Topology, Localization, and Quantum Information in Atomic, Molecular and Optical Systems

Μέγεθος: px
Εμφάνιση ξεκινά από τη σελίδα:

Download "Topology, Localization, and Quantum Information in Atomic, Molecular and Optical Systems"

Transcript

1 Topology, Localization, and Quantum Information in Atomic, Molecular and Optical Systems The Harvard community has made this article openly available. Please share how this access benefits you. Your story matters. Citation Accessed Citable Link Terms of Use Yao, Norman Ying Topology, Localization, and Quantum Information in Atomic, Molecular and Optical Systems. Doctoral dissertation, Harvard University. May 1, :52:49 AM EDT This article was downloaded from Harvard University's DASH repository, and is made available under the terms and conditions applicable to Other Posted Material, as set forth at (Article begins on next page)

2

3

4

5

6 W

7

8 40 87 G σσ LL (z) Gσσ LR (z) Πσσ LL (z) Πσσ LR (z)

9 ν =1/2

10 W

11

12

13

14

15

16

17

18 ß

19

20

21 Energy Site H = ij t ij c i c j + h.c. + i µ i n i t ij {µ i } µ i W W =0 t =0 W t t i + j t µ i µ j j

22 e r/ξ ξ H = α c αc α + α,β,γ,δ V α,β,γ,δ c αc β c γc δ.

23 ν =1/2 ν =1/2

24 1/r 2 H sd = 1 2N k,k,α,β J(k k )c k,α σ αβc k,β J(k k ) σ 1 2 J RKKY S 1 S 2 J RKKY r ξ J RKKY

25 J RKKY (r) = 1 8πr 3 J 2 exρ f ν 2 0 cos(2k f r) J ex ρ f ν o k f r ξ J SC RKKY (r) = 1 8πrξ 2 J 2 exρ f ν 2 0 cos(2k f r)e r/ξ. ξ 2 r ξ (ξ/r) 2 r 1 J RKKY H 0 = k,α ɛ k c k,α c k,α + k [c k c k + c k c k ]. Ψ k =(c k,c k,c k,c k )T H 0

26 H 0 = k Ψ k (ɛ kτ 3 + k τ 2 σ 2 )Ψ k. τ 1,2,3 σ 1,2,3 τ {c k,c k } σ G 0 (k, ω) = 1 iω ɛ k τ 3 k τ 2 σ 2. G(k, k,ω)=g 0 (k, ω)δ(k k )+G 0 (k, ω)t (k, k,ω)g 0 (k,ω) J(k k )=J ex T (ω) = 1 N (SJ ex /2) 2 g 0 (ω) 1 (g 0 (ω)sj ex /2) 2 g 0 (ω) =1/N k G 0(k, ω) S k ω + τ 2 σ 2 g 0 (ω) = πρ f 2 ω. 2 E b < ω = E b

27 2 ( 1+π 2 (JS/2) 2 ρ 2 f ) (ω + ωπ2 (JS/2) 2 ρ f ) 2 =0 E b = 1 π2 (JS/2) 2 ρ 2 f. 1+π 2 (JS/2) 2 ρ 2 f J ex 0 E b J ex E b S =1 0 e 0 e 1 e I =1/

28 H e,n = 0 S 2 z + µ e BS z + µ n BI z + AS z I z, 0 =2.87 µ e = 2.8 µ n = 0.43 A =3.0 ẑ

29

30 (a) R 1 R 2 R R1 (b) 2R 2,,,,,! 1 "! 2 " = =,,, Pair 1 Pair 2 R2 R R <r<2r t 1/r α α d d d =1

31 α = β β<α β<α d<α d<α d<α d<β d<β d<β+2 d<β/2 d< αβ α+β d<β/2 d<β/2 d<(β +2)/2 d< α(β+2) α+β+4 1/2 S z H = i ɛ i S z i ij t ij r ij α (S+ i S j + h.c.)+ ij V ij r ij β Sz i S z j ɛ i W α β β α ɛ i ɛ j t ij / r ij α R 1 < r ij < 2R 1 N 1 (R 1 ) (ρr d 1) t/rα 1 W ρ N 1 (R 1 ) R 1 d>α d = α δ t/r1 α

32 Dynamic Polarization (D) ν =1/2 α = β =1 α = β =2 (a) (c) α = β =3/2 α = β =3 (b) (d) Disorder Width (W ) d =1 t =1,V =2 α = β =1 α = β =3/2 α = β =2 α = β =3 W c 10 V S z R 2 δ 1,δ 2 >V(R 2 ) δ 1 δ 2 R 2 2R 2 N 2 (R 1,R 2 ) (n 1 (R 1 )R d 2) V/Rβ 2 t/r α 1, n 1 = ρn 1 N 2 R 2 R 1 R 2 d<β R 1 R 2 V/Rβ 2 t/r α 1 [1, V/Rβ 2 ] R t/r1 α 1 R β/α 2 d< αβ α+β

33 N 3 (R 1,R 2,R 3 ) (n 2 (R 1,R 2 )R d 3) V/Rβ 3 V/R β 2 n 2 = n 1 N 2 R 3 R 1,R 2 R 1 R 2 R 3 d<β/2 R 1 R β/α 2 R 2 R 3 t ij,v ij R 1 <R 2 V/R β 2 VR 2 1/R β+2 2 N 2 α = β d<β/2 α<β+4 α>β+4 α d c =1.5 α = β =3 d c =1.5 α =6 β =3 d c 2.3

34 d =1 α = β =1, 3/2, 2, 3 L =14 ν =1/2 W V ij = V =2 t ij = t =1 D ˆF = j Sz j e i2πj/l η k Dη k =1 k ˆF k k ˆF k k ˆF ˆF. k D Dη k L D 1 D 0 D α =2, 3 α =1 α =3 α =3/2 d =1 1 <α c < 3 α c =2

35 (a) (b) (c) z y x E (d) 1, 1 (e) 1, 1 1, 1 1, 1 1, 1 2B 0, 0 i t ij 1 r 6 ij ŷ ẑ ˆx = 1, 1, = 0, 0 α = β =3 α =6 j ŷ ẑ t ˆx α E d V d 2 /R 3 β =3 B W

36 V d 2 E 0 V 0 ν =1/2, 1/3, 1/4 L =16, 18, 20 H m = BJ 2 d z E = J =0,m j =0 = J =1,m j =1 H dd = 1 d i (1 3ˆr ij ˆr ij )d j 2 i j d rij 3 H dd {, } ɛ i = j i d s d a α = β =3 d rij 3 s,a = 1 dz 1 ± 0 d z 0 2 ɛ i W d sd a a ν(1 ν) 3 a0 0 d =1 d c =3/2 α = β =3

37 Dynamic Polarization (D) (a) β =3 V/t = ν =1/ Disorder Width (W ) Disorder Width (W ) (b) V/t =1 V/t =2 V/t =4 MBL Ergodic /4 1/3 1/2 Filling Fraction (ν) α β =3 W W c 1.4t V/t =1, 2, 4 d c =3 d =2 B = J =1,m j = 1 = J =1,m j =1 J z H = t 2 [ ij r (d i ij 6 + ) 2 (d j ) 2 +(d i ) 2 (d j +) 2] α =6 β =3 d =2 d c 2.3 d c =2.5 H NV = D 0 S 2 z + µ e BS z D 0

38 e Dk2t D D =0 T 1 k T 2 D T2 a 2 0/T 2 T 2 T 1 D T2 D e a 2 0/T a0 T a0 l (l 2 + Dt) d/2 e t/t T a0 10µ T T 1 25

39 2 3 T a0 1µ T 1,T 2 10

40 π/2 τ z i ±1

41 Ĥ = i h i τ z i + ij J ij τ z i τ z j + ijk J ijk τ z i τ z j τ z k +... J ij, J ijk,... ξ τi z τ z i Ĥ... + =( + )/ 2 h ( ) = h + j J jτ z j + j,k J jkτ z j τ z k +... π t/2 h ( ) t/2 t/2 π h ( )

42 spin echo DEER F(t) D(t) W L J z = J F (t) d ξ t/2 π π/2 τ z

43 ψ(t) = R π/2 e iĥ t 2 R π R π/2 e iĥ t 2 R π/2 ψ(0), Rr π/2 = j r (ˆ1 iˆσ y j )/ 2 Rr π =(Rr π/2 ) 2 D(t) ψ(t) ˆτ z ψ(t) = ( 1+e 2iJ Ij τ j ) t 2 j II N τ j j D(t) J Ij j I N J Ij t 1 J Ij t 1 1 D(t) D(t) D osc (t) D(t) = D(t)+D osc (t), D(t) =1/2 N (t), 1/2

44 J Ij exp( j I /ξ) j I ξ log(t) t t 0 /J Ik k = I + d N fast =0 t 0 t t 0 e N/ξ N (t) ξ log(t/t 0 ) D(t) t ξ ln 2 t t 0 e N/ξ D( ) 2 N (1 + t D(t) 2 /t 2 0) α/2 t t 0 e N/ξ =, 2 N t t 0 e N/ξ α = ξ ln 2 D osc (t) D osc (t) D osc (t)

45 1 0.8 fit J z =0.1J J z = J W =6J =3 d =3 N =7 Ĥ = J 2 (Ŝ+ i ij Ŝ j + Ŝ+ j Ŝ i )+J z Ŝi z Ŝz j + i ij h i Ŝ z i Ŝa j a {x, y, z} ±1/2 Ŝ± j = Ŝx j ± iŝy j h i [ W ; W ] J z =0 Ĥ W>0 J z 0 W/J S τ

46 fit J z =0.1J J z = J d =3 d =7 W =8J N =3 α W d = 7 N = 3 α = c 1 / ln(c 2 W ) ξ 1/ ln(w ) D( ) N J z D( ) J z N c/1.8 N L =12 S z ψ(0) D(0) = ψ(0) ˆσ z ψ(0) > 0 D(t) π/2 t =0

47 D(t) =A/(1 + t 2 /t 2 0) α/2 A D osc (t) D(t) D(t) d t 0 [ exp(d/ξ)] d α α α = c 1 / ln(c 2 W ) α = ξ ln 2 ξ 1/ ln(w ) N f(k) =c/1.8 N 1/2 N

48 J,J z T 1 1 J ξ ln(j 1 ) 6ξ 10 J /J J (1 10) 5µ 1 100µ 1 /J 1 (0.5 5) 10 3 J /J

49

50 1/2 H = ij JS + i S j + ij J σ + i σ j + i J z S z i σ z i + h.c S σ XY J J J z J 0 σ S {σ i } ±J z

51 J!" J #" J! J J J J J J z S J σ J N =20 Stot z = σtot z =0 σ J =1.0 J z 10.0 J J S ent = ρ A log ρ A = ρ B log ρ B A B D

52 N =8 Sent J =1.0 J =0.01 J z = 10.0 J =1.0 J =0.001 J z = 10.0 J =1.0 J = J z = 10.0 Fractional Polarization k =1 k =2 Time (1/J) Time (1/J) Time (1/J) Time (1/J) N = 16 Fractional Polarization Sent k =1 J =1.0 J =0.01 J z = 10.0 S-chain!!-chain" J =1.0 J =0.001 J z = 10.0 J =1.0 J = J z = 10.0 Time (1/J) Time (1/J) Time (1/J) N = 8 S z tot = σ z tot =0 k =1 S σ k =2 t int t d N = 16 k =1

53 ˆF = j Sz j e i2πj/l Sj z σj z k D k =1 k ˆF k k ˆF k k ˆF ˆF, k D D k N =8 J =10 2, 10 3, 10 4 J S ent t 1/J t 1/J J t 1/J J t int t>t int t>t d t int t d J S ent

54 t int 1/J t int 1/J 2 J, J J z t int J z t int J J S t d J z /J 2 J 2 /J z S t d D k=1 k =2 t e L k =1/L e Dk2t t d L 2 /D D J 2 /J z N =16 e 8 J t int t d

55 Sent J int J SPL quasi MBL thermal t L 2 /D t e L Short chain thermal t L 2 /D J d J 2 /J z SPL t e L Long chain 1/J 1/J int t (1/J) 1/J d 1/J 1/J int 1/J d t (1/J) ρ N = 8 N = 12 N = 16 d ρ/dw W/J S ent 1/J t>t int =1/J int J t>t d =1/J d D J 2 /J z S ent e L L 2 N

56 2.0 Sent J independent J 2 Time (1/J) t int J J 2 J L 2 t d 1/Dk 2 L 2 J =10 2 t int t d J J =10 5 S ent t d σ σ J

57 H W = i b z i S z i + i b z i σ z i b, b W H W ρ ψ = 1 N N i ψ Si+1 z Si z ψ H W ( ) N d ρ dw c. W/J d ρ dw N H T = H + H W J =1.0 J =0.01 J z = <W < N =8, N =16 ρ ψ J/3 ρ H W

58 H h = ij JS i S j + ij J σ i σ j + i J z S z i σ z i. J 0 t int 1/J J S J 2 J S ent J S σ JJ /J z 1/t d J 2 /J z

59

60 ẑ Θ 0, Φ 0 {X, Y, Z} ij {X, Y } H dd = 1 κ 2 i j R 3 ij [ ] d i d j 3(d i ˆR ij )(d j ˆR ij ), κ 1/4πɛ 0 µ 0 /4π R ij d i d j 0 ±1 ẑ

61 ±1 0 Ω + Ω Ω +, Ω 0 B = α( 1 + β 1 ) D = α ( β ) α =Ω + / Ω αβ =Ω / Ω Ω = Ω 2 + Ω + 2 E 0 = Ω 2 / E B = + Ω 2 / E D = d R 0 κd 2 /R0 3 Ω 2 / D 0 B 0 D B 0 a i = B 0 i n i = a i a i H B = ij t ij a i a j + 1 V ij n i n j, 2 i j t ij = B i 0 j H dd 0 i B j t ii = j i ( 0 i0 j H dd 0 i 0 j B i 0 j H dd B i 0 j ) V ij = B i B j H dd B i B j + 0 i 0 j H dd 0 i 0 j B i 0 j H dd B i 0 j 0 i B j H dd 0 i B j N i = i a i a i κd 2 /R0 3 H dd H B

62 J =1 Ω (r) Ω + (r) Ω ± δ J = 1 l p l t l t l W (p l) β π/n κ =1,i j t ij = d2 01 R 3 [ χ i (q 0 + [q 2 ]σ x + [q 2 ]σ y )χ j ], t ii = j i 2 q 0 R 3 (d0 d B i (d 0 ) 2 ), V ij =2 q 0 R 3 [ d B i d B j d 0 d B i d 0 d B j +(d 0 ) 2], d 0 d B ẑ 0 B d χ i = α i (1,β i ) T i q 0 = 1(1 2 3cos2 (Φ Φ 0 )sin 2 (Θ 0 )) q 2 = 3[cos(Φ Φ 2 0)cosΘ 0 i sin(φ Φ 0 )] 2 σ (R, Φ) R ij ij R Φ q 0 q 2 d B i

63 π/n (Θ 0, Φ 0 )=(sin 1 ( 2/3),π/4) q 0 =0 ˆX Ŷ H dd d + i d+ j d i d j d ± = (d x ± id y )/ 2 t ˆX ij = d2 01 χ R0 3 i tŷij = d2 01 χ R0 3 i [ 1 2 σx [ 1 2 σx σy 3 2 σy ] ] χ j, χ j. β =Ω /Ω + β β B 1 1 β Φ=π/4 W (p) = p t ij l Ψ l =arg[w (p l )] = arg[t 2 l t 2 l ] t l t l θ l =arg(t l ) θ l =arg(t l )= θ l+1 Ψ l =2θ l 2θ l+1 π/n θ l+1 = η l π η 2N β θ l β l+1 β l = sin( π 3 η + l π 2N ) sin( π 3 + η l π 2N ),

64 4N β 1 π/n t ij H B β = β 1,β 2 β 1 β 2 β 1 β 2 d B i ±1

65 β = β 1,β 2 Ψ, Ψ g 1 g 2 /R 3 β 2 β 1 c = 1 4π dkx dk y ( kx ˆd ky ˆd) ˆd H(k) = d(k) σ + f(k) ẑ H m = BJ 2 d z E + H D, B J d z ẑ E H D J =1 J, M E M J 0 ±1 ±1

66 d 1 d B i = d 1 V ij = 2 q 0 R 3 (d0 d 1 ) 2. V ij /t ij (d 0 d 1 ) 2 /d Ω H B N ν

67 (Θ 0, Φ 0 )=(0.46, 0.42) β 1 =3.6e 2.69i β 2 =5.8e 5.63i 1/27R 0 > 10 S(R, 0) = n(r)n(0) ν = 1/2 (Θ 0, Φ 0 )=(0.66,π/4) β 1 = 2.82e iφ 1 β 2 = 4.84e iφ 2 (d 0 d 1 ) 2 /d ĝ 1 ĝ 2 N s = 24 2π θ 1 4π θ 2

68 c = 1 ν =1/2 (d 0 d 1 ) 2 /d S(R, 0) = n(r)n(0) N s =32 ĝ 1 ĝ 2 φ 1 = φ 2 =0.1 k 2 =0,k 1 =0, 2π/3, 4π/3 ĝ 2 N s =24 ĝ 2

69

70 E H m = BJ 2 d z E E 0, 0 J =1 1, 1 1, 0 1, 1 J ẑ J, m E

71 φ {X, Y } ẑ Θ 0 Φ 0 {x, y, z} {X, Y, Z} J =0, 1 0, 0 M J = N s = 24 Θ 0 E =0 J =1

72 J =1 e 1 e 2 M H r = [ e 1 (Ω 1 1, 1 +Ω 2 1, 0 )+ e 2 (Ω 3 1, 0 +Ω 4 1, 1 )+ ] Ω i = 1 Ω(Ω 2 Ω 4 1, 1 Ω 1 Ω 4 1, 0 +Ω 1 Ω 3 1, 1 ) Ω H dd = 1 κ 2 i j R 3 ij [ ] d i d j 3(d i ˆR ij )(d j ˆR ij ), κ =1/(4πɛ 0 ) R ij i j d i d j d R 0 d κd 2 /R0 3 Ω i {, } 2B a i = i j i t ij = i j H dd i j

73 V ij = i j H dd i j + i j H dd i j i j H dd i j i j H dd i j H B = ij t ij a i a j + 1 V ij n i n j, 2 i j N = i a i a i t ii {a, b, A, B} J =1 a i t ij V ij g 1 g 2 C = 1 f 11.5 J =1 ν ν =1/2 N s =44 Θ 0 =cos 1 (1/ 3) ν =1/2

