for the cloned qubit." }, { "context": "(Due to De Huang)\n(i) Since $\\lvert 0 \\rangle \\lvert 0 \\rangle \\lvert 0 \\rangle$ and $\\lvert 1 \\rangle \\lvert 0 \\rangle \\lvert 0 \\rangle$ are orthogonal, we only need to verify that $U \\lvert 0 \\rangle \\lvert 0 \\rangle \\lvert 0 \\rangle$ and $U \\lvert 1 \\rangle \\lvert 0 \\rangle \\lvert 0 \\rangle$ are orthogonal. Indeed, note that $\\lvert 0 \\rangle \\lvert 0 \\rangle \\lvert 0 \\rangle$, $\\lvert 0 \\rangle \\lvert 0 \\rangle \\lvert 1 \\rangle$, $\\lvert 0 \\rangle \\lvert 1 \\rangle \\lvert 0 \\rangle$, $\\lvert 0 \\rangle \\lvert 1 \\rangle \\lvert 1 \\rangle$, $\\lvert 1 \\rangle \\lvert 0 \\rangle \\lvert 0 \\rangle$, $\\lvert 1 \\rangle \\lvert 0 \\rangle \\lvert 1 \\rangle$, $\\lvert 1 \\rangle \\lvert 1 \\rangle \\lvert 0 \\rangle$, $\\lvert 1 \\rangle \\lvert 1 \\rangle \\lvert 1 \\rangle$ are orthogonal to each other, since\n\n\\[\nU \\lvert 1 \\rangle \\lvert 0 \\rangle \\lvert 0 \\rangle = \\sqrt{\\frac{2}{3}} \\lvert 1 \\rangle \\lvert 1 \\rangle \\lvert 1 \\rangle + \\sqrt{\\frac{1}{6}} \\lvert 1 \\rangle \\lvert 0 \\rangle \\lvert 0 \\rangle + \\sqrt{\\frac{1}{6}} \\lvert 1 \\rangle \\lvert 1 \\rangle \\lvert 0 \\rangle,\n\\]\n\nwe have\n\n\\[\n\\langle 0 \\lvert \\langle 0 \\lvert \\langle 0 \\lvert U \\lvert 1 \\rangle \\lvert 0 \\rangle \\lvert 0 \\rangle \\rangle = \\langle 0 \\lvert \\langle 1 \\lvert \\langle 1 \\lvert U \\lvert 1 \\rangle \\lvert 0 \\rangle \\lvert 0 \\rangle \\rangle = \\langle 1 \\lvert \\langle 0 \\lvert \\langle 1 \\lvert U \\lvert 1 \\rangle \\lvert 0 \\rangle \\lvert 0 \\rangle \\rangle = 0.\n\\]\n\nAnd since\n\n\\[\nU \\lvert 0 \\rangle \\lvert 0 \\rangle \\lvert 0 \\rangle = \\sqrt{\\frac{2}{3}} \\lvert 0 \\rangle \\lvert 0 \\rangle \\lvert 0 \\rangle + \\sqrt{\\frac{1}{6}} \\lvert 0 \\rangle \\lvert 1 \\rangle \\lvert 1 \\rangle + \\sqrt{\\frac{1}{6}} \\lvert 1 \\rangle \\lvert 1 \\rangle \\lvert 1 \\rangle,\n\\]\n\nwe immediately have that $U \\lvert 0 \\rangle \\lvert 0 \\rangle \\lvert 0 \\rangle$ and $U \\lvert 1 \\rangle \\lvert 0 \\rangle \\lvert 0 \\rangle$ are orthogonal. Now we may extend $\\{ \\lvert 0 \\rangle \\lvert 0 \\rangle \\lvert 0 \\rangle, \\lvert 1 \\rangle \\lvert 0 \\rangle \\lvert 0 \\rangle \\}$ to\n\n\\[\n\\{ \\lvert 0 \\rangle \\lvert 0 \\rangle \\lvert 0 \\rangle, \\lvert 1 \\rangle \\lvert 0 \\rangle \\lvert 0 \\rangle, \\phi_3, \\phi_4, \\cdots, \\phi_8 \\}\n\\]\n\nas an orthogonal basis of all three-qubits, and also extend $\\{ U \\lvert 0 \\rangle \\lvert 0 \\rangle \\lvert 0 \\rangle, U \\lvert 1 \\rangle \\lvert 0 \\rangle \\lvert 0 \\rangle \\}$ to\n\n\\[\n\\{ U \\lvert 0 \\rangle \\lvert 0 \\rangle \\lvert 0 \\rangle, U \\lvert 1 \\rangle \\lvert 0 \\rangle \\lvert 0 \\rangle, \\psi_3, \\psi_4, \\cdots, \\psi_8 \\}\n\\]\n\nas another orthogonal basis of all three-qubits. Then one example of extending $U$ to a valid three-qubit unitary $\\tilde{U}$ would be\n\n\\[\n\\tilde{U} : \\lvert 0 \\rangle \\lvert 0 \\rangle \\lvert 0 \\rangle \\rightarrow U \\lvert 0 \\rangle \\lvert 0 \\rangle \\lvert 0 \\rangle, \\quad \\tilde{U} : \\lvert 1 \\rangle \\lvert 0 \\rangle \\lvert 0 \\rangle \\rightarrow U \\lvert 1 \\rangle \\lvert 0 \\rangle \\lvert 0 \\rangle,\n\\]\n\n\\[\n\\tilde{U} : \\phi_i \\rightarrow \\psi_i, \\quad i = 3, 4, \\cdots, 8.\n\\]\n\nIt’s easy to check that $\\tilde{U}$ is a valid three-qubit unitary because it linearly transforms an orthogonal basis to another orthogonal basis.\n\nThen the success probability is\n\n\\[\n|\\langle \\psi | \\langle T_2 | (\\langle \\psi | \\langle \\psi |) | \\psi \\rangle |^2 = | \\langle \\psi | \\langle \\psi | (\\alpha \\sqrt{\\frac{2}{3}} |0\\rangle |0\\rangle + \\beta \\sqrt{\\frac{1}{3}} | \\psi_+ \\rangle) (\\bar{\\alpha} \\sqrt{\\frac{2}{3}} \\langle 0 | \\langle 0 | + \\bar{\\beta} \\sqrt{\\frac{1}{3}} \\langle \\psi_+ |) | \\psi \\rangle |^2\n\\]\n\n\\[\n+ \\langle \\psi | \\langle \\psi | (\\beta \\sqrt{\\frac{2}{3}} |1\\rangle |1\\rangle + \\alpha \\sqrt{\\frac{1}{3}} | \\psi_+ \\rangle) (\\bar{\\beta} \\sqrt{\\frac{2}{3}} \\langle 1 | \\langle 1 | + \\bar{\\alpha} \\sqrt{\\frac{1}{3}} \\langle \\psi_+ |) | \\psi \\rangle |^2\n\\]\n\n\\[\n= | \\langle \\psi | \\langle \\psi | (\\alpha \\sqrt{\\frac{2}{3}} |0\\rangle |0\\rangle + \\beta \\sqrt{\\frac{1}{3}} | \\psi_+ \\rangle) |^2\n\\]\n\n\\[\n+ | \\langle \\psi | \\langle \\psi | (\\beta \\sqrt{\\frac{2}{3}} |1\\rangle |1\\rangle + \\alpha \\sqrt{\\frac{1}{3}} | \\psi_+ \\rangle) |^2\n\\]\n\n\\[\n= | \\alpha |^2 \\alpha \\sqrt{\\frac{2}{3}} + | \\beta |^2 \\alpha \\sqrt{\\frac{2}{3}} + | \\beta |^2 \\alpha \\sqrt{\\frac{2}{3}} + | \\alpha |^2 \\alpha \\sqrt{\\frac{2}{3}} |^2\n\\]\n\n\\[\n= \\frac{2}{3} | \\alpha |^2 + \\frac{2}{3} | \\beta |^2\n\\]\n\n\\[\n= \\frac{2}{3}.\n\\]", "question": "### (b) Let’s consider a second cloning map, which acts on the qubit input state together with two ancilla qubits as follows:\n\n\\[ \nU : |0\\rangle |0\\rangle |0\\rangle \\mapsto \\sqrt{\\frac{2}{3}} |0\\rangle |0\\rangle |0\\rangle + \\sqrt{\\frac{1}{6}} (|0\\rangle |1\\rangle + |1\\rangle |0\\rangle ) |1\\rangle , \n\\]\n\n\\[ \n|1\\rangle |0\\rangle |0\\rangle \\mapsto \\sqrt{\\frac{2}{3}} |1\\rangle |1\\rangle |1\\rangle + \\sqrt{\\frac{1}{6}} (|1\\rangle |0\\rangle + |0\\rangle |1\\rangle ) |0\\rangle . \n\\]\n\n(i) Verify that \\(U\\) can be extended into a valid three-qubit unitary.\n\nThe cloning map associated to \\(U\\) is the map \\(T_2\\) which first initializes two qubits to the \\(|0\\rangle |0\\rangle\\) state, then applies \\(U\\), and then traces out the third qubit.\n\n(ii) Evaluate the success probability of the map \\(U\\) on an arbitrary input pure state \\(\\rho = |\\psi\\rangle \\langle \\psi|\\)." }, { "context": "(Due to De Huang)\n(i) Note that\n\n\\[\nP_+^\\dagger = I^\\dagger - (| \\Psi_- \\rangle \\langle \\Psi_- |)^\\dagger = I - | \\Psi_- \\rangle \\langle \\Psi_- | = P_+,\n\\]\n\n\\[\nP_+ P_+ = (I - | \\Psi_- \\rangle \\langle \\Psi_- |) (I - | \\Psi_- \\rangle \\langle \\Psi_- |)\n\\]\n\n\\[\n= I - 2 | \\Psi_- \\rangle \\langle \\Psi_- | + | \\Psi_- \\rangle \\langle \\Psi_- | = I - | \\Psi_- \\rangle \\langle \\Psi_- |\n\\]\n\n\\[\n= P_+.\n\\]\n\nThen using the result of (a)(i), we have\n\n\\[\nT_3 (\\rho) = \\frac{2}{3} P_+ (\\rho \\otimes I) P_+\n\\]\n\n\\[\n= \\frac{4}{3} P_+ T_2 (\\rho) P_+\n\\]\n\n\\[\n= \\frac{4}{3} P_+ (A_1 \\rho A_1^\\dagger + A_2 \\rho A_2^\\dagger) P_+\n\\]\n\n\\[\n= (\\frac{2}{\\sqrt{3}} P_+ A_1) \\rho (\\frac{2}{\\sqrt{3}} P_+ A_1)^\\dagger + (\\frac{2}{\\sqrt{3}} P_+ A_2) \\rho (\\frac{2}{\\sqrt{3}} P_+ A_2)^\\dagger\n\\]\n\n\\[\n= V_1 \\rho V_1^\\dagger + V_2 \\rho V_2^\\dagger,\n\\]\n\nwhere \\( A_1, A_2 \\) are defined in (a)(i), and\n\n\\[\nV_1 = \\frac{2}{\\sqrt{3}} P_+ A_1, \\quad V_2 = \\frac{2}{\\sqrt{3}} P_+ A_2.\n\\]\n\nIf we see \\( P_+ = I - | \\Psi_- \\rangle \\langle \\Psi_- | \\) as a matrix in \\( \\mathbb{C}^{4 \\times 4} \\), then\n\n\\[\nP_+ = \\begin{pmatrix}\n1 & 0 & 0 & 0 \\\\\n0 & \\frac{1}{2} & \\frac{1}{2} & 0 \\\\\n0 & \\frac{1}{2} & \\frac{1}{2} & 0 \\\\\n0 & 0 & 0 & 1\n\\end{pmatrix}.\n\\]\n\nBy direct calculation, we can check that\n\n\\[ V_1^\\dagger V_1 + V_2^\\dagger V_2 = \\frac{4}{3} A_1^\\dagger P_+ P_+ A_1 + \\frac{4}{3} A_2^\\dagger P_+ P_+ A_2 = \\frac{4}{3} A_1^\\dagger P_+ A_1 + \\frac{4}{3} A_2^\\dagger P_+ A_2 = I. \\]\n\nTherefore \\( T_3 \\) is CPTP.