Tracking Quantum Circuits By Polynomials

Transcription

1 Tracking Quantum Circuits By Polynomials ACSS 2016, Kolkata Kenneth W. Regan 1 University at Buffalo (SUNY) 13 August, Includes joint work with Amlan Chakrabarti, U. Calcutta, and prospectively students...

2 Quantum Computers

3 Quantum Computers A Particular Wave of the Future...

4 Quantum Computers A Particular Wave of the Future... Apparently capable of out-performing current computers vastly at certain tasks...

5 Quantum Computers A Particular Wave of the Future... Apparently capable of out-performing current computers vastly at certain tasks most notably Factoring, Discete Logarithm, and simulating quantum systems.

6 Quantum Computers A Particular Wave of the Future... Apparently capable of out-performing current computers vastly at certain tasks most notably Factoring, Discete Logarithm, and simulating quantum systems. If, that is, we can build them scalably,

7 Quantum Computers A Particular Wave of the Future... Apparently capable of out-performing current computers vastly at certain tasks most notably Factoring, Discete Logarithm, and simulating quantum systems. If, that is, we can build them scalably, and if those tasks are not classically easy.

8 Quantum Computers A Particular Wave of the Future... Apparently capable of out-performing current computers vastly at certain tasks most notably Factoring, Discete Logarithm, and simulating quantum systems. If, that is, we can build them scalably, and if those tasks are not classically easy. Central notion: Quantum Circuits.

9 Quantum Computers A Particular Wave of the Future... Apparently capable of out-performing current computers vastly at certain tasks most notably Factoring, Discete Logarithm, and simulating quantum systems. If, that is, we can build them scalably, and if those tasks are not classically easy. Central notion: Quantum Circuits. Hardly the only player quantum adiabatic machines are closer to built and quantum communication systems are already deployed.

10 Boolean Circuits Have...

11 Boolean Circuits Have... n inputs x 1,..., x n {0, 1} n

12 Boolean Circuits Have... n inputs x 1,..., x n {0, 1} n r 1 outputs z 1,..., z r

13 Boolean Circuits Have... n inputs x 1,..., x n {0, 1} n r 1 outputs z 1,..., z r Maybe h-many nondeterministic inputs y 1,..., y h?

14 Boolean Circuits Have... n inputs x 1,..., x n {0, 1} n r 1 outputs z 1,..., z r Maybe h-many nondeterministic inputs y 1,..., y h? m gates g 1,..., g m (wlog. all NAND)

15 Boolean Circuits Have... n inputs x 1,..., x n {0, 1} n r 1 outputs z 1,..., z r Maybe h-many nondeterministic inputs y 1,..., y h? m gates g 1,..., g m (wlog. all NAND) Up to 2m + r wires (if fan-in 2)

16 Boolean Circuits Have... n inputs x 1,..., x n {0, 1} n r 1 outputs z 1,..., z r Maybe h-many nondeterministic inputs y 1,..., y h? m gates g 1,..., g m (wlog. all NAND) Up to 2m + r wires (if fan-in 2) Each wire has a definite 0-1 value.

17 Boolean Circuits Have... n inputs x 1,..., x n {0, 1} n r 1 outputs z 1,..., z r Maybe h-many nondeterministic inputs y 1,..., y h? m gates g 1,..., g m (wlog. all NAND) Up to 2m + r wires (if fan-in 2) Each wire has a definite 0-1 value. Bits have no common identity across wires, but we can create a universal layout for Boolean circuits in which they do retain identity...

18 Turing Cue Bits Space s, so n s ancillary cells.

19 Turing Cue Bits Space s, so n s ancillary cells. Can also be made reversible.

20 Quantum Circuits: similar picture, 90 flipped.

21 Quantum Circuits: similar picture, 90 flipped. Must be reversible. In certain circumstances values y 1, y 2,... can be copied to extra lines as f(x). Then reversing the gates re-creates the input x so the whole mapping is invertible ( copy-uncompute trick ).

22 Quantum Circuits Have...

23 Quantum Circuits Have... n input qubits x 1,..., x n {0, 1} n

24 Quantum Circuits Have... n input qubits x 1,..., x n {0, 1} n (What s a qubit?)

25 Quantum Circuits Have... n input qubits x 1,..., x n {0, 1} n (What s a qubit?) r 1 output qubits z 1,..., z r (think r = 1 or r = n)

26 Quantum Circuits Have... n input qubits x 1,..., x n {0, 1} n (What s a qubit?) r 1 output qubits z 1,..., z r (think r = 1 or r = n) s n ancilla qubits

27 Quantum Circuits Have... n input qubits x 1,..., x n {0, 1} n (What s a qubit?) r 1 output qubits z 1,..., z r (think r = 1 or r = n) s n ancilla qubits m-many quantum gates (arities can be 1,2,3)

28 Quantum Circuits Have... n input qubits x 1,..., x n {0, 1} n (What s a qubit?) r 1 output qubits z 1,..., z r (think r = 1 or r = n) s n ancilla qubits m-many quantum gates (arities can be 1,2,3) Maybe h of them are Hadamard gates, which supply nondeterminism.

29 Quantum Circuits Have... n input qubits x 1,..., x n {0, 1} n (What s a qubit?) r 1 output qubits z 1,..., z r (think r = 1 or r = n) s n ancilla qubits m-many quantum gates (arities can be 1,2,3) Maybe h of them are Hadamard gates, which supply nondeterminism. Qubits retain identity as wires transit gates.

30 Quantum Circuits Have... n input qubits x 1,..., x n {0, 1} n (What s a qubit?) r 1 output qubits z 1,..., z r (think r = 1 or r = n) s n ancilla qubits m-many quantum gates (arities can be 1,2,3) Maybe h of them are Hadamard gates, which supply nondeterminism. Qubits retain identity as wires transit gates. Each wire need not have a definite 0-1 value, owing to entanglement.