74 a b A B J =1 φ g 1 g 2 f 11.5 C = 1 {Θ 0, Φ 0 } = {0.68, 5.83} σ xy = 1 2π F (θx,θ y )dθ x dθ y = 0.5 {θ x,θ y } F (θ x,θ y )= ( Ψ θ y Ψ θ x Ψ θ x Ψ θ y ) ν =1/2 Q torus = ( N uc +1 N b N uc+1 2N b ) ( Nuc 1 N b N uc+1 2N b ) Nuc = N s /2 N b ν =1/2 Θ 0

75 σ xy ν =1/2 ν =1/2

76 d µ e 1 e 2 J,m = 2, ±2 v =41 (3) 1 Σ +

77 ν = 1/2 ν =1/2 N s = 24 N b =6 1/(3R 0 ) 3 (k x,k y )=(0, 0) (k x,k y )=( π, 0) ν =1/2 k x,k y N s = 16 N s = 44 N b =5 36 Q torus µ = E Nb +1 E Nb E Nb N b ν =1/2 ν =1/2 de/dθ 0 Θ 0 de/dθ 0 E =0.4 8

78 H lattice H hf Ω i H lattice H hf H hf 1 =160 E = B/d 0.5 Ω i E 1,0 E 1,1 M H lattice H hf Ω i

79

80 (a) (b) E b " η η " BCS ψ BCS E b r < ξ η η r ξ r r<ξ

81 r ξ H 0 =,σ ɛ c,σ c,σ + [c c + c c ]. ψ BCS J E b = 1 (πjsn 0/2) 2 β2 = 1 1+(πJSN 0 /2) 2 1+β 2 N 0 tan(δ) β = πjsn 0 /2 E b = cos(2δ) E b δ

82 T K exp ( 1/JN 0 ) T K β 1/ ln( /T K ) 1 φ sh ( ) 1 e /ξ sin(2δ) r ξ r<ξ η η η η r L r R ẑ H int = J σ d σ[s L f( L )c σ( )c σ ( )+S R f( R )c σ( )c σ ( )], S L(R) f( ) H int = J σ d d σ[s L e i( )r L f, + S R e i( )r R f, ]c σ, c σ, f

83 Ψ =(c,,c, ) H 0 = d Ψ [ɛ τ z + τ x ]Ψ τ H int = J d d Ψ [S Le i( )r L f, + S R e i( )r R f, ]Ψ + E 0 E 0 = J d f, [S L + S R ] H T = H 0 + H int d n = d (u n, ψ, + v n, ψ, ) H T = ε n d nd n 1 ε n = 2 n n n E tot = 1 ε n = E V n ε n (d nd n 1 2 ). dɛ ɛ δρ(ɛ) E V δρ(ɛ) I(r) I(r) =E, tot E, tot = 1 2 dɛ ɛ [δρ, (ɛ) δρ, (ɛ)]. δρ(ɛ) = 1 { [G π, (z) G(0) (z)]} z = ɛ + i0 + G (0) (z) =[z (ɛ τ z + τ x )] 1 G, (z)

84 T G, (z) =G (0) (z)+g(0) (z)t, G(0) (z), T, T T δρ(ɛ) = 1 π { [G(0) (z)t, G(0) (z)]} = 1 π { [JSΠ(1 JSG) 1 ]} Π G S 4 4 Π ll (z) = G ll (z) = d G (0) (z)g(0) (z)ei ( l l ) d G (0) (z)ei ( l l ) S ll = S l δ ll τ 0. τ 0 l l {L, R} J 1 [JSΠ(1 JSG) 1 ] [J 2 SΠSG] I( ) = E fβ 2 π(k f r) cos(2k fr)e 2r 3 ξ F1 [ 2r ξ ] + β2 (k f r) 2 sin2 (k f r)e 2r ξ F2 [ 2r ξ ]. k f r = L R

85 Interaction (khz) 400! 0! -400! 100! 150! R (nm) J YSR! RKKY E b ! E f = 11.7 k f = N 0 = 35 3 ξ =1.6µ J YSR E b 10 2 J YSR r ξ F 1 [α] =α dxe α( x ) F 0 2 [α] = 2 π x ) dx e α( 0 (x 2 +1) 1/r 2 /E f r E f /( k f ) ξ I( ) e 2r ξ O(J 2 )

86 J E b 0 η ɛ ɛ 0 [G, (z)] E b F (E b ) [1 SG(E b )] = 0. k f r 1 η η/e b F (E b )+η F (E b )=0

87 1 β η = 1 cos 2 (k f r) 1 β 2(k f e 2r r) 2 ξ. F (E b )+η F (E b )+ 1 2 η2 F (E b )+ 1 6 η3 F (E b )=0 η = 2 cos(k f r) (k f r) 2 e 2r ξ β 1 I(r) J YSR = η η E b 0 β 1 J YSR η J YSR = 1 cos 2 (k f r) 2r 1 β 2(k f r) 2 e ξ, 1 1 β U(r)c L, c R, U(r) =cos(k fr)/(k f r) 2E b β E b J YSR J YSR k f r> 1 1 β > ξ r.

88 r ξ J YSR r λ f ξ ξ / k f ξ J YSR = η η 1 r 2 β J YSR J YSR J YSR >J RKKY r J YSR

89 2 T K I I T K

90 ɛ = E b E b E b

91 ( /T K ) c S =1/2 /T K > ( /T K ) c S G =1/2 /T K < ( /T K ) c S G =0 H BCS = dk [ ξ (2π) 3 k c kσ c kσ + ( c k c k + h.c.)] σ 1/2 S 1 S 2 H int = J 2 S 1ψ 1σ ψ 1 + J 2 S 2ψ 2σ ψ 2. ψ 1 ψ 2 H T = H BCS + H int T K

92 H T P S 2 =0 SU c (2) Q x =(Q + + Q )/2 Q y =(Q + Q )/2i Q z = 1 dk 2 σ (c (2π) 3 kσ c kσ 1) 2 Q+ = dk c (2π) 3 k c π k Q = ( Q +) 0 U c (1) Q P S Q S 0 S =0 P = Q =0 T 0 S =1 P =+ Q =0 D + D S =1/2 P = ± Q =1 S 2 Q =2 P = π

93 (S, Q, P ) S 0 (0, 0, ) T 0 (1, 0, +) D ± ( 1, 1, ±) 2 S 2 (0, 2, ) /T K /T K 1 S S = sin(k F R) k F R R = R 1 R 2 k F S D ± I I J I/T K

94 S = 0.1 I/T K /T K S =1 S =1/2 S =0 S 2 S 0 /T K T 0 I<0 S 0 I>0 I T K, =0 S 2 I, T K Q =0 Q =2 S 0 S 2 =0 =0

95 /T K I/T K = 0.58 S = 0.1 S 2 D + T 0 S 0 T 0 I eff = E S0 E T0 I T K T K ω T K S =0 S =1/2 T K I 0 D + T K S =0 S =0 D ± /T K D ± D + S =1 1/2 /T K

96 S 2 φ ( ) 1 e /ζ sin(2δ) E = 1 β2 1+β 2 ζ β tan(δ) =JSN 0 π/2 N 0 T k F R β E sh =0 I <0 I >0 F 1 S 1 σψ 1 T 0 D + D 4

97 (a) 1.0 Esh/ Esh/ (b) I<0 I>0 Molecular Doublet Mol. Triplet Molecular Doublet Triplet Kondo Mol. Singlet β k F R 4.1 Kondo Singlet k F R 2.6 k F R 4.1 β = JNSπ/2 β 0.86 β 1.3 k F R Pb =1.55 T 20 mk T 0 S 0 E = hν S 0 S 0 D ± E S 0 di/dv S 0 T 0

98 T 0 D + D T 0 S 0 hν = E (D +,S 2 ) (D +,T 0 ) S 2 S 0

99

100 g κ

101 1/2 H = H 0 + H H 0 = N 1 i=1 κ(s+ i S i+1 + S i S+ i+1 ) H = g(s 0 + S1 + S + N+1 S N + ) S ± = S x ± is y S = σ / 2 σ =1 g κ H c i = e iπ i 1 0 S + j S j S i H 0 H 0 = N 1 i=1 κ(c i c i+1 + c i c i+1 ) f k = 1 N jkπ A j=1 sin N+1 c j k =1,,N A =(N+1 2 )1/2 H 0 = N k=1 E kf k f k E k =2κ cos kπ N+1 t k = g A H = N t k (c 0f k +( 1) k 1 c N+1 f k + ), k=1 kπ sin N N+1 k = z (N +1)/2 H t z g/a

102 g κ κ a b k = z g κ/ N E z E z±1 κ/n ( 1) z 1 c N+1 H eff = t z (c 0f z + c N+1 f z + ) H eff τ = π 2tz U eff = e iτh eff =( 1) f z f z (1 (c 0 + c N+1 )(c 0 + c N+1 )) {(1,c 0,c N+1,c 0c N+1 ) 00 0,N+1} U fermi eff =( 1) n 0+n N+1 +n z ( 1) n 0n N+1 0,N+1, n θ = f θ f θ H eff

103 0,N+1 =( 1) n 0n N+1 Φ i =(α + β ) 0 (α + β ) N+1 Ψ M,nz Ψ M = N j=1 S+ j S j n z N Φ f =( 0,j N+1,j ) 0,N+1 0,N+1 Φ i j=1 a a = a a = a a a b a b b b b F = i=1,2,3 [σi E(σ i )]

104 E e ikτ K N N H = i,j K i,js + i S j ɛ =1 F g/κ g κ ɛ ( ) 2 5 t k k z 3 E k [1 + ( 1) k+z cos(e k τ)] z = N+1 2 ( ) 10 t 2 k k z 3 E k N κ ɛ 0 g τ τ N t z t z κ/n τ 1/t z N/κ N N N E =0 H = (S0 z + SN+1 z ) N

105 =E k t k E =0 g L g R g L g R τ N/g

106 N =7 g/κ ɛ 0 = 10 3 τ 1/κ

107 g

108 T 1

109 L =12 N 1 H = κσi x σi+1 x + i=1 N i=1 Bσ z i κ B σi x σ ± i =(σi x ± iσ y i )/2 c i = σ+ i e iπ i 1 j=1 σ+ j σ j N 1 H JW = κ(c i c i+1 + c i c i+1 c ic i+1 c ic i+1 ) i=1 N B(c i c i c i c i ) i=1

110 H JW φ A φ φ =(c 1,c 2,...,c N,c 1,c 2,...,c N )T A ɛ ɛ Λ= 0 0 ɛ ɛ 2 ψ ψaψ T =Λ ±ɛ k d k = ψ 2k 1,j φ j d k = ψ 2k,jφ j k =1,,N H JW = N ɛ k (d k d k d k d k ), k=1 d ɛ k κ 2 + B 2 2Bκcos q k q k = kπ/(n +1) 0 N +1 g B H = g(σ x 0σ x 1 + σ x Nσ x N+1)+B (σ z 0 + σ z N+1).

111 H JW = g(c 0c 1 + c 0c 1 + c 1c 0 c 0 c 1 ) g(c N c N+1 + c N c N+1 + c N+1 c N c N c N+1 ) + B (c 0c 0 c 0 c 0 + c N+1 c N+1 c N+1 c N+1 ). B = ɛ z d z c i = N k=1 (ψt ) i,2k 1 d k + N k=1 (ψt ) i,2k d k c 1 c N d gψ 2z 1,1 = gψ 2z 1,N B, ɛ z ɛ z±1 H eff ɛ z (d zd z d z d z)+ɛ z (c 0c 0 c 0 c 0)+ɛ z (c N+1 c N+1 c N+1 c N+1 ) gψ 2z 1,1 (c 0d z + d zc 0 ) gψ 2z 1,N (c N+1 d z + d zc N+1 ). B<κ

112 g t "E g t "E >> gt (1) (2) g τ = π 2gψ2z 1,1 U eff = e iτh eff =( 1) n z ( 1) (c 0 +c N+1 )(c 0+c N+1 )/2 = ( 1) nz (1 (c 0 + c N+1 )(c 0 + c N+1 )), n z = d zd z U eff Ψ={ Ω,c 0 Ω,c N+1 Ω,c 0c N+1 Ω } Ω c 0 c N U eff Ψ=( 1) n z Ψ e iπ i 1 j=1 σ+ j σ j 2 =

113 a b = a b = a b a b 0

114 0 (QR) 1 2 N-1 N N+1 (QR) 0 N +1 X F = [ σ i E(σ i ) ], i=x,y,z E H = g(σ + 0 σ 1 + σ + N σ N+1 + )+ N 1 i=1 κ(σ+ i σ i+1 + ) U H t =2τ ρ DS ch

115 F DS = = = i=x,y,z i=x,y,z i=x,y,z [ σ i 0U(σ i 0 ρ DS ch )U ] [ U σ i 0U(σ i 0 ρ DS ch ) ] [ σ i 0(t)(σ i 0 ρ DS ch ) ], σ i 0(t) M = e ikt K (N +2) (N +2) H = N+1 i,j=0 K ijc i c j c m = i n K mnc n c m (t) = n M mnc n σ + 0 (t) =U σ + 0 U = U c 0U = i M 0ic i = i M 0iσ + i e iπσ+ l σ l, l<i σ0(t) z = 2c 0(t)c 0 (t) 1= 1+2 M0iM 0j c i c j ij = 1+2 M0iM 0j σ + i σ j e iπσ+ l σ l, ij i<l<j c 0 F DS σ ± =(σ x ± iσ y )/2 [σ0(t)(σ x 0 x ρ ch )] = [ (σ 0 + (t)+σ0 (t))((σ σ0 ) ρ ch ) ] σ 0 + (t)(σ0 ρ ch ) σ0 (t)(σ 0 + ρ ch ) i =0

116 [ σ + 0 (t)(σ0 ρ ch ) ] [ = ( ] M0iσ + i e iπσ+ l σ l )(σ 0 ρ ch ) i l<i = [ ] M00σ 0 + σ0 ρ ch = M 00. [ σ 0 (t)(σ + 0 ρ ch ) ] = M 00 σ z [σ0(t)(σ z 0 z ρ ch )] = [ σ0 z ρ ch ] [ + (2 ] M0iM 0j σ + i σ j e iπσ+ l σ l )(σ z 0 ρ ch ) ij i<l<j = [ 2M 00M 00 σ + 0 σ 0 σ z 0 ρ ch ] =2 M00 2, i = j i = j =0 [σ z 0]=0 F DS = (M 00 + M 00 + M 00 2 ). N +1 0 F SS = i=x,y,z [ σ i 0(t)(ρ SS ch σ i N+1) ],

117 ρ SS ch {0,,N} F SS σ x 0(t) = c 0(t)+c 0 (t) = i M 0ic i + M 0ic i = i i 1 [{ (M 0i )σi x + (M 0i )σ y i } ( σl z )]. l=0 i N +1 [σ x 0(t)(ρ ch σ x N+1)] = 2 (M 0,N+1 ) [ρ SS ch N ( σl z )]. l=0 σ y σ z [σ z 0(t)(ρ SS ch σz N+1 )] = 2 M 0,N+1 2 F SS = [2 (M 0,N+1) [ρ SS ch N ( σl z )] + M 0,N+1 2 ). l=0 F SS =1 M 0,N+1 =1 [ρ SS N ch l=0 ( σz l )] =1 P = N l=0 ( σz l ) = a b = a b {0 a, 0 b, 1,,N,(N +1) b, (N +1) a } U b {0 b, 1,,N,(N +1) b } U a

118 {0 a, 1,,N,(N +1) a } U = U b U a F enc = i=x,y,z [ σ i N+1(t)(σ i 0 ρ PP ch ρ N+1 ) ]. ρ PP ch {1,,N} σi 0 0 ρ N+1 (N +1) 0a,N+1 a 0b,N+1 b = P 2 = H Ua U a H Ua H Ua = g(c 0 a e iπn 0 b c 1 + c N eiπn (N+1) b c (N+1)a + ) F enc = 1 6 (2 M 0,N+1 2 [ M 2 0,N+1 M 0,0 M N+1,N+1 ] + M 0,N i M N+1,i M i,0 2 ) M

119 1/r 3 H B = N ωa i a i + i=1 N 1 i=1 κ(a i a i+1 + a i+1 a i). b k = 1 A A = (N +1)/2 k =1,,N H = k (ω + ɛ k)b k b k j sin jkπ N+1 a j ɛ k =2κcos( kπ ) N+1 H B = g(a 0a 1 + a N a N+1 + )+ω (a 0a 0 + a N+1 a N+1) g ω a 1 a N b k H B + H B = N t k (a 0b k +( 1) k 1 a N+1 b k + ) k=1 + ω (a 0a 0 + a N+1 a N+1)+ N (ω + ɛ k )b k b k, t k =(g/a)sin[kπ/(n +1)] b z ω = ω + ɛ z t z ɛ z ɛ z±1 H B eff = 2t z (η 0b z + b zη 0 ) η 0 =1/ 2(a 0 + a N+1 ) k=1

120 ξ ± =1/ 2(η 0 ± b z ) H B eff = 2t z (ξ +ξ + + ξ ξ ). H B eff τ B = π/( 2t z ) U B eff = e ihb eff τ B =( 1) ξ + ξ + ( 1) ξ ξ (U B eff ) ξ ± (U B eff )= ξ ± a 0 a N+1 a 0 (τ) (U B eff) a 0 (U B eff) = a N+1, a N+1 (τ) (U B eff) a N+1 (U B eff) = a 0, 0 1 a N+1 (τ) =M N+1,0 a 0 + ɛa ɛ ɛ =1 M N+1,0 2 g 2 a ɛ a i i =1,...,N +1 N +1 n N+1 (τ) =(1 ɛ) n 0 + ɛ n ɛ n i = a i a i n ɛ kt/ω > 1 g g ω/(kt)

121 40 Number of States T1 NV (units of ms) T1 NV (units of s) Participation Ratio Number of States Participation Ratio Number of States Participation Ratio Number of States Participation Ratio N = 11 T 1 d = 10 κ = 50 σ d σ d g L g R N = 51 N = 51 κ = 50

122 g g T 1 ɛ = L k z(g 2 ψ k,l 2 + g 2 R 2 k ψ k,r 2 2 k )+N t T 1, g L(R) ψ k,l(r) k z k N t T 1 N

