Solutions for Exercises 1 and 2
Exercise 1 Start with something simple: What is the stabilizer S such that is stabilized by S? G 1 = linspan ( 00, 11 )
Exercise 1 Start with something simple: What is the stabilizer S such that is stabilized by S? S = Z Z. G 1 = linspan ( 00, 11 )
Exercise 1 Start with something simple: What is the stabilizer S such that G 1 = linspan ( 00, 11 ) is stabilized by S? S = Z Z. Now what are the stabilizers S 1, S 2 such that G 2 = linspan ( 000, 111 ) is stabilized by S 1 and S 2?
Exercise 1 Start with something simple: What is the stabilizer S such that G 1 = linspan ( 00, 11 ) is stabilized by S? S = Z Z. Now what are the stabilizers S 1, S 2 such that G 2 = linspan ( 000, 111 ) is stabilized by S 1 and S 2? Ensure that qubits 1 and 2 are the same and also qubits 2 and 3 are the same. S 1 = Z Z I, S 2 = I Z Z, X = X X X
Exercise 1 Start with something simple: What is the stabilizer S such that G 1 = linspan ( 00, 11 ) is stabilized by S? S = Z Z. Now what are the stabilizers S 1, S 2 such that G 2 = linspan ( 000, 111 ) is stabilized by S 1 and S 2? Ensure that qubits 1 and 2 are the same and also qubits 2 and 3 are the same. S 1 = Z Z I, S 2 = I Z Z, X = X X X This is the 3 qubit code resistent to X errors.
Exercise 1 Similarly, the 3 qubit code resistent to Z errors will have stabilizers S 3, S 4 that stabilize the space Analagously, G 3 = linspan ( + + +, ). S 3 = X X I, S 4 = I X X, Z = Z Z Z We can see this by noticing G 3 = (H H H)G 2. So the stabilizer S 3 = (HHH)(ZZI )(HHH) = XXI
Exercise 1 This would also work if we worked with logical qubits instead of qubits. So, when we apply the 3 qubit Z-resistent code to the 3 qubit X -resistent code, S 3 =X X I = XXXXXXIII S 4 =I X X = IIIXXXXXX Z =ZZZ = ZZZZZZZZZ
Exercise 1 So now we need to add the stabilizers for each of the X -resistent codes to get the entire set of stabilizers. S 1 = ZZI III III S 2 = IZZ III III S 3 = III ZZI III S 4 = III IZZ III S 5 = III III ZZI S 6 = III III IZZ S 7 = XXX XXX III S 8 = III XXX XXX X = XXX XXX XXX Z = ZZZ ZZZ ZZZ
Exercise 2 Simulating one qubit measurements with constant size quantum circuit We have a depth-d quantum circuit, and we re interested in a measurement just on one qubit of the final state. How much of the circuit does this depend on? Idea: backward lightcone.
Exercise 2 Simulating one qubit measurements with constant size quantum circuit We have a depth-d quantum circuit, and we re interested in a measurement just on one qubit of the final state. How much of the circuit does this depend on? Idea: backward lightcone. x 0 x 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9
Exercise 2 Simulating one qubit measurements with constant size quantum circuit We have a depth-d quantum circuit, and we re interested in a measurement just on one qubit of the final state. How much of the circuit does this depend on? Idea: backward lightcone. x 0 x 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9
Exercise 2 Simulating one qubit measurements with constant size quantum circuit We have a depth-d quantum circuit, and we re interested in a measurement just on one qubit of the final state. How much of the circuit does this depend on? Idea: backward lightcone. x 0 x 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9
Exercise 2 Simulating one qubit measurements with constant size quantum circuit x 0 x 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9 We have a depth-d quantum circuit, and we re interested in a measurement just on one qubit of the final state. How much of the circuit does this depend on? Idea: backward lightcone. Answer: It depends on at most D = 2 d qubits.
Exercise 2 Simulating one qubit measurements with constant size quantum circuit x 0 x 1 x 2 x 3 x 4 x 5 We have a depth-d quantum circuit, and we re interested in a measurement just on one qubit of the final state. How much of the circuit does this depend on? Idea: backward lightcone. Answer: It depends on at most D = 2 d qubits. This circuit has depth d and acts on D qubits. It has size at most Dd. The measurement outcome has the same distribution as the same measurement in the original circuit.
Exercise 2 Characterizing a constant size quantum circuit with a constant depth classical circuit Our circuit of size Dd is fully described by a sequence of matrix multiplications, where each matrix has size 2 D 2 D. Notice that the depth of the new circuit 2 D = 2 2d is doubly exponential in the depth of the original circuit!
