3 filename=quantum-3dim110705a.tex ,2 [1],[2],[3] [3] U(x, y, z; t), p x ˆp x = h i x, p y ˆp y = h i y, p z ˆp z = h

R = Ar l B r l. A, B A, B.. r 2 R r = r2 [lar r l B r l2 ]=larl l B r l.2 r 2 R = [lar l l Br ] r r r = ll Ar l ll B = ll R rl.3 sin θ Θ = ll.4 Θsinθ

.3.2 3.3.2 Spherical Coorinates.5: Laplace 2 V = r 2 r 2 x = r cos φ sin θ, y = r sin φ sin θ, z = r cos θ.93 r 2 sin θ sin θ θ θ r 2 sin 2 θ 2 V =.94 2.94 z V φ Laplace r 2 r 2 r 2 sin θ.96.95 V r 2 R

sec13.dvi

13 13.1 O r F R = m d 2 r dt 2 m r m = F = m r M M d2 R dt 2 = m d 2 r dt 2 = F = F (13.1) F O L = r p = m r ṙ dl dt = m ṙ ṙ + m r r = r (m r ) = r F N. (13.2) N N = R F 13.2 O ˆn ω L O r u u = ω r 1 1:

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 v = mg + kv m v = mg k v v m v = mg + kv α = mg k v = α e rt + e rt m v = mg + kv v mg + kv = m v α + v = k m v (v α (v + α = k m ˆ ( v α ˆ αk v = m v + α ln v α v + α = αk m t + C v α v + α = e αk m

Gmech08.dvi

145 13 13.1 13.1.1 0 m mg S 13.1 F 13.1 F /m S F F 13.1 F mg S F F mg 13.1: m d2 r 2 = F + F = 0 (13.1) 146 13 F = F (13.2) S S S S S P r S P r r = r 0 + r (13.3) r 0 S S m d2 r 2 = F (13.4) (13.3) d 2

0 3 1 5 1.1.............................. 5 1. 3.................. 6 1.3.......................... 7 1.4.............................. 8 1.5.............................. 10 1.6.................................

ver F = i f i m r = F r = 0 F = 0 X = Y = Z = 0 (1) δr = (δx, δy, δz) F δw δw = F δr = Xδx + Y δy + Zδz = 0 (2) δr (2) 1 (1) (2 n (X i δx

ver. 1.0 18 6 20 F = f m r = F r = 0 F = 0 X = Y = Z = 0 (1 δr = (δx, δy, δz F δw δw = F δr = Xδx + Y δy + Zδz = 0 (2 δr (2 1 (1 (2 n (X δx + Y δy + Z δz = 0 (3 1 F F = (X, Y, Z δr = (δx, δy, δz S δr δw

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.

A 1. Boltzmann Planck u(ν, T )dν = 8πh ν 3 c 3 kt 1 dν h 6.63 10 34 J s Planck k 1.38 10 23 J K 1 Boltzmann u(ν, T ) T ν e hν c = 3 10 8 m s 1 2. Planck λ = c/ν Rayleigh-Jeans u(ν, T )dν = 8πν2 kt dν c

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( ) 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) (

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II No.1 [n/] [1]H n x) H n x) = 1) r n! r!n r)! x)n r r= []H n x) n,, H n x) = 1) n H n x) [3] H n x) = 1) n dn x e dx n e x [4] H n+1 x) = xh n x) nh n 1 x) ) d dx x H n x) = H n+1 x) d dx H nx) = nh

1 I 1.1 ± e = = - = C C MKSA [m], [Kg] [s] [A] 1C 1A 1 MKSA 1C 1C +q q +q q 1

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21 2 26 i 1 1 1.1............................ 1 1.2............................ 3 2 9 2.1................... 9 2.2.......... 9 2.3................... 11 2.4....................... 12 3 15 3.1..........