123 g L g R t z = g L ψ z,l = g R ψ z,r t = π/ 2t z ɛ = k z ( ψk,l 2 gl 2 2 k + ψ ) z,l 2 ψ k,r 2 Nπ + ψ z,r 2 2 k 2T1 g L ψ z,l, ( g L = Nπ 3 2 2T 1 ψ z,l k z ψ k,l 2 2 k ) + ψ 1 z,l 2 ψ k,r 2. ψ z,r 2 2 k 1/ T 1 10

124 {J i } J κ =1 g 0.7 T 1 N =11 N =51 1 ɛ 200 N =51 T 1

125 N PR = 1 N i=1 ψ i 4 N PR O(N) N PR N N PR σ κ σ κ 0.5κ < 2/3 T 1 5 gψ κ/n g = g M (N) κ N +2

126 g/κ N 1/6 N > 90% N = 100 J i = 1 2 (i +1)(N +1 i) H = N i=0 N+1 J i (σ + i σ i+1 + h.c.)+ i=0 h 2 σz i, h H = ij K ijc i c j K ij = J i δ j,i+1 + J j δ i,j+1 + hδ i,j H = N+1 k=0 ω kf k f k ω k = k + h N+1 2 c i (t) = j M ij(t)c i (0) h = N+1 t =2π M(2π) = 2 c i (2π) =c i (0) {J i } J i = J N i ψ H ψ ik =( 1) N+1+k ψ N+1 i,k

127 h = 3 (N +1) t = π 2 M ij = k ψ N+1 i,k ψ jk = δ N+1 i,j. {0, 1,...N} [ρ SS ch P ]=1 M 0,N+1 =1 F SS =1 h = 3(N +1) U 2 P = ( i) N+1 UP 2 = ( 1) N+1 F SS = [2 M 0,N+1 + M 0,N+1 2 ), F enc = [2 M 0,N+1 2 M 2 0,N+1 M 0,0 M N+1,N+1 + M 0,N i M N+1,i M i,0 2 ]. M 0,N+1 1 J i = 2 1 (i +1)(N +1 i) g κ J 0 = J N = g J 1 = J 2 =... = J N 1 = κ g/κ

128 1/T 1 M 0,N+1 1 N =2, 3 J i = 1 2 (i +1)(N +1 i) N>3 g = g M (N) ω k k N+1 2 h =0 F enc g M N 1/6 τ N 1/r 3 e iπ i 1 j=1 σ+ j σ j

129 XX N =12 90% N =10 98%

130 1/2 H N = γ e B S γn B I + A S z I z + A (S x I x + S y I y ), S 1/2 I A = A = N 1 H N = κ Si z Si+1 z + i=1 N (ω 0 + δ i )Si z, i=1 κ ω 0 δ i a 1 H eff = κsi z Si+1 z + JSNV z (Sa z + Sb z )+ i=1 N 1 i=b κs z i S z i+1, J a b

131 a)! 5-1%3-"617'01-"8234"53'. "!!!"#$%&'( "!!!!!!!!!!!! /&0010"(234 "!!!!)*"+,+-.( " b) J 9" 3 " % " * "!" c) - - N (1,N) (2,N 1) N +1 Q = H CP H CP Q M Q L Q M Q L

132 U eff = e ih eff T /2 S x NV e ih eff T /2 S x NV = e iκ S z i Sz i+1 T U local = e ih eff T e iκ S z i Sz i+1 T = e ijsz NV (Sz a+s z b )T κ(t + T )=2πm m U local N N th (N 1) st N +1 Q n+1 =( H i CP i ) n+1 U local Q M

133 Magnetic field gradient : Nitrogen impurity : Two-qubit NV Register Q L Q M Q L U directed = e ijsz NV Sz N T b H med = J(S z NV 1 + S z NV 2 )S z N b,

134 NV 1 NV 2 N b Q M ( H i CP i ) n+1

135 !"#$%&'()'*+#,-+%%&.'/-0",-1-2'!3/4'51,& '! "! "! "! #! "! $!!!! " "!!!! # #!!!! = %!! " 678'929%&: '! Q k m U p (k )! Q k #! 678'929%&: '! $ A B X U p = X 1 x π CP X 1 CP CP Q k =( H CP) k U (k) p = Q k U pq k n n +1 k = n 1 U swap = HU (k) p H XU (k) p Z HU (k) p H, X x Z z π

136

137 1/r 2 A B a R

138 %! $ " # A B a R λ Ω

139 1/r 6 1/2 H = 2 σi z + Ω 2 i σi x + i i<j C p r i r j p P i P j. Ω C p P i = =(1+σ z i )/2 m s =0 m s =1 H AB = i C p r A r i P p A P C p i + r B r i P p B P i. σa z σz B a R =[ζ(p)(p +1)C p / ] 1/p Ω=0

140 R max p r (r) p r (r) =n exp( nr) n R max = n 1 log N N = C p (a R R max ) p, L L A B E int = E E E + E, E αβ α A β B E int b 2 /L b

141 ε ε ε 1 + ε 2 ε 1 ε 2 ε 1 =exp[ 2 G t g/( λ)] G t g λ ε 1 =exp( c G t g / ) c G 1/L α = c G / = α 0 /L ε 2 = ε 2 (γt g ) γ t g ε 2 =1 exp[ (γt g ) δ ] (γt g ) δ δ

142 γ L γ = γ 0 L/L 0 γ 0 L 0 L 0 a R ε =exp( α 0 /Lt g )+(γ 0 L/L 0 t g ) δ. t opt α γ t opt = δl log[l 0 α 0 /(L 2 γ 0 )]/α 0 ε = ( δ L2 γ 0 log L ) δ 0α 0. L 0 α 0 L 2 γ 0 1/L 2 N a b ψ A,B =( A,B + A,B )/ 2

143 ψ SC = i i Ω(t) = ( ) Ω 0 sin 2 8t/t0 1+16t 2 /t 2 0 (t) = 0 [1 5exp( 4t/t 0 )], t 0 t = t 0 t π = π /E int π H H t g =2t 0 + t π F = ρ 2 AB ρ AB A B G t 0 F =1 cexp( d G t 0 ) c d ρ AB

144 F G t 0 G t 0 N = 34 t 0 Ω 0 p =3 C 3 = 100Ω 0 a 3 0 =2.3Ω 0 b =3a t 0 L/ a R F = 1 2 [1 c exp( d Gt 0 )] { 1+exp [ (γ 0 L/ a R t g ) 3]} δ =3 β t 0 G E int t 0 A B

145 S p =6 p =3 n =43 γ 0 = 10 KHz F =0.95 Ω 0 =2π 3.2 MHz 0 =2π 7 MHz C 3 n 3 =2π 320 MHz a =1µm γ 0 = 100 Hz Ω 0 =2π 80 KHz 0 =2π 170 KHz a =2nm T 2 m s =+1 m s = 1 T 1

146 γ 0

147 r a S =1 W

148 W 2.87 m s = 1 m s =1 m Ω m 0 1 H r = NΩ( 0 W +h.c.) 0 m s =0 W W = i.... N i NΩ/ H r = J W W J = NΩ 2 / m J J V dd W

149 m m>1 0 W Ω ext V dd J V dd Ω ext N =100 r =20nm m s =0, 1 σ α V ij = ( ) 1 3cos 2 µ 2 ϑ ij r i r j 3 { 1 4 [ 1+σ (i) z ][ 1+σ (j) z ] (i) σ + σ (j) } σ (i) σ (j) +, r i µ ϑ ij r i r j H = /2 i σ(i) z +Ω i σ(i) x + i<j V ij 0 m s =1 W H eff = N c µ 2 R 3 ( 1 q, 0 0 q,w +h.c.),

150 N c W 1 q, 0 m s =1 0 R =100nm m 2 m =3 φ ( N 2 N c = i...0 N φ ), i N c N Ω h 100 MHz N c 70 N Ω N c > 50 N c N c Ω NΩ/

151 N c Ω=0 Ω=h 110 MHz N c 70 = h 4 GHz) N c Ω 1 q, 0 m s =1 p q π t π R H eff R r V c t π 1/R 3

152 m s = 1 p q t π 600 µs V c 1/R 3 N N W N T 2 i W [ p W = p T2 1 W σ (i) z W 2] = 4 ( N p T ), N p T2 N

153 N T 2 W T eff 2 T 1 m s =1 m s =0 W m s =1 m s =0 p 1 0 T 1 p 0 1 T 1 p 1 0 T 1 0 m s =0 W 0 p W 0 = p T σ (i) 1 W 2 = p T1 1 0 N, N T W m s =1 T p 0 1 T 1 N p 0 1 T 1 T 1 T 2 T 1 /N T 2

154 R = 100 nm W T 2 T 1 1 T 1

155 t g t π T 1 /N T 2 ε =1 exp[ (4t π /T eff 2 ) 3 ] ε =10 2 T eff 2 =11ms 0 1 W W W 100 khz m s =1 A 2.14 MHz 14 N N N t π =70µs ε =10 2 T eff 2 =700µs ε =10 4 T eff 2 =3ms W

156

157

158

159 I =1/2 S =1 B Ω MW 0 e 1 e Ω RF 1 e t e n π 1 e n e τ π/2 n e e n n e S =1 0 e

160 0 e 1 e I =1/2 15 H e,n = 0 S 2 z + µ e BS z + µ n BI z + AS z I z, 0 =2.87 µ e = 2.8 µ n = 0.43 A =3.0 ẑ n e

161 0 e I z B z, µ 10µ

162 B z (y) = db z dy y + B z,0

163 6B;m`2 93, h?2 `+?Bi2+im`2 7Q` im`2 i2 [m MimK +QKTmi2`X U V irq@ /BK2MbBQM H?B2` `+?B+ H H iib+2 HHQrBM; 7Q` H2M;i?@b+ H2 # b2/ +QMi`QH- r?b+? 2M #H2b 7mHHv T ` H@ H2H QT2` ibqmbx i i?2 HQr2bi H2p2H- BM/BpB/m H TH [m2ii2b `2 QmiHBM2/ BM ;`2v M/ 2 +? +QMi BMb bbm;h2 +QKTmi ibqm H Lo `2;Bbi2`X i i?2 b2+qm/ H2p2H Q7?B2` `+?v- bmt2`@th [m2ii2- QmiHBM2/ BM r?bi2-2m+qkt bb2b H iib+2 Q7 TH [m2ii2bc 2 +? bmt2`@th [m2ii2 Bb b2t ` i2hv K MBTmH i2/ #v KB+`Q@bQH2MQB/ +QM}M2/ KB+`Qr p2 }2H/bX AM Q`/2` iq HHQr 7Q` [m MimK BM7Q`K ibqm i` Mb@ 72` +`Qbb #QmM/ `B2b Q7 bmt2`@th [m2ii2b- i?2`2 2tBbib /m H bmt2`@th [m2ii2 H iib+2 QmiHBM2/ BM `2/X U#V h?2 b+?2k ib+ Lo `2;Bbi2` BKTH Mi ibqm rbi?bm bmt2`@th [m2ii2x hrq `Qrb Q7 BM/B@ pb/m H TH [m2ii2b rbi?bm bmt2`@th [m2ii2 `2 b?qrmx Lo `2;Bbi2`b- +QMbBbiBM; Q7 M 2H2+i`QMB+ U;`22MV M/ Mm+H2 ` Uv2HHQrV btbm `2 /2TB+i2/ rbi?bm bi ;;2`2/ mt@bhqtbm; `` v r?b+? Bb `Qr@ `2T2iBiBp2X AM/BpB/m H `Qrb rbi?bm bbm;h2 TH [m2ii2 `2 bt2+b}2/ #v M BMi2;2` n rbi? n = 1 #2BM; i?2 #QiiQK `Qr M/ n = M #2BM; i?2 iqt `QrX hq +?B2p2 bi ;;2`2/ bi`m+im`2- r2 bt2+@ B7v mmb[m2 BKTH Mi ibqm `Qr rbi?bm 2 +? TH [m2ii2 r?2`2bm bbm;h2 BKTm`BiB2b `2 BKTH Mi2/ M/ bm#b2[m2mihv MM2 H2/X 6Q` ;Bp2M `Qr Q7 TH [m2ii2b- i?2 BKTH Mi ibqm `Qr +Q``2bTQM/BM; iq i?2 H27i@KQbi TH [m2ii2 Bb M 4 R- r?bh2 i?2 TH [m2ii2 iq i?2 BKK2/B i2 `B;?i? b BKTH Mi ibqm `Qr M 4 kc i?bb T ii2`m +QMiBMm2b mmibh i?2 }M H TH [m2ii2 BM ;Bp2M `Qr- r?b+? #v +QMbi`m+iBQM? b i?2?b;?2bi BKTH Mi ibqm `Qr MmK#2`X h?2 BKTH Mi ibqm T`Q+2bb Bb `2T2 i2/ 7Q` 2 +? `Qr Q7 TH [m2ii2b rbi?bm i?2 bmt2`@th [m2ii2 M/ +`2 i2b M `` v Q7 Lo `2;Bbi2`b- r?b+? 2 +? Q++mTv mmb[m2 `Qr BM i?2 bmt2`@th [m2ii2x abm+2 2 +? Lo `2;Bbi2` Q++mTB2b mmb[m2 `Qr rbi?bm i?2 bmt2`@th [m2ii2- i?2 K ;M2iB+ }2H/ ;` /B2Mi BM i?2 y /B`2+iBQM HHQrb 7Q` BM/BpB/m H bt2+i`qb+qtb+ //`2bbBM; Q7 bbm;h2 `2;Bbi2`bX *Q?2`2Mi +QmTHBM; Q7 bt ib HHv b2t ` i2/ Lo `2;Bbi2`b BM /D +2Mi TH [m2ii2b Bb K2/B i2/ #v / `F btbm +? BM / i #mb U.a*"V M/ Bb b+?2k ib+ HHv `2T`2b2Mi2/ #v i?2 +m`p2/ HBM2 +QMM2+iBM; BM/BpB/m H `2;Bbi2`bX h?2 b2+qm/ BKTH Mi ibqm bi2t +Q``2bTQM/b iq i?2 +`2 ibqm Q7 i?2b2?q`bxqmi H M/ p2`ib+ H / `F btbm +? BMbX R98

164 1/2 H int =4κSzS 1 z 2 + (ω 0 + δ i )Sz, i κ ω 0 δ i H drive = i=1,2 2Ω isx i cos[(ω 0 + δ i )t] (x, y, z) (z, y, x) i=1,2 H int = κ(s + 1 S 2 + S 1 S + 2 )+Ω 1 S 1 z +Ω 2 S 2 z. H int Ω 1 Ω 2 κ Ω 1 Ω 2 κ Ω 1 Ω 2 H int = i κ(s+ i S i+1 + S i S+ i+1 )+ i Ω isz i

165 κ 1/2 κ i,i+1 Ω i Ω i+1 H FFST = g(s + NV 1 S 1 N 1 + S + NV 2 S N + )+ i=1 κ(s + i S i+1 + S i S+ i+1 )

166 Ω i Ω j g κ g Ω

167 H FFST g B z (y)

168 N = T1 NV N = T1 NV T N 1 T N T NV 1

169 p SS err N(p SS off + p adia + p dip + p SS T 1 + p SS T 2). ( ) p SS off Ω 2 i g Ω i g p adia ( ) κ 2 p dip Ω i p SS T 1 T 1 p SS T 2 N N N p FFST err p FFST off + p fermi + p g + p FFST T 1 + p FFST T 2. p SS err p FFST err p fermi g/ N κ/n ( g/ N κ/n )2 p g g p SS err t FFST

170 T NV T 1 T 1 10 T 1 50 ɛ 1.4%

171

172 g H = ( ) Ω( ) Ω N S N x +4κS NV z S N z, Ω Ω N B = D = B + 2Ω 0 0 2Ω B Ω H = D D ( + 2Ω2 ) Ω2 1 2 Ω N( + N + N ) +2κ( B D + D B )( + N + N + ), ± N = N ± 2 N S N x 2κ { ( + D + D + ) 2Ω ( D + D ) } ( + N N + N + N ). { D, } g κ Ω

173 κ + κ 2 / 2 0 (α + β ) (α + β D ) π D D D Ω D

174

175

176 1/2

177 6B;m`2 8R, a+?2k ib+ `2T`2b2Mi ibqm Q7 iqtqhq;b+ HHv T`Qi2+i2/ bi i2 i` Mb72` h?2 ;`2v /`QTH2i `2T`2b2Mib k. `` v Q7 BMi2` +ibm; btbmb imm2/ BMiQ i?2 *ag" T? b2x Zm MimK btbm@`2;bbi2`b +QKTQb2/ Q7 i` Mb72` [m#bi U;`22MV M/ K2KQ`v [m#bi U;QH/V `2 `` M;2/ `QmM/ i?2 2/;2 Q7 i?2 k. /`QTH2i M/ +QmTHBM; #2ir22M i?2k Q++m`b i?`qm;? i?2 +?B` H 2/;2 KQ/2X URV "v K T@ TBM; i?2 [m MimK BM7Q`K ibqm QMiQ 72`KBQMB+ r p2@t +F2i U#Hm2V i` p2hbm; HQM; i?2 2/;2- i?2 [m MimK bi i2 + M #2 i` Mb72``2/ iq `2KQi2 `2;Bbi2`X h?2 r p2t +F2i i` p2hb QMHv BM i?2 /B`2+@ ibqm Q7 i?2 #Hm2 ``Qrc i?bb +?B` HBiv T`2p2Mib KQ/2 HQ+ HBx ibqm M/ /2bi`m+iBp2 # +Fb+ ii2`bm;x i bt2+b}2/ ibk2 i i?2 `2KQi2 `2;Bbi2` HQ+ ibqm- i?2 +QmTHBM; Bb im`m2/ QM M/ i?2 r p2t +F2i Bb + Tim`2/ UkVX :Bp2M M M+BHH `v K2KQ`v [m#bi M/ HQ+ H `2;Bbi2` K MBTmH ibqmb- irq@[m#bi ; i2 UjV + M #2 T2`7Q`K2/ #27Q`2 i?2 [m MimK bi i2 Bb i` Mb72``2/ # +F iq i?2 Q`B;BM H `2;Bbi2` M/ biq`2/ U9@8VX h?bb HHQrb 7Q` mmbp2`b H +QKTmi ibqm #2ir22M i?2 K2KQ`v [m#bib Q7 bt ib HHv b2t ` i2/ `2;Bbi2`bX?QM2v+QK# H iib+2 b /2TB+i2/ BM 6B;X 8k (jkd)x h?2 bbq+b i2/ > KBHiQMB M M im` HHv ;2M2` HBx2b i?2 EBi 2p KQ/2H (jrn) M/ 72 im`2b +?B` H btbm HB[mB/ ;`QmM/ bi i2 U*aG" T? b2vh0 = 1! x x 1! y y 1! z z κσi σj + κσi σj + κσi σj, 2 x,x! 2 y,y! 2 z,z! links links UdXRV links r?2`2 1σ `2 S mhb btbm QT2` iq`b U! = 1VX h?2 KQ/2H K v #2 bqhp2/ #v BMi`Q/m+BM; 7Qm` J DQ` M QT2` iq`b- {γ 0, γ 1, γ 2, γ 3 } 7Q` 2 +? btbm- b b?qrm b+?2k ib+ HHv BM 6B;X 8k M/ #v `2T`2b2MiBM; i?2 btbm H;2#` b, σ x = iγ 1 γ 0 - σ y = iγ 2 γ 0 - σ z = iγ 3 γ 0 (jk8- jkd)x h?2 J DQ` M QT2` iq`b `2 >2`KBiB M M/ b ibb7v i?2 bi M/ `/ MiB+QKKmi ibqm `2H ibqm {γ l, γ m } = 2δlm X h?2 >BH#2`i bt +2 bbq+b i2/ rbi? i?2 T?vbB+ H btbm Bb irq@/bk2mbbqm H bm#bt +2 Q7 i?2 2ti2M/2/ 7Qm`@/BK2MbBQM H J DQ` M >BH#2`i bt +2c i?mb- r2 Kmbi BKTQb2 i?2 ; m;2 T`QD2+iBQM- P = 1+D 2 r?2`2 D = γ 1 γ 2 γ 3 γ 0 (jk8)x R8N

