QCD ( ), ( ) 8 23 Typeset by FoilTEX

QCD ( ), ( ) 8 23 Typeset by FoilTEX Introduction:QCD Phase diagram of QCD system Critical end point Quark Gluon Plasma T c Temperature 1st order phase transition line Hadron 0 CSC 0 Chemical potential

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 ±

7 7. ( ) SU() SU() 9 ( MeV) p 98.8 π + π 0 n 99.57 9.57 97.4 497.70 δm m 0.4%.% 0.% 0.8% π 9.57 4.96 Σ + Σ 0 Σ 89.6 9.46 K + K 0 49.67 (7.) p p = αp + βn, n n = γp + δn (7.a) [ ] p ψ ψ = Uψ, U = n [ α

meiji_resume_1.PDF

β β β (q 1,q,..., q n ; p 1, p,..., p n ) H(q 1,q,..., q n ; p 1, p,..., p n ) Hψ = εψ ε k = k +1/ ε k = k(k 1) (x, y, z; p x, p y, p z ) (r; p r ), (θ; p θ ), (ϕ; p ϕ ) ε k = 1/ k p i dq i E total = E

20 4 20 i 1 1 1.1............................ 1 1.2............................ 4 2 11 2.1................... 11 2.2......................... 11 2.3....................... 19 3 25 3.1.............................

July 28, H H 0 H int = H H 0 H int = H int (x)d 3 x Schrödinger Picture Ψ(t) S =e iht Ψ H O S Heisenberg Picture Ψ H O H (t) =e iht O S e i

July 8, 4. H H H int H H H int H int (x)d 3 x Schrödinger Picture Ψ(t) S e iht Ψ H O S Heisenberg Picture Ψ H O H (t) e iht O S e iht Interaction Picture Ψ(t) D e iht Ψ(t) S O D (t) e iht O S e ih t (Dirac

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)

TOP URL 1

TOP URL http://amonphys.web.fc2.com/ 1 30 3 30.1.............. 3 30.2........................... 4 30.3...................... 5 30.4........................ 6 30.5.................................. 8 30.6...............................

H 0 H = H 0 + V (t), V (t) = gµ B S α qb e e iωt i t Ψ(t) = [H 0 + V (t)]ψ(t) Φ(t) Ψ(t) = e ih0t Φ(t) H 0 e ih0t Φ(t) + ie ih0t t Φ(t) = [

3 3. 3.. H H = H + V (t), V (t) = gµ B α B e e iωt i t Ψ(t) = [H + V (t)]ψ(t) Φ(t) Ψ(t) = e iht Φ(t) H e iht Φ(t) + ie iht t Φ(t) = [H + V (t)]e iht Φ(t) Φ(t) i t Φ(t) = V H(t)Φ(t), V H (t) = e iht V (t)e

,,..,. 1

016 9 3 6 0 016 1 0 1 10 1 1 17 1..,,..,. 1 1 c = h = G = ε 0 = 1. 1.1 L L T V 1.1. T, V. d dt L q i L q i = 0 1.. q i t L q i, q i, t L ϕ, ϕ, x µ x µ 1.3. ϕ x µ, L. S, L, L S = Ld 4 x 1.4 = Ld 3 xdt 1.5

) ] [ 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. k λ ν ω T v p v g k = π λ ω = πν = π T v p = λν = ω k v g = dω dk 1) ) 3) 4). p = hk = h λ 5) E = hν = hω 6) h = h π 7) h =6.6618 1 34 J sec) hc=197.3 MeV fm = 197.3 kev pm= 197.3 ev nm = 1.97 1 3 ev

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

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

2005 9/8-11 2 2.2 ( 2-5) γ ( ) γ cos θ 2πr πρhr 2 g h = 2γ cos θ ρgr (2.1) γ = ρgrh (2.2) 2 cos θ θ cos θ = 1 (2.2) γ = 1 ρgrh (2.) 2 2. p p ρgh p ( ) p p = p ρgh (2.) h p p = 2γ r 1 1 (Berry,1975) 2-6

d ϕ i) t d )t0 d ϕi) ϕ i) t x j t d ) ϕ t0 t α dx j d ) ϕ i) t dx t0 j x j d ϕ i) ) t x j dx t0 j f i x j ξ j dx i + ξ i x j dx j f i ξ i x j dx j d )

23 M R M ϕ : R M M ϕt, x) ϕ t x) ϕ s ϕ t ϕ s+t, ϕ 0 id M M ϕ t M ξ ξ ϕ t d ϕ tx) ξϕ t x)) U, x 1,...,x n )) ϕ t x) ϕ 1) t x),...,ϕ n) t x)), ξx) ξ i x) d ϕi) t x) ξ i ϕ t x)) M f ϕ t f)x) f ϕ t )x) fϕ

Part () () Γ Part ,

Contents a 6 6 6 6 6 6 6 7 7. 8.. 8.. 8.3. 8 Part. 9. 9.. 9.. 3. 3.. 3.. 3 4. 5 4.. 5 4.. 9 4.3. 3 Part. 6 5. () 6 5.. () 7 5.. 9 5.3. Γ 3 6. 3 6.. 3 6.. 3 6.3. 33 Part 3. 34 7. 34 7.. 34 7.. 34 8. 35