\n\n(ii) For any single-state \\( |\\psi\\rangle \\), we have\n\n\\[ \\langle \\psi | \\langle \\psi | \\Psi_- \\rangle = \\frac{1}{\\sqrt{2}} (\\langle \\psi | 0 \\rangle \\langle \\psi | 1 \\rangle - \\langle \\psi | 1 \\rangle \\langle \\psi | 0 \\rangle) = 0, \\]\n\n\\[ \\langle \\psi | \\langle \\psi | P_+ = \\langle \\psi | \\langle \\psi | - \\langle \\psi | \\langle \\psi | \\Psi_- \\rangle \\langle \\Psi_- | = \\langle \\psi | \\langle \\psi |, \\]\n\n\\[ P_+ |\\psi\\rangle |\\psi\\rangle = |\\psi\\rangle |\\psi\\rangle - |\\Psi_- \\rangle \\langle \\Psi_- | |\\psi\\rangle |\\psi\\rangle = |\\psi\\rangle |\\psi\\rangle, \\]\n\nthus the success probability of \\( T_3 \\) is\n\n\\[ |\\langle \\psi | \\langle \\psi | T_3 (|\\psi\\rangle |\\psi\\rangle) | \\langle \\psi | \\langle \\psi | \\rangle = \\frac{2}{3} \\langle \\psi | \\langle \\psi | P_+ (|\\psi\\rangle |\\psi\\rangle \\langle \\psi | \\langle \\psi | \\otimes I) P_+ |\\psi\\rangle |\\psi\\rangle \\]\n\n\\[ = \\frac{2}{3} \\langle \\psi | \\langle \\psi | (|\\psi\\rangle \\langle \\psi | \\otimes I) |\\psi\\rangle |\\psi\\rangle \\]\n\n\\[ = \\frac{2}{3} (\\langle \\psi | \\langle \\psi | |\\psi\\rangle |\\psi\\rangle) (\\langle \\psi | I | \\psi\\rangle) \\]\n\n\\[ = \\frac{2}{3}. \\]\n\n(iii) We can see that for any single-state \\( |\\psi\\rangle \\),\n\n\\[ |\\langle \\psi | \\langle \\psi | T_2 (|\\psi\\rangle |\\psi\\rangle) | \\langle \\psi | \\langle \\psi | \\rangle = |\\langle \\psi | \\langle \\psi | T_3 (|\\psi\\rangle |\\psi\\rangle) | \\langle \\psi | \\langle \\psi | \\rangle = \\frac{2}{3}, \\]\n\nthat is, the map \\( T_2 \\) and \\( T_3 \\) have the same success probability. The essential reason for this result is that we actually have\n\n\\[ T_2 (|\\psi\\rangle \\langle \\psi |) = T_3 (|\\psi\\rangle \\langle \\psi |) \\]\n\nfor any single-state \\( |\\psi\\rangle \\). To see this, we first rewrite \\( U |\\psi\\rangle |0\\rangle |0\\rangle \\) as\n\n\\[ U |\\psi\\rangle |0\\rangle |0\\rangle = \\alpha U |0\\rangle |0\\rangle |0\\rangle + \\beta U |1\\rangle |0\\rangle |0\\rangle \\]\n\n\\[ = \\alpha \\left( \\sqrt{\\frac{2}{3}} |0\\rangle |0\\rangle |0\\rangle + \\sqrt{\\frac{1}{6}} (|0\\rangle |1\\rangle + |1\\rangle |0\\rangle) |1\\rangle \\right) \\]\n\n\\[ + \\beta \\left( \\sqrt{\\frac{2}{3}} |1\\rangle |1\\rangle |1\\rangle + \\sqrt{\\frac{1}{6}} (|1\\rangle |0\\rangle + |0\\rangle |1\\rangle) |0\\rangle \\right) \\]\n\n\\[ = \\frac{1}{\\sqrt{3}} \\left[|\\Phi_+\\rangle \\langle \\alpha |0\\rangle + \\beta |1\\rangle | + \\frac{1}{\\sqrt{3}} \\left[|\\Phi_-\\rangle \\langle \\alpha |0\\rangle - \\beta |1\\rangle | + \\frac{1}{\\sqrt{3}} \\left[|\\Psi_+\\rangle \\langle \\alpha |1\\rangle + \\beta |0\\rangle | \\right]\n\\]\n\nHere \\(|\\Phi_+\\rangle, |\\Phi_-\\rangle, |\\Psi_+\\rangle\\) together with \\(|\\Psi_-\\rangle\\) are the Bell basis, i.e.\n\n\\[\n|\\Phi_+\\rangle = \\frac{1}{\\sqrt{2}} (|00\\rangle + |11\\rangle), \\quad |\\Phi_-\\rangle = \\frac{1}{\\sqrt{2}} (|00\\rangle - |11\\rangle),\n\\]\n\n\\[\n|\\Psi_+\\rangle = \\frac{1}{\\sqrt{2}} (|01\\rangle + |10\\rangle), \\quad |\\Psi_-\\rangle = \\frac{1}{\\sqrt{2}} (|01\\rangle - |10\\rangle).\n\\]\n\nThen we have\n\n\\[\nT_2(|\\psi\\rangle \\langle \\psi|) = \\text{tr}_3(U(|\\psi\\rangle \\langle \\psi| \\otimes |0\\rangle \\langle 0|)U^\\dagger)\n\\]\n\n\\[\n= \\frac{1}{3} \\left( \\text{tr}(|\\psi\\rangle \\langle \\psi|) |\\Phi_+\\rangle \\langle \\Phi_+| + \\text{tr}(Z|\\psi\\rangle \\langle \\psi|Z) |\\Phi_-\\rangle \\langle \\Phi_-| + \\text{tr}(X|\\psi\\rangle \\langle \\psi|X) |\\Psi_+\\rangle \\langle \\Psi_+| + \\text{tr}(Z|\\psi\\rangle \\langle \\psi|X) |\\Psi_-\\rangle \\langle \\Psi_-| \\right)\n\\]\n\n\\[\n= \\frac{1}{3} \\left( \\langle \\psi| \\psi \\rangle |\\Phi_+\\rangle \\langle \\Phi_+| + \\langle \\psi| \\psi \\rangle |\\Phi_-\\rangle \\langle \\Phi_-| + \\langle \\psi| \\psi \\rangle |\\Psi_+\\rangle \\langle \\Psi_+| + \\langle \\psi| \\psi \\rangle |\\Psi_-\\rangle \\langle \\Psi_-| \\right)\n\\]\n\nOn the other hand, since\n\n\\[\n|\\Phi_+\\rangle \\langle \\Phi_+| + |\\Phi_-\\rangle \\langle \\Phi_-| + |\\Psi_+\\rangle \\langle \\Psi_+| + |\\Psi_-\\rangle \\langle \\Psi_-| = I,\n\\]\n\nwe have\n\n\\[\nI - |\\Psi_-\\rangle \\langle \\Psi_-| = |\\Phi_+\\rangle \\langle \\Phi_+| + |\\Phi_-\\rangle \\langle \\Phi_-| + |\\Psi_+\\rangle \\langle \\Psi_+|.\n\\]\n\nThus\n\n\\[\nT_3(|\\psi\\rangle \\langle \\psi|)\n\\]\n\n\\[\n= \\frac{2}{3} (I - |\\Psi_-\\rangle \\langle \\Psi_-|) (|\\psi\\rangle \\langle \\psi| \\otimes I) (I - |\\Psi_-\\rangle \\langle \\Psi_-|)\n\\]\n\n\\[\n= \\frac{2}{3} \\left( |\\Phi_+\\rangle \\langle \\Phi_+| + |\\Phi_-\\rangle \\langle \\Phi_-| + |\\Psi_+\\rangle \\langle \\Psi_+| \\right) (|\\psi\\rangle \\langle \\psi| \\otimes I) \\left( |\\Phi_+\\rangle \\langle \\Phi_+| + |\\Phi_-\\rangle \\langle \\Phi_-| + |\\Psi_+\\rangle \\langle \\Psi_+| \\right)\n\\]\n\n\\[\n= \\frac{2}{3} \\left( |\\Phi_+\\rangle \\langle \\Phi_+| (|\\psi\\rangle \\langle \\psi| \\otimes I) |\\Phi_+\\rangle \\langle \\Phi_+| + |\\Phi_-\\rangle \\langle \\Phi_-| (|\\psi\\rangle \\langle \\psi| \\otimes I) |\\Phi_-\\rangle \\langle \\Phi_-| + |\\Psi_+\\rangle \\langle \\Psi_+| (|\\psi\\rangle \\langle \\psi| \\otimes I) |\\Psi_+\\rangle \\langle \\Psi_+| \\right).\n\\]\n\nNote that\n\n\\[\n\\langle \\Phi_+ | (|\\psi\\rangle \\langle \\psi| \\otimes I) |\\Phi_+\\rangle = \\frac{1}{2} \\left( \\langle 0| \\langle 0| + \\langle 1| \\langle 1| \\right) (|\\psi\\rangle \\langle \\psi|) = \\frac{1}{2} \\langle \\psi| \\psi \\rangle,\n\\]\n\n\\[\n\\begin{aligned}\n\\langle \\Phi_- | (|\\psi\\rangle \\langle \\psi| \\otimes I) | \\Phi_- \\rangle &= \\frac{1}{2} \\langle \\psi | (|0\\rangle \\langle 0| + |1\\rangle \\langle 1|) |\\psi \\rangle = \\frac{1}{2} \\langle \\psi | \\psi \\rangle, \\\\\n\\langle \\Psi_+ | (|\\psi\\rangle \\langle \\psi| \\otimes I) | \\Psi_+ \\rangle &= \\frac{1}{2} \\langle \\psi | (|0\\rangle \\langle 0| + |1\\rangle \\langle 1|) |\\psi \\rangle = \\frac{1}{2} \\langle \\psi | \\psi \\rangle, \\\\\n\\langle \\Phi_+ | (|\\psi\\rangle \\langle \\psi| \\otimes I) | \\Phi_- \\rangle &= \\frac{1}{2} \\langle \\psi | (|0\\rangle \\langle 0| - |1\\rangle \\langle 1|) |\\psi \\rangle = \\frac{1}{2} \\langle \\psi | Z | \\psi \\rangle, \\\\\n\\langle \\Phi_+ | (|\\psi\\rangle \\langle \\psi| \\otimes I) | \\Psi_+ \\rangle &= \\frac{1}{2} \\langle \\psi | (|1\\rangle \\langle 0| + |0\\rangle \\langle 1|) |\\psi \\rangle = \\frac{1}{2} \\langle \\psi | X | \\psi \\rangle, \\\\\n\\langle \\Phi_- | (|\\psi\\rangle \\langle \\psi| \\otimes I) | \\Psi_+ \\rangle &= \\frac{1}{2} \\langle \\psi | (|1\\rangle \\langle 0| - |0\\rangle \\langle 1|) |\\psi \\rangle = \\frac{1}{2} \\langle \\psi | X Z | \\psi \\rangle.\n\\end{aligned}\n\\]\n\nTherefore\n\n\\[\n\\begin{aligned}\nT_3 (|\\psi \\rangle \\langle \\psi |) &= \\frac{2}{3} \\left( \\langle \\psi | \\psi \\rangle |\\Phi_+\\rangle \\langle \\Phi_+| + \\langle \\psi | \\psi \\rangle |\\Phi_-\\rangle \\langle \\Phi_-| + \\langle \\psi | \\psi \\rangle |\\Psi_+\\rangle \\langle \\Psi_+| \\right. \\\\\n& \\quad + \\langle \\psi | Z | \\psi \\rangle |\\Phi_+\\rangle \\langle \\Phi_-| + \\langle \\psi | X | \\psi \\rangle |\\Phi_-\\rangle \\langle \\Psi_+| \\right) \\\\\n&= T_2 (|\\psi \\rangle \\langle \\psi |)\n\\end{aligned}\n\\]\n\nIt's done.", "question": "### (c) Consider a third cloning map \\(T_3\\) defined as \\(T_3(\\rho) = \\frac{2}{3} P_+ (\\rho \\otimes I) P_+\\), where \\(P_+ = I - |\\Psi_-\\rangle \\langle \\Psi_-|\\) and \\(|\\Psi_-\\rangle = \\frac{1}{\\sqrt{2}}(|0\\rangle |1\\rangle - |1\\rangle |0\\rangle)\\).\n(i) Verify that \\(T_3\\) is a CPTP map.\n(ii) Evaluate its success probability as a universal cloning map.\n(iii) Is this a coincidence — is there a relationship between the three maps you have considered?" } ]