178 H γ = i 4 κ i,j Û i,j γ 0 i γ 0 j, Ûi,j = iγ α i γ α j α ij ij Ûi,j Û i,j Ûl,m {U i,j = ±1} {U i,j } γ 0 U i,j 2 w(p) = ij p U i,j p p ij U i,j =+1 γ 0 π/2 w(p) =+1 π π/2 π/2 w(p) = 1 v Q κ i,j Q k,i(iu i,j )Q k,j = δ kk ɛ k H γ = 1 N/2 2 k= N/2 ɛ kc k c k c k = 1 2 j Q k,jγj 0 N k

179 ɛ k = ɛ k c k = c k k>0 H γ = k>0 ɛ k (c k c k 1 2 ), c k c k b U i,j H T = H 0 + H int H int L R H int = S 2 (σz L + σ z R)+g L σ β L σβ a + g R σ η R ση b. S β,η g L g R

180 Ûi,j γ 0 i π π/2 U ij w(p) = 1

181 a b H int = S 2 (iγl 3γ0 L + iγ3 R γ0 R )+g LγL 1γ1 aγl 0γ0 a + g R γr 1 γ1 b γ0 R γ0 b σ x σ x U i,j U L,a U R,b H T = H + H int = k>0 ɛ k (c k c k 1 2 ) + S (c L c L 1 2 )+ S(c R c R 1 2 ) g L U L,a (c L + c L ) i 2 ( k Q k,ac k + k Q k,a c k ) g R U R,b (c R + c R ) i 2 ( k Q k,bc k + k Q k,b c k ), c L,R =1/2(γL,R 0 iγ L,R 3 ) c L,R =1/2(γ L,R 0 + iγ L,R 3 ) σ z

182 L R S > 0 g L,g R < S c k S k ɛ k H eff = i g L Q k,a c L 2 c k i g R Q k,b c R 2 c k + h.c. τ ɛ κ/l l τ l/κ

183 61 40 y k y a = π b =0.46κ 0.14κ 0.17κ S g L (t) g L (t) = vf(t), dt t f(t ) 2 f(t) v g R (t) S l

184 T N v n p e v/t n p π N v π ξ a a π

185 v H p H T H p H e = v dp 2π pc pc p = v dx γ(x)(i )γ(x),

186 c p = c p {c k } p {γ(x),γ(y)} = δ(x y) c p H e = λ dx γ(x)(i )γ(x)(i ) 2 γ(x)(i ) 3 γ(x), λ Γ int p ɛ p k B T Γ int p Γ int p λ2 p 13 + λ2 p 11 T 2 + O(T 4 ). v v κ λ κ( κ κ )2 a 7 Γ int p κ2 S ( κ κ )4 (ap) 14 S = vp p<1/a Γ dec p σ α i Γ dec p S(ω)

187 e d/ξ d i S(ω 0 ) e ω 0/k B T ω 0 =2 v S l Γ dec p e S/k B T + le V /k B T. S(ω) 1 ω 2 +1/t 2 c t c Γ dec p 1/ 2 v v 1/t c

188 e S/k B T S l

189 L R R L

190

191

192 N ρ = i ρ i ρ i Q = { +,, + i, i, 0, 1 } ρ F tol ρ i M i

193 F i =1/ Q ρ i Tr[ρ i M i (ρ i )] p h = 1 Q Tr[P acc M(ρ)] 1 e ND(F exp F tol ), ρ Q Q = Q N P acc M = i M i F exp =1/N i F i D F tol F tol N F tol N F tol N (2F tol 1)N 2F tol 1 p d = 1 Q ρ Q Tr [ P 2 acct (ρ) ] e ND(2F tol 1 2/3), T 2F tol 1 > 2/3 5/6

194 a b original qticket:... cloned qticket:... F tol N " 1 " 2 " 3 minimum overlap " N #2 F tol N " 1 " 2 " 3 " N #1 " N 1" " N #2 " N #1 " N "# +" 0" " Z challenge questions " X F tol F tol N (2F tol 1)N Ftol cv = 1+1/ 2 2

195 X Z { 0, +, 0,, 1, +, 1,, +, 0,, 0, +, 1,, 1 }. n r n r 2 X Z Ftol cv r n F exp >Ftol cv n

196 r p cv h ( rd(fexp F cv 1 e )) n tol. F cv tol > 1+1/ / Ftol cv p cv d ( ) 2 v ( 2 1/2+e rd(f tol 1+1/ 2 2 )) n, v F dishonest <F tol <F exp

197 , 0 0, + 1, + 0, + 0, + +, 1, 0 1, + Z X 0, 0 0, 1 1, 1 0, 1, + +,, + +, n = 4 r =2 8 F tol =3/4 F tol =3/4 F tol > 1/2+1/ 8

198 c

199 c (c+1)(c+2) X 1,...,X n {0, 1} δ i S {1,...,n} Pr [ i S X ] i i S δ i Pr[ n X i γn] e nd(γ δ) i=1 δ := n 1 N i=1 δ i γ δ γ 1 D(p q) =p ln p q +(1 p)ln 1 p 1 q (X =1)=p (X =1)=q

200 P ρ acc N F tol ρ = N i=1 ρ i 0 F tol 1 P ρ acc = N ( bi ρ i + b i ρ i ). b: b 1 F tol N i=1 b {0, 1} N N b 1 = N i=1 b i b i =1 b i ρ i = ρ i b 1 F tol N 1 b M i F i F exp =1/N i F i F exp >F tol p v 1 e ND(F tol F exp). X =(X 1,...,X N ) Pr[ X = b]= 1 Q = N i=1 [ Tr M(ρ) ρ Q N ( bi ρ i + b ) ] i ρ i i=1 1 Tr [ M i (ρ i )(b i ρ i + 6 b i ρ i ) ] ρ i Q

201 X i Pr[X i ]=F i 1 Q ρ Q Tr[P accm(ρ)] ρ = Pr[ N i=1 X i F tol N] L 1 µ

202 (a) (c) (b) (d) ±1 { +1, 1 } S =1 m s =0, ±1 E B ( =1) H NV =(D 0 + d E z )S 2 z + µ B g s S B d [ Ex (S x S y + S y S x )+E y (S 2 x S 2 y) ],

203 D 0 /2π 2.88 g s 2 µ B d d ±1 B = g s µ B B z / E ±1 E = E 0 (a + a ) a ω m E 0 ±1 0 = B ω m D 0 ±1 0 H i = g ( σ + i a + a σ i ) σ ± i = ±1 i 1 i g J z = 1 2 i 1 i 1 1 i 1 J ± = J x ± ij y = i σ± i H = ω m a a + B J z + g ( a J + aj + ), g

204 L w, h ( ) 1/2 g 2π 180 L 3 w GHz, ρe ρ E (L, w, h) =(1, 0.1, 0.1) µ ω m /2π 1 g/2π 1 g e /2π 10 T 2 η = g2 T 2 γ n th γ = ω m /Q n th =(e ω m/k B T 1) 1 T Q =10 6 T 2 =10 T =4 η 0.8 g = B ω m H ( ) e R He R R = g a J aj + (g/ ) 2 H eff = ω m a a + ( B + λa a ) J z + λ 2 J +J, λ =2g 2 / J + J = J 2 J 2 z + J z J

205 J 2 z ψ 0 x J x ψ 0 = J ψ 0 J 2 y = J 2 z = J/2 J 2 z z J z J 2 min = 1 2 ( V + V 2 + V 2 yz ξ 2 = 2J J min 2 J x 2, ) V ± = J 2 y ± J 2 z Vyz = J y J z + J z J y /2 ξ 2 < 1 H eff [ ρ = i λ 2 J z 2 + ( B + λa a ) ] J z,ρ + 1 D[σ 2T z]ρ i 2 +Γ γ ( n th +1)D[J ]+Γ γ n th D[J + ], i D[c]ρ = cρc 1 2 ( c cρ + ρc c ) T 1 2 λ/2 B T 1

206 Γ γ = γg 2 / 2 n = a a n ψ 0 N =100 n th T 1 2 Γ γ ξ 2 Γ γ T 2 N J 1 n th 1 ξ 2 4Γ γ n th Jλ 2 t + t T 2. t ξ 2 opt 2 Jη, t opt = T 2 / Jη J ξopt 2 J 2/3 n = a a J z

207 1 0.5 ξ (a) 10 n th = J λt ξ 2 opt 0.1 (b) N N N N = 100 T 2 = 10 n th n th =1 T 2 =1 ω m /2π =1 g/2π =1 Q = 10 6 T 2 H int (t) =λj z f(t)δn(t). f(t) J z J z π δn(t) =n(t) n n

208 n S n (ω) = dte iωt δn(t)δn(0) J + (t) = e χ e iµ(jz 1/2) J + (0), J 2 +(t) J + (t)j z (t) χ µ dω χ = λ 2 F (ωτ) 2π µ = λ 2 dω 2π ω 2 Sn (ω), K(ωτ) A ω 2 n (ω), S n (ω) =(S n (ω)+s n ( ω)) /2 A n (ω) =(S n (ω) S n ( ω)) /2 F (ωτ) = ω2 dte iωt f(t) 2 τ 2 π K(ωτ) µ F K F M =4 S n (ω) =2γ n th ( n th +1)/(ω 2 + γ 2 ) ω =0 χ 0 (t) = 1λ2 n 2 2 th t2 n th 1 T2 2/λ n th t = t opt n th > J M π

209 t χ th λ 2 γ n 2 th t3 /M 2 Γ γ T 1 2 M n th γt2 ω dr = ω m + δ S n (ω) ω = ±δ n dr n th n dr ( n th +1) n dr t/m =2π/δ π χ dr ( λ δ ) 2 ndr n th γt µ λ2 δ n drt n dr n th χ th Γ γ n dr M =4 g J eff <J J 1 g i i g i/ i g2 i i g2 i / i g4 i g i

210 Sn(ω) f (t) t/τ F (ωτ) ξ (a) 0.1 (b) K (ωτ) δ 0 δ ω J λt ξ 2 min (c) F K M =4 f(t) M =4 n th = 10 n dr = 10 3, , 10 6 n th = 50, 10 M =4 g/2π =1 T 2 = 10 N = 100 ω m /2π =1 Q = 10 6 n dr (1, 0.1, 0.1) µ N N eff 100 η 1 H int = λ ( σ 1 + σ2 +h.c. )

211 4

212 σ y Q = ν/ ν ν ν ν

213 Excitation Laser Si Photodiode Diamond Film Bragg Mirror Microwave Line µ > 0 ±

214 H gs =(D gs + d σ z )S 2 z + gµ b S B + d σ x (S x S y + S y S x )+d σ y (S 2 x S 2 y) d, C 3v D gs µ b S k k = {x, y, z} σ D gs S z ω 0 ±1 ω D gs ω D gs B 1 =2b 1 cos(2πωt)ˆx H gs V = e 2πiωtS2 z H gs =(D gs + d σ z ω)s 2 z + gµ b B z S z + gµ b b 1 S x

215 ρ = 1 i [H gs,ρ]+ k L k ρl k 1 2 L k L kρ 1 2 ρl k L k ρ L k r k I Ω=gµ b b 1 =D gs ω F (I,Ω, ) = γρ ss 22 + γ2 κ+γ ρss 33 ρ ss 22 ρ ss 33 F σ y (τ) = πQ (S/N) τ τ S/N C 1

216 A V 3 E 3 A I D es γ orbital states electron spin sublevels D gs κ λ 1 E 1 A1 S B 9.5 x 105 I 0 9 F(!) (photons/sec) C ν Detuning from resonance,! (Hz) 3 E 3 A 1 E 1 A I γ κ λ S z 0 ±1 C 1 T2 =88 σ y (τ) = τ 1/2 Q S/N

217 D gs T δω δωt T 1 T T c T c T c T 2 m f =0 π T T 2 π ) 2 x ) τ π x τ π ) 2 x θ ) φ θ φ t p θ = gµ b b 1 t p Ω 1 gµ b b 1 (D gs Ω),gµ b B z

218 U echo = e i π 2 Sx e i(δωs2 z +bs z )τ e i π 2 Sx e i π 2 Sx e i π 2 Sx e i(δωs2 z +bs z )τ e i π 2 Sx = WW W W = e i π 2 Sx e i(δωs2 z +bs z)τ e i π 2 Sx = e i(δω ˆX bs y )τ, W W = e i π 2 S x e i(δωs2 z +bs z )τ e i π 2 S x = e i(δω ˆX+bS y)τ, ˆX ˆX = [ ˆX,S y ]=0 W W U echo = e 2i(D gs Ω)τ ˆX ˆX 0 π 4 ˆx U 45 = e i π 4 S y U 45 0 = 1 2 ( 0 i + + ±1 S z U echo U 45 0 = 1 ( ) ie iδωτ + + e iδωτ 0. 2 Sz 2 δν S z B A S z I z π/2 π/4 S z

219 Optical Pumping Z π π π 4 T T 4 TF } S 2 z X Y S z S z S 2 z S z S 2 z Sz 2 2T S z ψ 0 = m s =0 U echo ψ f = U echo U 45 ψ 0 = 1 2 sin(φ) cos(φ)( ) φ =(D gs ω)t = δωt ˆM ˆM = a b ( ) a b ˆM

220 ψ f δω ω 0 = 1 ω 0 ˆM ˆM / ω ˆM 2 = ˆM 2 ˆM 2 2a 3b ˆM M τ = M T δω ω 0 M = ξ D gs Tτ ξ 5 b 0 a λ/γ T = T 2 1 D gs =2870 δω/ω 0 = / τ 0.2% N 1/ N N µ 3 3 1/ N / τ / τ 100µ 1 2 σ pulsed y / τ

221 T 1/2 2T e (2T/T 2) n n 3 T 2 T e n =1 2 S/N T e D gs 1µ

222 T 2, 10 20µ D gs dd gs /dt = 74.2(7) dd gs /dt 100 dd gs /dt =100 D gs dd gs /dt 75 dd gs /dt

223 δ D gs V ij S i S j = κ 2 3ˆri ˆr j δ ij r 3 S i S j S i κ a r (1) = aŷ r (2) = 3a ˆx a 2 2ŷ r(3) = 3a a ˆx 2 2ŷ H = V ij S i S j = 3κ 4a 3 (S2 z 2/3). ɛ kl H = S i S j V ij = κ 2 ( ) 3ˆr i ˆr j δ ij ɛ r k r 3 kl ˆr l S i S j = 3κ 2a 3 [ 3 4 (ɛ xx + ɛ yy )(S 2 z 2/3) (ɛ xx ɛ yy )(S 2 x S 2 y) (ɛ xy + ɛ yx )(S x S y + S y S x ) 1 2 ɛ zx(s x S z + S z S x ) 1 2 ɛ zy(s y S z + S z S y )]. α = 3 4 (ɛ xx + ɛ yy ) 3κ 2a 3 3κ 4a 3 a =2.38 α = 4.32(ɛ xx + ɛ yy ) =2.88

224 (a) "(T) (b) area A!d, Ed Diamond 1 Diamond 2 "0 clock 1 clock 2 (c)!1, E1 clamp1 Diamond clamp2 z T0 T!2, E2 Pz=Fz/A=Y! #T dd gs /dt η 1,2 E 1,2 ɛ xx = ɛ yy = ɛ zz dd gs dt ( ) =( ) 15, 1K 5 δ 3 ddgs dt

225 E 1 η c1 E 2 η c2 <η c1 T ɛ 1,2 η d (1 + η c1,2 E c1,2 /E d ) T T T 0 ω 0 T T

226 ω 1 (T ) = ω 0 + β 1 T ω 2 (T ) = ω 0 + β 2 T β 1,2 = dɛ dd gs dt dɛ = η d (1 + η c1,2 E c1,2 /E d )(dd gs /dɛ) τ =0 T 0 T 0 t φ 1,2 (t) =ω 0 t + t 0 β 1,2 T (t )dt ± φ 0, φ 0 = ξ t/ T 2,N φ(t) =φ 2 (t) φ 1 (t) = t 0 β 1,2 T (t )dt ± 2 φ 0 β 1,2 β 2 β 1 φ 1 (t) φ 2 (t) t φ 1(t) = φ 1 (t) β 1 T (t )dt 0 ( t ) = φ 1 (t) β 1,2 T (t )dt 0 = φ 1 (t) β 1 β 1,2 ( φ(t) 2 φ 0 ). β 1 β 1,2 t t ω 1 ω 1 = = ( ξ β1 (1+2 T2 NtD gs β 1,2 ) 2 ) 1/2 ( ) ( ( ) ) 2 1/2 ω β ω T =T 0 β 1,2

227 D ZFS (khz) (a) (b) (c) Brass Tungsten khz, 6.92 mk Temperature (mk) D ZFS (khz) T=5 mk T=10 mk Position (mm) diamond d clamp D gs β 1,2 β 1 ν beat 10 Q 10 6 D gs β 1,2 η d (1 + η c1,2 E c1,2 /E d )/( 75 T 0.01 dd gs /dt