notekiso1_09.dvi

39 3 3.1 2 Ax 1,y 1 Bx 2,y 2 x y fx, y z fx, y x 1,y 1, 0 x 1,y 1,fx 1,y 1 x 2,y 2, 0 x 2,y 2,fx 2,y 2 A s I fx, yds lim fx i,y i Δs. 3.1.1 Δs 0 x i,y i N Δs 1 I lim Δx 2 +Δy 2 0 x 1 fx i,y i Δx i 2 +Δy

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)

1 16 10 5 1 2 2.1 a a a 1 1 1 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) 4 2 3 4 2 5 2.4 x y (x,y) l a x = l cot h cos a, (3) y = l cot h sin a (4) h a

2. 2 P M A 2 F = mmg AP AP 2 AP (G > : ) AP/ AP A P P j M j F = n j=1 mm j G AP j AP j 2 AP j 3 P ψ(p) j ψ(p j ) j (P j j ) A F = n j=1 mgψ(p j ) j AP

1. 1 213 1 6 1 3 1: ( ) 2: 3: SF 1 2 3 1: 3 2 A m 2. 2 P M A 2 F = mmg AP AP 2 AP (G > : ) AP/ AP A P P j M j F = n j=1 mm j G AP j AP j 2 AP j 3 P ψ(p) j ψ(p j ) j (P j j ) A F = n j=1 mgψ(p j ) j AP

W u = u(x, t) u tt = a 2 u xx, a > 0 (1) D := {(x, t) : 0 x l, t 0} u (0, t) = 0, u (l, t) = 0, t 0 (2)

3 215 4 27 1 1 u u(x, t) u tt a 2 u xx, a > (1) D : {(x, t) : x, t } u (, t), u (, t), t (2) u(x, ) f(x), u(x, ) t 2, x (3) u(x, t) X(x)T (t) u (1) 1 T (t) a 2 T (t) X (x) X(x) α (2) T (t) αa 2 T (t) (4)

Gmech08.dvi

51 5 5.1 5.1.1 P r P z θ P P P z e r e, z ) r, θ, ) 5.1 z r e θ,, z r, θ, = r sin θ cos = r sin θ sin 5.1) e θ e z = r cos θ r, θ, 5.1: 0 r

18 2 F 12 r 2 r 1 (3) Coulomb km Coulomb M = kg F G = ( ) ( ) ( ) 2 = [N]. Coulomb

r 1 r 2 r 1 r 2 2 Coulomb Gauss Coulomb 2.1 Coulomb 1 2 r 1 r 2 1 2 F 12 2 1 F 21 F 12 = F 21 = 1 4πε 0 1 2 r 1 r 2 2 r 1 r 2 r 1 r 2 (2.1) Coulomb ε 0 = 107 4πc 2 =8.854 187 817 10 12 C 2 N 1 m 2 (2.2)

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 (

9 1. (Ti:Al 2 O 3 ) (DCM) (Cr:Al 2 O 3 ) (Cr:BeAl 2 O 4 ) 2. 2.1 Ĥ ψ n (r) ω n Schrödinger Ĥ ψ n (r) = ω n ψ n (r), (1) ω i ψ (r, t) = [Ĥ + Ĥint (t)] ψ (r, t), (2) Ĥ int (t) = eˆxe cos ωt ˆdE cos ωt, (3)

50 2 I SI MKSA r q r q F F = 1 qq 4πε 0 r r 2 r r r r (2.2 ε 0 = 1 c 2 µ 0 c = m/s q 2.1 r q' F r = 0 µ 0 = 4π 10 7 N/A 2 k = 1/(4πε 0 qq

49 2 I II 2.1 3 e e = 1.602 10 19 A s (2.1 50 2 I SI MKSA 2.1.1 r q r q F F = 1 qq 4πε 0 r r 2 r r r r (2.2 ε 0 = 1 c 2 µ 0 c = 3 10 8 m/s q 2.1 r q' F r = 0 µ 0 = 4π 10 7 N/A 2 k = 1/(4πε 0 qq F = k r

