吸収分光.PDF

Similar documents
飽和分光

LLG-R8.Nisus.pdf

PDF

Microsoft Word - 学士論文(表紙).doc

1 7 ω ω ω 7.1 0, ( ) Q, 7.2 ( Q ) 7.1 ω Z = R +jx Z 1/ Z 7.2 ω 7.2 Abs. admittance (x10-3 S) RLC Series Circuit Y R = 20 Ω L = 100


LD

( ) sin 1 x, cos 1 x, tan 1 x sin x, cos x, tan x, arcsin x, arccos x, arctan x. π 2 sin 1 x π 2, 0 cos 1 x π, π 2 < tan 1 x < π 2 1 (1) (

9 1. (Ti:Al 2 O 3 ) (DCM) (Cr:Al 2 O 3 ) (Cr:BeAl 2 O 4 ) Ĥ0 ψ n (r) ω n Schrödinger Ĥ 0 ψ n (r) = ω n ψ n (r), (1) ω i ψ (r, t) = [Ĥ0 + Ĥint (

(Compton Scattering) Beaming 1 exp [i (k x ωt)] k λ k = 2π/λ ω = 2πν k = ω/c k x ωt ( ω ) k α c, k k x ωt η αβ k α x β diag( + ++) x β = (ct, x) O O x

() x + y + y + x dy dx = 0 () dy + xy = x dx y + x y ( 5) ( s55906) 0.7. (). 5 (). ( 6) ( s6590) 0.8 m n. 0.9 n n A. ( 6) ( s6590) f A (λ) = det(a λi)

N cos s s cos ψ e e e e 3 3 e e 3 e 3 e



ω 0 m(ẍ + γẋ + ω0x) 2 = ee (2.118) e iωt x = e 1 m ω0 2 E(ω). (2.119) ω2 iωγ Z N P(ω) = χ(ω)e = exzn (2.120) ϵ = ϵ 0 (1 + χ) ϵ(ω) ϵ 0 = 1 +

V(x) m e V 0 cos x π x π V(x) = x < π, x > π V 0 (i) x = 0 (V(x) V 0 (1 x 2 /2)) n n d 2 f dξ 2ξ d f 2 dξ + 2n f = 0 H n (ξ) (ii) H

1 No.1 5 C 1 I III F 1 F 2 F 1 F 2 2 Φ 2 (t) = Φ 1 (t) Φ 1 (t t). = Φ 1(t) t = ( 1.5e 0.5t 2.4e 4t 2e 10t ) τ < 0 t > τ Φ 2 (t) < 0 lim t Φ 2 (t) = 0

2.2 h h l L h L = l cot h (1) (1) L l L l l = L tan h (2) (2) L l 2 l 3 h 2.3 a h a h (a, h)

m dv = mg + kv2 dt m dv dt = mg k v v m dv dt = mg + kv2 α = mg k v = α 1 e rt 1 + e rt m dv dt = mg + kv2 dv mg + kv 2 = dt m dv α 2 + v 2 = k m dt d

m(ẍ + γẋ + ω 0 x) = ee (2.118) e iωt P(ω) = χ(ω)e = ex = e2 E(ω) m ω0 2 ω2 iωγ (2.119) Z N ϵ(ω) ϵ 0 = 1 + Ne2 m j f j ω 2 j ω2 iωγ j (2.120)

x E E E e i ω = t + ikx 0 k λ λ 2π k 2π/λ k ω/v v n v c/n k = nω c c ω/2π λ k 2πn/λ 2π/(λ/n) κ n n κ N n iκ k = Nω c iωt + inωx c iωt + i( n+ iκ ) ωx

. ev=,604k m 3 Debye ɛ 0 kt e λ D = n e n e Ze 4 ln Λ ν ei = 5.6π / ɛ 0 m/ e kt e /3 ν ei v e H + +e H ev Saha x x = 3/ πme kt g i g e n