228 D gs σ ɛ F ijkl S i S j ɛ kl F Sz 2 (D gs + A 1 (ɛ xx + ɛ yy )+A 2 ɛ zz )(S 2 z 2/3) A 1 A 2 ɛ xx = ɛ yy = ɛ zz dd gs /dt =2A 1 + A 2 T T 0 L d L = L d η d T P = E d L d /L = E d ε d = E d Tη d L T 0.01K

229 Al + ion clock Ensemble NV Echo (0.01ppb x 1mm 3 ) TXCO Commercial Rb Single NV Echo Rb Chip Clock Ensemble NV CW SAW Oscillator Allan Deviation after 1s averaging Single NV CW / f N 2

230 σ y = τ 1/2 9 /

231 200 m s =0 m s = ±1 d /dt = (2π)77 / N

232 " #!!"#$%&'( )%*%&+( Temperatureaccuracy Kelvin CdSe QD SThM Nano Diamond Raman Liquid crystal Infrared Green FluorescentProtein Seebeck! "!! "!,-./0'( )%*%&+( #! " " ProjectedNano Diamond Bulk Diamond Sensor size um ±1 0 (T ) δ

233 ! 1.1 " !"#$!" "!" Normalized fluorescence population τ (us) T ( o C) 2τ 50 2π 2τ = 250 µ 2τ = 50 µ 2τ 2τ m s =0 m s = 1 η = C d /dt 1 T Nt, T t C T C /

234 ω 1 2 ( 0 + B ) B = 1 2 ( ) τ 2π +1 1 τ ± % τ η =(9± 1.8) / 2τ =250µ δt =1.8 ± 0.3 2τ (2d /dt 2τ) 1 2τ < 2τ 1 µ

235 N 500 = nm 2.5

236 " # Normalized fluorescence data max. slope meas. freq. y ( µm) ! $ " #!"#$ % & ω (GHz) x ( µm)! 4 $ off AU Fit on AU Fit " theory NV data T (K) T (K) 5! # $ % & laser power ( µw) distance ( µm) µ

237 0.8 ± 0.1 µ 0.8 µ δt =(44± 10) T (r) = Q 4πκr Q κ r 72 ± µ 1 T =( 20 ± 50)

238 12 µ 0.5 ± µ 3.9 ± µ /

239 y ( µm) " x ( µm) # $ +$'()*!"#$$$$$$$!"%$$$$$$$$!#$$$$$$$$%$$$$$$$$$#$$$$$$$$"%$$$$$$$"#!" &'()#$!"$ &'()#$ &'()#%!"%!"#$$$$$$$!"%$$$$$$$$!#$$$$$$$$%$$$$$$$$$#$$$$$$$$"%$$$$$$$"# & '()* T K Pos 1 Pos 2 y ( µm) &'()! T (K) !"#$ Au Fluorescence (kcps) *#+#$%-. /#+#,15#3,1%#4 y ( µm) !"% *#+#$%,-. /#+# 60 /#+#012#3,1$# x ( µm) x ( µm) /

240 50 µ 5 =2.87 f 1,2 f (ω )+ f ω ω f 3,4 f (ω + )+ f ω ω+ ( ) δω + δb + δt d dt ( ) δω δb + δt d dt δt = δω (f 1 + f 2 ) (f 3 + f 4 ) d /dt (f 1 f 2 )+(f 3 f 4 ), ω ± δω δb

241 C 0.03 δt σ δt = σ/(c d dt 2τ) c 2τ t<30 s t η = δt t m 1 δt = N 1 ΣN i=1 (T i mp i ) 2 T i P i σ (δt) =δt 1 2 Γ 2 (n/2) Γ( ) N 1 Γ 2 ((n 1)/2)

242 9 / 250 µ 13 T 1 / µ / n 200 µ B µ 50 µ

243 2

244 S 1,S 2 R H 12 = ɛ 1 S z 1 + ɛ 2 S z 2 + t 12 (S + 1 S 2 + S 1 S + 2 )+V 12 S z 1S z 2 ɛ i W t 12,V 12 R R t 12,V 12 ɛ i W t 12 = = S z =0 δ a = ɛ 1 ɛ 2 δ a t a = t 12 τ α a, S z =0

245 (a), (b) r 1 r =, =, t R 2 = (c) a δ a + δ b + V ab,,, δ b δ a V δ ab a δ b V ab, = 3 r 3 r 4 4 b, (δ a + δ b )+V ab R 1 S 1 S 2 a S 3 S 4 b t a,t b H ab H a = δ a τ z a + t a τ x a τ x a V 12 S z 1S z 2 S z =0 a =12 b =34 R 1 R 2 t(r 2 ) ɛ i W a b S z a =0 S z b =0 H ab = δ a τ z a + t a τ x a + δ b τ z b + t b τ x b + V ab τ z a τ z b V ab = V 13 V 14 V 23 + V 24 τa z = S1 z = S2 z H ab H ab

246 R 1,R 2 V ab 0 t a/b 0 τ z δ a/b t a/b t 2 a + δ 2 a, t 2 b + δ2 b V ab t a t b µ α µ z H ab τ µ O(1) τ x τ z H = H ab + H cd + H int = ab µ z ab + cd µ z cd + V αβ µ α abµ β cd, α,β {x,z} H int = V ac τ z a τ z c + V ad τ z a τ z d + V bcτ z b τ z c + V bd τ z b τ z d µ V (r) 1/r 1/r 3

247 V (r) 1/r β V ab = V 13 V 14 V 23 + V 24 ( 1 = 1 ) ( ) R β 13 R β 14 R β 24 R β 23 ( ) ( ) 1 = R 2 + r 4 r β R r 3 r 1 β R 2 + r 3 r 2 1 β R 2 + r 4 r 2 β 1 ( ) β 2r3 r 1 +2r 4 r 2 2r 4 r 1 2r 3 r 2 R2 1. R β 2 R R β+2 2 R 1 <R 2 /2 R 2 1 R ij V ab = V 13 V 14 V 23 + V 24 a b V ab R 2 1/R 5 2 V ab N 2 (R 1,R 2 ) V/R β 2 VR 2 1/R β+2 2

248 R 1 R 2 N 2 (R 1,R 2 ) (n 1 (R 1 )R2) d VR2 1/R β+2 2 t/r1 α R d (β+2) 2, n 1 = ρn 1 n 1 (R 1 )R d 2 R 2 2R 2 VR2 1 /Rβ+2 2 t/r1 α R 1 R 2 d>β+2 R 2 V (R 2 ) t(r 1 ) V (R 2 ) t(r 1 ) R2 1 /Rβ+2 2 1/R1 α 1/10 R 2 R α+2 β+2 1 N 2 (R 1,R 2 ) R1 d+2 (R α+2 β+2 1 ) d (β+2) α+2 d α+d β+2 = R1. d> α(β+2) α+β+4

249 N 3 (R 1,R 2,R 3 ) (n 2 (R 1,R 2 )R d 3) Ṽ/Rβ 3 Ṽ/R β 2 =(n 2 (R 1,R 2 )R3) d ṼR2 1/R β+2 3 ṼR1/R 2 β+2 2 = R 2d α+2 1 R d 2 R d β 2 3 n 2 = n 1 N 2 R 3 R 1,R 2 R 1 R 2 R 3 d>(β +2)/2 R 1 R (β+2)/(α+2) 2 R 2 R 3 N 3 V σ z α>β+4 R 1 R 2 R 3 α>β R 2 R 1 R β+2/α+2 2 R 1 R β/α 2 R <R 2 R 2 O(1) R 1 R <R 2

250 R N 2 (R 1, R) ρn 1 (R 1 ) R d R d α 1 R d. O(1) R R α/d 1 1 R 2 V (R 2 ) t(r 1 ) R >R 2 R = R 2 R 1 R = R 2 V R <R 2 R 2 R R 2 R α/d 1 1 R α β 1 d c = αβ α+β d< αβ α+β R 2 O(1) N 2 (R 2 ) 1 R 2

251 R 2 N 2 (R 1,R 2 ) R 2 α = β R 1 R 2 R 2 N 2 (R 2 ) 1 1 N 2 (R 2 )=ρ 2 R 2d β 2 V W = R2d β2 = 1 W ρ 2 V = 1 1 ρ 2 a β 0 W V/a β 0 D = W V/a β 0 ρ 1/a d 0 R 2 a 0 D 1/(2d β), d = β =3 R 2 a 0 3 D d =2 R 2 a 0 D α = β =3 d c =1.5 D>1 d c α = β =3 d z i d z j

252 i<j d z i d z j r 3 ij = i<j (d s i + d a i σ z i )(d s j + d a j σ z j ) r 3 ij = i<j (d s ) 2 r 3 ij + i<j d a d s (σ z rij 3 i + σj z )+ i<j (d a ) 2 σ z rij 3 i σj z, d z = d s + d a d z = d a d s ɛ i = i j d a d s i ɛ rij 3 i = j i d a d s Q rij 3 j = ν da d s a 3 0 l lat 1 ν l 3 Q j =1 0 δɛ 2 i = ɛ2 i ɛ i 2 ɛ 2 i = ( )( d a d s d j i k i Q a d s rij 3 j Q rik 3 k ) Q j Q k = νδ jk + ν 2 (1 δ jk ) ( ) 2 [ ɛ 2 i = (ν ν 2 ) 1 l lat + ( ν ) 1 2 l 6 l lat ] l 3 d a d s a 3 0 W = ɛ ɛ 2i = 2i ɛ i 2 = da d s (ν ν a 3 2 ) 1 0 l, 6 l lat ν 1 W ν da d s ρ ν/a 3 0 W = da d s (ν ν a 2 ) 1 3 l lat 0 l 6 a N 2 (R 2 )=ρ 2 R 2d β 2 V W = R2d 32 = 1 W ρ 2 V = 1 1 ρ 2 a 3 0 d a d s (ν ν a 2 ) 3 l lat 0 V/a l 6.

253 V (d a ) 2 ( d s 1 R 2 a 0 (ν ν d a ν 2 ) ) 1 2d l 6 l lat ν 1 ( ) 1/3 d d =3 R 2 s a0 d a / ν d =2 R 2 ds d a a 0 /ν 3/2

254 H B = ij t ija i a j i j V ijn i n j X Y (Θ 0, Φ 0 ) = s 1, 1 + v 1, 1 + w 1, 0 s =Ω 2 Ω 4 / Ω v =Ω 1 Ω 3 / Ω w = Ω 1 Ω 4 / Ω i j R =(R, θ, φ) {x, y, z} H = 1 6 4πɛ 0 R 3 2 ( 1) q C q(θ, 2 φ)tq 2 (d (i), d (j) ), q= 2 C k q (θ, φ) k z q T 2 2 ( ) T±2(d 2 (i), d (j) )=d (i) ± d (j) ± T±1(d 2 (i), d (j) )= d (i) z d (j) ± + d (i) ± d (j) z / 2

255 ( T0 2 (d (i), d (j) )= d (i) d (j) + +2d (i) z d (j) z ) + d (i) + d (j) / 6 d ± = (d x ± id y )/ 2 1, 0 1, ±1 T 2 ±1 t ij i j T 2 0 i j = 2 3 [d2 00w i w j 1 2 d2 01(v i v j + s i s j )], i j T 2 +2 i j = d 2 01(v i s j ), i j T 2 2 i j = d 2 01(s i v j ), d 00 = 1, 0 d z 0, 0 d 01 = 1, ±1 d ± 0, 0 t ij V ij V ij = i j H dd i j + i j H dd i j i j H dd i j i j H dd i j i d i = d 1 ( s i 2 + v i 2 )+µ 0 w i 2 d 1 = 1, ±1 d z 1, ±1 µ 0 = 1, 0 d z 1, 0 1, 0 V ij

256 ij d i j d z d z (d +d + d d + ) i j = d 2 0, i j d z d z (d +d + d d + ) i j = d i d 0, i j d z d z (d +d + d d + ) i j = d 0 d j, i j d z d z (d +d + d d + ) i j = d i d j 1 2 µ2 01(s i wi w j s j + w i vi v j wj + ) i j d + d + i j = µ 2 01(s i wi w j vj + w i vi s j wj ), i j d d i j = µ 2 01(w i s i v j wj + v i wi w j s j), d 0 = 0, 0 d z 0, 0 µ 01 = 1, ±1 d ± 1, 0 1, 0 1, ±1 H dd t ii = j i ( i j H dd i j i j H dd i j ) t ii 2B {a, b, A, B} a A b B t ij V ij g 1, g 2 w t ij V ij a A b B g 2 w a/b = w A/B w a/b = w A/B

257 I 1 =4 I 2 =3/2 H Q , ±1 T±2 2 H dd Ω M H hf {a, b, A, B} A H hf A = a H hf a = B H hf B = b H hf b 10 3

258 A B t ii E(R, t) =E(R)e iωt + ˆX H lattice = E(R) α(ω)e(r) E(R) = E(R) p β p(r)e p e p α(ω) H lattice = E 2 (R) [ 2α α 3 +(α α ) p C 2 pγ p ] α α γ 0 = β 0 2 1/3 γ ±1 =1/ 3(β0β ± β β 0 ) γ ±2 = 2/3β β ± [ ] H lattice = E 2 2α α (R) +(α α ) 0, 0 C0 2 0, 0 γ 0 3 H lattice = E 2 (R)[ 2α α 3 +(α α ){γ 0 ( s 2 1, 1 C 2 0 1, 1 + v 2 1, 1 C 2 0 1, 1 + w 2 1, 0 C 2 0 1, 0 )+γ 2 sv 1, 1 C 2 2 1, 1 + γ 2 s v 1, 1 C 2 2 1, 1 }]. δe = H lattice H lattice A B {s, v, w} t ii

259 σ + π σ γ ±2 γ ±1 =E 1,0 E 1,1 H lattice ˆx ŷ ẑ λ 0 {a, b, A, B} λ L = R 0 λ 0 λ L λ L k ˆX Ŷ λ L k ( ˆX ± Ŷ ) 2λ L ˆk ẑ Ω 2 Ω 3 Ω 1 =Ω 4 =0 Ω 1 =Ω 4 Ω 2 =Ω 3 =0 M Θ 0 =0.68, Φ 0 =5.83

260 k x k y Θ 0 =0.05 E 32 Θ 0 =1.05 E 28 Θ 0 =0.68 E 36 Θ 0 =0.25 E 40 {θ a,θ b,φ a,φ b,α a,α b,γ a,γ b } = {0.53, 0.97, 1.36, 3.49, 2.84, 2.03, 4.26, 3.84} s i =sin(α i )sin(θ i ) v i =sin(α i )cos(θ i )e iφ i w i =cos(α i )e iγ i ν =1/2 N s =24 {Θ 0, Φ 0,θ a,θ b,φ a,φ b,α a,α b,γ a,γ b } = {0.65, 3.68, 2.4, 2.97, 6.06, 4.1, 0.97, 2.74, 3.44, 1.74} f 7 A B a b γ A = π + γ a γ B = π + γ b w a/b = w A/B E 8 ν =1/2 Θ 0

261 40 87 Θ 0 A B d 00 = d 01 1, ±1 s i s i d 00 /d 01 v i v i d 00 /d 01 (0, 0) ( π, 0) E < 4 A B

262

263 I(r) = β2 4π 0 [G 0 (r; z)g 0 (r; z)]dx G 0 (r; z) = d 3 k z+ɛτ z+ τ x z 2 ɛ 2 2 G 0 (r; z = ix) = 2πρ 0 k f r e 2 +x 2 /v f r (cos(k f r) ) 2 + x 2 τ z +sin(k f r)[ix + τ x ]. 2 + x 2

264 [G 0 (r; z)g 0 (r; z)] I(r) = = k f r 0 x dx e 0 β 2 π(k f r 0 ) 2 E f β 2 [ 2 + x 2 cos(2k f r 0 )] π(k f r 0 ) 2 (x ) [ cos(2k f r 0 ) 0 dx e k f r 0 x E f ] +2 2 sin 2 (k f r) k f r 0 x dx e E f 0 (x ) x x I(r) = β 2 cos(2k π(k f r 0 ) 2 f r 0 ) 0 dx [ e k f r 0 x 2 +1 E f ] +2sin 2 (k f r) 0 k f r 0 x 2 +1 E f dx e (x 2 +1) dx [e k f r 0 ] x E f = E k f r 0 f 1 k f r 0 e E f 0 dx [ e k f r 0 x 2 +1 ] E f = E f k f r 0 e k f r 0 E f [ ] kf r 0 F 1 E f F 1 F 1 [α] =α 0 dxe α( x ). k f r 0 x 2 +1 dx e E f 0 (x 2 +1) = π k f r 0 [ ] 2 e E kf f r 0 F 2 E f

265 F 2 F 2 [α] = 2 π 0 x dx e α( ). (x 2 +1) I(r) = E fβ 2 π(k f r 0 ) cos(2k fr 3 0 )e k f r 0 [ ] E kf f r 0 F 1 + β2 = E [ fβ 2 π(k f r 0 ) cos(2k fr 3 0 )e 2r 0 2r0 ξ F1 ξ k f r 0 E f (k f r 0 ) 2 sin2 (k f r)e E f ] [ + β2 (k f r 0 ) 2 sin2 (k f r)e 2r 0 2r0 ξ F2 ξ [ ] kf r 0 F 2 E f ] F 1 F 2 F 1 [α] =α 0 dxe α( x ) 1.25(α +0.65) 1/2 F 2 [α] = 2 π 0 x dx e α( ) (x 2 +1) 0.8. (α +0.65) 1/2 [G, (z) G (0) (z)] = [SΠ+SΠSG + SΠSGSG +...]= [SΠ(z)(1 SG(0) (z)) 1 ]. S S

266 SC Interaction (khz) Interaction Strength Hz 6! ! Full SC Correction Perturbative SC Correction β =0.01 0! 3! 5! 7! Distance between spins nm R (nm) β =0.01 =0 S [SΠ(z)(1 SG (0) (z)) 1 ] [SΠ(z)SG (0) (z)]. Π(z) G(z) [SΠ(z)SG (0) (z)] = n,m Ψ n S Ψ m Ψ m S Ψ n. (z ε m ) 2 (z ε n ) E tot +iλ 0 iλ = = iλ iλ iλ dɛ 4π 2Re[ iɛ (iɛ ε m ) 2 (iɛ ε n ) ] ɛdɛ ɛ 4πi (iɛ ε m ) 2 (iɛ ε n ) dɛ 4π ɛ (ɛ + iε m ) 2 (ɛ + iε n ). ±i iε m,n iλ