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d dt P = d ( ) dv G M vg = F M = F (4.1) dt dt M v G P = M v G F (4.1) d dt H G = M G (4.2) H G M G Z K O I z R R O J x k i O P r! j Y y O -

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量子力学 問題

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II (1 4 ) 1. p.13 1 (x, y) (a, b) ε(x, y; a, b) f (x, y) f (a, b) A, B (6.5) y = b f (x, b) f (a, b) x a = A + ε(x, b; a, b) x a x a A = f x (a, b) y x 3 3y 3 (x, y) (, ) f (x, y) = x + y (x, y) = (, )

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II A A441 : October 02, 2014 Version : Kawahira, Tomoki TA (Kondo, Hirotaka )

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

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7 -a 7 -a February 4, 2007 1. 2. 3. 4. 1. 2. 3. 1 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 z

2 Chapter 4 (f4a). 2. (f4cone) ( θ) () g M. 2. (f4b) T M L P a θ (f4eki) ρ H A a g. v ( ) 2. H(t) ( )

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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)) ( )

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液晶の物理１：連続体理論（弾性，粘性）

The Physics of Liquid Crystals P. G. de Gennes and J. Prost (Oxford University Press, 1993) Liquid crystals are beautiful and mysterious; I am fond of them for both reasons. My hope is that some readers

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1 1 sin cos P (primary) S (secondly) 2 P S A sin(ω2πt + α) A ω 1 ω α V T m T m 1 100Hz m 2 36km 500Hz. 36km 1

sin cos P (primary) S (secondly) 2 P S A sin(ω2πt + α) A ω ω α 3 3 2 2V 3 33+.6T m T 5 34m Hz. 34 3.4m 2 36km 5Hz. 36km m 34 m 5 34 + m 5 33 5 =.66m 34m 34 x =.66 55Hz, 35 5 =.7 485.7Hz 2 V 5Hz.5V.5V V

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I 1 m 2 l k 2 x = 0 x 1 x 1 2 x 2 g x x 2 x 1 m k m 1-1. L x 1, x 2, ẋ 1, ẋ 2 ẋ 1 x = 0 1-2. 2 Q = x 1 + x 2 2 q = x 2 x 1 l L Q, q, Q, q M = 2m µ = m 2 1-3. Q q 1-4. 2 x 2 = h 1 x 1 t = 0 2 1 t x 1 (t)

p = mv p x > h/4π λ = h p m v Ψ 2 Ψ

II p = mv p x > h/4π λ = h p m v Ψ 2 Ψ Ψ Ψ 2 0 x P'(x) m d 2 x = mω 2 x = kx = F(x) dt 2 x = cos(ωt + φ) mω 2 = k ω = m k v = dx = -ωsin(ωt + φ) dt = d 2 x dt 2 0 y v θ P(x,y) θ = ωt + φ ν = ω [Hz] 2π

1 variation 1.1 imension unit L m M kg T s Q C QT 1 A = C s 1 MKSA F = ma N N = kg m s 1.1 J E = 1 mv W = F x J = kg m s 1 = N m 1.

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米田 戸倉川月 7 限 193~21 西 5-19 応用数学 A 積分定理 Gaussの定理 divbd = B nds Stokesの定理 E bds = E dr Green の定理 g x f y dxdy = fdx + gdy = f e i + ge j dr Gauss の発散定理 S n FdS = Fd 1777-1855 ドイツ Johann arl Friedrich Gauss

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SO(3) 71 5.7 5.7.1 1 ħ L k l k l k = iϵ kij x i j (5.117) l k SO(3) l z l ± = l 1 ± il = i(y z z y ) ± (z x x z ) = ( x iy) z ± z( x ± i y ) = X ± z ± z (5.118) l z = i(x y y x ) = 1 [(x + iy)( x i y )

9 2 1 f(x, y) = xy sin x cos y x y cos y y x sin x d (x, y) = y cos y (x sin x) = y cos y(sin x + x cos x) x dx d (x, y) = x sin x (y cos y) = x sin x