[ { "context": "The most general way in which Bob can try to ascertain Alice’s classical bits is by creating a POVM with four operators, each corresponding to a single guess for Alice’s pair of qubits. Thus, we want a POVM\n\n\\[\n\\{M_0, M_1, M_+, M_-\\}\n\\]\n\nWhich maximizes the value\n\n\\[\nP(0|0) + P(1|1) + P(+|+) + P(-|-)\n\\]\n\n\\[\n= \\frac{4}{4} tr(M_0|0\\rangle \\langle 0|) + tr(M_1|1\\rangle \\langle 1|) + tr(M_+|+\\rangle \\langle +|) + tr(M_-|- \\rangle \\langle -|)\n\\]\n\n\\[\n= \\frac{\\langle 0|M_0|0\\rangle + \\langle 1|M_1|1\\rangle + \\langle +|M_+|+\\rangle + \\langle -|M_-|-\rangle}{4}\n\\]\n\nAnd we want to know this maximal value. Observe that\n\n\\[\n\\langle 0|M_0|0\\rangle + \\langle 1|M_1|1\\rangle \\leq \\langle 0|M_0|0\\rangle + \\langle 1|M_0|1\\rangle + \\langle 0|M_1|0\\rangle + \\langle 1|M_1|1\\rangle = tr(M_0 + M_1)\n\\]\n\nFrom the positive definiteness of these matrices, and similarly\n\n\\[\n\\langle +|M_+|+\\rangle + \\langle -|M_-|-\rangle \\leq \\langle +|M_+|+\\rangle + \\langle -|M_+|-\rangle + \\langle +|M_-|+\\rangle + \\langle -|M_-|-\rangle = tr(M_+ + M_-)\n\\]\n\nAnd we therefore have\n\n\\[\n\\langle 0|M_0|0\\rangle + \\langle 1|M_1|1\\rangle + \\langle +|M_+|+\\rangle + \\langle -|M_-|-\rangle \\leq tr(M_0 + M_1) + tr(M_+ + M_-)\n\\]\n\n\\[\n= tr(M_0 + M_1 + M_+ + M_-)\n\\]\n\n\\[\n= tr(I)\n\\]\n\n\\[\n= 2\n\\]\n\nAnd so\n\n\\[\n\\frac{P(0|0) + P(1|1) + P(+|+) + P(-|-)}{4}\n\\]\n\\[\n= \\frac{tr(M_0|0\\rangle \\langle 0|) + tr(M_1|1\\rangle \\langle 1|) + tr(M_+|+\\rangle \\langle +|) + tr(M_-|- \\rangle \\langle -|)}{4}\n\\]\n\\[\n= \\frac{\\langle 0|M_0|0\\rangle + \\langle 1|M_1|1\\rangle + \\langle +|M_+|+\\rangle + \\langle -|M_-|- \\rangle}{4}\n\\]\n\\[\n\\leq \\frac{2}{4}\n\\]\n\\[\n= \\frac{1}{2}\n\\]\n\nAnd so \\(\\frac{1}{2}\\) is an upper bound on the probability of success. We can attain this upper bound with\n\\[\nM_0 = |0\\rangle \\langle 0|\n\\]\n\\[\nM_1 = |1\\rangle \\langle 1|\n\\]\n\\[\nM_+ = 0\n\\]\n\\[\nM_- = 0\n\\]\nWhich measures in the standard basis. Thus, \\(\\frac{1}{2}\\) is the maximum value with which Bob can correctly guess both of Alice's two classical bits.", "question": "### (a) The first idea she has is to encode her two classical bits into her preparation of one of four states in \\(\\{ |0\\rangle, |1\\rangle, |+\\rangle, |-\\rangle \\}\\), and then send this qubit to Bob. Suppose that the a priori distribution of Alice’s two classical bits is uniform. What is the maximum probability with which Bob can correctly guess both of Alice’s two classical bits?" }, { "context": "Initially, the Alice-Bob qubit pair is maximally entangled:\n\n\\[\n\\frac{1}{\\sqrt{2}} (|0\\rangle_A|0\\rangle_B + |1\\rangle_A|1\\rangle_B)\n\\]\n\nSuppose Alice applies one of the four unitary transformations\n\n\\[\n\\{I, X, Z, ZX\\}\n\\]\n\nto her qubit and then sends it to Bob. Now, Bob has the qubit pair state\n\n\\[\n\\frac{1}{\\sqrt{2}} ((Z^{k_1}X^{k_2}|0\\rangle_A) \\otimes |0\\rangle_B + (Z^{k_1}X^{k_2}|1\\rangle_A) \\otimes |1\\rangle_B)\n\\]\n\nNow, consider the possible values for this pair\n\n\\[\n|\\psi\\rangle_{00} = \\frac{1}{\\sqrt{2}} (|0\\rangle_A|0\\rangle_B + |1\\rangle_A|1\\rangle_B)\n\\]\n\n\\[\n|\\psi\\rangle_{01} = \\frac{1}{\\sqrt{2}} (|1\\rangle_A|0\\rangle_B + |0\\rangle_A|1\\rangle_B)\n\\]\n\n\\[\n|\\psi\\rangle_{10} = \\frac{1}{\\sqrt{2}} (|0\\rangle_A|0\\rangle_B - |1\\rangle_A|1\\rangle_B)\n\\]\n\n\\[\n|\\psi\\rangle_{11} = \\frac{1}{\\sqrt{2}} (|1\\rangle_A|0\\rangle_B - |0\\rangle_A|1\\rangle_B)\n\\]\n\nThis quadruple constitutes a basis. We can see this since all the states are normalized, and the two pairs of states which share components in the standard basis, their inner products evaluate to 0. Thus, all Bob has to do is to measure his state in this basis, thereby ascertaining which of the four states he has.", "question": "### (b) Suppose that Alice and Bob share a maximally entangled pair of qubits. Alice thinks that it is a good idea to start by performing one of four unitary transformations on the qubit in her possession depending on the value of the two classical bits that she wishes to communicate and send her qubit to Bob. What next? Help Alice (and Bob) devise a scheme that achieves the desired task with certainty." }, { "context": "No, Eve cannot recover any information about the classical bits Alice is sharing with Bob. To see this, we trace out Bob’s bit from the two-qubit state, leaving only the intercepted qubit. Letting \\( U_A \\) be the unitary applied by Alice, the density matrix for Alice and Bob’s state is:\n\n\\[\n\\begin{aligned}\n&\\frac{1}{2} (U_A|0\\rangle_A)(\\langle0|U_A^\\dagger) \\otimes |0\\rangle_B\\langle0|_B + \\\\\n&\\frac{1}{2} (U_A|0\\rangle_A)(\\langle1|U_A^\\dagger) \\otimes |0\\rangle_B\\langle1|_B + \\\\\n&\\frac{1}{2} (U_A|1\\rangle_A)(\\langle0|U_A^\\dagger) \\otimes |1\\rangle_B\\langle0|_B + \\\\\n&\\frac{1}{2} (U_A|1\\rangle_A)(\\langle1|U_A^\\dagger) \\otimes |1\\rangle_B\\langle1|_B\n\\end{aligned}\n\\]\n\nAnd so if we trace out the second bit, we get\n\n\\[\n\\begin{aligned}\n&\\frac{1}{2} (U_A|0\\rangle_A)(\\langle0|U_A^\\dagger) + (U_A|1\\rangle_A)(\\langle1|U_A^\\dagger) \\\\\n&= U_A (\\frac{1}{2} (|0\\rangle\\langle0| + |1\\rangle\\langle1|)) U_A^\\dagger \\\\\n&= U_A \\frac{I}{2} U_A^\\dagger\n\\end{aligned}\n\\]\n\nNow, neither the application of \\( X \\) or \\( Z \\) changes the value of the maximally mixed state \\( \\frac{1}{2} \\), so whatever Alice’s bits are, Eve’s state is maximally mixed, and so she learns nothing.", "question": "### (c) After all the thought that Alice and Bob put into coming up with a working scheme, they finally decide to employ it. Unfortunately, the tireless eavesdropper Eve has heard of their new scheme, and as soon as Alice and Bob use it, she intercepts the qubit as it’s sent from Alice to Bob. Can Eve recover information about the two confidential classical bits that Alice intended to share with Bob?" } ]

[ { "context": "We first prove a Lemma: If \\( X, Y \\in M_d(\\mathbb{C}), X \\geq 0, Y \\geq 0 \\), then \\( \\text{tr}(XY) \\geq 0 \\).\n\n**Proof:** Consider the eigenvalue decomposition of \\( X \\),\n\n\\[ X = Q \\Lambda Q^\\dagger \\]\n\nwhere \\( Q \\) is unitary, and \\( \\Lambda \\) is a diagonal matrix with diagonal elements \\( \\lambda_1 \\geq \\lambda_2 \\geq \\cdots \\geq \\lambda_d \\geq 0 \\). Then\n\n\\[ \\text{tr}(XY) = \\text{tr}(Q \\Lambda Q^\\dagger Y) = \\text{tr}(\\Lambda Q^\\dagger Y Q).\\]\n\nLet \\( s_1, s_2, \\ldots, s_d \\) be the diagonal elements of \\( Q^†YQ \\). Since \\( Y \\ge 0 \\), we have \\( Q^†YQ \\ge 0 \\), and thus \\( s_i \\ge 0, \\, i = 1, 2, \\ldots, d \\). Therefore\n\n\\[ \\text{tr}(XY) = \\text{tr}(\\Lambda Q^†YQ) = \\sum_{i=1}^{d} \\lambda_i s_i \\ge 0.\\]\n\nLet \\( \\Omega_1 = \\{ X \\in M_d(\\mathbb{C}) : X \\ge 0, \\Phi(X) = B \\} \\), \\( \\Omega_2 = \\{ Y \\in M_d(\\mathbb{C}) : \\Phi^*(Y) \\ge A, Y = Y^† \\} \\). Now given any \\( X \\in \\Omega_1, Y \\in \\Omega_2 \\), we have\n\n\\[ \\text{tr}(BY) = \\text{tr}(\\Phi(X)Y) \\]\n\n\\[= \\text{tr}\\left( \\sum_{i=1}^{k} K_i X K_i^† Y \\right)\\]\n\n\\[= \\text{tr}\\left( \\sum_{i=1}^{k} X K_i^† Y K_i \\right)\\]\n\n\\[= \\text{tr}(X \\Phi^*(Y)).\\]\n\nSince \\( \\Phi^*(Y) \\ge A \\), i.e. \\( \\Phi^*(Y) - A \\ge 0 \\), using the Lemma we have\n\n\\[ \\text{tr}(X(\\Phi^*(Y) - A)) \\ge 0,\\]\n\n\\[\\Rightarrow \\text{tr}(BY) = \\text{tr}(X \\Phi^*(Y)) \\ge \\text{tr}(XA) = \\text{tr}(AX).\\]\n\nSince \\( X, Y \\) are arbitrary in \\( \\Omega_1, \\Omega_2 \\), we immediately have\n\n\\[ \\beta = \\min_{Y \\in \\Omega_2} \\text{tr}(BY) \\ge \\max_{X \\in \\Omega_1} \\text{tr}(AX) = \\alpha.\\]", "question": "### (a) Show that it is always the case that \\(\\alpha \\leq \\beta\\). This condition is called **weak duality**." }, { "context": "Consider the eigenvalue decomposition of \\( M \\),\n\n\\[ M = U \\Lambda U^†, \\]\n\nwhere \\( U \\) is unitary, and \\( \\Lambda \\) is a diagonal matrix with diagonal elements \\( \\lambda_1 \\ge \\lambda_2 \\ge \\ldots \\ge \\lambda_d \\). We have\n\n\\[ \\lambda I \\ge M \\Rightarrow \\lambda I - M \\ge 0 \\Rightarrow U^†(\\lambda I - M)U \\ge 0 \\Rightarrow \\lambda I - \\Lambda \\ge 0.