A note on adiabatic theorem for Markov chains and adiabatic quantum computation. Yevgeniy Kovchegov Oregon State University

Transcription

1 A note on adiabatic theorem for Markov chains and adiabatic quantum computation Yevgeniy Kovchegov Oregon State University

2 Introduction Max Born and Vladimir Fock in 1928: a physical system remains in its instantaneous eigenstate if a given perturbation is acting on it slowly enough and if there is a gap between the eigenvalue and the rest of the Hamiltonian s spectrum (see Wikipedia page on adiabatic theorem).

3 Introduction A way of expressing Hamiltonians via reversible Markov chains Corresponding theorem for Markov chains.

4 Quantum adiabatic theorem Given two Hamiltonians, H initial and H final, acting on a quantum system. Let H(s) = (1 s)h initial + sh final System evolves according to H(t/T ) on t : 0 T (adiabatic evolution).

5 Quantum adiabatic theorem H(s) = (1 s)h initial + sh final System evolves according to H(t/T ) on t : 0 T (adiabatic evolution). Adiabatic thm of quantum mechanics: For T large enough, the final state of the system will be close to the ground state of H final, i.e. ɛ close in l 2 norm whenever T C ɛβ 3, where β is the least spectral gap of H(s) over all s [0, 1].

6 Mixing and relaxation times Definition. If µ and ν are two probability distributions over Ω, the total variation distance is µ ν T V = 1 2 x Ω µ(x) ν(x) = sup µ(a) ν(a) A Ω Observe that the total variation distance measures the coincidence between the distributions on a scale from zero to one.

7 Mixing and relaxation times Definition. Suppose P is an irreducible and aperiodic Markov chain with stationary distribution π, i.e. πp = π. Given an ɛ > 0, the mixing time t mix (ɛ) is defined as t mix (ɛ) = inf { t : νp t π T V ɛ, for all ν }

8 Mixing and relaxation ( ) times Suppose P = p(x, y) is reversible: x,y π(x)p(x, y) = π(y)p(y, x) x, y Ω P is reversible P is self-adjoint w.r.t. π All real eigenvalues. β = 1 λ 2 is the spectral gap of P. The relaxation time is defined as τ rlx = 1 β

9 Mixing and relaxation times Theorem. If P is a reversible, irreducible and aperiodic Markov chain. Then (τ rlx 1) log(2ɛ) 1 t mix (ɛ) τ rlx log(ɛ min x Ω π(x)) 1

10 Adiabatic time Given P initial and P final, where P final is irreducible and aperiodic. Let P s = (1 s)p initial + sp final π f is the stationary distribution for P final. Definition. Given ɛ > 0, a time T ɛ is called the adiabatic time if it is the least T such that max ν νp 1T P 2T P T 1 P 1 π f T V ɛ, T where the maximum is taken over all probability distributions ν over Ω.

11 Adiabatic theorem for Markov chains Theorem A (K.2008) Let t mix denote the mixing time for P final. Then the adiabatic time ( tmix (ɛ/2) 2 ) T ɛ = O ɛ

12 Proof: νp 1T P 2T P T 1 P 1 = T! T where u N = νp 1T P 2T P NT of the terms. N!T T N u NP T N final + E, and E is the rest Hence, by triangle inequality, max ν νp 1T P 2T P T 1 P 1 π f T V T max ν νp T N final π f T V where 0 S N 1 T! N!T T N. T! N!T T N + S N,

13 Proof: Let T = Kt mix (ɛ/2) and N = (K 1)t mix (ɛ/2) Then and max ν simplifies to νp T N final π f T V ɛ/2 e T N log xdx+(t N) log T ( K 1 e ) K 1 t mix (ɛ/2) T! N!T T N T! N!T T N 1

14 Proof: max ν νp 1T P 2T P T 1 P 1 π f T V T max ν νp T N final π f T V where 0 S N 1 T! N!T T N 0 S N 1 T! 1 N!T T N T! N!T T N + S N, and ( K 1 e ) K 1 t mix (ɛ/2)

15 Proof: Need the least K such that 1 ( K 1 e ) K 1 t mix (ɛ/2) ɛ/2 and since log(1 + x) = x x2 2 + O(x3 ), the least such K is approximated as follows K t mix (ɛ/2) 2 log (1 ɛ/2) t mix(ɛ/2) ɛ Thus for T = Kt mix (ɛ/2) t mix(ɛ/2) 2 ɛ, max ν νp 1T P 2T P T 1 P 1 π f T V ɛ T