267 iλ iε n iλ dɛ ɛ iλ 4π (ɛ + iε m ) 2 (ɛ + iε n ) = 1 ε n (ε n Λ). 2 (ε n ε m ) 2 m iλ dɛ ɛ iλ 4π (ɛ + iε m ) 2 (ɛ + iε n ) = 1 ε n (ε n Λ) + ε m (ε m Λ) 4 (ε n ε m ) 2 = 1 (ε n Λ) 4 (ε n ε m ) = 1 4 iλ iλ dɛ 1 2π (ɛ + iε m )(ɛ + iε n ). 4δE tot =2 +iλ 0 0 dɛ 4π 2 [i [SG(0) (iɛ)sg (0) (iɛ)]] dɛ 4π 2 [i [SG(0) (iɛ)sg (0) (iɛ)]] iλ 0 dɛ 4π 2 [i [SG(0) (iɛ)sg (0) (iɛ)]], G σσ LL (z) Gσσ LR (z) Πσσ LL (z) Πσσ LR (z) G σσ LL (z) Gσσ LR (z) Πσσ LL (z) Πσσ LR (z) L R G σσ LL (z) =Gσσ RR (z)

268 G σσ LR (z) =Gσσ RL (z) G σσ LL (z) = d 3 1 k z 2 2 ɛ 2 z ɛ k k z + ɛ k = ρ 0 2 z 2 z z G σσ LR (z) G σσ LR(z) = d 3 1 k z 2 2 ɛ 2 z ɛ k k z + ɛ k e ik r 0, r 0 = r L r R G σσ LR (z) = 2 z 2 r v 0 f ρ 0 e k f r 0 2 z 2 z sin(k fr 0 )+ 2 z 2 cos(k f r 0 ) sin(k f r 0 ) sin(k f r 0 ) z sin(k f r 0 ) 2 z 2 cos(k f r 0 ). Π σσ LL (z) Π σσ LL (z) = ρ 0 ( 2 z 2 ) 3/2 z z. k f r 1

REPORT DOCUMENTATION PAGE

REPORT DOCUMENTATION PAGE REPORT DOCUMENTATION PAGE Form Approved OMB NO. 0704-0188 The public reporting burden for this collection of information is estimated to average 1 hour per response, including the time for reviewing instructions,

Διαβάστε περισσότερα

A Classical Perspective on Non-Diffractive Disorder

A Classical Perspective on Non-Diffractive Disorder A Classical Perspective on Non-Diffractive Disorder The Harvard community has made this article openly available. Please share how this access benefits you. Your story matters. Citation Accessed Citable

Διαβάστε περισσότερα

Defects in Hard-Sphere Colloidal Crystals

Defects in Hard-Sphere Colloidal Crystals Defects in Hard-Sphere Colloidal Crystals The Harvard community has made this article openly available. Please share how this access benefits you. Your story matters. Citation Accessed Citable Link Terms

Διαβάστε περισσότερα

Gradient Descent for Optimization Problems With Sparse Solutions

Gradient Descent for Optimization Problems With Sparse Solutions Gradient Descent for Optimization Problems With Sparse Solutions The Harvard community has made this article openly available. Please share how this access benefits you. Your story matters Citation Chen,

Διαβάστε περισσότερα

Diamond platforms for nanoscale photonics and metrology

Diamond platforms for nanoscale photonics and metrology Diamond platforms for nanoscale photonics and metrology The Harvard community has made this article openly available. Please share how this access benefits you. Your story matters. Citation Accessed Citable

Διαβάστε περισσότερα

C 1 D 1. AB = a, AD = b, AA1 = c. a, b, c : (1) AC 1 ; : (1) AB + BC + CC1, AC 1 = BC = AD, CC1 = AA 1, AC 1 = a + b + c. (2) BD 1 = BD + DD 1,

C 1 D 1. AB = a, AD = b, AA1 = c. a, b, c : (1) AC 1 ; : (1) AB + BC + CC1, AC 1 = BC = AD, CC1 = AA 1, AC 1 = a + b + c. (2) BD 1 = BD + DD 1, 1 1., BD 1 B 1 1 D 1, E F B 1 D 1. B = a, D = b, 1 = c. a, b, c : (1) 1 ; () BD 1 ; () F; D 1 F 1 (4) EF. : (1) B = D, D c b 1 E a B 1 1 = 1, B1 1 = B + B + 1, 1 = a + b + c. () BD 1 = BD + DD 1, BD =

Διαβάστε περισσότερα

m i N 1 F i = j i F ij + F x

m i N 1 F i = j i F ij + F x N m i i = 1,..., N m i Fi x N 1 F ij, j = 1, 2,... i 1, i + 1,..., N m i F i = j i F ij + F x i mi Fi j Fj i mj O P i = F i = j i F ij + F x i, i = 1,..., N P = i F i = N F ij + i j i N i F x i, i = 1,...,

Διαβάστε περισσότερα

Solutions - Chapter 4

Solutions - Chapter 4 Solutions - Chapter Kevin S. Huang Problem.1 Unitary: Ût = 1 ī hĥt Û tût = 1 Neglect t term: 1 + hĥ ī t 1 īhĥt = 1 + hĥ ī t ī hĥt = 1 Ĥ = Ĥ Problem. Ût = lim 1 ī ] n hĥ1t 1 ī ] hĥt... 1 ī ] hĥnt 1 ī ]

Διαβάστε περισσότερα

Ax = b. 7x = 21. x = 21 7 = 3.

Ax = b. 7x = 21. x = 21 7 = 3. 3 s st 3 r 3 t r 3 3 t s st t 3t s 3 3 r 3 3 st t t r 3 s t t r r r t st t rr 3t r t 3 3 rt3 3 t 3 3 r st 3 t 3 tr 3 r t3 t 3 s st t Ax = b. s t 3 t 3 3 r r t n r A tr 3 rr t 3 t n ts b 3 t t r r t x 3

Διαβάστε περισσότερα

k k ΚΕΦΑΛΑΙΟ 1 G = (V, E) V E V V V G E G e = {v, u} E v u e v u G G V (G) E(G) n(g) = V (G) m(g) = E(G) G S V (G) S G N G (S) = {u V (G)\S v S : {v, u} E(G)} G v S v V (G) N G (v) = N G ({v}) x V (G)

Διαβάστε περισσότερα

k k ΚΕΦΑΛΑΙΟ 1 G = (V, E) V E V V V G E G e = {v, u} E v u e v u G G V (G) E(G) n(g) = V (G) m(g) = E(G) G S V (G) S G N G (S) = {u V (G)\S v S : {v, u} E(G)} G v S v V (G) N G (v) = N G ({v}) x V (G)

Διαβάστε περισσότερα

Parts Manual. Trio Mobile Surgery Platform. Model 1033

Parts Manual. Trio Mobile Surgery Platform. Model 1033 Trio Mobile Surgery Platform Model 1033 Parts Manual For parts or technical assistance: Pour pièces de service ou assistance technique : Für Teile oder technische Unterstützung Anruf: Voor delen of technische

Διαβάστε περισσότερα

Estimation of grain boundary segregation enthalpy and its role in stable nanocrystalline alloy design

Estimation of grain boundary segregation enthalpy and its role in stable nanocrystalline alloy design Supplemental Material for Estimation of grain boundary segregation enthalpy and its role in stable nanocrystalline alloy design By H. A. Murdoch and C.A. Schuh Miedema model RKM model ΔH mix ΔH seg ΔH

Διαβάστε περισσότερα

Το άτομο του Υδρογόνου

Το άτομο του Υδρογόνου Το άτομο του Υδρογόνου Δυναμικό Coulomb Εξίσωση Schrödinger h e (, r, ) (, r, ) E (, r, ) m ψ θφ r ψ θφ = ψ θφ Συνθήκες ψ(, r θφ, ) = πεπερασμένη ψ( r ) = 0 ψ(, r θφ, ) =ψ(, r θφ+, ) π Επιτρεπτές ενέργειες

Διαβάστε περισσότερα

(... )..!, ".. (! ) # - $ % % $ & % 2007

(... )..!, .. (! ) # - $ % % $ & % 2007 (! ), "! ( ) # $ % & % $ % 007 500 ' 67905:5394!33 : (! ) $, -, * +,'; ), -, *! ' - " #!, $ & % $ ( % %): /!, " ; - : - +', 007 5 ISBN 978-5-7596-0766-3 % % - $, $ &- % $ % %, * $ % - % % # $ $,, % % #-

Διαβάστε περισσότερα

Dissertation for the degree philosophiae doctor (PhD) at the University of Bergen

Dissertation for the degree philosophiae doctor (PhD) at the University of Bergen Dissertation for the degree philosophiae doctor (PhD) at the University of Bergen Dissertation date: GF F GF F SLE GF F D Ĉ = C { } Ĉ \ D D D = {z : z < 1} f : D D D D = D D, D = D D f f : D D

Διαβάστε περισσότερα

u(x, y) =f(x, y) Ω=(0, 1) (0, 1)

u(x, y) =f(x, y) Ω=(0, 1) (0, 1) u(x, y) =f(x, y) Ω=(0, 1) (0, 1) u(x, y) =g(x, y) Γ=δΩ ={0, 1} {0, 1} Ω Ω Ω h Ω h h ˆ Ω ˆ u v = fv Ω u = f in Ω v V H 1 (Ω) V V h V h ψ 1,ψ 2,...,ψ N, ˆ ˆ u v = Ω Ω fv v V ˆ ˆ u v = Ω ˆ ˆ u ψ i = Ω Ω Ω

Διαβάστε περισσότερα

F (x) = kx. F (x )dx. F = kx. U(x) = U(0) kx2

F (x) = kx. F (x )dx. F = kx. U(x) = U(0) kx2 F (x) = kx x k F = F (x) U(0) U(x) = x F = kx 0 F (x )dx U(x) = U(0) + 1 2 kx2 x U(0) = 0 U(x) = 1 2 kx2 U(x) x 0 = 0 x 1 U(x) U(0) + U (0) x + 1 2 U (0) x 2 U (0) = 0 U(x) U(0) + 1 2 U (0) x 2 U(0) =

Διαβάστε περισσότερα

W τ R W j N H = 2 F obj b q N F aug F obj b q Ψ F aug Ψ ( ) ϱ t + + p = 0 = 0 Ω f = Γ Γ b ϱ = (, t) = (, t) Ω f Γ b ( ) ϱ t + + p = V max 4 3 2 1 0-1 -2-3 -4-4 -3-2 -1 0 1 2 3 4 x 4 x 1 V mn V max

Διαβάστε περισσότερα

ϕ n n n n = 1,..., N n n {X I, Y I } {X r, Y r } (x c, y c ) q r = x a y a θ X r = [x r, y r, θ r ] X I = [x I, y I, θ I ] X I = R(θ)X r R(θ) R(θ) = cosθ sinθ 0 sinθ cosθ 0 0 0 1 Ẋ I = R(θ)Ẋr y r ẏa r

Διαβάστε περισσότερα

ΠΕΡΙΟΔΙΚΟΣ ΠΙΝΑΚΑΣ ΣΤΟΙΧΕΙΩΝ

ΠΕΡΙΟΔΙΚΟΣ ΠΙΝΑΚΑΣ ΣΤΟΙΧΕΙΩΝ ΠΕΡΙΟΔΙΚΟΣ ΠΙΝΑΚΑΣ ΣΤΟΙΧΕΙΩΝ Περίοδοι περιοδικού πίνακα Ο περιοδικός πίνακας αποτελείται από 7 περιόδους. Ο αριθμός των στοιχείων που περιλαμβάνει κάθε περίοδος δεν είναι σταθερός, δηλ. η περιοδικότητα

Διαβάστε περισσότερα

φ(t) TE 0 φ(z) φ(z) φ(z) φ(z) η(λ) G(z,λ) λ φ(z) η(λ) η(λ) = t CIGS 0 G(z,λ)φ(z)dz t CIGS η(λ) φ(z) 0 z

Διαβάστε περισσότερα

Νόµοςπεριοδικότητας του Moseley:Η χηµική συµπεριφορά (οι ιδιότητες) των στοιχείων είναι περιοδική συνάρτηση του ατοµικού τους αριθµού.

Νόµοςπεριοδικότητας του Moseley:Η χηµική συµπεριφορά (οι ιδιότητες) των στοιχείων είναι περιοδική συνάρτηση του ατοµικού τους αριθµού. Νόµοςπεριοδικότητας του Moseley:Η χηµική συµπεριφορά (οι ιδιότητες) των στοιχείων είναι περιοδική συνάρτηση του ατοµικού τους αριθµού. Περιοδικός πίνακας: α. Είναι µια ταξινόµηση των στοιχείων κατά αύξοντα

Διαβάστε περισσότερα

!"#$ % &# &%#'()(! $ * +

!#$ % &# &%#'()(! $ * + ,!"#$ % &# &%#'()(! $ * + ,!"#$ % &# &%#'()(! $ * + 6 7 57 : - - / :!", # $ % & :'!(), 5 ( -, * + :! ",, # $ %, ) #, '(#,!# $$,',#-, 4 "- /,#-," -$ '# &",,#- "-&)'#45)')6 5! 6 5 4 "- /,#-7 ",',8##! -#9,!"))

Διαβάστε περισσότερα

!!" #7 $39 %" (07) ..,..,.. $ 39. ) :. :, «(», «%», «%», «%» «%». & ,. ). & :..,. '.. ( () #*. );..,..'. + (# ).

!! #7 $39 % (07) ..,..,.. $ 39. ) :. :, «(», «%», «%», «%» «%». & ,. ). & :..,. '.. ( () #*. );..,..'. + (# ). 1 00 3 !!" 344#7 $39 %" 6181001 63(07) & : ' ( () #* ); ' + (# ) $ 39 ) : : 00 %" 6181001 63(07)!!" 344#7 «(» «%» «%» «%» «%» & ) 4 )&-%/0 +- «)» * «1» «1» «)» ) «(» «%» «%» + ) 30 «%» «%» )1+ / + : +3

Διαβάστε περισσότερα

ΓΗ ΚΑΙ ΣΥΜΠΑΝ. Εικόνα 1. Φωτογραφία του γαλαξία μας (από αρχείο της NASA)

ΓΗ ΚΑΙ ΣΥΜΠΑΝ. Εικόνα 1. Φωτογραφία του γαλαξία μας (από αρχείο της NASA) ΓΗ ΚΑΙ ΣΥΜΠΑΝ Φύση του σύμπαντος Η γη είναι μία μονάδα μέσα στο ηλιακό μας σύστημα, το οποίο αποτελείται από τον ήλιο, τους πλανήτες μαζί με τους δορυφόρους τους, τους κομήτες, τα αστεροειδή και τους μετεωρίτες.

Διαβάστε περισσότερα

Ι ΙΟΤΗΤΕΣ ΤΩΝ ΑΤΟΜΩΝ. Παππάς Χρήστος Επίκουρος Καθηγητής

Ι ΙΟΤΗΤΕΣ ΤΩΝ ΑΤΟΜΩΝ. Παππάς Χρήστος Επίκουρος Καθηγητής ΗΛΕΚΤΡΟΝΙΚΗ ΟΜΗ ΚΑΙ Ι ΙΟΤΗΤΕΣ ΤΩΝ ΑΤΟΜΩΝ Παππάς Χρήστος Επίκουρος Καθηγητής ΤΟ ΜΕΓΕΘΟΣ ΤΩΝ ΑΤΟΜΩΝ Ατομική ακτίνα (r) : ½ της απόστασης μεταξύ δύο ομοιοπυρηνικών ατόμων, ενωμένων με απλό ομοιοπολικό δεσμό.

Διαβάστε περισσότερα

ITU-R P (2012/02) &' (

ITU-R P (2012/02) &' ( ITU-R P.530-4 (0/0) $ % " "#! &' ( P ITU-R P. 530-4 ii.. (IPR) (ITU-T/ITU-R/ISO/IEC).ITU-R http://www.itu.int/itu-r/go/patents/en. ITU-T/ITU-R/ISO/IEC (http://www.itu.int/publ/r-rec/en ) () ( ) BO BR BS

Διαβάστε περισσότερα

ΝΟΜΟΣ ΤΗΣ ΠΕΡΙΟ ΙΚΟΤΗΤΑΣ : Οι ιδιότητες των χηµικών στοιχείων είναι περιοδική συνάρτηση του ατοµικού τους αριθµού.

ΝΟΜΟΣ ΤΗΣ ΠΕΡΙΟ ΙΚΟΤΗΤΑΣ : Οι ιδιότητες των χηµικών στοιχείων είναι περιοδική συνάρτηση του ατοµικού τους αριθµού. 1. Ο ΠΕΡΙΟ ΙΚΟΣ ΠΙΝΑΚΑΣ Οι άνθρωποι από την φύση τους θέλουν να πετυχαίνουν σπουδαία αποτελέσµατα καταναλώνοντας το λιγότερο δυνατό κόπο και χρόνο. Για το σκοπό αυτό προσπαθούν να οµαδοποιούν τα πράγµατα

Διαβάστε περισσότερα

τροχιακά Η στιβάδα καθορίζεται από τον κύριο κβαντικό αριθµό (n) Η υποστιβάδα καθορίζεται από τους δύο πρώτους κβαντικούς αριθµούς (n, l)

τροχιακά Η στιβάδα καθορίζεται από τον κύριο κβαντικό αριθµό (n) Η υποστιβάδα καθορίζεται από τους δύο πρώτους κβαντικούς αριθµούς (n, l) ΑΤΟΜΙΚΑ ΤΡΟΧΙΑΚΑ Σχέση κβαντικών αριθµών µε στιβάδες υποστιβάδες - τροχιακά Η στιβάδα καθορίζεται από τον κύριο κβαντικό αριθµό (n) Η υποστιβάδα καθορίζεται από τους δύο πρώτους κβαντικούς αριθµούς (n,

Διαβάστε περισσότερα

... 5 A.. RS-232C ( ) RS-232C ( ) RS-232C-LK & RS-232C-MK RS-232C-JK & RS-232C-KK

... 5 A.. RS-232C ( ) RS-232C ( ) RS-232C-LK & RS-232C-MK RS-232C-JK & RS-232C-KK RS-3C WIWM050 014.1.9 P1 :8... 1... 014.0.1 1 A... 014.0. 1... RS-3C()...01.08.03 A.. RS-3C()...01.08.03 3... RS-3C()... 003.11.5 4... RS-3C ()... 00.10.01 5... RS-3C().008.07.16 5 A.. RS-3C().0 1.08.