31 Quantum Circuits Have... n input qubits x 1,..., x n {0, 1} n (What s a qubit?) r 1 output qubits z 1,..., z r (think r = 1 or r = n) s n ancilla qubits m-many quantum gates (arities can be 1,2,3) Maybe h of them are Hadamard gates, which supply nondeterminism. Qubits retain identity as wires transit gates. Each wire need not have a definite 0-1 value, owing to entanglement. Under the hood are S = 2 s complex entries of a unit state vector.

32 A Qubit

33 A Qubit A qubit is a physical entity that combines as described by a 2-vector (a, b) = ae 0 + be 1 over C and yields two observations: e 0 with probability a 2 or e 1 with probability b 2.

34 A Qubit A qubit is a physical entity that combines as described by a 2-vector (a, b) = ae 0 + be 1 over C and yields two observations: e 0 with probability a 2 or e 1 with probability b 2. A qutrit yields 3 results and is (a, b, c) where a 2 + b 2 + c 2 = 1.

35 Quantum Gates

36 Quantum Gates A k-ary gate can be represented by a K K unitary matrix, K = 2 k.

37 Quantum Gates A k-ary gate can be represented by a K K unitary matrix, K = 2 k. Common gates for k = 1, K = 2:

38 Quantum Gates A k-ary gate can be represented by a K K unitary matrix, K = 2 k. Common gates for k = 1, K = 2: [ ] 1 0 I = identity 0 1

39 Quantum Gates A k-ary gate can be represented by a K K unitary matrix, K = 2 k. Common gates for k = 1, K = 2: [ ] 1 0 I = identity 0 1 [ ] 0 1 X = negation, aka. NOT 1 0

40 Quantum Gates A k-ary gate can be represented by a K K unitary matrix, K = 2 k. Common gates for k = 1, K = 2: [ ] 1 0 I = identity 0 1 [ ] 0 1 X = negation, aka. NOT 1 0 [ ] H = Hadamard gate. 1 1

41 Quantum Gates A k-ary gate can be represented by a K K unitary matrix, K = 2 k. Common gates for k = 1, K = 2: [ ] 1 0 I = identity 0 1 [ ] 0 1 X = negation, aka. NOT 1 0 [ ] H = Hadamard gate. 1 1 Only non-permutation gate needed for universality.

42 Quantum Gates A k-ary gate can be represented by a K K unitary matrix, K = 2 k. Common gates for k = 1, K = 2: [ ] 1 0 I = identity 0 1 [ ] 0 1 X = negation, aka. NOT 1 0 [ ] H = Hadamard gate. 1 1 Only non-permutation gate needed for universality. [ ] [ ] [ ] 0 i But also common: Y =, Z =, S =. i i

43 Binary Gates With k = 2 qubits, K = 4. Controlled Not showing quantum coordinates: CNOT =

44 Binary Gates With k = 2 qubits, K = 4. Controlled Not showing quantum coordinates: CNOT = Permutation ( ),

45 Binary Gates With k = 2 qubits, K = 4. Controlled Not showing quantum coordinates: CNOT = Permutation ( ), swap (3 4).

46 Binary Gates With k = 2 qubits, K = 4. Controlled Not showing quantum coordinates: CNOT = Permutation ( ), swap (3 4). Also called CX.

47 Binary Gates With k = 2 qubits, K = 4. Controlled Not showing quantum coordinates: CNOT = Permutation ( ), swap (3 4). Also called CX CNOT (H I) =

48 Binary Gates With k = 2 qubits, K = 4. Controlled Not showing quantum coordinates: CNOT = Permutation ( ), swap (3 4). Also called CX CNOT (H I) = Applied to e 00 = (1, 0, 0, 0) T gives 1 2 (e 00 + e 11 ).

49 Binary Gates With k = 2 qubits, K = 4. Controlled Not showing quantum coordinates: CNOT = Permutation ( ), swap (3 4). Also called CX CNOT (H I) = Applied to e 00 = (1, 0, 0, 0) T gives 1 2 (e 00 + e 11 ). EPR Entanglement.

50 Ternary Toffoli Gate: K = 8

51 Ternary Toffoli Gate: K = 8 TOF = diag(1, 1, 1, 1, 1, 1), then [ ]

52 Ternary Toffoli Gate: K = 8 TOF = diag(1, 1, 1, 1, 1, 1), then [ ] Fixes 000,..., 101; swaps

53 Ternary Toffoli Gate: K = 8 TOF = diag(1, 1, 1, 1, 1, 1), then [ ] Fixes 000,..., 101; swaps Control-Control-NOT, hence also called CCX.

54 Ternary Toffoli Gate: K = 8 TOF = diag(1, 1, 1, 1, 1, 1), then [ ] Fixes 000,..., 101; swaps Control-Control-NOT, hence also called CCX. TOF(a, b, 1) = (,, a NAND b), Thus TOF is classically universal.

55 Ternary Toffoli Gate: K = 8 TOF = diag(1, 1, 1, 1, 1, 1), then [ ] Fixes 000,..., 101; swaps Control-Control-NOT, hence also called CCX. TOF(a, b, 1) = (,, a NAND b), Thus TOF is classically universal. H + TOF is quantum universal.

56 Ternary Toffoli Gate: K = 8 TOF = diag(1, 1, 1, 1, 1, 1), then [ ] Fixes 000,..., 101; swaps Control-Control-NOT, hence also called CCX. TOF(a, b, 1) = (,, a NAND b), Thus TOF is classically universal. H + TOF is quantum universal. H + CNOT is not quantum universal; it recognizes a proper subclass of P.

57 Example Quantum Circuit

58 Example Quantum Circuit 1 H I H I (s 3).

59 Example Quantum Circuit 1 H I H I (s 3). 2 CNOT I (s 2). First three lines have CXI.

60 Example Quantum Circuit 1 H I H I (s 3). 2 CNOT I (s 2). First three lines have CXI. 3 CIX semantically but not syntactically of I and CNOT.

61 Example Quantum Circuit 1 H I H I (s 3). 2 CNOT I (s 2). First three lines have CXI. 3 CIX semantically but not syntactically of I and CNOT. 4 After the S in stage 4, a TOF with controls on 1,3 and target on 2. The whole C computes a unitary U C.