\\]\n\nNotice that \\( \\lambda I - \\Lambda \\) is a diagonal matrix. Thus we have \\( \\lambda - \\lambda_i \\ge 0, \\, i = 1, 2, \\ldots, d \\), which means all eigenvalues of \\( M \\) are less than or equal to \\( \\lambda \\).", "question": "### (b) Remember that the inequality \\(M \\geq N\\), for \\(M, N\\) Hermitian \\(d \\times d\\) matrices, is always taken to mean \\(M - N \\geq 0\\), or equivalently all eigenvalues of \\((M - N)\\) are non-negative. What does the condition \\(M \\leq \\lambda I\\), for some fixed Hermitian \\(M \\in M_d(\\mathbb{C})\\) and \\(\\lambda \\in \\mathbb{R}\\), mean on the eigenvalues of \\(M\\)?" }, { "context": "We can choose that \\( A = M \\in M_d(\\mathbb{C}), B = 1 \\in \\mathbb{C} \\), and each \\( K_i, i = 1, 2, \\ldots, d \\) is a \\( 1 \\times d \\) vector such that the \\( i \\)-th element is 1 and the other elements are 0. Then for any \\( X \\in M_d(\\mathbb{C}) \\) and any \\( y \\in \\mathbb{C} \\), we have\n\n\\[ \\Phi(X) = \\sum_{i=1}^{d} K_i X K_i^\\dagger = \\sum_{i=1}^{d} X_{ii} = \\text{tr}(X), \\]\n\n\\[ \\Phi^*(y) = \\sum_{i=1}^{d} K_i^\\dagger y K_i = yI. \\]\n\nNow we have \\( tr(By) = y \\), and \\( y = y^{\\dagger} \\Rightarrow y \\in \\mathbb{R} \\). Then the dual problem is just\n\n\\[ \\beta = \\min_{y \\in \\mathbb{R}} y \\\\\n\\text{s.t.} \\quad yI \\geq M. \\]\n\nUsing the result of (b), it's easy to see that \\( \\beta = \\lambda_1(M) \\). The primal problem is\n\n\\[ \\alpha = \\max_{X} tr(MX) \\\\\n\\text{s.t.} \\quad tr(X) = 1, \\\\\nX \\geq 0. \\]\n\nGiven \\( X \\geq 0 \\), we can always find the root decomposition of \\( X \\),\n\n\\[ X = PP^{\\dagger}. \\]\n\nLet \\( p_i \\) denote the \\( i \\)-th column of \\( P \\), then we have\n\n\\[ tr(MX) = tr(MPP^{\\dagger}) = tr(P^{\\dagger}MP) = \\sum_{i=1}^{d} p_i^{\\dagger}Mp_i, \\]\n\nand\n\n\\[ 1 = tr(X) = tr(PP^{\\dagger}) = tr(P^{\\dagger}P) \\Rightarrow \\sum_{i=1}^{d} \\|p_i\\|_2^2 = 1. \\]\n\nThus the primal problem is equivalent to\n\n\\[ \\alpha = \\max_{p_i, i=1,2,\\ldots,d} \\sum_{i=1}^{d} p_i^{\\dagger}Mp_i \\\\\n\\text{s.t.} \\quad \\sum_{i=1}^{d} \\|p_i\\|_2^2 = 1. \\]\n\nSince given that \\( M \\) is Hermitian, for any feasible \\( p_i, i = 1, 2, \\ldots, d \\), we have\n\n\\[ \\sum_{i=1}^{d} p_i^{\\dagger}Mp_i \\leq \\sum_{i=1}^{d} \\lambda_1(M)p_i^{\\dagger}p_i = \\lambda_1(M) \\sum_{i=1}^{d} \\|p_i\\|_2^2 = \\lambda_1(M). \\]\n\nThus \\( \\alpha \\leq \\lambda_1(M) \\). In particular, if we choose \\( p_1 \\) to be the normalized eigenvector of \\( M \\) associated with \\( \\lambda_1(M) \\), and \\( p_i = 0, i = 2, 3, \\ldots, d \\), then\n\n\\[ \\sum_{i=1}^{d} \\|p_i\\|_2^2 = 1, \\quad \\sum_{i=1}^{d} p_i^{\\dagger}Mp_i = p_1^{\\dagger}Mp_1 = \\lambda_1(M). \\]\n\nTherefore we have \\( \\alpha = \\lambda_1(M) = \\beta \\).", "question": "### (c) Express the problem of computing the largest eigenvalue \\(\\lambda_1(M)\\) of a given \\(d \\times d\\) Hermitian matrix \\(M\\) in the form of a **dual problem** as above. That is, specify the map \\(\\Phi\\) (via the matrices \\(K_i\\)) and the matrices \\(A\\) and \\(B\\) such that \\(\\beta = \\lambda_1(M)\\). Write the primal problem. Show that, in this case, its optimum \\(\\alpha = \\beta\\)." }, { "context": "Recall that in class we have shown that\n\n\\[\\| \\rho - \\sigma \\|_{tr} = \\max_{E_1, E_2} \\text{tr}(\\rho E_1) + \\text{tr}(\\sigma E_2) - 1\\]\n\ns.t. \\( E_1 \\geq 0, \\, E_2 \\geq 0, \\, E_1 + E_2 = I \\),\n\ngiven that \\(\\rho, \\sigma\\) are density matrices. Let\n\n\\[ A = \\begin{pmatrix} \\rho & 0 \\\\ 0 & \\sigma \\end{pmatrix} \\in M_{2d}(C), \\quad B = I \\in M_d(C),\\]\n\n\\[ K_1 = \\begin{pmatrix} I & 0 \\\\ 0 & 0 \\end{pmatrix} \\in M_{d \\times 2d}(C), \\quad K_2 = \\begin{pmatrix} 0 & I \\\\ 0 & 0 \\end{pmatrix} \\in M_{d \\times 2d}(C).\\]\n\nConsider the two sets \\(\\Omega_1 = \\{(E_1, E_2) \\in (M_d(C), M_d(C)) : E_1 \\geq 0, E_2 \\geq 0, E_1 + E_2 = I\\}, \\Omega_2 = \\{X \\in M_{2d}(C) : X \\geq 0, \\Phi(X) = I\\}\\). For any matrix\n\n\\[X = \\begin{pmatrix} X_1 & X_3 \\\\ X_3^\\dagger & X_2 \\end{pmatrix} \\in \\Omega_2,\\]\n\nlet \\(E_1 = X_1, E_2 = X_2\\), then we have\n\n\\[\\text{tr}(AX) = \\text{tr}(\\rho E_1) + \\text{tr}(\\sigma E_2),\\]\n\nand \\((E_1, E_2) \\in \\Omega_1\\), because\n\n\\[X \\geq 0 \\implies E_1 \\geq 0, \\, E_2 \\geq 0,\\]\n\n\\[\\Phi(X) = B = I \\implies K_1 X K_1^\\dagger + K_2 X K_2^\\dagger = E_1 + E_2 = I.\\]\n\nConversely, for any \\((E_1, E_2) \\in \\Omega_1\\), let\n\n\\[X = \\begin{pmatrix} E_1 & 0 \\\\ 0 & E_2 \\end{pmatrix},\\]\n\nthen we have\n\n\\[\\text{tr}(AX) = \\text{tr}(\\rho E_1) + \\text{tr}(\\sigma E_2),\\]\n\nand \\(X \\in \\Omega_2\\), because\n\n\\[E_1 \\geq 0, \\, E_2 \\geq 0 \\implies X \\geq 0,\\]\n\n\\[E_1 + E_2 = I \\implies \\Phi(X) = K_1 X K_1^\\dagger + K_2 X K_2^\\dagger = I = B.\\]\n\nTherefore if we consider the primal problem\n\n\\[\\alpha = \\max_X \\text{tr}(AX) - 1\\]\ns.t. \\(\\Phi(X) = B, \\, X \\geq 0\\),\nit's easy to see that \\(\\alpha = \\| \\rho - \\sigma \\|_{tr} = \\| M \\|_{tr}\\). Notice that this primal problem with a ‘-1’ in the objective function is a little different from the original form above, but we can fix this\n\nby simply adding one more dimension to the problem, which would have more complicated expressions for \\(A, B, K_1, K_2\\). For convenience, we just stick to this modified primal problem.\n\nNow the modified dual problem\n\n\\[\\beta = \\min_Y \\, \\text{tr}(BY) - 1\\]\n\\[\\text{s.t.} \\quad \\Phi^*(Y) \\geq A,\\]\n\\[Y = Y^\\dagger,\\]\nis equivalent to\n\n\\[\\beta = \\min_Y \\, \\text{tr}(Y) - 1\\]\n\\[\\text{s.t.} \\quad Y \\geq \\rho, \\quad Y \\geq \\sigma,\\]\n\\[Y = Y^\\dagger,\\]\nwhich can be easily verified by substituting the explicit expressions of \\(A, B, K_1, K_2\\) into the modified dual problem.\n\nNext we need to prove in this case \\(\\beta = \\alpha = \\|\\rho - \\sigma\\|_{\\text{tr}}\\). Since in (a) we already showed that \\(\\alpha \\leq \\beta\\), we only need to show that there exists a feasible \\(Y\\) such that \\(\\text{tr}(Y) - 1 = \\alpha = \\|\\rho - \\sigma\\|_{\\text{tr}}\\). Consider the eigenvalue decomposition of \\(\\rho - \\sigma\\),\n\n\\[\\rho - \\sigma = Q \\Lambda Q^\\dagger,\\]\nwhere \\(Q\\) is unitary, and \\(\\Lambda\\) is a diagonal matrix with diagonal elements\n\n\\[\\lambda_1 \\geq \\lambda_2 \\geq \\lambda_r \\geq 0 \\geq \\lambda_{r+1} \\geq \\ldots \\geq \\lambda_d.\\]\n\nSince \\(\\rho, \\sigma\\) are density matrices, we have\n\n\\[\\|\\rho - \\sigma\\|_{\\text{tr}} = \\sum_{i=1}^r \\lambda_i = \\frac{1}{2} \\sum_{i=1}^d |\\lambda_i| = \\frac{1}{2} \\sum_{i=1}^r \\lambda_i - \\frac{1}{2} \\sum_{i=r+1}^d \\lambda_i.\\]\n\nLet \\(q_i\\) denote the \\(i\\)-th column of \\(Q\\). Now define\n\n\\[s_i = q_i^\\dagger \\rho q_i, \\quad i = 1, 2, \\ldots, d,\\]\n\\[t_i = q_i^\\dagger \\sigma q_i, \\quad i = 1, 2, \\ldots, d.\\]\n\nWe have\n\n\\[s_i - t_i = q_i^\\dagger (\\rho - \\sigma) q_i = \\lambda_i \\geq 0, \\quad i = 1, 2, \\ldots, r,\\]\n\\[t_i - s_i = -q_i^\\dagger (\\rho - \\sigma) q_i = -\\lambda_i \\geq 0, \\quad i = r+1, r+2, \\ldots, d.\\]\n\nLet \\(S, T, \\Sigma\\) be three diagonal matrices such that their diagonal vectors are \\((s_1, s_2, \\ldots, s_d)\\), \\((t_1, t_2, \\ldots, t_d)\\) and \\((s_1, s_2, \\ldots, s_r, t_{r+1}, t_{r+2}, \\ldots, t_d)\\) respectively. Then\n\n\\[\\Sigma - S = \\text{diag}(0, 0, \\ldots, 0, t_{r+1} - s_{r+1}, \\ldots, t_d - s_d) \\geq 0,\\]\n\\[\\Sigma - T = \\text{diag}(s_1 - t_1, s_2 - t_2, \\ldots, s_r - t_r, 0, \\ldots, 0) \\geq 0,\\]\n\nwhere diag(v) denotes the diagonal matrix with diagonal vector v. Notice that \\( S \\) and \\( T \\) are the diagonal parts of \\( Q^\\dagger \\rho Q \\) and \\( Q^\\dagger \\sigma Q \\) respectively. We should also notice that\n\n\\[ q_i^\\dagger (\\rho - \\sigma) q_j = 0, \\quad i \\neq j, \\]\nwhich means the non-diagonal part of \\( Q^\\dagger \\rho Q \\) and \\( Q^\\dagger \\sigma Q \\) are the same, i.e.\n\n\\[ Q^\\dagger \\rho Q - S = Q^\\dagger \\sigma Q - T \\triangleq L.\\]\n\nNow we have\n\n\\[ Q^\\dagger \\rho Q = S + L, \\quad Q^\\dagger \\sigma Q = T + L.\\]\n\nLet\n\n\\[ Y = Q(\\Sigma + L)Q^\\dagger.