Διαβάστε περισσότερα

ITU-R P (2009/10)

ITU-R P (2009/10) ITU-R.45-4 (9/) % # GHz,!"# $$ # ITU-R.45-4.. (IR) (ITU-T/ITU-R/ISO/IEC).ITU-R http://www.tu.t/itu-r/go/patets/e. (http://www.tu.t/publ/r-rec/e ) () ( ) BO BR BS BT F M RA S RS SA SF SM SNG TF V.ITU-R

Διαβάστε περισσότερα

a; b 2 R; a < b; f : [a; b] R! R y 2 R: y : [a; b]! R; ( y (t) = f t; y(t) ; a t b; y(a) = y : f (t; y) 2 [a; b]r: f 2 C ([a; b]r): y 2 C [a; b]; y(a) = y ; f y ỹ ỹ y ; jy ỹ j ky ỹk [a; b]; f y; ( y (t)

Διαβάστε περισσότερα

J J l 2 J T l 1 J T J T l 2 l 1 J J l 1 c 0 J J J J J l 2 l 2 J J J T J T l 1 J J T J T J T J {e n } n N {e n } n N x X {λ n } n N R x = λ n e n {e n } n N {e n : n N} e n 0 n N k 1, k 2,..., k n N λ

Διαβάστε περισσότερα

SUPPLEMENTAL INFORMATION. Fully Automated Total Metals and Chromium Speciation Single Platform Introduction System for ICP-MS

SUPPLEMENTAL INFORMATION. Fully Automated Total Metals and Chromium Speciation Single Platform Introduction System for ICP-MS Electronic Supplementary Material (ESI) for Journal of Analytical Atomic Spectrometry. This journal is The Royal Society of Chemistry 2018 SUPPLEMENTAL INFORMATION Fully Automated Total Metals and Chromium

Διαβάστε περισσότερα

Note: Please use the actual date you accessed this material in your citation.

Note: Please use the actual date you accessed this material in your citation. MIT OpenCourseWare http://ocw.mit.edu 6.03/ESD.03J Electromagnetics and Applications, Fall 005 Please use the following citation format: Markus Zahn, 6.03/ESD.03J Electromagnetics and Applications, Fall

Διαβάστε περισσότερα

Q π (/) ^ ^ ^ Η φ. <f) c>o. ^ ο. ö ê ω Q. Ο. o 'c. _o _) o U 03. ,,, ω ^ ^ -g'^ ο 0) f ο. Ε. ιη ο Φ. ο 0) κ. ο 03.,Ο. g 2< οο"" ο φ.

Q π (/) ^ ^ ^ Η φ. <f) c>o. ^ ο. ö ê ω Q. Ο. o 'c. _o _) o U 03. ,,, ω ^ ^ -g'^ ο 0) f ο. Ε. ιη ο Φ. ο 0) κ. ο 03.,Ο. g 2< οο ο φ. II 4»» «i p û»7'' s V -Ζ G -7 y 1 X s? ' (/) Ζ L. - =! i- Ζ ) Η f) " i L. Û - 1 1 Ι û ( - " - ' t - ' t/î " ι-8. Ι -. : wî ' j 1 Τ J en " il-' - - ö ê., t= ' -; '9 ',,, ) Τ '.,/,. - ϊζ L - (- - s.1 ai

Διαβάστε περισσότερα

r r t r r t t r t P s r t r P s r s r r rs tr t r r t s ss r P s s t r t t tr r r t t r t r r t t s r t rr t Ü rs t 3 r r r 3 rträ 3 röÿ r t

r r t r r t t r t P s r t r P s r s r r rs tr t r r t s ss r P s s t r t t tr r r t t r t r r t t s r t rr t Ü rs t 3 r r r 3 rträ 3 röÿ r t r t t r t ts r3 s r r t r r t t r t P s r t r P s r s r P s r 1 s r rs tr t r r t s ss r P s s t r t t tr r 2s s r t t r t r r t t s r t rr t Ü rs t 3 r t r 3 s3 Ü rs t 3 r r r 3 rträ 3 röÿ r t r r r rs

Διαβάστε περισσότερα

Microscopie photothermique et endommagement laser

Microscopie photothermique et endommagement laser Microscopie photothermique et endommagement laser Annelise During To cite this version: Annelise During. Microscopie photothermique et endommagement laser. Physique Atomique [physics.atom-ph]. Université

Διαβάστε περισσότερα

a; b 2 R; a < b; f : [a; b] R! R y 2 R: y : [a; b]! R; ( y (t) = f t; y(t) ; a t b; y(a) = y : f (t; y) 2 [a; b]r: f 2 C ([a; b]r): y 2 C [a; b]; y(a) = y ; f y ỹ ỹ y ; jy ỹ j ky ỹk [a; b]; f y; ( y (t)

Διαβάστε περισσότερα

γ 1 6 M = 0.05 F M = 0.05 F M = 0.2 F M = 0.2 F M = 0.05 F M = 0.05 F M = 0.05 F M = 0.2 F M = 0.05 F 2 2 λ τ M = 6000 M = 10000 M = 15000 M = 6000 M = 10000 M = 15000 1 6 τ = 36 1 6 τ = 102 1 6 M = 5000

Διαβάστε περισσότερα

! "# $ % $&'& () *+ (,-. / 0 1(,21(,*) (3 4 5 "$ 6, ::: ;"<$& = = 7 + > + 5 $?"# 46(A *( / A 6 ( 1,*1 B"',CD77E *+ *),*,*) F? $G'& 0/ (,.

! # $ % $&'& () *+ (,-. / 0 1(,21(,*) (3 4 5 $ 6, ::: ;<$& = = 7 + > + 5 $?# 46(A *( / A 6 ( 1,*1 B',CD77E *+ *),*,*) F? $G'& 0/ (,. ! " #$%&'()' *('+$,&'-. /0 1$23(/%/4. 1$)('%%'($( )/,)$5)/6%6 7$85,-9$(- /0 :/986-$, ;2'$(2$ 1'$-/-$)('')5( /&5&-/ 5(< =(4'($$,'(4 1$%$2/996('25-'/(& ;/0->5,$ 1'$-/%'')$(($/3?$%9'&-/?$( 5(< @6%-'9$

Διαβάστε περισσότερα

Errata (Includes critical corrections only for the 1 st & 2 nd reprint)

Errata (Includes critical corrections only for the 1 st & 2 nd reprint) Wedesday, May 5, 3 Erraa (Icludes criical correcios oly for he s & d repri) Advaced Egieerig Mahemaics, 7e Peer V O eil ISB: 978474 Page # Descripio 38 ie 4: chage "w v a v " "w v a v " 46 ie : chage "y

Διαβάστε περισσότερα

Jeux d inondation dans les graphes

Jeux d inondation dans les graphes Jeux d inondation dans les graphes Aurélie Lagoutte To cite this version: Aurélie Lagoutte. Jeux d inondation dans les graphes. 2010. HAL Id: hal-00509488 https://hal.archives-ouvertes.fr/hal-00509488

Διαβάστε περισσότερα

!"! #!"!!$ #$! %!"&' & (%!' #!% #" *! *$' *.!! )#/'.0! )#/.*!$,)# * % $ %!!#!!%#'!)$! #,# #!%# ##& )$&# 11!!#2!

!! #!!!$ #$! %!&' & (%!' #!% # *! *$' *.!! )#/'.0! )#/.*!$,)# * % $ %!!#!!%#'!)$! #,# #!%# ##& )$&# 11!!#2! # $ #$ % (% # )*%%# )# )$ % # * *$ * #,##%#)#% *-. )#/###%. )#/.0 )#/.* $,)# )#/ * % $ % # %# )$ #,# # %# ## )$# 11 #2 #**##%% $#%34 5 # %## * 6 7(%#)%%%, #, # ## # *% #$# 8# )####, 7 9%%# 0 * #,, :;

Διαβάστε περισσότερα

ITU-R P (2012/02)

ITU-R P (2012/02) ITU-R P.56- (0/0 P ITU-R P.56- ii.. (IPR (ITU-T/ITU-R/ISO/IEC.ITU-R ttp://www.itu.int/itu-r/go/patents/en. (ttp://www.itu.int/publ/r-rec/en ( ( BO BR BS BT F M P RA RS S SA SF SM SNG TF V 0.ITU-R ITU 0..(ITU

Διαβάστε περισσότερα

m 1, m 2 F 12, F 21 F12 = F 21

m 1, m 2 F 12, F 21 F12 = F 21 m 1, m 2 F 12, F 21 F12 = F 21 r 1, r 2 r = r 1 r 2 = r 1 r 2 ê r = rê r F 12 = f(r)ê r F 21 = f(r)ê r f(r) f(r) < 0 f(r) > 0 m 1 r1 = f(r)ê r m 2 r2 = f(r)ê r r = r 1 r 2 r 1 = 1 m 1 f(r)ê r r 2 = 1 m

Διαβάστε περισσότερα

Łs t r t rs tø r P r s tø PrØ rø rs tø P r s r t t r s t Ø t q s P r s tr. 2stŁ s q t q s t rt r s t s t ss s Ø r s t r t. Łs t r t t Ø t q s

Łs t r t rs tø r P r s tø PrØ rø rs tø P r s r t t r s t Ø t q s P r s tr. 2stŁ s q t q s t rt r s t s t ss s Ø r s t r t. Łs t r t t Ø t q s Łs t r t rs tø r P r s tø PrØ rø rs tø P r s r t t r s t Ø t q s P r s tr st t t t Ø t q s ss P r s P 2stŁ s q t q s t rt r s t s t ss s Ø r s t r t P r røs r Łs t r t t Ø t q s r Ø r t t r t q t rs tø

Διαβάστε περισσότερα

HONDA. Έτος κατασκευής

HONDA. Έτος κατασκευής Accord + Coupe IV 2.0 16V (CB3) F20A2-A3 81 110 01/90-09/93 0800-0175 11,00 2.0 16V (CB3) F20A6 66 90 01/90-09/93 0800-0175 11,00 2.0i 16V (CB3-CC9) F20A8 98 133 01/90-09/93 0802-9205M 237,40 2.0i 16V

Διαβάστε περισσότερα

ITU-R P (2012/02) khz 150

ITU-R P (2012/02) khz 150 (0/0) khz 0 P ii (IPR) (ITU-T/ITU-R/ISO/IEC) ITU-R http://www.itu.int/itu-r/go/patents/en http://www.itu.int/publ/r-rec/en BO BR BS BT F M P RA RS S SA SF SM SNG TF V ITU-R 0 ITU 0 (ITU) khz 0 (0-009-00-003-00-994-990)

Διαβάστε περισσότερα

Between Square and Circle

Between Square and Circle DOCTORAL T H E SIS Between Square and Circle A Study on the Behaviour of Polygonal Steel Profiles Under Compression Panagiotis Manoleas Steel Structures Printed by Luleå University of Technology, Graphic

Διαβάστε περισσότερα

rs r r â t át r st tíst Ó P ã t r r r â

rs r r â t át r st tíst Ó P ã t r r r â rs r r â t át r st tíst P Ó P ã t r r r â ã t r r P Ó P r sã rs r s t à r çã rs r st tíst r q s t r r t çã r r st tíst r t r ú r s r ú r â rs r r â t át r çã rs r st tíst 1 r r 1 ss rt q çã st tr sã

Διαβάστε περισσότερα

! " #! $ %&! '( #)!' * +#, " -! %&! "!! ! " #$ % # " &' &'... ()* ( +, # ' -. + &', - + &' / # ' -. + &' (, % # , 2**.

!  #! $ %&! '( #)!' * +#,  -! %&! !! !  #$ % #  &' &'... ()* ( +, # ' -. + &', - + &' / # ' -. + &' (, % # , 2**. ! " #! $ %&! '( #)!' * +#, " -! %&! "!!! " #$ % # " &' &'... ()* ( +, # ' -. + &', - + &' / 0123 4 # ' -. + &' (, % #. -5 0126, 2**., 2, + &' %., 0, $!, 3,. 7 8 ', $$, 9, # / 3:*,*2;

Διαβάστε περισσότερα

A 1 A 2 A 3 B 1 B 2 B 3

A 1 A 2 A 3 B 1 B 2 B 3 16 0 17 0 17 0 18 0 18 0 19 0 20 A A = A 1 î + A 2 ĵ + A 3ˆk A (x, y, z) r = xî + yĵ + zˆk A B A B B A = A 1 B 1 + A 2 B 2 + A 3 B 3 = A B θ θ A B = ˆn A B θ A B î ĵ ˆk = A 1 A 2 A 3 B 1 B 2 B 3 W = F

Διαβάστε περισσότερα

d dx x 2 = 2x d dx x 3 = 3x 2 d dx x n = nx n 1

d dx x 2 = 2x d dx x 3 = 3x 2 d dx x n = nx n 1 d dx x 2 = 2x d dx x 3 = 3x 2 d dx x n = nx n1 x dx = 1 2 b2 1 2 a2 a b b x 2 dx = 1 a 3 b3 1 3 a3 b x n dx = 1 a n +1 bn +1 1 n +1 an +1 d dx d dx f (x) = 0 f (ax) = a f (ax) lim d dx f (ax) = lim 0 =

Διαβάστε περισσότερα

μ μ dω I ν S da cos θ da λ λ Γ α/β MJ Capítulo 1 % βpic ɛ Eridani V ega β P ic F ormalhaut 10 9 15% 70 Virgem 47 Ursa Maior Debris Disk Debris Disk μ 90% L ac = GM M ac R L ac R M M ac L J T

Διαβάστε περισσότερα

Homework 8 Model Solution Section

Homework 8 Model Solution Section MATH 004 Homework Solution Homework 8 Model Solution Section 14.5 14.6. 14.5. Use the Chain Rule to find dz where z cosx + 4y), x 5t 4, y 1 t. dz dx + dy y sinx + 4y)0t + 4) sinx + 4y) 1t ) 0t + 4t ) sinx

Διαβάστε περισσότερα

Molekulare Ebene (biochemische Messungen) Zelluläre Ebene (Elektrophysiologie, Imaging-Verfahren) Netzwerk Ebene (Multielektrodensysteme) Areale (MRT, EEG...) Gene Neuronen Synaptische Kopplung kleine

Διαβάστε περισσότερα

())*+,-./0-1+*)*2, *67()(,01-+4(-8 9 0:,*2./0 30 ;+-7 3* *),+*< 7+)0 3* (=24(-) 04(-() 18(4-3-) 3-2(>*+)(3-3*

())*+,-./0-1+*)*2, *67()(,01-+4(-8 9 0:,*2./0 30 ;+-7 3* *),+*< 7+)0 3* (=24(-) 04(-() 18(4-3-) 3-2(>*+)(3-3* ! " # $ $ %&&' % $ $! " # ())*+,-./0-1+*)*2,-3-4050+*67()(,01-+4(-8 9 0:,*2./0 30 ;+-7 3* *),+*< 7+)0 3* *),+-30 *5 35(2(),+-./0 30 *,0+ 3* (=24(-) 04(-() 18(4-3-) 3-2(>*+)(3-3* *3*+-830-+-2?< +(*2,-30+

Διαβάστε περισσότερα

!"!# ""$ %%"" %$" &" %" "!'! " #$!

!!# $ %% %$ & % !'!  #$! " "" %%"" %" &" %" " " " % ((((( ((( ((((( " %%%% & ) * ((( "* ( + ) (((( (, (() (((((* ( - )((((( )((((((& + )(((((((((( +. ) ) /(((( +( ),(, ((((((( +, 0 )/ (((((+ ++, ((((() & "( %%%%%%%%%%%%%%%%%%%(

Διαβάστε περισσότερα

V r,k j F k m N k+1 N k N k+1 H j n = 7 n = 16 Ṽ r ñ,ñ j Ṽ Ṽ j x / Ṽ W 2r V r D N T T 2r 2r N k F k N 2r Ω R 2 n Ω I n = { N: n} n N R 2 x R 2, I n Ω R 2 u R 2, I n x k+1 = x k + u k, u, x R 2,

Διαβάστε περισσότερα

Teor imov r. ta matem. statist. Vip. 94, 2016, stor

Teor imov r. ta matem. statist. Vip. 94, 2016, stor eor imov r. ta matem. statist. Vip. 94, 6, stor. 93 5 Abstract. e article is devoted to models of financial markets wit stocastic volatility, wic is defined by a functional of Ornstein-Ulenbeck process

Διαβάστε περισσότερα

01 π π 4 1 2 I(t) C V C L V L C L Q(t) Φ(t) (L, C) Q(t) V C + V L =0 Q(t) C + Q(t) Ld2 =0 dt 2 d 2 Q(t) + Q(t) dt 2 LC =0 d 2 Q(t) + ω dt 0Q(t) 2 =0 2 Q(t) ω0 2 = 1 LC V L + V C =0 d 2 Φ(t)

Διαβάστε περισσότερα

l 1 p r i = ρ ij α j + w i j=1 ρ ij λ α j j p w i p α j = 1, α j 0, j = 1,..., p j=1 R B B B m j [ρ 1j, ρ 2j,..., ρ Bj ] T = }{{} α + [,,..., ] R B p p α [α 1,..., α p ] [w 1,..., w p ] M m 1 m 2,

Διαβάστε περισσότερα

Περιεχόμενα. A(x 1, x 2 )

Περιεχόμενα. A(x 1, x 2 ) Περιεχόμενα A(x 1, x 2 7 Ολοκληρώματα της Μαγνητοϋδροδυναμικής και Μαγνητοϋδροδυναμικά Κύματα Σχήμα 7.1: Οι τριδιάστατες ελικοειδείς μαγνητικές γραμμές στις οποίες εφάπτεται το διάνυσμα του μαγνητικού

Διαβάστε περισσότερα

ECE Spring Prof. David R. Jackson ECE Dept. Notes 2

ECE Spring Prof. David R. Jackson ECE Dept. Notes 2 ECE 634 Spring 6 Prof. David R. Jackson ECE Dept. Notes Fields in a Source-Free Region Example: Radiation from an aperture y PEC E t x Aperture Assume the following choice of vector potentials: A F = =

Διαβάστε περισσότερα

ΠΕΡΙΟΔΙΚΟ ΣΥΣΤΗΜΑ ΤΩΝ ΣΤΟΙΧΕΙΩΝ (1) Ηλία Σκαλτσά ΠΕ ο Γυμνάσιο Αγ. Παρασκευής

ΠΕΡΙΟΔΙΚΟ ΣΥΣΤΗΜΑ ΤΩΝ ΣΤΟΙΧΕΙΩΝ (1) Ηλία Σκαλτσά ΠΕ ο Γυμνάσιο Αγ. Παρασκευής ΠΕΡΙΟΔΙΚΟ ΣΥΣΤΗΜΑ ΤΩΝ ΣΤΟΙΧΕΙΩΝ (1) Ηλία Σκαλτσά ΠΕ04.01 5 ο Γυμνάσιο Αγ. Παρασκευής Όπως συμβαίνει στη φύση έτσι και ο άνθρωπος θέλει να πετυχαίνει σπουδαία αποτελέσματα καταναλώνοντας το λιγότερο δυνατό

Διαβάστε περισσότερα

! "#" "" $ "%& ' %$(%& % &'(!!")!*!&+ ,! %$( - .$'!"