62 Input-Output and Measurement

63 Input-Output and Measurement Input: E x = e x0 s n

64 Input-Output and Measurement Input: E x = e x0 s n = e x1 e x2 e xn e (s n) 0.

65 Input-Output and Measurement Input: E x = e x0 s n = e x1 e x2 e xn e (s n) 0. Output: A state vector (z 0,..., z S 1 ), S = 2 s.

66 Input-Output and Measurement Input: E x = e x0 s n = e x1 e x2 e xn e (s n) 0. Output: A state vector (z 0,..., z S 1 ), S = 2 s. Measure all lines: For any outcome b {0, 1} s, Pr[C(x) b] = z b 2 = E x U C e b 2.

67 Input-Output and Measurement Input: E x = e x0 s n = e x1 e x2 e xn e (s n) 0. Output: A state vector (z 0,..., z S 1 ), S = 2 s. Measure all lines: For any outcome b {0, 1} s, Pr[C(x) b] = z b 2 = E x U C e b 2. (Show how DavyW applet does this.)

68 Input-Output and Measurement Input: E x = e x0 s n = e x1 e x2 e xn e (s n) 0. Output: A state vector (z 0,..., z S 1 ), S = 2 s. Measure all lines: For any outcome b {0, 1} s, Pr[C(x) b] = z b 2 = E x U C e b 2. (Show how DavyW applet does this.) For outcome d {0, 1} r on r-many designated qubit lines, Pr[C(x) d] = b d z b 2.

69 Input-Output and Measurement Input: E x = e x0 s n = e x1 e x2 e xn e (s n) 0. Output: A state vector (z 0,..., z S 1 ), S = 2 s. Measure all lines: For any outcome b {0, 1} s, Pr[C(x) b] = z b 2 = E x U C e b 2. (Show how DavyW applet does this.) For outcome d {0, 1} r on r-many designated qubit lines, Pr[C(x) d] = b d z b 2. Can project as amplitudes: C(x) (z 0,..., z 2 r 1 ) where z d 2 is the probability of outcome d {0, 1} r.

70 Input-Output and Measurement Input: E x = e x0 s n = e x1 e x2 e xn e (s n) 0. Output: A state vector (z 0,..., z S 1 ), S = 2 s. Measure all lines: For any outcome b {0, 1} s, Pr[C(x) b] = z b 2 = E x U C e b 2. (Show how DavyW applet does this.) For outcome d {0, 1} r on r-many designated qubit lines, Pr[C(x) d] = b d z b 2. Can project as amplitudes: C(x) (z 0,..., z 2 r 1 ) where z d 2 is the probability of outcome d {0, 1} r. Call this amplitude z d as A[C(x) d].

71 BQP Definition A language L belongs to BQP if there are poly-time uniform quantum circuits C n for each n such that forall n and inputs x {0, 1} n, designating qubit 1 for yes/no output: x L = Pr[C n (x) 1] > 3 4, x / L = Pr[C n (x) 1] < 1 4,

72 BQP Definition A language L belongs to BQP if there are poly-time uniform quantum circuits C n for each n such that forall n and inputs x {0, 1} n, designating qubit 1 for yes/no output: x L = Pr[C n (x) 1] > 3 4, x / L = Pr[C n (x) 1] < 1 4, The language L = {(x, w) : w is an initial part of the unique prime factorization of x} captures the task of factoring x.

73 BQP Definition A language L belongs to BQP if there are poly-time uniform quantum circuits C n for each n such that forall n and inputs x {0, 1} n, designating qubit 1 for yes/no output: x L = Pr[C n (x) 1] > 3 4, x / L = Pr[C n (x) 1] < 1 4, The language L = {(x, w) : w is an initial part of the unique prime factorization of x} captures the task of factoring x. So the fact that L BQP says that QCs can efficiently factor.

74 BQP Definition A language L belongs to BQP if there are poly-time uniform quantum circuits C n for each n such that forall n and inputs x {0, 1} n, designating qubit 1 for yes/no output: x L = Pr[C n (x) 1] > 3 4, x / L = Pr[C n (x) 1] < 1 4, The language L = {(x, w) : w is an initial part of the unique prime factorization of x} captures the task of factoring x. So the fact that L BQP says that QCs can efficiently factor. Goal: calculate the amplitude of C n (x) 1,

75 BQP Definition A language L belongs to BQP if there are poly-time uniform quantum circuits C n for each n such that forall n and inputs x {0, 1} n, designating qubit 1 for yes/no output: x L = Pr[C n (x) 1] > 3 4, x / L = Pr[C n (x) 1] < 1 4, The language L = {(x, w) : w is an initial part of the unique prime factorization of x} captures the task of factoring x. So the fact that L BQP says that QCs can efficiently factor. Goal: calculate the amplitude of C n (x) 1, explicitly or implicitly.

76 BQP Definition A language L belongs to BQP if there are poly-time uniform quantum circuits C n for each n such that forall n and inputs x {0, 1} n, designating qubit 1 for yes/no output: x L = Pr[C n (x) 1] > 3 4, x / L = Pr[C n (x) 1] < 1 4, The language L = {(x, w) : w is an initial part of the unique prime factorization of x} captures the task of factoring x. So the fact that L BQP says that QCs can efficiently factor. Goal: calculate the amplitude of C n (x) 1, explicitly or implicitly. But no classical way known without paying 2 n time overhead as for factoring.

77 Points of the Polynomial Simulation Existing general simulations pay 2 n m or 2 s m overhead right away.

78 Points of the Polynomial Simulation Existing general simulations pay 2 n m or 2 s m overhead right away. Idea: Paying only O(s m) we can build a polynomial P = P C that encodes all info of C algebraically.

79 Points of the Polynomial Simulation Existing general simulations pay 2 n m or 2 s m overhead right away. Idea: Paying only O(s m) we can build a polynomial P = P C that encodes all info of C algebraically. The amplitude(s) we seek are calculable from the numbers of solutions to certain equations P (y ; ˆx) = ẑ.