µµ InGaAs/GaAs PIN InGaAs PbS/PbSe InSb InAs/InSb MCT (HgCdTe)

I ( ) 1 de Broglie 1 (de Broglie) p λ k h Planck ( Js) p = h λ = k (1) h 2π : Dirac k B Boltzmann ( J/K) T U = 3 2 k BT

64 3 g=9.85 m/s 2 g=9.791 m/s 2 36, km ( ) 1 () 2 () m/s : : a) b) kg/m kg/m k

c 2009 i

05Mar2001_tune.dvi

keisoku01.dvi

II ( ) (7/31) II ( [ (3.4)] Navier Stokes [ (6/29)] Navier Stokes 3 [ (6/19)] Re

SFGÇÃÉXÉyÉNÉgÉãå`.pdf

ELECTRONIC IMAGING IN ASTRONOMY Detectors and Instrumentation 5 Instrumentation and detectors


N/m f x x L dl U 1 du = T ds pdv + fdl (2.1)

1 (Berry,1975) 2-6 p (S πr 2 )p πr 2 p 2πRγ p p = 2γ R (2.5).1-1 : : : : ( ).2 α, β α, β () X S = X X α X β (.1) 1 2

4. ϵ(ν, T ) = c 4 u(ν, T ) ϵ(ν, T ) T ν π4 Planck dx = 0 e x 1 15 U(T ) x 3 U(T ) = σt 4 Stefan-Boltzmann σ 2π5 k 4 15c 2 h 3 = W m 2 K 4 5.


meiji_resume_1.PDF

Gmech08.dvi

Microsoft Word - 章末問題

The Physics of Atmospheres CAPTER :

ii 3.,. 4. F. (), ,,. 8.,. 1. (75%) (25%) =7 20, =7 21 (. ). 1.,, (). 3.,. 1. ().,.,.,.,.,. () (12 )., (), 0. 2., 1., 0,.

85 4

1. (8) (1) (x + y) + (x + y) = 0 () (x + y ) 5xy = 0 (3) (x y + 3y 3 ) (x 3 + xy ) = 0 (4) x tan y x y + x = 0 (5) x = y + x + y (6) = x + y 1 x y 3 (

80 4 r ˆρ i (r, t) δ(r x i (t)) (4.1) x i (t) ρ i ˆρ i t = 0 i r 0 t(> 0) j r 0 + r < δ(r 0 x i (0))δ(r 0 + r x j (t)) > (4.2) r r 0 G i j (r, t) dr 0

( ) e + e ( ) ( ) e + e () ( ) e e Τ ( ) e e ( ) ( ) () () ( ) ( ) ( ) ( )

Microsoft PowerPoint - 山形大高野send ppt [互換モード]

201711grade1ouyou.pdf

ma22-9 u ( v w) = u v w sin θê = v w sin θ u cos φ = = 2.3 ( a b) ( c d) = ( a c)( b d) ( a d)( b c) ( a b) ( c d) = (a 2 b 3 a 3 b 2 )(c 2 d 3 c 3 d

, 3, 6 = 3, 3,,,, 3,, 9, 3, 9, 3, 3, 4, 43, 4, 3, 9, 6, 6,, 0 p, p, p 3,..., p n N = p p p 3 p n + N p n N p p p, p 3,..., p n p, p,..., p n N, 3,,,,

Untitled


r d 2r d l d (a) (b) (c) 1: I(x,t) I(x+ x,t) I(0,t) I(l,t) V in V(x,t) V(x+ x,t) V(0,t) l V(l,t) 2: 0 x x+ x 3: V in 3 V in x V (x, t) I(x, t

.2 ρ dv dt = ρk grad p + 3 η grad (divv) + η 2 v.3 divh = 0, rote + c H t = 0 dive = ρ, H = 0, E = ρ, roth c E t = c ρv E + H c t = 0 H c E t = c ρv T