Exercise 2 Characterizing a constant size quantum circuit with a constant depth classical circuit Our circuit of size Dd is fully described by a sequence of matrix multiplications, where each matrix has size 2 D 2 D. Notice that the depth of the new circuit 2 D = 2 2d is doubly exponential in the depth of the original circuit! Let be the unitary describing the circuit. Then we can compute matrix entries of in constant time using the above.
Exercise 2 Characterizing a constant size quantum circuit with a constant depth classical circuit Our circuit of size Dd is fully described by a sequence of matrix multiplications, where each matrix has size 2 D 2 D. Notice that the depth of the new circuit 2 D = 2 2d is doubly exponential in the depth of the original circuit! Let be the unitary describing the circuit. Then we can compute matrix entries of in constant time using the above. We can compute the probabilities of the measurement outcome like so: ( 0 I )C x D 2 = 0z C x D 2. (1) z {0,1} D 1 Each complex number 0z C x D can be computed in constant time, and we need to compute only a constant number of them. So we can compute the probability of measuring 0.
Exercise 2 Classical constant depth circuits compute the same functions as quantum ones Suppose that computes f most of the time, in the sense that f (x) x 2 1 2.
Exercise 2 Classical constant depth circuits compute the same functions as quantum ones Suppose that computes f most of the time, in the sense that f (x) x 2 1 2. For the i th bit, the probability that measuring the i th bit of x gives the i th bit of f (x) is at least 1 2.
Exercise 2 Classical constant depth circuits compute the same functions as quantum ones Suppose that computes f most of the time, in the sense that f (x) x 2 1 2. For the i th bit, the probability that measuring the i th bit of x gives the i th bit of f (x) is at least 1 2. Therefore, we can extract each bit of the output of f.
Exercise 2 Classical constant depth circuits compute the same functions as quantum ones Suppose that computes f most of the time, in the sense that f (x) x 2 1 2. For the i th bit, the probability that measuring the i th bit of x gives the i th bit of f (x) is at least 1 2. Therefore, we can extract each bit of the output of f. Theorem If quantum circuit of constant depth computes f with probability 1 2, then there is a constant-depth classical circuit computing f.
Exercise 7 Computational lemmas Let φ D = D 1/2 i i i. φ D A B φ D = (D 1/2 i i i )A B(D 1/2 j j j )
Exercise 7 Computational lemmas Let φ D = D 1/2 i i i. φ D A B φ D = (D 1/2 i i i )A B(D 1/2 j j j ) = D 1 ij i i A B j j
Exercise 7 Computational lemmas Let φ D = D 1/2 i i i. φ D A B φ D = (D 1/2 i i i )A B(D 1/2 j j j ) = D 1 ij = D 1 ij i i A B j j i A j i B j
Exercise 7 Computational lemmas Let φ D = D 1/2 i i i. φ D A B φ D = (D 1/2 i i i )A B(D 1/2 j j j ) = D 1 ij = D 1 ij = D 1 ij i i A B j j i A j i B j i A j j B T i
Exercise 7 Computational lemmas Let φ D = D 1/2 i i i. φ D A B φ D = (D 1/2 i i i )A B(D 1/2 j j j ) = D 1 ij = D 1 ij = D 1 ij = D 1 i i i A B j j i A j i B j i A j j B T i i A j j B T i j
Exercise 7 Computational lemmas Let φ D = D 1/2 i i i. φ D A B φ D = (D 1/2 i i i )A B(D 1/2 j j j ) = D 1 ij = D 1 ij = D 1 ij = D 1 i i i A B j j i A j i B j i A j j B T i i A j j B T i j = D 1 i i AB T i = D 1 Tr AB T.
Exercise 7 Computational lemmas Suppose AB = BA. Then also AB T = B T A.
Exercise 7 Computational lemmas Suppose AB = BA. Then also AB T = B T A. By cyclicity, Tr AB = Tr BA. But by linearity, Tr AB = Tr BA. Therefore, Tr AB = 0.
Exercise 7 Computational lemmas Suppose AB = BA. Then also AB T = B T A. By cyclicity, Tr AB = Tr BA. But by linearity, Tr AB = Tr BA. Therefore, Tr AB = 0. We also note that AB = BA iff ABA 1 B 1 = I. The quantity ABA 1 B 1 is known as the group commutator of A and B.