80 Points of the Polynomial Simulation Existing general simulations pay 2 n m or 2 s m overhead right away. Idea: Paying only O(s m) we can build a polynomial P = P C that encodes all info of C algebraically. The amplitude(s) we seek are calculable from the numbers of solutions to certain equations P (y ; ˆx) = ẑ. Call that number e.g. N P,x,z [1].

81 Points of the Polynomial Simulation Existing general simulations pay 2 n m or 2 s m overhead right away. Idea: Paying only O(s m) we can build a polynomial P = P C that encodes all info of C algebraically. The amplitude(s) we seek are calculable from the numbers of solutions to certain equations P (y ; ˆx) = ẑ. Call that number e.g. N P,x,z [1]. Opposite: N P,x,z [0]

82 Points of the Polynomial Simulation Existing general simulations pay 2 n m or 2 s m overhead right away. Idea: Paying only O(s m) we can build a polynomial P = P C that encodes all info of C algebraically. The amplitude(s) we seek are calculable from the numbers of solutions to certain equations P (y ; ˆx) = ẑ. Call that number e.g. N P,x,z [1]. Opposite: N P,x,z [0] For general P this number is NP-hard to compute. No Free Lunch.

83 Points of the Polynomial Simulation Existing general simulations pay 2 n m or 2 s m overhead right away. Idea: Paying only O(s m) we can build a polynomial P = P C that encodes all info of C algebraically. The amplitude(s) we seek are calculable from the numbers of solutions to certain equations P (y ; ˆx) = ẑ. Call that number e.g. N P,x,z [1]. Opposite: N P,x,z [0] For general P this number is NP-hard to compute. No Free Lunch. It is complete for the solution-counting class #P.

84 Points of the Polynomial Simulation Existing general simulations pay 2 n m or 2 s m overhead right away. Idea: Paying only O(s m) we can build a polynomial P = P C that encodes all info of C algebraically. The amplitude(s) we seek are calculable from the numbers of solutions to certain equations P (y ; ˆx) = ẑ. Call that number e.g. N P,x,z [1]. Opposite: N P,x,z [0] For general P this number is NP-hard to compute. No Free Lunch. It is complete for the solution-counting class #P. But in many cases, #SAT solvers may be effective.

85 Points of the Polynomial Simulation Existing general simulations pay 2 n m or 2 s m overhead right away. Idea: Paying only O(s m) we can build a polynomial P = P C that encodes all info of C algebraically. The amplitude(s) we seek are calculable from the numbers of solutions to certain equations P (y ; ˆx) = ẑ. Call that number e.g. N P,x,z [1]. Opposite: N P,x,z [0] For general P this number is NP-hard to compute. No Free Lunch. It is complete for the solution-counting class #P. But in many cases, #SAT solvers may be effective. Computation by solver more offline than managing 2 n -sized arrays.

86 Points of the Polynomial Simulation Existing general simulations pay 2 n m or 2 s m overhead right away. Idea: Paying only O(s m) we can build a polynomial P = P C that encodes all info of C algebraically. The amplitude(s) we seek are calculable from the numbers of solutions to certain equations P (y ; ˆx) = ẑ. Call that number e.g. N P,x,z [1]. Opposite: N P,x,z [0] For general P this number is NP-hard to compute. No Free Lunch. It is complete for the solution-counting class #P. But in many cases, #SAT solvers may be effective. Computation by solver more offline than managing 2 n -sized arrays. P C may yield other algebraic and physical info about C, including about entanglement.

87 The Basic Theorem Theorem (Dawson et al , implicitly before?) Given C built from TOF gates and h-many H gates, we can efficiently compute a polynomial P C (x 1,..., x n, y 1,..., y h, z 1,..., z r ) and a constant R (here, R = 2 h ) such that for all x {0, 1} n and z {0, 1} r, A[C(x) z] = 1 R (N P,x,z[1] N P,x,z [0]).

88 The Basic Theorem Theorem (Dawson et al , implicitly before?) Given C built from TOF gates and h-many H gates, we can efficiently compute a polynomial P C (x 1,..., x n, y 1,..., y h, z 1,..., z r ) and a constant R (here, R = 2 h ) such that for all x {0, 1} n and z {0, 1} r, A[C(x) z] = 1 R (N P,x,z[1] N P,x,z [0]). Thus BQP reduces to the difference between two #P functions.

89 The Basic Theorem Theorem (Dawson et al , implicitly before?) Given C built from TOF gates and h-many H gates, we can efficiently compute a polynomial P C (x 1,..., x n, y 1,..., y h, z 1,..., z r ) and a constant R (here, R = 2 h ) such that for all x {0, 1} n and z {0, 1} r, A[C(x) z] = 1 R (N P,x,z[1] N P,x,z [0]). Thus BQP reduces to the difference between two #P functions. Note heavy promise: 0 N[1] N[0] R = 2 h.

90 The Basic Theorem Theorem (Dawson et al , implicitly before?) Given C built from TOF gates and h-many H gates, we can efficiently compute a polynomial P C (x 1,..., x n, y 1,..., y h, z 1,..., z r ) and a constant R (here, R = 2 h ) such that for all x {0, 1} n and z {0, 1} r, A[C(x) z] = 1 R (N P,x,z[1] N P,x,z [0]). Thus BQP reduces to the difference between two #P functions. Note heavy promise: 0 N[1] N[0] R = 2 h. Means all but a trace of pairs y, y {0, 1} h cancel.

91 The Basic Theorem Theorem (Dawson et al , implicitly before?) Given C built from TOF gates and h-many H gates, we can efficiently compute a polynomial P C (x 1,..., x n, y 1,..., y h, z 1,..., z r ) and a constant R (here, R = 2 h ) such that for all x {0, 1} n and z {0, 1} r, A[C(x) z] = 1 R (N P,x,z[1] N P,x,z [0]). Thus BQP reduces to the difference between two #P functions. Note heavy promise: 0 N[1] N[0] R = 2 h. Means all but a trace of pairs y, y {0, 1} h cancel. Hence cannot approximate the difference by approximating each term. Need exact solution counting.