2009 9 6 16 7 1 7.1 1 1 1 9 2 1 f(x, y) = xy sin x cos y x y cos y y x sin x d (x, y) = y cos y (x sin x) = y cos y(sin x + x cos x) x dx d (x, y) = x sin x (y cos y) = x sin x(cos y y sin y) y dy 1 sin

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No δs δs = r + δr r = δr (3) δs δs = r r = δr + u(r + δr, t) u(r, t) (4) δr = (δx, δy, δz) u i (r + δr, t) u i (r, t) = u i x j δx j (5) δs 2

No.2 1 2 2 δs δs = r + δr r = δr (3) δs δs = r r = δr + u(r + δr, t) u(r, t) (4) δr = (δx, δy, δz) u i (r + δr, t) u i (r, t) = u i δx j (5) δs 2 = δx i δx i + 2 u i δx i δx j = δs 2 + 2s ij δx i δx j

18 I ( ) (1) I-1,I-2,I-3 (2) (3) I-1 ( ) (100 ) θ ϕ θ ϕ m m l l θ ϕ θ ϕ 2 g (1) (2) 0 (3) θ ϕ (4) (3) θ(t) = A 1 cos(ω 1 t + α 1 ) + A 2 cos(ω 2 t + α

18 I ( ) (1) I-1,I-2,I-3 (2) (3) I-1 ( ) (100 ) θ ϕ θ ϕ m m l l θ ϕ θ ϕ 2 g (1) (2) 0 (3) θ ϕ (4) (3) θ(t) = A 1 cos(ω 1 t + α 1 ) + A 2 cos(ω 2 t + α 2 ), ϕ(t) = B 1 cos(ω 1 t + α 1 ) + B 2 cos(ω 2 t

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数学演習：微分方程式

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6 2 2 x y x y t P P = P t P = I P P P ( ) ( ) ,, ( ) ( ) cos θ sin θ cos θ sin θ, sin θ cos θ sin θ cos θ y x θ x θ P

6 x x 6.1 t P P = P t P = I P P P 1 0 1 0,, 0 1 0 1 cos θ sin θ cos θ sin θ, sin θ cos θ sin θ cos θ x θ x θ P x P x, P ) = t P x)p ) = t x t P P ) = t x = x, ) 6.1) x = Figure 6.1 Px = x, P=, θ = θ P

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( : December 27, 2015) CONTENTS I. 1 II. 2 III. 2 IV. 3 V. 5 VI. 6 VII. 7 VIII. 9 I. 1 f(x) f (x) y = f(x) x ϕ(r) (gradient) ϕ(r) (gradϕ(r) ) ( ) ϕ(r)

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(1.2) T D = 0 T = D = 30 kn 1.2 (1.4) 2F W = 0 F = W/2 = 300 kn/2 = 150 kn 1.3 (1.9) R = W 1 + W 2 = = 1100 N. (1.9) W 2 b W 1 a = 0

1 1 1.1 1.) T D = T = D = kn 1. 1.4) F W = F = W/ = kn/ = 15 kn 1. 1.9) R = W 1 + W = 6 + 5 = 11 N. 1.9) W b W 1 a = a = W /W 1 )b = 5/6) = 5 cm 1.4 AB AC P 1, P x, y x, y y x 1.4.) P sin 6 + P 1 sin 45

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

I (008 4 0 de Broglie (de Broglie p λ k h Planck ( 6.63 0 34 Js p = h λ = k ( h π : Dirac k B Boltzmann (.38 0 3 J/K T U = 3 k BT ( = λ m k B T h m = 0.067m 0 m 0 = 9. 0 3 kg GaAs( a T = 300 K 3 fg 07345

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II 9 7 8 HP lecture 017 II 6/13 7/18 I I 0 (004) L. D. Landau and E. M. Lifshitz, Quantum Mechanics Pergamon Press (1991) Steven Weinberg Lectures on Quantum Mechanics Cambridge University Press (015)

2000年度『数学展望 I』講義録

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