( ) ,

重力方向に基づくコントローラの向き決定方法

1 1.1 ( ). z = a + bi, a, b R 0 a, b 0 a 2 + b 2 0 z = a + bi = ( ) a 2 + b 2 a a 2 + b + b 2 a 2 + b i 2 r = a 2 + b 2 θ cos θ = a a 2 + b 2, sin θ =


II Karel Švadlenka * [1] 1.1* 5 23 m d2 x dt 2 = cdx kx + mg dt. c, g, k, m 1.2* u = au + bv v = cu + dv v u a, b, c, d R

#A A A F, F d F P + F P = d P F, F y P F F x A.1 ( α, 0), (α, 0) α > 0) (x, y) (x + α) 2 + y 2, (x α) 2 + y 2 d (x + α)2 + y 2 + (x α) 2 + y 2 =

( ) ( 40 )+( 60 ) Schrödinger 3. (a) (b) (c) yoshioka/education-09.html pdf 1

基礎数学I

Gmech08.dvi

: , 2.0, 3.0, 2.0, (%) ( 2.

[ ] [ ] [ ] [ ] [ ] [ ] ADC


B

7 π L int = gψ(x)ψ(x)φ(x) + (7.4) [ ] p ψ N = n (7.5) π (π +,π 0,π ) ψ (σ, σ, σ )ψ ( A) σ τ ( L int = gψψφ g N τ ) N π * ) (7.6) π π = (π, π, π ) π ±

main.dvi


本文/目次(裏白)

untitled

微分積分 サンプルページ この本の定価 判型などは, 以下の URL からご覧いただけます. このサンプルページの内容は, 初版 1 刷発行時のものです.

I

4‐E ) キュリー温度を利用した消磁:熱消磁

Z: Q: R: C: sin 6 5 ζ a, b

Hanbury-Brown Twiss (ver. 2.0) van Cittert - Zernike mutual coherence

grad φ(p ) φ P grad φ(p ) p P p φ P p l t φ l t = 0 g (0) g (0) (31) grad φ(p ) p grad φ φ (P, φ(p )) xy (x, y) = (ξ(t), η(t)) ( )

untitled


) ] [ h m x + y + + V x) φ = Eφ 1) z E = i h t 13) x << 1) N n n= = N N + 1) 14) N n n= = N N + 1)N + 1) 6 15) N n 3 n= = 1 4 N N + 1) 16) N n 4

1. z dr er r sinθ dϕ eϕ r dθ eθ dr θ dr dθ r x 0 ϕ r sinθ dϕ r sinθ dϕ y dr dr er r dθ eθ r sinθ dϕ eϕ 2. (r, θ, φ) 2 dr 1 h r dr 1 e r h θ dθ 1 e θ h

5 c P 5 kn n t π (.5 P 7 MP π (.5 n t n cos π. MP 6 4 t sin π 6 cos π 6.7 MP 4 P P N i i i i N i j F j ii N i i ii F j i i N ii li i F j i ij li i i i

1 Visible spectroscopy for student Spectrometer and optical spectrum phys/ishikawa/class/index.html

08-Note2-web

<4D F736F F D B B83578B6594BB2D834A836F815B82D082C88C602E646F63>

C el = 3 2 Nk B (2.14) c el = 3k B C el = 3 2 Nk B

第1章 微分方程式と近似解法

128 3 II S 1, S 2 Φ 1, Φ 2 Φ 1 = { B( r) n( r)}ds S 1 Φ 2 = { B( r) n( r)}ds (3.3) S 2 S S 1 +S 2 { B( r) n( r)}ds = 0 (3.4) S 1, S 2 { B( r) n( r)}ds

untitled

I-2 (100 ) (1) y(x) y dy dx y d2 y dx 2 (a) y + 2y 3y = 9e 2x (b) x 2 y 6y = 5x 4 (2) Bernoulli B n (n = 0, 1, 2,...) x e x 1 = n=0 B 0 B 1 B 2 (3) co

untitled

量子力学 問題

数学の基礎訓練I

1 [ 1] (1) MKS? (2) MKS? [ 2] (1) (42.195k) k 2 (2) (3) k/hr [ 3] t = 0 10 ( 1 velocity [/s] 8 4 O