92 My Extensions

93 My Extensions Say a gate is balanced if all nonzero entries re iθ of its matrix have equal magnitude r.

94 My Extensions Say a gate is balanced if all nonzero entries re iθ of its matrix have equal magnitude r. A circuit C is balanced if every gate in C is balanced.

95 My Extensions Say a gate is balanced if all nonzero entries re iθ of its matrix have equal magnitude r. A circuit C is balanced if every gate in C is balanced. K(C) = the least K such that all θ in entries of gates in C are multiples of 2π/K. Min-Phase

96 My Extensions Say a gate is balanced if all nonzero entries re iθ of its matrix have equal magnitude r. A circuit C is balanced if every gate in C is balanced. K(C) = the least K such that all θ in entries of gates in C are multiples of 2π/K. Min-Phase Let G be a field or ring such that G embeds the K-th roots of unity ω j by a multiplicative homomorphism e(ω j ). Theorem Can arrange P C = gates g P g such that for all x and z, A[C(x) z] = 1 R K 1 j=0 ω j N PC,x,z[e(ω j )]

97 My Extensions Say a gate is balanced if all nonzero entries re iθ of its matrix have equal magnitude r. A circuit C is balanced if every gate in C is balanced. K(C) = the least K such that all θ in entries of gates in C are multiples of 2π/K. Min-Phase Let G be a field or ring such that G embeds the K-th roots of unity ω j by a multiplicative homomorphism e(ω j ). Theorem Can arrange P C = gates g P g such that for all x and z, A[C(x) z] = 1 R K 1 j=0 ω j N PC,x,z[e(ω j )] = 1 ω PC(x,y,z). R y

98 How it Works: The Quantum Grille Consider a general 2 2 matrix A. Assign an indicator variable u to its input and y to its output: (1 y) y (1 u) a 11 a 12 u a 21 a 22

99 How it Works: The Quantum Grille Consider a general 2 2 matrix A. Assign an indicator variable u to its input and y to its output: (1 y) y (1 u) a 11 a 12 u a 21 a 22 P A = a 11 + (a 21 a 11 )u + (a 12 a 11 )y + (a 11 a 12 a 21 + a 22 )uy.

100 How it Works: The Quantum Grille Consider a general 2 2 matrix A. Assign an indicator variable u to its input and y to its output: (1 y) y (1 u) a 11 a 12 u a 21 a 22 P A = a 11 + (a 21 a 11 )u + (a 12 a 11 )y + (a 11 a 12 a 21 + a 22 )uy. Given u, y {0, 1}, only one path is allowed others zeroed out.

101 How it Works: The Quantum Grille Consider a general 2 2 matrix A. Assign an indicator variable u to its input and y to its output: (1 y) y (1 u) a 11 a 12 u a 21 a 22 P A = a 11 + (a 21 a 11 )u + (a 12 a 11 )y + (a 11 a 12 a 21 + a 22 )uy. Given u, y {0, 1}, only one path is allowed others zeroed out. Carries out Feynman s Sum Over Paths construction.

102 How it Works: The Quantum Grille Consider a general 2 2 matrix A. Assign an indicator variable u to its input and y to its output: (1 y) y (1 u) a 11 a 12 u a 21 a 22 P A = a 11 + (a 21 a 11 )u + (a 12 a 11 )y + (a 11 a 12 a 21 + a 22 )uy. Given u, y {0, 1}, only one path is allowed others zeroed out. Carries out Feynman s Sum Over Paths construction. Works over any field or ring that embeds 0, 1, 1.

103 How it Works: The Quantum Grille Consider a general 2 2 matrix A. Assign an indicator variable u to its input and y to its output: (1 y) y (1 u) a 11 a 12 u a 21 a 22 P A = a 11 + (a 21 a 11 )u + (a 12 a 11 )y + (a 11 a 12 a 21 + a 22 )uy. Given u, y {0, 1}, only one path is allowed others zeroed out. Carries out Feynman s Sum Over Paths construction. Works over any field or ring that embeds 0, 1, 1. If gate is deterministic, can substitute y by expression in u.

104 How it Works: The Quantum Grille Consider a general 2 2 matrix A. Assign an indicator variable u to its input and y to its output: (1 y) y (1 u) a 11 a 12 u a 21 a 22 P A = a 11 + (a 21 a 11 )u + (a 12 a 11 )y + (a 11 a 12 a 21 + a 22 )uy. Given u, y {0, 1}, only one path is allowed others zeroed out. Carries out Feynman s Sum Over Paths construction. Works over any field or ring that embeds 0, 1, 1. If gate is deterministic, can substitute y by expression in u. Every qubit at every stage has a well-defined local algebraic value.

105 How it Works: Two-Qubit Gate Now let A be a general 4 4 matrix. Assign indicator variables u 1, u 2 to the two incoming qubits and y 1, y 2 to their outgoing selves: (1 y 1 )(1 y 2 ) (1 y 1 )y 2 y 1 (1 y 2 ) y 1 y 2 (1 u 1 )(1 u 2 ) a 11 a 12 a 13 a 14 (1 u 1 )u 2 a 21 a 22 a 23 a 24 u 1 (1 u 2 ) a 31 a 32 a 33 a 34 u 1 u 2 a 41 a 42 a 43 a 44

106 How it Works: Two-Qubit Gate Now let A be a general 4 4 matrix. Assign indicator variables u 1, u 2 to the two incoming qubits and y 1, y 2 to their outgoing selves: (1 y 1 )(1 y 2 ) (1 y 1 )y 2 y 1 (1 y 2 ) y 1 y 2 (1 u 1 )(1 u 2 ) a 11 a 12 a 13 a 14 (1 u 1 )u 2 a 21 a 22 a 23 a 24 u 1 (1 u 2 ) a 31 a 32 a 33 a 34 u 1 u 2 a 41 a 42 a 43 a 44 Similar for 8 8 etc. Initially there are a lot of terms.