Exercise 7 Computational lemmas Suppose AB = BA. Then also AB T = B T A. By cyclicity, Tr AB = Tr BA. But by linearity, Tr AB = Tr BA. Therefore, Tr AB = 0. We also note that AB = BA iff ABA 1 B 1 = I. The quantity ABA 1 B 1 is known as the group commutator of A and B. Notice that the group commutator factorizes across the tensor product. More precisely, suppose A = A 0 A 1, B = B 0 B 1. Then ABA 1 B 1 = A 0 B 0 A 1 0 B 1 0 A 1 B 1 A 1 1 B 1 1. (2)
Exercise 7 Constructing the operators Consider the matrices C 1 = Z I I I C 2 = X Z I I C 3 = X X Z I. C d = X X X Z
Exercise 7 Constructing the operators Consider the matrices C 1 = Z I I I C 2 = X Z I I C 3 = X X Z I. C d = X X X Z We claim that C i C j = C j C i when i j.
Exercise 7 Constructing the operators Consider the matrices C 1 = Z I I I C 2 = X Z I I C 3 = X X Z I. C d = X X X Z We claim that C i C j = C j C i when i j. To see this, consider the group commutator term-by-term in the tensor product.
Exercise 7 Constructing the operators Consider the matrices C 1 = Z I I I C 2 = X Z I I C 3 = X X Z I. C d = X X X Z We claim that C i C j = C j C i when i j. To see this, consider the group commutator term-by-term in the tensor product. Exactly one group commutator is between an X and Z. The rest are between X and X or between I and something.
Exercise 7 Constructing the operators Consider the matrices C 1 = Z I I I C 2 = X Z I I C 3 = X X Z I. C d = X X X Z We claim that C i C j = C j C i when i j. To see this, consider the group commutator term-by-term in the tensor product. Exactly one group commutator is between an X and Z. The rest are between X and X or between I and something. The commutators of the latter type are all equal to I. The X, Z commutator is equal to I.
Exercise 7 Constructing the operators Consider the matrices C 1 = Z I I I C 2 = X Z I I C 3 = X X Z I. C d = X X X Z We claim that C i C j = C j C i when i j. To see this, consider the group commutator term-by-term in the tensor product. Exactly one group commutator is between an X and Z. The rest are between X and X or between I and something. The commutators of the latter type are all equal to I. The X, Z commutator is equal to I. Therefore, the whole commutator is I.
Exercise 7 Analyzing the operators Given vector u R d, let C(u) = i u ic i. Let s compute φ d C(u) C(v) φ d.
Exercise 7 Analyzing the operators Given vector u R d, let C(u) = i u ic i. Let s compute φ d C(u) C(v) φ d. By our first computational lemma, this is equal to 1 D Tr C(u)C(v)T.
Exercise 7 Analyzing the operators Given vector u R d, let C(u) = i u ic i. Let s compute φ d C(u) C(v) φ d. By our first computational lemma, this is equal to 1 D Tr C(u)C(v)T. By linearity, this is equal to 1 D ij Tr u iv j C i Cj T.
Exercise 7 Analyzing the operators Given vector u R d, let C(u) = i u ic i. Let s compute φ d C(u) C(v) φ d. By our first computational lemma, this is equal to 1 D Tr C(u)C(v)T. By linearity, this is equal to 1 D ij Tr u iv j C i C T By our second computational lemma, Tr C i C T j also 1 D Tr C ic T i = 1. j. = 0 for i j. But
Exercise 7 Analyzing the operators Given vector u R d, let C(u) = i u ic i. Let s compute φ d C(u) C(v) φ d. By our first computational lemma, this is equal to 1 D Tr C(u)C(v)T. By linearity, this is equal to 1 D ij Tr u iv j C i C T By our second computational lemma, Tr C i C T j also 1 D Tr C ic T i = 1. Therefore, φ D C(u) C(v) φ D = i u iv i = u v. j. = 0 for i j. But
Exercise 7 Analyzing the operators Given vector u R d, let C(u) = i u ic i. Let s compute φ d C(u) C(v) φ d. By our first computational lemma, this is equal to 1 D Tr C(u)C(v)T. By linearity, this is equal to 1 D ij Tr u iv j C i C T By our second computational lemma, Tr C i Cj T = 0 for i j. But also 1 D Tr C ici T = 1. Therefore, φ D C(u) C(v) φ D = i u iv i = u v. To conclude: any correlation achieved by inner products of vectors is also achieved by making measurements on a state. j.