\\]\n\nObviously \\( Y = Y^\\dagger \\), and we have\n\n\\[ Q^\\dagger (Y - \\rho) Q = \\Sigma + L - S - L = \\Sigma - S \\geq 0 \\quad \\Rightarrow \\quad Y \\geq \\rho,\\]\n\\[ Q^\\dagger (Y - \\sigma) Q = \\Sigma + L - T - L = \\Sigma - T \\geq 0 \\quad \\Rightarrow \\quad Y \\geq \\sigma,\\]\nthus \\( Y \\) is a feasible solution to the dual problem. Moreover, notice that we have\n\n\\[ \\sum_{i=1}^d s_i = \\text{tr}(Q^\\dagger \\rho Q) = \\text{tr}(\\rho) = 1, \\quad \\sum_{i=1}^d t_i = \\text{tr}(Q^\\dagger \\sigma Q) = \\text{tr}(\\sigma) = 1,\\]\ntherefore\n\n\\[ \\text{tr}(Y) = \\text{tr}(\\Sigma) = \\sum_{i=1}^r s_i + \\sum_{i=r+1}^d t_i = \\sum_{i=1}^r (s_i - t_i) + \\sum_{i=1}^d t_i = \\sum_{i=1}^r \\lambda_i + 1 = \\|\\rho - \\sigma\\|_{tr} + 1,\\]\nthat is \\( \\beta \\leq \\text{tr}(Y) - 1 = \\|\\rho - \\sigma\\|_{tr} = \\alpha \\). In all we have \\( \\beta = \\alpha = \\|\\rho - \\sigma\\|_{tr} \\).", "question": "### (d) Suppose given a Hermitian matrix \\(M\\) that is the difference of two density matrices, \\(M = \\rho - \\sigma\\). Express the problem of computing \\(\\|M\\|_{\\text{tr}} = \\frac{1}{2} \\|M\\|_1\\) in the form of a primal problem as above. That is, specify the map \\(\\Phi\\) and the matrices \\(A\\) and \\(B\\) such that \\(\\alpha = \\|M\\|_{\\text{tr}}\\). [Hint: recall the operational interpretation of the trace distance as optimal distinguishing probability.] Write the dual problem. Show that, in this case, its optimum \\(\\beta = \\alpha\\)." }, { "context": "The success probability of distinguishing with a POVM \\( \\{M_i\\} \\) is\n\n\\[ \\sum_{i=1}^k p_i \\Pr(M_i | \\rho_i) = \\sum_{i=1}^k p_i \\text{tr}(M_i \\rho_i). \\]\n\nDefine\n\n\\[ A = \\begin{pmatrix}\np_1 \\rho_1 & p_2 \\rho_2 & \\cdots \\\\\n& \\ddots \\\\\n& & p_k \\rho_k\n\\end{pmatrix} \\in M_{kd}(C), \\quad B = I \\in M_{d}(C), \\]\n\n\\[ K_i = (0, \\ldots, 0, I, 0, \\ldots, 0) \\in M_{d \\times kd}(C), \\quad i = 1, 2, \\ldots, k. \\]\n\nthen using the similar argument in (d), we can see that solving the optimization problem\n\n\\[ \\alpha = \\max_{M_i} \\sum_{i=1}^k p_i \\text{tr}(M_i \\rho_i) \\]\ns.t. \\quad M_i \\geq 0, \\quad i = 1, 2, \\ldots, k,\n\\sum_{i=1}^{k} M_i = I,\nis equivalent to solving the primal problem\n\n\\[\\alpha = \\max_X \\, \\text{tr}(AX)\\]\ns.t. \\quad \\Phi(X) = B,\n\\quad \\quad \\quad X \\geq 0,\nand given an optimal solution \\(X^*\\) for the primal problem, we can recover an optimal solution \\(\\{M_i^*\\}_{i \\geq k}\\) for the first problem by taking \\(\\{M_i^*\\}_{i \\geq k}\\) to be the diagonal blocks of \\(X^*\\). Indeed, let \\(\\Omega_1\\) and \\(\\Omega_2\\) be the feasible sets of the two problems above respectively. For any \\(\\{M_i\\}_{i \\geq k} \\in \\Omega_1\\), let\n\n\\[X = \\begin{pmatrix}\nM_1 & M_2 & \\cdots & M_k\n\\end{pmatrix},\\]\nthen we have\n\n\\[\\text{tr}(AX) = \\sum_{i=1}^{k} p_i \\text{tr}(M_i \\rho_i), \\quad K_i X K_i^\\dagger = M_i, \\quad i = 1, 2, \\ldots, k,\\]\nand \\(X \\in \\Omega_2\\) because\n\n\\[\\sum_{i=1}^{k} M_i = I \\implies \\Phi(X) = B = I,\\]\n\\[M_i \\geq 0, \\quad i = 1, 2, \\ldots, k \\implies X \\geq 0.\\]\n\nConversely, for any \\(X \\in \\Omega_2\\), let \\(M_1, M_2, \\ldots, M_k\\) be the diagonal blocks of \\(X\\), then we have\n\n\\[\\sum_{i=1}^{k} p_i \\text{tr}(M_i \\rho_i) = \\text{tr}(AX), \\quad K_i X K_i^\\dagger = M_i, \\quad i = 1, 2, \\ldots, k,\\]\nand \\(\\{M_i\\}_{i \\geq k} \\in \\Omega_1\\) because\n\n\\[\\Phi(X) = B = I \\implies \\sum_{i=1}^{k} M_i = I,\\]\n\\[X \\geq 0 \\implies M_i \\geq 0, \\quad i = 1, 2, \\ldots, k.\\]", "question": "### (e) Suppose you are given one of \\(k\\) possible density matrices, \\(\\rho_1, ..., \\rho_k\\), each with a priori probability \\(p_1, ..., p_k\\) respectively. Your goal is to find the optimal guessing measurement: this is the \\(k\\)-outcome POVM which maximizes your chances of producing the index \\(j \\in \\{1, ..., k\\}\\), given one copy of \\(\\rho_j\\) (which is assumed to occur with probability \\(p_j\\)). First write a formula that expresses the success probability of distinguishing with a POVM \\(\\{M_x\\}\\). Show that the problem of optimizing this quantity can be expressed as a semidefinite program in primal or dual form (whichever you find most convenient).\n\nIt turns out that in many cases (essentially all “well-behaved” cases) the optimum of the primal problem of a semidefinite program equals the optimum of the dual problem. This is useful for several reasons. First of all, note how the primal is a maximization problem, while the dual is a minimization problem. Therefore any feasible solution (a candidate solution that satisfies all the constraints) to the primal provides a lower bound on the optimum, while a feasible solution to the dual provides an upper bound. The fact that they are equal shows that one can get tight bounds in this way. In addition, formulating a problem in, say, primal form, and then looking at the dual formulation, can provide useful insights on the problem. We will see examples of this later on in the course, when we discuss the relation between “guessing probability” and “conditional min-entropy”." } ]

[ { "context": "We have a maximally entangled pair of qubits\n\n\\[ |\\Phi^+\\rangle = \\sum_{0 \\leq i \\leq d-1} \\frac{1}{\\sqrt{d}} |i\\rangle_A \\otimes |i\\rangle_B \\]\n\nTo find the reduced state on \\( A \\) we trace out the \\( B \\) system\n\n\\[ Tr_B(|\\Phi^+\\rangle \\langle \\Phi^+|) = Tr_B \\left( \\sum_{0 \\leq i,j \\leq d-1} \\frac{1}{d} (|i\\rangle_A \\otimes |i\\rangle_B)(\\langle j|_A \\otimes \\langle j|_B) \\right) \\]\n\n\\[ = \\sum_{0 \\leq i,j \\leq d-1} \\frac{1}{d} Tr_B \\left( (|i\\rangle_A \\otimes |i\\rangle_B)(\\langle j|_A \\otimes \\langle j|_B) \\right) \\]\n\n\\[ = \\sum_{0 \\leq i,j \\leq d-1} \\frac{1}{d} Tr_B \\left( (|i\\rangle_A \\langle j|_A \\otimes |i\\rangle_B \\langle j|_B) \\right) \\]\n\n\\[ = \\sum_{0 \\leq i,j \\leq d-1} \\frac{1}{d} |i\\rangle_A \\langle j|_A \\]\n\nSo we get the maximally mixed state on the \\( A \\) system.", "question": "### (i) What is the reduced state on subsystem A?" }, { "context": "\\( M \\otimes I \\) and \\( I \\otimes M^T \\) are both \\( d^2 \\times d^2 \\) matrices, and \\( |\\Phi^+\\rangle \\) is a vector of length \\( d^2 \\), so \\( M \\otimes I |\\Phi^+\\rangle \\) and \\( I \\otimes M^T |\\Phi^+\\rangle \\) are vectors of length \\( d^2 \\). To see these are equal, we must show their components are equal. That is, for each \\( 0 \\leq k, l \\leq d-1 \\), we must show\n\n\\[ ((\\langle k| \\otimes \\langle l|) M \\otimes I |\\Phi^+\\rangle) = ((\\langle k| \\otimes \\langle l|) I \\otimes M^T |\\Phi^+\\rangle) \\]\n\nNow see\n\n\\[ ((\\langle k| \\otimes \\langle l|) M \\otimes I |\\Phi^+\\rangle) = ((\\langle k| \\otimes \\langle l|) M \\otimes I \\left( \\sum_{0 \\leq i \\leq d-1} \\frac{1}{\\sqrt{d}} |i\\rangle_A \\otimes |i\\rangle_B \\right)) \\]\n\n\\[ = \\sum_{0 \\leq i \\leq d-1} \\frac{1}{\\sqrt{d}} ((\\langle k| \\otimes \\langle l|) M \\otimes I (|i\\rangle_A \\otimes |i\\rangle_B)) \\]\n\n\\[ = \\sum_{0 \\leq i \\leq d-1} \\frac{1}{\\sqrt{d}} \\langle k| M |i\\rangle \\otimes \\langle l| I |i\\rangle \\]\n\n\\[ = \\sum_{0 \\leq i \\leq d-1} \\frac{1}{\\sqrt{d}} \\langle k| M |i\\rangle \\otimes \\langle l| i\\rangle \\]\n\n\\[ = \\frac{1}{\\sqrt{d}} \\langle k| M |l\\rangle \\]\n\n\\[ = \\frac{1}{\\sqrt{d}} M_{kl} \\]\n\nAnd\n\n\\[ (\\langle k| \\otimes \\langle l|) \\otimes M^T \\Phi^+ = (\\langle k| \\otimes \\langle l|) \\otimes M^T \\left( \\sum_{0 \\leq i \\leq d-1} \\frac{1}{\\sqrt{d}} |i\\rangle_A \\otimes |i\\rangle_B \\right) \\]\n\n\\[ = \\sum_{0 \\leq i \\leq d-1} \\frac{1}{\\sqrt{d}} (\\langle k| \\otimes \\langle l|) \\otimes M^T (|i\\rangle_A \\otimes |i\\rangle_B) \\]\n\n\\[ = \\sum_{0 \\leq i \\leq d-1} \\frac{1}{\\sqrt{d}} \\langle k|i\\rangle \\otimes \\langle l|M^T |i\\rangle \\]\n\n\\[ = \\sum_{0 \\leq i \\leq d-1} \\frac{1}{\\sqrt{d}} \\langle k|i\\rangle \\otimes \\langle l|M^T |i\\rangle \\]\n\n\\[ = \\frac{1}{\\sqrt{d}} M^T_{lk} \\]\n\n\\[ = \\frac{1}{\\sqrt{d}} M_{kl} \\]\n\nSo these two are indeed equal.", "question": "### (ii) Let M ∈ M_d(C). Show that M ⊗ I|φ⁺⟩ = I ⊗ Mᵀ|φ⁺⟩." } ]