107 How it Works: Two-Qubit Gate Now let A be a general 4 4 matrix. Assign indicator variables u 1, u 2 to the two incoming qubits and y 1, y 2 to their outgoing selves: (1 y 1 )(1 y 2 ) (1 y 1 )y 2 y 1 (1 y 2 ) y 1 y 2 (1 u 1 )(1 u 2 ) a 11 a 12 a 13 a 14 (1 u 1 )u 2 a 21 a 22 a 23 a 24 u 1 (1 u 2 ) a 31 a 32 a 33 a 34 u 1 u 2 a 41 a 42 a 43 a 44 Similar for 8 8 etc. Initially there are a lot of terms. However, with substitution for permutation gates, the entire polynomial collapses to the constant 1

108 How it Works: Two-Qubit Gate Now let A be a general 4 4 matrix. Assign indicator variables u 1, u 2 to the two incoming qubits and y 1, y 2 to their outgoing selves: (1 y 1 )(1 y 2 ) (1 y 1 )y 2 y 1 (1 y 2 ) y 1 y 2 (1 u 1 )(1 u 2 ) a 11 a 12 a 13 a 14 (1 u 1 )u 2 a 21 a 22 a 23 a 24 u 1 (1 u 2 ) a 31 a 32 a 33 a 34 u 1 u 2 a 41 a 42 a 43 a 44 Similar for 8 8 etc. Initially there are a lot of terms. However, with substitution for permutation gates, the entire polynomial collapses to the constant 1!

109 How it Works: Two-Qubit Gate Now let A be a general 4 4 matrix. Assign indicator variables u 1, u 2 to the two incoming qubits and y 1, y 2 to their outgoing selves: (1 y 1 )(1 y 2 ) (1 y 1 )y 2 y 1 (1 y 2 ) y 1 y 2 (1 u 1 )(1 u 2 ) a 11 a 12 a 13 a 14 (1 u 1 )u 2 a 21 a 22 a 23 a 24 u 1 (1 u 2 ) a 31 a 32 a 33 a 34 u 1 u 2 a 41 a 42 a 43 a 44 Similar for 8 8 etc. Initially there are a lot of terms. However, with substitution for permutation gates, the entire polynomial collapses to the constant 1! The effect on P C is then how the substituted terms are input to subsequent Hadamard and other kinds of gates.

110 How it Works: Qutrit Case Let qutrit values be 0, 1, 1 indexed by the basis vectors e 0 = (1, 0, 0) T, e 1 = (0, 1, 0) T, and e 2 = (0, 0, 1) T.

111 How it Works: Qutrit Case Let qutrit values be 0, 1, 1 indexed by the basis vectors e 0 = (1, 0, 0) T, e 1 = (0, 1, 0) T, and e 2 = (0, 0, 1) T. We need indicator polynomials in u and y that have the same nonzero value (here, 2) only when u, y have the corresponding values.

112 How it Works: Qutrit Case Let qutrit values be 0, 1, 1 indexed by the basis vectors e 0 = (1, 0, 0) T, e 1 = (0, 1, 0) T, and e 2 = (0, 0, 1) T. We need indicator polynomials in u and y that have the same nonzero value (here, 2) only when u, y have the corresponding values. Given a general 3 3 matrix A, it goes: Now P A has 36 terms. 2 2y 2 y 2 + y y 2 y 2 2u 2 a 11 a 12 a 13 u 2 + u a 21 a 22 a 23 u 2 u a 31 a 32 a 33

113 How it Works: Qutrit Case Let qutrit values be 0, 1, 1 indexed by the basis vectors e 0 = (1, 0, 0) T, e 1 = (0, 1, 0) T, and e 2 = (0, 0, 1) T. We need indicator polynomials in u and y that have the same nonzero value (here, 2) only when u, y have the corresponding values. Given a general 3 3 matrix A, it goes: 2 2y 2 y 2 + y y 2 y 2 2u 2 a 11 a 12 a 13 u 2 + u a 21 a 22 a 23 u 2 u a 31 a 32 a 33 Now P A has 36 terms. But it simplifies for certain matrices: A = ω ω 2 ; P A = iuy + 2u 2 y ω 2 ω Here ω = 1 2 ( 1 + 3i), and the H 3 multiplier is 2/ 3.

114 Multiplicative and Additive Cases P C is simply the product of P g over all gates g,

115 Multiplicative and Additive Cases P C is simply the product of P g over all gates g, possibly after substitutions for input and output variables and in-between.

116 Multiplicative and Additive Cases P C is simply the product of P g over all gates g, possibly after substitutions for input and output variables and in-between. For min-phase K, works over any ring that embeds (0, e 2πij/K ).

117 Multiplicative and Additive Cases P C is simply the product of P g over all gates g, possibly after substitutions for input and output variables and in-between. For min-phase K, works over any ring that embeds (0, e 2πij/K ). However, degree is high: Θ(s).

118 Multiplicative and Additive Cases P C is simply the product of P g over all gates g, possibly after substitutions for input and output variables and in-between. For min-phase K, works over any ring that embeds (0, e 2πij/K ). However, degree is high: Θ(s). We would prefer to add terms P g and work e.g. over Z K.

119 Multiplicative and Additive Cases P C is simply the product of P g over all gates g, possibly after substitutions for input and output variables and in-between. For min-phase K, works over any ring that embeds (0, e 2πij/K ). However, degree is high: Θ(s). We would prefer to add terms P g and work e.g. over Z K. But where can the zero-values for invalid paths go?

120 Multiplicative and Additive Cases P C is simply the product of P g over all gates g, possibly after substitutions for input and output variables and in-between. For min-phase K, works over any ring that embeds (0, e 2πij/K ). However, degree is high: Θ(s). We would prefer to add terms P g and work e.g. over Z K. But where can the zero-values for invalid paths go? My trick: Allocate new validity indicator variables w,...

121 Multiplicative and Additive Cases P C is simply the product of P g over all gates g, possibly after substitutions for input and output variables and in-between. For min-phase K, works over any ring that embeds (0, e 2πij/K ). However, degree is high: Θ(s). We would prefer to add terms P g and work e.g. over Z K. But where can the zero-values for invalid paths go? My trick: Allocate new validity indicator variables w,... Given a constraint c with values 0 = OK, 1 = fail, add q c = w 0 c + 2w 1 c + 4w 2 c k 1 w k 1 c.