16 Mixing and relaxation times Theorem. If P is a reversible, irreducible and aperiodic Markov chain. Then (τ rlx 1) log(2ɛ) 1 t mix (ɛ) τ rlx log(ɛ min x Ω π(x)) 1 Theorem A (K.2008) Let t mix denote the mixing time for P final. Then the adiabatic time ( tmix (ɛ/2) 2 ) T ɛ = O ɛ

17 A Corollary If P final is reversible, irreducible and aperiodic with its spectral gap β > 0, then T ɛ = O log 2 ɛ + log 1 min x Ω π f (x) ɛβ 2

18 Continuous time Markov processes Suppose Q is a bounded Markov generator for a continuous time Markov chain P (t), and λ max i Ω j:j i q(i, j) Uniformization: P (t) = n=0 (λt) n e λt Pλ n n!, where P λ = I+ 1 λ Q

19 Mixing times: continuous time Definition. P (t) is an irreducible continuous time Markov chain with stationary distribution π. Given an ɛ > 0, the mixing time t mix (ɛ) is defined as t mix (ɛ) = inf {t : νp (t) π T V ɛ, all ν}

20 Continuous time Suppose Q initial and Q final are bounded Markov generators, and π f is the only stationary distribution for Q final. Let Q[s] = (1 s)q initial + sq final be a time non-homogeneous generator. Given T > 0, let P T (s, t) (0 s t T ) denote a matrix of transition probabilities of a Markov process generated by Q [ t T ].

21 Adiabatic time: continuous time Definition. Given ɛ > 0, a time T ɛ is called the adiabatic time if it is the least T such that max νp ν T (0, T ) π f T V ɛ, where the maximum is taken over all probability distributions ν over Ω.

22 Adiabatic theorem for Markov processes. Let λ max i Ω j:j i q initial (i, j) and λ max i Ω j:j i q final (i, j) q initial (i, j) and q final (i, j) are the rates in Q initial and Q final respectively.

23 Adiabatic theorem for Markov processes. λ max i Ω Then j:j i q initial (i, j), λ, max i Ω j:j i q final (i, j) Theorem B.(K. 2008) Let t mix denote the mixing time for Q final. Then adiabatic time T ɛ λt mix(ɛ/2) 2 ɛ

24 Proof: λ max i Ω j:j i q t (i, j), where q t (i, j) are rates in Q [ t T ] (0 t T ). Let T = Kt mix (ɛ/2) and N = (K 1)t mix (ɛ/2), so that max ν νp final (T N) π f T V ɛ/2, where P final (t) = e tq final, transition probability generated by Q final. Let P 0 = I + 1 λ Q initial and P 1 = I + 1 λ Q final.

25 Proof: νp T (0, T ) = u N n=0 where u N = νp T (0, N) and (λ(t N) n e λ(t N) I n n!, I n = n! (T N) n [(1 x1 /T )P 0 + (x 1 /T )P 1 ] [(1 x n /T )P 0 +(x n /T )P 1 ]dx 1... dx n, over N < x 1 < x 2 < < x n < T.

26 Proof: Hence νp T (0, T ) = u N ( n=0 (λ(t N) n e λ(t N) (T N) n T n n! P n 1 TN x 1... x n dx 1... dx n ) +E = e λt mix(ɛ/2) u N ( n=0 λ n [(1 1 2K )t mix(ɛ/2)] n n! P n 1 +E = e λt mix(ɛ/2) u N P final ( λ ( 1 1 2K ) tmix (ɛ/2) ) +E where E is the rest of the terms. )

27 Proof: νp T (0, T ) = e λt mix(ɛ/2) u N P final ( λ ( 1 1 2K ) tmix (ɛ/2) ) +E where E is the rest of the terms. Thus, the total variation distance, max νp ν T (0, T ) π f T V e λtmix(ɛ/2) ɛ/2+s N whenever λ ( 1 2K 1 ) 1, i.e. K λ 2(λ 1), easy condition.

28 Proof: max νp T (0, T ) π f T V e λtmix(ɛ/2) ɛ/2+s N ν Here S N = E π f T V 1 e λt mix(ɛ/2) n=0 λ n [(1 1 2K )t mix(ɛ/2)] n n! = 1 e λt mix (ɛ/2) 2K ɛ/2 whenever K λt mix(ɛ/2) ɛ, and therefore T ɛ λt mix(ɛ/2) 2 ɛ