[ { "context": "Assume that (2) is true, i.e.\n\n\\[T(X) = \\sum_s K_s X K_s^\\dagger, \\quad \\forall X \\in M_d(\\mathbb{C}).\\]\n\nFor any \\( d'' \\geq 0 \\) and any \\( X \\otimes Y \\in M_d(\\mathbb{C}) \\otimes M_{d''}(\\mathbb{C}) \\) such that \\( X \\otimes Y \\geq 0 \\), we have\n\n\\[T \\otimes id_{d''} (X \\otimes Y) = T(X) \\otimes Y = \\sum_s K_s X K_s^\\dagger \\otimes Y.\\]\n\nThen for any joint state\n\n\\[|\\Phi\\rangle = \\sum_{0 \\leq i \\leq d-1} \\sum_{0 \\leq j \\leq d''-1} a_{ij} |i\\rangle \\otimes |j\\rangle \\in \\text{span}\\{\\mathbb{C}^d \\otimes \\mathbb{C}^{d''}\\},\\]\n\nwe have\n\\[\\langle \\Phi | T \\otimes id_{d''} (X \\otimes Y) | \\Phi \\rangle = \\sum_s \\langle \\Phi | (K_s X K_s^\\dagger \\otimes Y) | \\Phi \\rangle\\]\n\n\\[= \\sum_s \\sum_{i,j} \\sum_{k,l} a_{ij} a_{kl} \\langle i | K_s X K_s^\\dagger | k \\rangle \\langle j | Y | l \\rangle\\]\n\n\\[= \\sum_s \\left( \\sum_{i,j} a_{ij} \\langle i | K_s \\rangle \\otimes \\langle j | \\right) (X \\otimes Y) \\left( \\sum_{k,l} a_{kl} \\langle k | K_s^\\dagger \\rangle \\otimes \\langle l | \\right)\\]\n\n\\[= \\sum_s \\left( \\langle \\Phi | K_s \\otimes \\mathbb{I} \\right) (X \\otimes Y) \\left( K_s^\\dagger \\otimes \\mathbb{I} | \\Phi \\rangle \\right)\\]\n\n\\[\\geq 0.\\]\nSince \\( | \\Phi \\rangle \\) is arbitrary, we have \\( T \\otimes id_{d''} (X \\otimes Y) \\geq 0 \\). And since \\( d'' \\) and \\( X \\otimes Y \\) are arbitrary, we can conclude that \\( T \\) is completely positive.", "question": "### (a) Show that (2) \\(\\Rightarrow\\) (3), i.e. that if \\( T \\) is such that \\( T(X) = \\sum_j K_j X K_j^\\dagger \\) for \\( X \\in M_d(\\mathbb{C}) \\), then \\( T \\) is completely positive." }, { "context": "(3) \\(\\Rightarrow\\) (1) is trivial. If \\( T \\) is completely positive, then by definition, in the case \\( d'' = d \\), \\( T \\otimes id_d \\) is positive. Since \\( \\Phi^+ = | \\Phi^+ \\rangle \\langle \\Phi^+ | \\) is positive semidefinite, we immediately have that\n\\[J(T) = T \\otimes id_d (\\Phi^+)\\]\nis positive semidefinite.", "question": "### (b) Explain why (3) \\(\\Rightarrow\\) (1)." }, { "context": "We may always assume that\n\\[X | j \\rangle = \\sum_{0 \\leq j \\leq d-1} x_{ij} | i \\rangle,\\]\nthen we can write \\( X \\) as\n\\[X = \\sum_{0 \\leq i,j \\leq d-1} x_{ij} | i \\rangle \\langle j |,\\]\nand we have\n\\[T(X) = \\sum_{0 \\leq i,j \\leq d-1} x_{ij} T(| i \\rangle \\langle j |).\\]\nOn the other hand, we have\n\\[(id_d \\otimes t_d(J(T))) (\\mathbb{I} \\otimes X) = \\left( \\sum_{0 \\leq i,j \\leq d-1} T(| i \\rangle \\langle j |) \\otimes t_d(| i \\rangle \\langle j |) \\right) (\\mathbb{I} \\otimes X)\\]\n\\[= \\sum_{0 \\leq i,j \\leq d-1} (T(| i \\rangle \\langle j |) \\otimes | j \\rangle \\langle i |) (\\mathbb{I} \\otimes X)\\]\n\\[= \\sum_{0 \\leq i,j \\leq d-1} T(| i \\rangle \\langle j |) \\otimes (| j \\rangle \\langle i | X)\\]\n\\[= \\sum_{0 \\leq i,j \\leq d-1} T(| i \\rangle \\langle j |) \\otimes (| j \\rangle \\langle i | (\\sum_{0 \\leq k,l \\leq d-1} x_{kl} | k \\rangle \\langle l |))\\]\n\\[= \\sum_{0 \\leq i,j \\leq d-1} T(| i \\rangle \\langle j |) \\otimes (\\sum_{0 \\leq k,l \\leq d-1} x_{kl} | j \\rangle \\langle l |)\\]\n\\[= \\sum_{0 \\leq i,j \\leq d-1} x_{il} T(| i \\rangle \\langle j |) \\otimes | j \\rangle \\langle l |,\\]\n\\[= \\sum_{0 \\leq i,j \\leq d-1} x_{il} T(| i \\rangle \\langle j |) \\otimes | j \\rangle \\langle l |,\\]\nand thus\n\\[tr_A \\left( (id_{d'} \\otimes td(J(T))) (\\lvert B \\rangle \\otimes X_A) \\right) = tr_A \\left( \\sum_{0 \\leq i,j \\leq d-1} x_{i} T(\\lvert i \\rangle \\langle j \\rvert) \\lvert B \\rangle \\otimes \\lvert j \\rangle \\langle i \\rvert \\right)\\]\n\\[= \\sum_{0 \\leq i,j \\leq d-1} x_{ij} T(\\lvert i \\rangle \\langle j \\rvert)\\]\n\\[= T(X).\\]\nNow we can define a bidirectional map between linear map \\( T \\) from \\( M_d(\\mathbb{C}) \\) to \\( M_{d'}(\\mathbb{C}) \\) and their operator representation \\( J(T) \\) in \\( M_{d'}(\\mathbb{C}) \\otimes M_d(\\mathbb{C}) \\). One direction is\n\\[T \\rightarrow J(T),\\]\nand the other direction is\n\\[J(T) \\rightarrow tr_A \\left( (id_{d'} \\otimes td(J(T))) (\\lvert B \\rangle \\otimes \\cdot \\lambda_A) \\right) = T(\\cdot).\\]\nThese two directions are inverse to each other, so this is a one-to-one map. Let's define \\( J^{-1}(J(T)) = T \\) for future use. Also we can see that both \\( J \\) and \\( J^{-1} \\) are linear.", "question": "### (c) Let \\( t_d : M_d(\\mathbb{C}) \\rightarrow M_d(\\mathbb{C}) \\) be the linear map defined by \\( X \\mapsto X^T \\). \\( t_d \\) transposes the matrix but we take care to define it in this way as we could be taking the transpose just on a single subsystem. Show that the action of \\( T \\) on \\( X \\in M_d(\\mathbb{C}) \\) can be written in terms of its Choi-Jamiolkowski representation \\( J(T) \\) as\n\\[ T(X) = \\text{Tr}_{\\Lambda}((\\text{id}_d \\otimes t_d(J(T)))(\\mathbb{I}_d \\otimes X_{\\Lambda})). \\tag{2} \\]\nand deduce that there is a one-to-one correspondence between linear maps from \\( M_d(\\mathbb{C}) \\) to \\( M_{d^2}(\\mathbb{C}) \\) and their operator representations in \\( M_{d^2}(\\mathbb{C}) \\)." }, { "context": "For an arbitrary \\( Z \\in M_{d' \\times d} \\),\n\\[Z = \\sum_{0 \\leq i \\leq d' - 1} \\sum_{0 \\leq j \\leq d - 1} a_{ij} \\lvert i \\rangle \\lvert j \\rangle,\\]\nwe have\n\\[vec(Z) = \\sum_{0 \\leq i \\leq d' - 1} \\sum_{0 \\leq j \\leq d - 1} a_{ij} vec(\\lvert i \\rangle \\lvert j \\rangle)\\]\n\\[= \\sum_{0 \\leq i \\leq d' - 1} \\sum_{0 \\leq j \\leq d - 1} a_{ij} \\lvert i \\rangle \\otimes \\lvert j \\rangle\\]\n\\[= \\sum_{0 \\leq j \\leq d - 1} \\left( \\sum_{0 \\leq i \\leq d' - 1} a_{ij} \\lvert i \\rangle \\right) \\otimes \\lvert j \\rangle\\]\n\\[= \\sum_{0 \\leq j \\leq d - 1} Z \\lvert j \\rangle \\otimes \\lvert j \\rangle,\\]\nwhere we have used the fact that\n\\[Z \\lvert j \\rangle = \\sum_{0 \\leq i \\leq d' - 1} a_{ij} \\lvert i \\rangle, \\quad 0 \\leq j \\leq d - 1.\\]\nTherefore we have\n\\[J(T) = T \\otimes id_d(\\Phi^+)\\]\n\\[= \\sum_{0 \\leq i,j \\leq d-1} T(|i\\rangle \\langle j|) \\otimes |i\\rangle \\langle j|\\]\n\\[= \\sum_{0 \\leq i,j \\leq d-1} (Z|i\\rangle \\langle j|Z^\\dagger) \\otimes |i\\rangle \\langle j|\\]\n\\[= \\sum_{0 \\leq i,j \\leq d-1} ((Z|i\\rangle) \\otimes |i\\rangle)((\\langle j|Z^\\dagger) \\otimes \\langle j|)\\]\n\\[= (\\sum_{0 \\leq i,j \\leq d-1} (Z|i\\rangle) \\otimes |i\\rangle) (\\sum_{0 \\leq i,j \\leq d-1} (\\langle j|Z^\\dagger) \\otimes \\langle j|)\\]\n\\[= |\\zeta\\rangle \\langle \\zeta|,\\]\nwith \\(|\\zeta\\rangle = vec(Z)\\).", "question": "### (d) Define the linear map \\( \\text{vec} : M_{d^2}(\\mathbb{C}) \\rightarrow M_d(\\mathbb{C}) \\otimes M_d(\\mathbb{C}) \\) by its action on the standard basis, \\( \\text{vec} : |i\\rangle \\langle j| \\mapsto |i\\rangle \\otimes |j\\rangle \\) for \\( 0 \\leq i < d \\) and \\( 0 \\leq j < d \\). Let \\( Z \\in M_{d^2}(\\mathbb{C}) \\). Show that a map of the form \\( T : X \\mapsto Z X Z^\\dagger \\) has Choi-Jamiolkowski representation \\( |\\zeta\\rangle \\langle \\zeta| \\) where \\( |\\zeta\\rangle = \\text{vec}(Z) \\)." }, { "context": "If (1) is true, \\(J(T)\\) is positive semidefinite, we should be able to write \\(J(T)\\) in form of its eigenvalue decomposition\n\\[J(T) = \\sum_s \\lambda_s |\\zeta_s\\rangle \\langle \\zeta_s|,\\]\nwhere \\(\\lambda_s > 0\\) for each \\(s\\), and each\n\\[|\\zeta_s\\rangle = \\sum_{0 \\leq i,j \\leq d-1} c_{ij}^s |i\\rangle \\otimes |j\\rangle \\in span\\{C^d \\otimes C^d\\}\\]\nis an normalized eigenstate of \\(J(T)\\). Let's define\n\\[Z_s = \\sum_{0 \\leq i,j \\leq d-1} c_{ij}^s |i\\rangle \\langle j|, \\quad T_s = J^{-1}(|\\zeta_s\\rangle \\langle \\zeta_s|),\\]\nwhere the notation \\(J^{-1}\\) has been defined in (c). Then it's easy to check that\n\\[vec(Z_s) = |\\zeta_s\\rangle, \\quad J(T_s) = |\\zeta_s\\rangle \\langle \\zeta_s|,\\]\nand using the result of (c) and (d) we have\n\\[T_s(X) = Z_s X Z_s^\\dagger, \\quad \\forall X \\in M_d(C).\\]\nNow we have\n\\[J(T) = \\sum_s \\lambda_s |\\zeta_s\\rangle \\langle \\zeta_s| = \\sum_s \\lambda_s J(T_s),\\]\nthen by linearity we have\n\\[T = J^{-1}(J(T)) = \\sum_s \\lambda_s J^{-1}(J(T_s)) = \\sum_s \\lambda_s T_s.\\]\nFurther, since \\(\\lambda_s > 0\\) for each \\(s\\), we can define\n\\[K_s = \\sqrt{\\lambda_s} Z_s,\\]\nthen we have\n\\[T(X) = \\sum_s \\lambda_s T_s(X) = \\sum_s \\lambda_s Z_s X Z_s^\\dagger = \\sum_s K_s X K_s^\\dagger, \\quad \\forall X \\in M_d(C),\\]\nwhich means (2) is true.", "question": "### (e) Show that (1) \\(\\Rightarrow\\) (2), i.e. suppose that a map \\( T : M_d(\\mathbb{C}) \\rightarrow M_{d^2}(\\mathbb{C}) \\) has a positive semidefinite Choi-Jamiolkowski representation. Construct a set of maps \\( \\{K_j \\in M_{d^2}(\\mathbb{C})\\} \\) such that \\( T(X) = \\sum_j K_j X K_j^\\dagger \\) for any \\( X \\in M_d(\\mathbb{C}) \\). [**Hint:** You might find calculations from part (d) helpful.]" } ]