Gauss Gauss ɛ 0 E ds = Q (1) xy σ (x, y, z) (2) a ρ(x, y, z) = x 2 + y 2 (r, θ, φ) (1) xy A Gauss ɛ 0 E ds = ɛ 0 EA Q = ρa ɛ 0 EA = ρea E = (ρ/ɛ 0 )e

SO(2)

OHO.dvi


q π =0 Ez,t =ε σ {e ikz ωt e ikz ωt } i/ = ε σ sinkz ωt 5.6 x σ σ *105 q π =1 Ez,t = 1 ε σ + ε π {e ikz ωt e ikz ωt } i/ = 1 ε σ + ε π sinkz ωt 5.7 σ

x () g(x) = f(t) dt f(x), F (x) 3x () g(x) g (x) f(x), F (x) (3) h(x) = x 3x tf(t) dt.9 = {(x, y) ; x, y, x + y } f(x, y) = xy( x y). h (x) f(x), F (x

Transcription:

3 Rb 1

1 4 1.1 4 1. 4 5.1 5. 5 3 8 3.1 8 4 1 4.1 External Cavity Laser Diode: ECLD 1 4. 1 4.3 Polarization Beam Splitter: PBS 13 4.4 Photo Diode: PD 13 4.5 13 4.6 13 5 Rb 14 6 15 6.1 ECLD 15 6. 15 6.3 16 6.4 16

7 17 A 18 B C LS 1 D Hyperfine Structure E 4 3

1 1.1 197 1. Rb 4

.1 J =, ± 1 F =, ± 1 J J = F F = g = e g = e. 1 Torr 5Torr 3 E t E t h (..1) (..1) E E = h (..) (..1) (..) 1 (..3) π t 5

τ γ τ 1 = L( ) 1 γ π L ( ) = (..4) γ 4π ( ) + = Full Width of Half Maximum: FWHM γ = (..5) π θ 1 + cosθ (..6) c (3..6) c = cosθ (..7) 1 = cosθ f ) m z + d z z z ( z 6

f ( = z ) π m k B T 1/ m z exp kbt (..8) k B T z (..8) m G( ) = exp c (..9) k BT = FWHM 1/ kbt ln = (..1) c m 7

3 3.1 MHz ~ GHz MHz ~ GHz Rb Fig3.1 Fig3.1 1 8

Fig3. Fig3.. Fig3. 1 N1 N N N Fig3.3 Fig3.3 9

4 4.1 External Cavity Laser Diode: ECLD Laser Diode: LD GaAs Fig4.1 h W g < h < W W (4.1.1) fc fv Wg W fc W fv W < W W (4.1.1) h g fc fv h ' Fig4.1 1

ECLD ECLD LD Littrow LD Littman-Metcalf Fig4. Fig4.3 Littrow Littman-Metcalf θ θ m d λ 1 ( θ θ ) mλ d sin 1 + sin = (4.1.) Fig4. Fig4.3 Littrow Fig4. Littrow LD 78nm d=1/18mm Fig4.4 78nm (4.1.) Littrow 1 45 1 LD PZT LD LD 11

LD PZT LD 8% mw khz Fig4.4 4. / /4 / /4 Fig4.5 Fig4.5 x y E E lin_ red lin_ blue = E xˆ sin( ωt + kz+ φ = E yˆ sin( ωt + kz+ φ lin_ red ) lin_ blue ) (4..1) Fig4.5 k xˆ yˆ 1