! #  $ %& ' %$(%& % &'(!!)!*!&+ ,! %$( - .$'! ! "#" "" $ "%& ' %$(%&!"#$ % &'(!!")!*!&+,! %$( -.$'!" /01&$23& &4+ $$ /$ & & / ( #(&4&4!"#$ %40 &'(!"!!&+ 5,! %$( - &$ $$$".$'!" 4(02&$ 4 067 4 $$*&(089 - (0:;

Διαβάστε περισσότερα

Ταλαντώσεις 6.1 Απλή Αρµονική Ταλάντωση σε µία ιάσταση Ελατήριο σε οριζόντιο επίπεδο Σχήµα 6.1

Ταλαντώσεις 6.1 Απλή Αρµονική Ταλάντωση σε µία ιάσταση Ελατήριο σε οριζόντιο επίπεδο Σχήµα 6.1 6 Ταλαντώσεις 6.1 Απλή Αρµονική Ταλάντωση σε µία ιάσταση 6.1.1 Ελατήριο σε οριζόντιο επίπεδο Υποθέτουµε ότι το ελατήριο έχει αρχικό µήκος µηδέν, ιδανικό ελατήριο. F=-kx x K M x Σχήµα 6.1 ιαστάσεις µεγεθών

Διαβάστε περισσότερα

m r = F m r = F ( r) m r = F ( v) F = F (x) m dv dt = F (x) vdv = F (x)dx d dt = dx dv dt dx = v dv dx

m r = F m r = F ( r) m r = F ( v) F = F (x) m dv dt = F (x) vdv = F (x)dx d dt = dx dv dt dx = v dv dx m r = F m r = F ( r) m r = F ( v) x F = F (x) m dv dt = F (x) d dt = dx dv dt dx = v dv dx vdv = F (x)dx 2 mv2 x 2 mv2 0 = F (x )dx x 0 K = 2 mv2 W x0 x = x x 0 F (x)dx K K 0 = W x0 x x, x 2 x K 2 K =

Διαβάστε περισσότερα

f RF f LO f RF ±f LO Ιδανικός μείκτης RF Είσοδος f RF f RF ± f LO IF Έξοδος f LO LO Είσοδος f RF f LO (ω RF t) (ω LO t) = 1 2 [(ω RF + ω LO )t + (ω RF ω LO )t] RF LO IF f RF ± f LO 0 180 +1 RF IF 1 LO

Διαβάστε περισσότερα

MICROMASTER Vector MIDIMASTER Vector

MICROMASTER Vector MIDIMASTER Vector s MICROMASTER Vector MIDIMASTER Vector... 2 1.... 4 2. -MICROMASTER VECTOR... 5 3. -MIDIMASTER VECTOR... 16 4.... 24 5.... 28 6.... 32 7.... 54 8.... 56 9.... 61 Siemens plc 1998 G85139-H1751-U553B 1.

Διαβάστε περισσότερα

!"#!"!"# $ "# '()!* '+!*, -"*!" $ "#. /01 023 43 56789:3 4 ;8< = 7 >/? 44= 7 @ 90A 98BB8: ;4B0C BD :0 E D:84F3 B8: ;4BG H ;8

Διαβάστε περισσότερα

(x y) = (X = x Y = y) = (Y = y) (x y) = f X,Y (x, y) x f X

(x y) = (X = x Y = y) = (Y = y) (x y) = f X,Y (x, y) x f X X, Y f X,Y x, y X x, Y y f X Y x y X x Y y X x, Y y Y y f X,Y x, y f Y y f X Y x y x y X Y f X,Y x, y f X Y x y f X,Y x, y f Y y x y X : Ω R Y : Ω E X < y Y Y y 0 X Y y x R x f X Y x y gy X Y gy gy : Ω

Διαβάστε περισσότερα

Mesh Parameterization: Theory and Practice

Mesh Parameterization: Theory and Practice Mesh Parameterization: Theory and Practice Kai Hormann, Bruno Lévy, Alla Sheffer To cite this version: Kai Hormann, Bruno Lévy, Alla Sheffer. Mesh Parameterization: Theory and Practice. This document is

Διαβάστε περισσότερα

Déformation et quantification par groupoïde des variétés toriques

Déformation et quantification par groupoïde des variétés toriques Défomation et uantification pa goupoïde de vaiété toiue Fédéic Cadet To cite thi veion: Fédéic Cadet. Défomation et uantification pa goupoïde de vaiété toiue. Mathématiue [math]. Univeité d Oléan, 200.

Διαβάστε περισσότερα

Physique des réacteurs à eau lourde ou légère en cycle thorium : étude par simulation des performances de conversion et de sûreté

Physique des réacteurs à eau lourde ou légère en cycle thorium : étude par simulation des performances de conversion et de sûreté Physique des réacteurs à eau lourde ou légère en cycle thorium : étude par simulation des performances de conversion et de sûreté Alexis Nuttin To cite this version: Alexis Nuttin. Physique des réacteurs

Διαβάστε περισσότερα

AC 1 = AB + BC + CC 1, DD 1 = AA 1. D 1 C 1 = 1 D 1 F = 1. AF = 1 a + b + ( ( (((

AC 1 = AB + BC + CC 1, DD 1 = AA 1. D 1 C 1 = 1 D 1 F = 1. AF = 1 a + b + ( ( ((( ? / / / o/ / / / o/ / / / 1 1 1., D 1 1 1 D 1, E F 1 D 1. = a, D = b, 1 = c. a, b, c : #$ #$ #$ 1) 1 ; : 1)!" ) D 1 ; ) F ; = D, )!" D 1 = D + DD 1, % ) F = D + DD 1 + D 1 F, % 4) EF. 1 = 1, 1 = a + b

Διαβάστε περισσότερα

Studies in Magnetism and Superconductivity under Extreme Pressure

Studies in Magnetism and Superconductivity under Extreme Pressure Washington University in St. Louis Washington University Open Scholarship All Theses and Dissertations (ETDs) 1-1-2011 Studies in Magnetism and Superconductivity under Extreme Pressure Wenli Bi Washington

Διαβάστε περισσότερα

! " #$% & '()()*+.,/0.

!  #$% & '()()*+.,/0. ! " #$% & '()()*+,),--+.,/0. 1!!" "!! 21 # " $%!%!! &'($ ) "! % " % *! 3 %,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,0 %%4,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,5

Διαβάστε περισσότερα

(product-operator) I I cos ω ( t sin ω ( t x x ) + Iy )

(product-operator) I I cos ω ( t sin ω ( t x x ) + Iy ) (product-operator) I I cos( t) + I sin( t) x x y z 2π (rad) y 1 y t x = 2πν x t (rad) sin t Iy# cos t t Ix# Ix# (t ) z Ix# Iy# Ix# (t ) z Ix cos (t ) + Iy sin (t ) -x -y t y I-y# I-y# (t ) z (t ) z x I-y#

Διαβάστε περισσότερα

J! "#$ %"& ( ) ) ) " *+, -./0-, *- /! /!+12, ,. 6 /72-, 0,,3-8 / ',913-51:-*/;+ 5/<3/ +15;+ 5/<3=9 -!.1!-9 +17/> ) ) &

J! #$ %& ( ) ) )  *+, -./0-, *- /! /!+12, ,. 6 /72-, 0,,3-8 / ',913-51:-*/;+ 5/<3/ +15;+ 5/<3=9 -!.1!-9 +17/> ) ) & J! "#$ %"& J ' ( ) ) ) " *+, -./0-, L *- /! /!+12,3-4 % +15,. 6 /72-, 0,,3-8 / ',913-51:-*/;+ 5/01 ',913-51:--

Διαβάστε περισσότερα

γ n ϑ n n ψ T 8 Q 6 j, k, m, n, p, r, r t, x, y f m (x) (f(x)) m / a/b (f g)(x) = f(g(x)) n f f n I J α β I = α + βj N, Z, Q ϕ Εὐκλείδης ὁ Ἀλεξανδρεύς Στοιχεῖα ἄκρος καὶ μέσος λόγος ὕδωρ αἰθήρ ϕ φ Φ τ

Διαβάστε περισσότερα

Κεφάλαιο 1 Πραγματικοί Αριθμοί 1.1 Σύνολα

Κεφάλαιο 1 Πραγματικοί Αριθμοί 1.1 Σύνολα x + = 0 N = {,, 3....}, Z Q, b, b N c, d c, d N + b = c, b = d. N = =. < > P n P (n) P () n = P (n) P (n + ) n n + P (n) n P (n) n P n P (n) P (m) P (n) n m P (n + ) P (n) n m P n P (n) P () P (), P (),...,

Διαβάστε περισσότερα

1951 {0, 1} N = N \ {0} n m M n, m N F x i = (x i 1,..., xi m) x j = (x 1 j,..., xn j ) i j M M i j x i j m n M M M M T f : F m F f(m) f M (f(x 1 1,..., x1 m),..., f(x n 1,..., xn m)) T R F M R M R x

Διαβάστε περισσότερα

ΜΕΤΑΠΤΥΧΙΑΚΗ ΙΠΛΩΜΑΤΙΚΗ ΕΡΓΑΣΙΑ. Ελευθερίου Β. Χρυσούλα. Επιβλέπων: Νικόλαος Καραμπετάκης Καθηγητής Α.Π.Θ.

ΜΕΤΑΠΤΥΧΙΑΚΗ ΙΠΛΩΜΑΤΙΚΗ ΕΡΓΑΣΙΑ. Ελευθερίου Β. Χρυσούλα. Επιβλέπων: Νικόλαος Καραμπετάκης Καθηγητής Α.Π.Θ. ΑΡΙΣΤΟΤΕΛΕΙΟ ΠΑΝΕΠΙΣΤΗΜΙΟ ΘΕΣΣΑΛΟΝΙΚΗΣ ΤΜΗΜΑ ΜΑΘΗΜΑΤΙΚΩΝ ΜΕΤΑΠΤΥΧΙΑΚΟ ΠΡΟΓΡΑΜΜΑ ΣΠΟΥ ΩΝ ΘΕΩΡΗΤΙΚΗ ΠΛΗΡΟΦΟΡΙΚΗ ΚΑΙ ΘΕΩΡΙΑ ΣΥΣΤΗΜΑΤΩΝ ΚΑΙ ΕΛΕΓΧΟΥ Αναγνώριση συστημάτων με δεδομένη συνεχή και κρουστική συμπεριφορά

Διαβάστε περισσότερα

Ó³ Ÿ , º 1(130).. 7Ä ±μ. Ñ Ò É ÉÊÉ Ö ÒÌ ² μ, Ê

Ó³ Ÿ , º 1(130).. 7Ä ±μ. Ñ Ò É ÉÊÉ Ö ÒÌ ² μ, Ê Ó³ Ÿ. 006.. 3, º 1(130).. 7Ä16 Š 530.145 ˆ ƒ ˆ ˆŒ ˆŸ Š ƒ.. ±μ Ñ Ò É ÉÊÉ Ö ÒÌ ² μ, Ê É μ ² Ö Ó μ μ Ö μ μ²õ μ É μ ÌÉ ±ÊÎ É ² ³ É μ - Î ±μ μ ÊÌ ±μ Ëμ ³ μ- ±² μ ÒÌ ³μ ²ÖÌ Ê ±. ³ É ÔÉμ μ μ μ Ö, Ö ²ÖÖ Ó ±μ³

Διαβάστε περισσότερα

!"#!$% &' ( )*+*,% $ &$ -.&01#(2$#3 4-$ #35667

!#!$% &' ( )*+*,% $ &$ -.&01#(2$#3 4-$ #35667 !"#!$% & &' ( )*+*,% $ -*(-$ -.*/% $- &$ -.&01#(2$#3 4-$ #35667 5051 & 00000000000000000000000000000000000000000000000000000000000000000000000000000 9 508&:;&& 0000000000000000000000000000000000000000000000000

Διαβάστε περισσότερα

f H f H ψ n( x) α = 0.01 n( x) α = 1 n( x) α = 3 n( x) α = 10 n( x) α = 30 ū i ( x) α = 1 ū i ( x) α = 3 ū i ( x) α = 10 ū i ( x) α = 30 δū ij ( x) α = 1 δū ij ( x) α = 3 δū ij ( x) α = 10 δū ij ( x)

Διαβάστε περισσότερα

a(z) = k 0 1 z k = k 0 2 k z k = k 0 z k = (1 + z) n. k

a(z) = k 0 1 z k = k 0 2 k z k = k 0 z k = (1 + z) n. k !" #$%% $&$'$ # %( $)%*&%' '+ &'&% ! " " # $ " " % " & ' # () *+ (, *,-.$ / " " " * $ 0 * " # " $ * $ 0 # % " & ', # ' * # " & #! " # %& *%& $ % & ' " ( z D log! ) * (% % (+, ) " " -. // 0 ', % 0 ', %

Διαβάστε περισσότερα

Multi-GPU numerical simulation of electromagnetic waves

Multi-GPU numerical simulation of electromagnetic waves Multi-GPU numerical simulation of electromagnetic waves Philippe Helluy, Thomas Strub To cite this version: Philippe Helluy, Thomas Strub. Multi-GPU numerical simulation of electromagnetic waves. ESAIM:

Διαβάστε περισσότερα

ITU-R P (2009/10)

ITU-R P (2009/10) ITU-R.38-6 (009/0 $% #! " #( ' * & ' /0,-. # GHz 00 MHz 900 ITU-R.38-6 ii.. (IR (ITU-T/ITU-R/ISO/IEC.ITU-R http://www.itu.int/itu-r/go/patents/en. (http://www.itu.int/publ/r-rec/en ( ( BO BR BS BT F M

Διαβάστε περισσότερα

Formulas of Agrawal s Fiber-Optic Communication Systems NA n 2 ; n n. NA( )=n1 a

Formulas of Agrawal s Fiber-Optic Communication Systems NA n 2 ; n n. NA( )=n1 a Formula o grawal Fiber-Oti Communiation Sytem Chater (ntroution) 8 / max m M / E nh N h M m 4 6.66. J e 9.6 / m log /mw SN / / /, NZ SN log / Z max N E Chater (Otial Fiber) Setion - (Geometrial Oti erition)

Διαβάστε περισσότερα

Answers to practice exercises

Answers to practice exercises Answers to practice exercises Chapter Exercise (Page 5). 9 kg 2. 479 mm. 66 4. 565 5. 225 6. 26 7. 07,70 8. 4 9. 487 0. 70872. $5, Exercise 2 (Page 6). (a) 468 (b) 868 2. (a) 827 (b) 458. (a) 86 kg (b)

Διαβάστε περισσότερα

ITU-R P ITU-R P (ITU-R 204/3 ( )

ITU-R P ITU-R P (ITU-R 204/3 ( ) 1 ITU-R P.530-1 ITU-R P.530-1 (ITU-R 04/3 ) (007-005-001-1999-1997-1995-1994-199-1990-1986-198-1978)... ( ( ( 1 1. 1 : - - ) - ( 1 ITU-R P.530-1..... 6.3. :. ITU-R P.45 -. ITU-R P.619 -. ) (ITU-R P.55

Διαβάστε περισσότερα

Answers - Worksheet A ALGEBRA PMT. 1 a = 7 b = 11 c = 1 3. e = 0.1 f = 0.3 g = 2 h = 10 i = 3 j = d = k = 3 1. = 1 or 0.5 l =

Answers - Worksheet A ALGEBRA PMT. 1 a = 7 b = 11 c = 1 3. e = 0.1 f = 0.3 g = 2 h = 10 i = 3 j = d = k = 3 1. = 1 or 0.5 l = C ALGEBRA Answers - Worksheet A a 7 b c d e 0. f 0. g h 0 i j k 6 8 or 0. l or 8 a 7 b 0 c 7 d 6 e f g 6 h 8 8 i 6 j k 6 l a 9 b c d 9 7 e 00 0 f 8 9 a b 7 7 c 6 d 9 e 6 6 f 6 8 g 9 h 0 0 i j 6 7 7 k 9

Διαβάστε περισσότερα

Transfert sécurisé d Images par combinaison de techniques de compression, cryptage et de marquage

Transfert sécurisé d Images par combinaison de techniques de compression, cryptage et de marquage Transfert sécurisé d Images par combinaison de techniques de compression, cryptage et de marquage José Marconi Rodrigues To cite this version: José Marconi Rodrigues. Transfert sécurisé d Images par combinaison

Διαβάστε περισσότερα

χ 2 1 N =0 1 1 2 3 npn 1 2 1 9 N =0 1 1 1 1 2 6 6 4 9 B V 70 100 10 1 2 2 2 2 a 1 a 2 δ 1, δ 2 δ 3. b 1 b 2 Γ, K, K M K K A B a 1 = ( ) ( ) 3a 2, a 3a, a 2 2 = 2, a, 2 a = a 1 = a 2 2.46 ( ) (

Διαβάστε περισσότερα

Inflation and Reheating in Spontaneously Generated Gravity

Inflation and Reheating in Spontaneously Generated Gravity Univesità di Bologna Inflation and Reheating in Spontaneously Geneated Gavity (A. Ceioni, F. Finelli, A. Tonconi, G. Ventui) Phys.Rev.D81:123505,2010 Motivations Inflation (FTV Phys.Lett.B681:383-386,2009)

Διαβάστε περισσότερα

L. F avart. CLAS12 Workshop Genova th of Feb CLAS12 workshop Feb L.Favart p.1/28

L. F avart. CLAS12 Workshop Genova th of Feb CLAS12 workshop Feb L.Favart p.1/28 L. F avart I.I.H.E. Université Libre de Bruxelles H Collaboration HERA at DESY CLAS Workshop Genova - 4-8 th of Feb. 9 CLAS workshop Feb. 9 - L.Favart p./8 e p Integrated luminosity 96- + 3-7 (high energy)

Διαβάστε περισσότερα