[ { "context": "The success probability is\n\n\\[\n\\Pr(\\text{success}) = \\frac{1}{4} \\sum_{x, \\theta \\in \\{0,1\\}} \\text{tr}(T(|x\\rangle \\langle x| \\otimes |x\\rangle \\langle x| \\otimes |x\\rangle \\langle x|)).\n\\]", "question": "### (a) Write out the formula which expresses the success probability of an attack specified by a CPTP map \\( T \\)." }, { "context": "Notice that for any matrices \\( N \\in M_4(\\mathbb{C}) \\) and \\( M \\in M_2(\\mathbb{C}) \\), we have\n\n\\[\n\\begin{aligned}\n\\text{tr}(J(T)(N \\otimes M)) &= \\sum_{i,j \\in \\{0,1\\}} \\text{tr}((T(|i\\rangle \\langle j|) \\otimes |i\\rangle \\langle j|)(N \\otimes M)) \\\\\n&= \\sum_{i,j \\in \\{0,1\\}} \\text{tr}(T(|i\\rangle \\langle j|)N \\otimes |i\\rangle \\langle j|M) \\\\\n&= \\sum_{i,j \\in \\{0,1\\}} \\text{tr}(T(|i\\rangle \\langle j|)N) \\text{tr}(|i\\rangle \\langle j|M) \\\\\n&= \\sum_{i,j \\in \\{0,1\\}} \\text{tr}(T(|i\\rangle \\langle j|)N) m_{ij} \\\\\n&= \\text{tr}(T(\\sum_{i,j \\in \\{0,1\\}} m_{ij} |i\\rangle \\langle j|)N) \\\\\n&= \\text{tr}(T(M)N),\n\\end{aligned}\n\\]\n\nwhere we have used that\n\n\\[\nM = \\sum_{i,j \\in \\{0,1\\}} m_{ij} |i\\rangle \\langle j|,\n\\]\n\n\\[\nm_{ij} = \\langle i|M|j\\rangle = \\text{tr}(|i\\rangle \\langle j|M), \\quad i, j \\in \\{0, 1\\}.\n\\]\n\nThen using this result by taking \\( N = |x\\rangle \\langle x| \\otimes |x\\rangle \\langle x|, M = |x\\rangle \\langle x| \\) for each pair of \\( (x, \\theta) \\), we have\n\n\\[\n\\begin{aligned}\n\\text{tr}(J(T)Q) &= \\sum_{x, \\theta \\in \\{0,1\\}} \\text{tr}(J(T)(|x\\rangle \\langle x| \\otimes |x\\rangle \\langle x| \\otimes |x\\rangle \\langle x|)) \\\\\n&= \\sum_{x, \\theta \\in \\{0,1\\}} \\text{tr}(T(|x\\rangle \\langle x| \\otimes |x\\rangle \\langle x| \\otimes |x\\rangle \\langle x|)) \\\\\n&= 4 \\Pr(\\text{success}),\n\\end{aligned}\n\\]\n\nthat is\n\n\\[\n\\Pr(\\text{success}) = \\frac{1}{4} \\text{tr}(J(T)Q).\n\\]", "question": "### (b) Let \\( J(T) = \\sum_{i,j \\in \\{0,1\\}} T(|i\\rangle \\langle j|) \\otimes |i\\rangle \\langle j| \\) be the Choi-Jamiolkowski representation of the map \\( T \\). Consider the matrix \\( Q = \\sum_{x, \\theta \\in \\{0,1\\}} |x\\rangle \\otimes |\\theta\\rangle \\langle x| \\otimes \\langle \\theta| \\). Write the success probability of the map \\( T \\) as a simple expression involving \\( J(T) \\) and \\( Q \\)." }, { "context": "If \\( T \\) is trace-preserving, then we have\n\n\\[\n\\text{tr}_1(J(T)) = \\sum_{0 \\leq i,j \\leq d-1} \\text{tr}(T(|i\\rangle \\langle j|)) |i\\rangle \\langle j|\n\\]\n\n\\[\n= \\sum_{0 \\leq i,j \\leq d-1} \\delta_{ij} |i\\rangle \\langle j|\n\\]\n\n\\[\n= \\sum_{0 \\leq i \\leq d-1} |i\\rangle \\langle i|\n\\]\n\n\\[\n= I_d.\n\\]\n\nConversely, if we have\n\n\\[\n\\sum_{0 \\leq i,j \\leq d-1} \\text{tr}(T(|i\\rangle \\langle j|)) |i\\rangle \\langle j| = I_d,\n\\]\n\nthen\n\n\\[\n\\delta_{ij} = \\langle i | I_d | j \\rangle = \\langle i | \\left( \\sum_{0 \\leq k,l \\leq d-1} \\text{tr}(T(|k\\rangle \\langle l|)) |k\\rangle \\langle l| \\right) | j \\rangle = \\text{tr}(T(|i\\rangle \\langle j|)), \\quad \\forall 0 \\leq i, j \\leq d-1.\n\\]\n\nTherefore for any\n\n\\[\nX = \\sum_{0 \\leq i,j \\leq d-1} x_{ij} |i\\rangle \\langle j| \\in M_d(\\mathbb{C}),\n\\]\n\nwe have\n\n\\[\n\\text{tr}(T(X)) = \\text{tr} \\left( \\sum_{0 \\leq i,j \\leq d-1} x_{ij} T(|i\\rangle \\langle j|) \\right)\n\\]\n\n\\[\n= \\sum_{0 \\leq i,j \\leq d-1} x_{ij} \\text{tr}(T(|i\\rangle \\langle j|))\n\\]\n\n\\[\n= \\sum_{0 \\leq i \\leq d-1} x_{ii}\n\\]\n\n\\[\n= \\text{tr}(X).\n\\]\n\nSince \\( X \\) is arbitrary, we may conclude that \\( T \\) is trace-preserving.", "question": "### (c) Show that the condition that the CP map \\( T \\) is trace-preserving can be expressed as the condition that its Choi-Jamiolkowski representation \\( J(T) \\) satisfies\n\\[ \\text{Tr}_1 (J(T)) = \\sum_{0 \\leq i,j \\leq d-1} \\text{Tr} \\left( T(|i\\rangle \\langle j|) \\right) |i\\rangle \\langle j| = I_d. \\]" }, { "context": "Recall in (b) we have shown that\n\n\\[\n\\text{Pr}(\\text{success}(T)) = \\frac{1}{4} \\text{tr}(Q J(T)),\n\\]\n\nso we may take \\( A = \\frac{1}{4} Q \\), and the variable \\( X = J(T) \\). After obtaining the optimal \\( X^* \\), we may recover the optimal \\( T^* \\) as \\( T^* = J^{-1}(X^*) \\), where \\( J^{-1} \\) is defined in problem 4 (c) as\n\n\\[\nJ^{-1}(X)( \\cdot ) = \\text{tr}_A \\left( (id_A \\otimes t_2(X))(I_B \\otimes ( \\cdot ) \\otimes \\lambda) \\right).\n\\]\n\nDefine\n\n\\[\nK_{ij} = \\langle i | \\langle j | \\otimes I_{t_2}, \\quad i, j \\in \\{0, 1\\},\n\\]\n\n\\[\n\\Phi(X) = \\sum_{i,j \\in \\{0,1\\}} K_{ij} X K_{ij}^{\\dagger},\n\\]\n\nthen for any matrices \\( N \\in M_4(\\mathbb{C}) \\) and \\( M \\in M_2(\\mathbb{C}) \\) we have\n\n\\[\n\\Phi(N \\otimes M) = \\sum_{i,j \\in \\{0,1\\}} K_{ij} (N \\otimes M) K_{ij}^{\\dagger}\n= \\sum_{i,j \\in \\{0,1\\}} (\\langle i| \\langle j| N |i \\rangle |j \\rangle) \\otimes M\n= \\text{tr}(N) M.\n\\]\n\nThen\n\n\\[\n\\Phi(J(T)) = \\sum_{0 \\leq i,j \\leq d-1} \\Phi(T(|i \\rangle \\langle j|)) \\otimes |i \\rangle \\langle j| = \\sum_{0 \\leq i,j \\leq d-1} \\text{tr}(T(|i \\rangle \\langle j|)) |i \\rangle \\langle j| = \\text{tr}(J(T)).\n\\]\n\nNow if we take \\( B = I_2 \\), then the condition\n\n\\[\n\\Phi(J(T)) = \\Phi(X) = B = I_2\n\\]\n\nensures that \\( T \\) is trace-preserving, by the result of (c). Moreover, the condition\n\n\\[\nJ(T) = X \\geq 0\n\\]\n\nensures that \\( T \\) is completely positive, by the result of problem 4. Then finally, we can obtain the optimal success probability of attack based on CPTP map by solving the primal problem\n\n\\[\n\\alpha = \\max_X \\text{tr} \\left( \\frac{1}{4} QX \\right)\n\\]\nsubject to\n\n\\[\n\\Phi(X) = I_2,\nX \\geq 0,\n\\]\nwith the optimal success probability equal to \\( \\alpha \\) and the optimal CPTP map \\( T^* = J^{-1}(X^*) \\).\n\nThe dual problem is\n\n\\[\n\\beta = \\min_Y \\text{tr}(Y)\n\\]\nsubject to\n\n\\[\n\\Phi^*(Y) \\geq \\frac{1}{4} Q,\nY = Y^{\\dagger}.\n\\]\n\nNotice that\n\n\\[\n\\Phi^*(Y) = \\sum_{i,j \\in \\{0,1\\}} K_{ij}^{\\dagger} Y K_{ij} = \\sum_{i,j \\in \\{0,1\\}} \\langle i| Y |j \\rangle \\langle i| j \\rangle \\otimes Y = I_4 \\otimes Y,\n\\]\nthe dual problem can have a more explicit form\n\n\\[\n\\beta = \\min_Y \\text{tr}(Y)\n\\]\nsubject to\n\n\\[\nI_4 \\otimes Y \\geq \\frac{1}{4} Q,\nY = Y^{\\dagger}.\n\\]", "question": "### (d) Find a semidefinite program in primal form (see problem 2.) whose optimum is the success probability of an arbitrary attack on the single-qubit Wiesner quantum money scheme. *[Hint: recall the characterization of CP maps from their Choi-Jamiolkowski representation given in the previous problem, and use the previous question as well]* Write down the dual semidefinite program." }, { "context": "Let's solve the primal problem in (d):\n\n\\[\n\\alpha = \\max_X \\, \\text{tr}(AX) \\\\\n\\text{s.t.} \\quad \\Phi(X) = I_2, \\\\\nX \\geq 0.