Fig4.5 E E lin_ red lin_ blue = E xˆ sin( ωt kz+ φ) + E = E xˆ sin( ωt + kz+ φ) + E yˆ sin( ωt kz+ φ + π / ) yˆ sin( ωt + kz+ φ π / ) (4..) x y 4.3 Polarization Beam Splitter: PBS PBS Fig4.6 4.4 Photo Diode: PD Fig4.6 LD PD 4.5 85Rb 87 Rb Rb Rb 4.6 1 LD 6~7 13

5 Rb Fig5.1 S+1 L =,1,,3, L = S, P, D,... n L Fig5.1 Rb 795nm D1 78nm D J Fig5. 14

6 6.1 ECLD ECLD LD PZT 1 1mA LD PZT 1V PZT 6. Fig6.1 PZT 87Rb 78.47nm 78.34nm 85 Rb 78.44nm 78.38nm Fig6.1 15

6.3 FWHM ECLD 6.4 78.47nm 6.4 5V 5 6.3 16

7 ECLD 3 17

A + t x( t) γ x( t) + ω x( ) = (A.1) ω k m γ t ( t x t) = x e cos( ω ) (A.) ω I(ω) (A.) 1 i ω e ω t x( t) = A dω π ( ) (A.3) 1 + iωt 1 A( ω) = x( t) e dt = π π = x 8π i 1 + ( ω ω ) + γ i( ω ω ) 1 + + γ x e γ t cos( ω t) e iωt dt (A.4) ( ) I(ω) A( ω) A * ( ω ) ω ω << ω ω + ω C I ( ω ω ) = (A.5) ( ω ω ) + ( γ ) C 18

L( ω ω I L ( ω ω ) dω 1 ) = I( ω ω) = 1 γ L ( ω ω) = (A.6) π ( ω ω ) + ( γ ) γ = (A.7) π (3..5) τ τ γ ~ h (A.3) 19

B L = r p ( i ) L = r h (B.1) L 1 1 L = h sin θ (B.) sin θ θ θ sin θ ϕ Y L Y ( L +1) h Y = L (B.3) ( ) L L L +1 L L = L( L +1)h (B.4)

C LS LS J L S J = L + S (C.1) L S L S J J J = L + S, L + S 1,, L S (C.) + 1 S ( S L) 1 + L ( S > L) H ˆ = ξ( L S) (C.3) FS ξ L S J = L + S ˆ J S L H FS = ξ (C.4) J J Ĥ (.1.3) J, L, S > fs ˆ ξ h < J, L, S H FS J, L, S >= L [ J ( J + 1) S( S + 1) L( + 1) ] (C.5) 1

D Hyperfine Structure F J I F = J + I (D.1) J I J I F F F = J + I, J + I 1,, J I (D.) + 1 I ( I J ) +1 J ( I > J ) k Hˆ ( ) HFS = k Hˆ HFS (D.3) k k F, J, I > k < F J I Hˆ HFS F J I >= < F J I Hˆ ( ),,,,,, F, J, I > HFS k (D.4) k = 1 I 1 J 1 ˆ ) (1 H HFS = A( I J) (D.5)

A k = I 1 1 J ˆ () H HFS 6 = B ( I J ) + 3I J I( I + 1) J( J + 1) I(I 1)J (J 1) (D.6) B,... k = 3,4,5 k 3 ˆ 6K + 3K I ( I + 1) J( J + 1) < F, J, I HHFS F, J, I >= AK + B I (I 1) J(J 1) (D.7) 1 K = [ F ( F + 1) I( I + 1) J ( J + 1) ] (D.8) (D.7) 1 I 1 J 1 I J =1 1 3

E M J M J = J, J 1,, J (E.1) J +1 W ze FS ξ h = W FS µ J H ex [( J( J + 1) L( L + 1) S( S + 1) )] + µ B M J g J H ex (E.) µ M g g B g J ( J + 1) + S( S + 1) L( L + 1) J( J + 1) J J g J = 1+ (E.3) J H ex µ J µ J = µ B g J J (E.4) 4