122 Multiplicative and Additive Cases P C is simply the product of P g over all gates g, possibly after substitutions for input and output variables and in-between. For min-phase K, works over any ring that embeds (0, e 2πij/K ). However, degree is high: Θ(s). We would prefer to add terms P g and work e.g. over Z K. But where can the zero-values for invalid paths go? My trick: Allocate new validity indicator variables w,... Given a constraint c with values 0 = OK, 1 = fail, add q c = w 0 c + 2w 1 c + 4w 2 c k 1 w k 1 c. Then c = 1 = binary assignments to w 0,..., w k 1 run through all K values = the entire net contribution over u, y, w cancels.

123 Multiplicative and Additive Cases P C is simply the product of P g over all gates g, possibly after substitutions for input and output variables and in-between. For min-phase K, works over any ring that embeds (0, e 2πij/K ). However, degree is high: Θ(s). We would prefer to add terms P g and work e.g. over Z K. But where can the zero-values for invalid paths go? My trick: Allocate new validity indicator variables w,... Given a constraint c with values 0 = OK, 1 = fail, add q c = w 0 c + 2w 1 c + 4w 2 c k 1 w k 1 c. Then c = 1 = binary assignments to w 0,..., w k 1 run through all K values = the entire net contribution over u, y, w cancels. Whereas c = 0 zeroes all such terms, so he only effect is to inflate R.

124 Additive Extension Theorem Given any C of minphase K, we can efficiently compute a polynomial Q C (x 1,..., x n, y 1,..., y h, z 1,..., z r, w 1,..., w t ) over Z K and a constant R such that for all x {0, 1} n and z {0, 1} r, A[C(x) z] = 1 R K 1 j=0 ω j N QC,x,z[j]

125 Additive Extension Theorem Given any C of minphase K, we can efficiently compute a polynomial Q C (x 1,..., x n, y 1,..., y h, z 1,..., z r, w 1,..., w t ) over Z K and a constant R such that for all x {0, 1} n and z {0, 1} r, A[C(x) z] = 1 R K 1 j=0 ω j N QC,x,z[j] = 1 ω QC(x,y,z,w), R where Q C = gates g q g + constraints c q c has bounded degree. Thus we can do all calculations using Q C over Z K. y,w

126 Computing the Polynomials Annotate every juncture of qubit i with variable y j or term u i.

127 Computing the Polynomials Annotate every juncture of qubit i with variable y j or term u i. Initially x i and 0 terms, P C = 1, Q C = 0.

128 Computing the Polynomials Annotate every juncture of qubit i with variable y j or term u i. Initially x i and 0 terms, P C = 1, Q C = 0. u i H : new variable y j, P C = (1 u i y j ) Q C + = 2 k 1 u i y j.

129 Computing the Polynomials Annotate every juncture of qubit i with variable y j or term u i. Initially x i and 0 terms, P C = 1, Q C = 0. u i H : new variable y j, P C = (1 u i y j ) Q C + = 2 k 1 u i y j. CNOT with incoming terms u i on control, u j on target: u i stays, u j := 2u i u j u i u j.

130 Computing the Polynomials Annotate every juncture of qubit i with variable y j or term u i. Initially x i and 0 terms, P C = 1, Q C = 0. u i H : new variable y j, P C = (1 u i y j ) Q C + = 2 k 1 u i y j. CNOT with incoming terms u i on control, u j on target: u i stays, u j := 2u i u j u i u j. No change to P C or Q C, as with any permutation gate.

131 Computing the Polynomials Annotate every juncture of qubit i with variable y j or term u i. Initially x i and 0 terms, P C = 1, Q C = 0. u i H : new variable y j, P C = (1 u i y j ) Q C + = 2 k 1 u i y j. CNOT with incoming terms u i on control, u j on target: u i stays, u j := 2u i u j u i u j. No change to P C or Q C, as with any permutation gate. In characteristic 2, linearity is preserved.

132 Computing the Polynomials Annotate every juncture of qubit i with variable y j or term u i. Initially x i and 0 terms, P C = 1, Q C = 0. u i H : new variable y j, P C = (1 u i y j ) Q C + = 2 k 1 u i y j. CNOT with incoming terms u i on control, u j on target: u i stays, u j := 2u i u j u i u j. No change to P C or Q C, as with any permutation gate. In characteristic 2, linearity is preserved. TOF: controls u i, u j stay, target u k changes to 2u i u j u k u i u j u k.

133 Computing the Polynomials Annotate every juncture of qubit i with variable y j or term u i. Initially x i and 0 terms, P C = 1, Q C = 0. u i H : new variable y j, P C = (1 u i y j ) Q C + = 2 k 1 u i y j. CNOT with incoming terms u i on control, u j on target: u i stays, u j := 2u i u j u i u j. No change to P C or Q C, as with any permutation gate. In characteristic 2, linearity is preserved. TOF: controls u i, u j stay, target u k changes to 2u i u j u k u i u j u k. Linearity not preserved.

134 Computing the Polynomials Annotate every juncture of qubit i with variable y j or term u i. Initially x i and 0 terms, P C = 1, Q C = 0. u i H : new variable y j, P C = (1 u i y j ) Q C + = 2 k 1 u i y j. CNOT with incoming terms u i on control, u j on target: u i stays, u j := 2u i u j u i u j. No change to P C or Q C, as with any permutation gate. In characteristic 2, linearity is preserved. TOF: controls u i, u j stay, target u k changes to 2u i u j u k u i u j u k. Linearity not preserved. Similar considerations in paper by Bacon-van Dam-Russell, 2008 (unpub., morphed into least action talk... ).

135 Equality Constraints To enforce a desired output value z i on qubit i with final term u i : P C = (1 + 2u i z i u i z i ) Q C += w j (u i + z i 2u i z i ).

136 Equality Constraints To enforce a desired output value z i on qubit i with final term u i : P C = (1 + 2u i z i u i z i ) Q C += w j (u i + z i 2u i z i ). In characteristic 2, Q C remains quadratic.