\n\\]\n\nFor Wiesner's scheme, we have\n\n\\[\nA = \\frac{1}{4} Q = \\frac{1}{4} (|\\psi_1\\rangle \\langle \\psi_1| + |\\psi_2\\rangle \\langle \\psi_2| + |\\psi_3\\rangle \\langle \\psi_3| + |\\psi_4\\rangle \\langle \\psi_4|),\n\\]\n\nwhere\n\n\\[\n|\\psi_1\\rangle = |0\\rangle |0\\rangle |0\\rangle, \\quad |\\psi_2\\rangle = |1\\rangle |1\\rangle |1\\rangle, \\quad |\\psi_3\\rangle = |+\\rangle |+\\rangle |+\\rangle, \\quad |\\psi_4\\rangle = |\\-\\rangle |\\-\\rangle |\\-\\rangle.\n\\]\n\nWith help of Matlab, we can easily find the eigenvalue decomposition of \\(Q\\),\n\n\\[\nQ = U \\Lambda U^\\dagger,\n\\]\n\nwhere \\(U\\) is unitary, and\n\n\\[\n\\Lambda = \\text{diag}\\left(\\frac{3}{8}, \\frac{3}{8}, \\frac{1}{8}, \\frac{1}{8}, 0, 0, 0, 0\\right).\n\\]\n\nThat is, all eigenvalues of \\(A\\) are\n\n\\[\n\\lambda_1 = \\lambda_2 = \\frac{3}{8}, \\quad \\lambda_3 = \\lambda_4 = \\frac{1}{8}, \\quad \\lambda_5 = \\lambda_6 = \\lambda_7 = \\lambda_8 = 0.\n\\]\n\nThen we can immediately obtain an upper bound for our objective function given that \\(X\\) is a feasible solution,\n\n\\[\n\\text{tr}(AX) = \\text{tr}(U \\Lambda U^\\dagger X) \\leq \\lambda_1(A) \\text{tr}(U^\\dagger X U) = \\lambda_1(A) \\text{tr}(X) = 2 \\lambda_1(A) = \\frac{3}{4}.\n\\]\n\nTherefore if we can achieve this upper bound with some feasible \\(X\\), then the problem is solved. Indeed, to make the inequality to become equality in the formula above, i.e.\n\n\\[\n\\text{tr}(U^\\dagger X U) = \\lambda_1(A) \\text{tr}(U^\\dagger X U),\n\\]\n\nwe need the diagonal entries of \\(U^\\dagger X U\\) to focus on the first two entries which are associated with \\(\\lambda_1(A), \\lambda_2(A)\\). Recall that \\(\\text{tr}(U^\\dagger X U) = \\text{tr}(X) = 2\\), a natural guess would be\n\n\\[\nU^\\dagger X^* U = \\text{diag}(1, 1, 0, 0, 0, 0, 0, 0),\n\\]\n\nand we have\n\n\\[\nX^* = U \\text{diag}(1, 1, 0, 0, 0, 0, 0, 0) U^\\dagger = \n\\begin{pmatrix}\n3/4 & 0 & 0 & 1/4 & 0 & 1/4 & 1/4 & 0 \\\\\n0 & 1/12 & 1/12 & 0 & 1/12 & 0 & 0 & 1/12 \\\\\n0 & 1/12 & 1/12 & 0 & 1/12 & 0 & 0 & 1/12 \\\\\n1/4 & 0 & 0 & 1/12 & 0 & 1/12 & 1/12 & 0 \\\\\n0 & 1/12 & 1/12 & 0 & 1/12 & 0 & 0 & 1/12 \\\\\n1/4 & 0 & 0 & 1/12 & 0 & 1/12 & 1/12 & 0 \\\\\n1/4 & 0 & 0 & 1/12 & 0 & 1/12 & 1/12 & 0 \\\\\n0 & 1/4 & 1/4 & 0 & 1/4 & 0 & 0 & 3/4\n\\end{pmatrix}\n\\]\n\n\\[\n= |\\zeta_1\\rangle \\langle \\zeta_1| + |\\zeta_2\\rangle \\langle \\zeta_2|,\n\\]\n\nwhere \\( |\\zeta_1\\rangle, |\\zeta_2\\rangle \\) are the eigenstates of \\( A \\) corresponding to eigenvalues \\( \\lambda_1, \\lambda_2 \\),\n\n\\[ |\\zeta_1\\rangle = \\frac{1}{\\sqrt{12}} (3|0\\rangle|0\\rangle|0\\rangle + |0\\rangle|1\\rangle|1\\rangle + 1|1\\rangle|0\\rangle|1\\rangle + 1|1\\rangle|1\\rangle|0\\rangle), \\]\n\n\\[ |\\zeta_2\\rangle = \\frac{1}{\\sqrt{12}} (|0\\rangle|0\\rangle|1\\rangle + |0\\rangle|1\\rangle|0\\rangle + 1|1\\rangle|0\\rangle|0\\rangle + 3|1\\rangle|1\\rangle|1\\rangle). \\]\n\nIt's easy to check that \\( X^* \\) is a feasible solution, i.e.\n\n\\[ \\Phi(X^*) = I_2, \\quad X^* \\geq 0, \\]\n\nthus we have \\( \\alpha = \\frac{3}{4} \\), the optimal success probability is \\( \\frac{3}{4} \\).\n\nOur next mission is to recover \\( T^* \\) from \\( X^* \\). Now we can make use of the useful results in problem 4. Let\n\n\\[ Z_1 = \\frac{1}{\\sqrt{12}} (3|0\\rangle|0\\rangle|0\\rangle + |0\\rangle|1\\rangle|1\\rangle|0\\rangle + 1|1\\rangle|0\\rangle|1\\rangle|0\\rangle), \\]\n\n\\[ Z_2 = \\frac{1}{\\sqrt{12}} (|0\\rangle|0\\rangle|1\\rangle + |0\\rangle|1\\rangle|0\\rangle + 1|1\\rangle|0\\rangle|0\\rangle + 3|1\\rangle|1\\rangle|1\\rangle), \\]\n\nthen we have\n\n\\[ \\text{vec}(Z_1) = |\\zeta_1\\rangle, \\quad \\text{vec}(Z_2) = |\\zeta_2\\rangle. \\]\n\nDefine\n\n\\[ T_1(\\rho) = Z_1 \\rho Z_1^\\dagger, \\quad \\forall \\rho \\in M_2(\\mathbb{C}), \\]\n\n\\[ T_2(\\rho) = Z_2 \\rho Z_2^\\dagger, \\quad \\forall \\rho \\in M_2(\\mathbb{C}). \\]\n\nBy the result of problem 4(d), we have\n\n\\[ J^{-1}(|\\zeta_1\\rangle \\langle \\zeta_1|) = T_1, \\quad J^{-1}(|\\zeta_2\\rangle \\langle \\zeta_2|) = T_2. \\]\n\nFinally we have\n\n\\[ T^* = J^{-1}(X^*) = J^{-1}(|\\zeta_1\\rangle \\langle \\zeta_1|) + J^{-1}(|\\zeta_2\\rangle \\langle \\zeta_2|) = T_1 + T_2, \\]\n\nthat is we have\n\n\\[ T^*(\\rho) = Z_1 \\rho Z_1^\\dagger + Z_2 \\rho Z_2^\\dagger \\quad \\forall \\rho \\in M_2(\\mathbb{C}). \\]", "question": "### (e) Solve the semidefinite program! That is, give an explicit matrix which achieves the optimum, together with the value of the optimum. *[Hint: I will allow you to google — but if you do so, state your source. Serious bonus points for solving the problem yourself, either by hand (explain your reasoning) or using Matlab or any other program (print out your code).*" }, { "context": "Consider a linear map \\( U \\) such that\n\n\\[ U : |0\\rangle|0\\rangle|0\\rangle \\longrightarrow \\frac{1}{\\sqrt{12}} \\left( (3|0\\rangle|0\\rangle + 1|1\\rangle) \\otimes |0\\rangle + (|0\\rangle|1\\rangle + 1|1\\rangle) \\otimes |1\\rangle \\right), \\]\n\n\\[ U : |1\\rangle|0\\rangle|0\\rangle \\longrightarrow \\frac{1}{\\sqrt{12}} \\left( (|0\\rangle|1\\rangle + 1|1\\rangle) \\otimes |0\\rangle + (|0\\rangle|0\\rangle + 3|1\\rangle|1\\rangle) \\otimes |1\\rangle \\right). \\]\n\nBy direct calculation, we can check that\n\n\\[ |U|0\\rangle|0\\rangle|0\\rangle| = |U|1\\rangle|0\\rangle|0\\rangle| = 1, \\quad (U|0\\rangle|0\\rangle|0\\rangle)^{\\dagger} (U|1\\rangle|0\\rangle|0\\rangle) = 0, \\]\n\ntherefore we can extend \\( U \\) to be a unitary operator for all three-qubits (use a similar argument for HW2 problem 6(b)). We still denote this extended unitary operator as \\( U \\). Notice that\n\n\\[\nZ_1|0\\rangle = \\frac{1}{\\sqrt{12}} (3|0\\rangle|0\\rangle + |1\\rangle|1\\rangle), \\quad Z_1|1\\rangle = \\frac{1}{\\sqrt{12}} (|0\\rangle|1\\rangle + 3|1\\rangle|0\\rangle),\n\\]\n\n\\[\nZ_2|0\\rangle = \\frac{1}{\\sqrt{12}} (|0\\rangle|1\\rangle + |1\\rangle|0\\rangle), \\quad Z_2|1\\rangle = \\frac{1}{\\sqrt{12}} (|0\\rangle|0\\rangle + 3|1\\rangle|1\\rangle),\n\\]\n\nthus we have\n\n\\[\nU(|0\\rangle|0\\rangle|0\\rangle) = (Z_1|0\\rangle) \\otimes |0\\rangle + (Z_2|0\\rangle) \\otimes |1\\rangle,\n\\]\n\n\\[\nU(|1\\rangle|0\\rangle|0\\rangle) = (Z_1|1\\rangle) \\otimes |0\\rangle + (Z_2|1\\rangle) \\otimes |1\\rangle.\n\\]\n\nThen for any single qubit \\( |\\phi\\rangle \\), by linearity, we always have\n\n\\[\nU(|\\phi\\rangle|0\\rangle|0\\rangle) = (Z_1|\\phi\\rangle) \\otimes |0\\rangle + (Z_2|\\phi\\rangle) \\otimes |1\\rangle.\n\\]\n\nNotice that we have\n\n\\[\ntr_3(U|\\phi\\rangle|0\\rangle\\langle 0| \\langle \\phi|U^{\\dagger}) = tr_3 \\left( Z_1|\\phi\\rangle \\langle \\phi|Z_1^{\\dagger} \\otimes |0\\rangle \\langle 0| + Z_2|\\phi\\rangle \\langle \\phi|Z_2^{\\dagger} \\otimes |1\\rangle \\langle 1| + Z_1|\\phi\\rangle \\langle \\phi|Z_2^{\\dagger} \\otimes |0\\rangle \\langle 1| + Z_2|\\phi\\rangle \\langle \\phi|Z_1^{\\dagger} \\otimes |1\\rangle \\langle 0| \\right)\n\\]\n\n\\[\n= Z_1|\\phi\\rangle \\langle \\phi|Z_1^{\\dagger} + Z_2|\\phi\\rangle \\langle \\phi|Z_2^{\\dagger} = T^*(|\\phi\\rangle \\langle \\phi|).\n\\]\n\nNow we can clarify our optimal attack found in (e). Given a money qubit \\( |\\phi\\rangle \\), the attack steps are\n\n(i) **Operation**: Append \\( |0\\rangle|0\\rangle \\) to \\( |\\phi\\rangle \\). **Outcome**: \\( |\\phi\\rangle|0\\rangle|0\\rangle \\).\n\n(ii) **Operation**: Apply \\( U \\) to \\( |\\phi\\rangle|0\\rangle|0\\rangle \\). **Outcome**: \\( (Z_1|\\phi\\rangle) \\otimes |0\\rangle + (Z_2|\\phi\\rangle) \\otimes |1\\rangle \\).\n\n(iii) **Operation**: Trace out the third qubit. **Outcome**: \\( T^*(|\\phi\\rangle \\langle \\phi|) \\).", "question": "### (f) Give an explicit representation of the attack you found in (e) as a sequence of three operations: (i) appending some auxiliary qubits in state |0⟩; (ii) applying a unitary transformation on all qubits; (iii) performing a partial trace or measurement map on some of the qubits. *[If you weren’t able to solve (e), you can ignore this question. It will only count for 1 point.]*" } ]