137 Equality Constraints To enforce a desired output value z i on qubit i with final term u i : P C = (1 + 2u i z i u i z i ) Q C += w j (u i + z i 2u i z i ). In characteristic 2, Q C remains quadratic. Theorem (Cai-Chen-Lipton-Lu 2010, after Grigoriev-Karpinski (et al.)) For quadratic p(x 1,..., x n ) over Z K, and all a < K, N p [a] is computable in poly(nk) time.

138 Equality Constraints To enforce a desired output value z i on qubit i with final term u i : P C = (1 + 2u i z i u i z i ) Q C += w j (u i + z i 2u i z i ). In characteristic 2, Q C remains quadratic. Theorem (Cai-Chen-Lipton-Lu 2010, after Grigoriev-Karpinski (et al.)) For quadratic p(x 1,..., x n ) over Z K, and all a < K, N p [a] is computable in poly(nk) time. Open: replace K by log K in the time? Affirmative for A[C(x) z].

139 Gottesman-Knill: alternative methodology

140 Gottesman-Knill: alternative methodology To represent u i S we need K = 4.

141 Gottesman-Knill: alternative methodology To represent u i S we need K = 4. H gives Q C += 2u i y j.

142 Gottesman-Knill: alternative methodology To represent u i S we need K = 4. H gives Q C += 2u i y j. CNOT: Nonlinear term has a 2 which will cancel the 2 from Hadamard.

143 Gottesman-Knill: alternative methodology To represent u i S we need K = 4. H gives Q C += 2u i y j. CNOT: Nonlinear term has a 2 which will cancel the 2 from Hadamard. Equality constraint w j (u i + z i 2u i z i ): OK with [G-K], [CCLL] because w j appears only here.

144 Gottesman-Knill: alternative methodology To represent u i S we need K = 4. H gives Q C += 2u i y j. CNOT: Nonlinear term has a 2 which will cancel the 2 from Hadamard. Equality constraint w j (u i + z i 2u i z i ): OK with [G-K], [CCLL] because w j appears only here. S: u i left alone but Q C += u 2 i.

145 Gottesman-Knill: alternative methodology To represent u i S we need K = 4. H gives Q C += 2u i y j. CNOT: Nonlinear term has a 2 which will cancel the 2 from Hadamard. Equality constraint w j (u i + z i 2u i z i ): OK with [G-K], [CCLL] because w j appears only here. S: u i left alone but Q C += u 2 i. Inductively every term in Q C has form y 2 j or 2y iy j.

146 Gottesman-Knill: alternative methodology To represent u i S we need K = 4. H gives Q C += 2u i y j. CNOT: Nonlinear term has a 2 which will cancel the 2 from Hadamard. Equality constraint w j (u i + z i 2u i z i ): OK with [G-K], [CCLL] because w j appears only here. S: u i left alone but Q C += u 2 i. Inductively every term in Q C has form y 2 j or 2y iy j. These terms are invariant under 0 2, 1 3.

147 Gottesman-Knill: alternative methodology To represent u i S we need K = 4. H gives Q C += 2u i y j. CNOT: Nonlinear term has a 2 which will cancel the 2 from Hadamard. Equality constraint w j (u i + z i 2u i z i ): OK with [G-K], [CCLL] because w j appears only here. S: u i left alone but Q C += u 2 i. Inductively every term in Q C has form y 2 j or 2y iy j. These terms are invariant under 0 2, 1 3. Hence [CCLL] gives poly-time simulation by soution counting in Z 4.

148 Open Questions

149 Open Questions Solution counting is #P-complete for degree 3 over Z K in general.

150 Open Questions Solution counting is #P-complete for degree 3 over Z K in general. Are some structured subcases of degree 3 tractable? Can they come from families of QC s?

151 Open Questions Solution counting is #P-complete for degree 3 over Z K in general. Are some structured subcases of degree 3 tractable? Can they come from families of QC s? What else is (physically!) meaningful about polynomials in the circuit s partition function?

152 Open Questions Solution counting is #P-complete for degree 3 over Z K in general. Are some structured subcases of degree 3 tractable? Can they come from families of QC s? What else is (physically!) meaningful about polynomials in the circuit s partition function? Invariants based on Strassen s geometric degree γ(f) concept, others?

153 Open Questions Solution counting is #P-complete for degree 3 over Z K in general. Are some structured subcases of degree 3 tractable? Can they come from families of QC s? What else is (physically!) meaningful about polynomials in the circuit s partition function? Invariants based on Strassen s geometric degree γ(f) concept, others? Baur-Strassen showed that log 2 γ(f) lower-bounds the arithmetical complexity of f, indeed the number of binary multiplication gates.

154 Open Questions Solution counting is #P-complete for degree 3 over Z K in general. Are some structured subcases of degree 3 tractable? Can they come from families of QC s? What else is (physically!) meaningful about polynomials in the circuit s partition function? Invariants based on Strassen s geometric degree γ(f) concept, others? Baur-Strassen showed that log 2 γ(f) lower-bounds the arithmetical complexity of f, indeed the number of binary multiplication gates. Relevance to complexity of quantum circuits C...

155 Open Questions Solution counting is #P-complete for degree 3 over Z K in general. Are some structured subcases of degree 3 tractable? Can they come from families of QC s? What else is (physically!) meaningful about polynomials in the circuit s partition function? Invariants based on Strassen s geometric degree γ(f) concept, others? Baur-Strassen showed that log 2 γ(f) lower-bounds the arithmetical complexity of f, indeed the number of binary multiplication gates. Relevance to complexity of quantum circuits C... Possibly quantify the entangling capacity of C?

156 Open Questions Solution counting is #P-complete for degree 3 over Z K in general. Are some structured subcases of degree 3 tractable? Can they come from families of QC s? What else is (physically!) meaningful about polynomials in the circuit s partition function? Invariants based on Strassen s geometric degree γ(f) concept, others? Baur-Strassen showed that log 2 γ(f) lower-bounds the arithmetical complexity of f, indeed the number of binary multiplication gates. Relevance to complexity of quantum circuits C... Possibly quantify the entangling capacity of C? Delineate algebraic properties of qutrit circuits...