( ) ) ) ) 5) 1 J = σe 2 6) ) 9) 1955 Statistical-Mechanical Theory of Irreversible Processes )

( 3 7 4 ) 2 2 ) 8 2 954 2) 955 3) 5) J = σe 2 6) 955 7) 9) 955 Statistical-Mechanical Theory of Irreversible Processes 957 ) 3 4 2 A B H (t) = Ae iωt B(t) = B(ω)e iωt B(ω) = [ Φ R (ω) Φ R () ] iω Φ R (t)

6 2 T γ T B (6.4) (6.1) [( d nm + 3 ] 2 nt B )a 3 + nt B da 3 = 0 (6.9) na 3 = T B V 3/2 = T B V γ 1 = const. or T B a 2 = const. (6.10) H 2 = 8π kc2

1 6 6.1 (??) (P = ρ rad /3) ρ rad T 4 d(ρv ) + PdV = 0 (6.1) dρ rad ρ rad + 4 da a = 0 (6.2) dt T + da a = 0 T 1 a (6.3) ( ) n ρ m = n (m + 12 ) m v2 = n (m + 32 ) T, P = nt (6.4) (6.1) d [(nm + 32 ] )a

SO(3) 7 = = 1 ( r ) + 1 r r r r ( l ) (5.17) l = 1 ( sin θ ) + sin θ θ θ ϕ (5.18) χ(r)ψ(θ, ϕ) l ψ = αψ (5.19) l 1 = i(sin ϕ θ l = i( cos ϕ θ l 3 = i ϕ

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 )

TOP URL 1

TOP URL http://amonphys.web.fc.com/ 1 19 3 19.1................... 3 19.............................. 4 19.3............................... 6 19.4.............................. 8 19.5.............................

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..........

( )

7..-8..8.......................................................................... 4.................................... 3...................................... 3..3.................................. 4.3....................................

n (1.6) i j=1 1 n a ij x j = b i (1.7) (1.7) (1.4) (1.5) (1.4) (1.7) u, v, w ε x, ε y, ε x, γ yz, γ zx, γ xy (1.8) ε x = u x ε y = v y ε z = w z γ yz

1 2 (a 1, a 2, a n ) (b 1, b 2, b n ) A (1.1) A = a 1 b 1 + a 2 b 2 + + a n b n (1.1) n A = a i b i (1.2) i=1 n i 1 n i=1 a i b i n i=1 A = a i b i (1.3) (1.3) (1.3) (1.1) (ummation convention) a 11 x

QCD 1 QCD GeV 2014 QCD 2015 QCD SU(3) QCD A µ g µν QCD 1

QCD 1 QCD GeV 2014 QCD 2015 QCD SU(3) QCD A µ g µν QCD 1 (vierbein) QCD QCD 1 1: QCD QCD Γ ρ µν A µ R σ µνρ F µν g µν A µ Lagrangian gr TrFµν F µν No. Yes. Yes. No. No! Yes! [1] Nash & Sen [2] Riemann

SO(2)

TOP URL http://amonphys.web.fc2.com/ 1 12 3 12.1.................................. 3 12.2.......................... 4 12.3............................. 5 12.4 SO(2).................................. 6

1 1.1 H = µc i c i + c i t ijc j + 1 c i c j V ijklc k c l (1) V ijkl = V jikl = V ijlk = V jilk () t ij = t ji, V ijkl = V lkji (3) (1) V 0 H mf = µc

013 6 30 BCS 1 1.1........................ 1................................ 3 1.3............................ 3 1.4............................... 5 1.5.................................... 5 6 3 7 4 8

Dirac 38 5 Dirac 4 4 γ µ p µ p µ + m 2 = ( p µ γ µ + m)(p ν γ ν + m) (5.1) γ = p µ p ν γ µ γ ν p µ γ µ m + mp ν γ ν + m 2 = 1 2 p µp ν {γ µ, γ ν } + m

Dirac 38 5 Dirac 4 4 γ µ p µ p µ + m 2 p µ γ µ + mp ν γ ν + m 5.1 γ p µ p ν γ µ γ ν p µ γ µ m + mp ν γ ν + m 2 1 2 p µp ν {γ µ, γ ν } + m 2 5.2 p m p p µ γ µ {, } 10 γ {γ µ, γ ν } 2η µν 5.3 p µ γ µ + mp

nakayama.key

2017/11/1@Flavor Physics Workshop 2017 Contents CP 1932 10 4 p + p! p + p + p + p P. Blasi, 1311.7346 d d. 10 Gpc 10 0 Cohen, De Rujula, Glashow (1997) d B0 Flux [photons cm -2 s -1 MeV -1 sr -1

K E N Z U 2012 7 16 HP M. 1 1 4 1.1 3.......................... 4 1.2................................... 4 1.2.1..................................... 4 1.2.2.................................... 5................................

1 A A.1 G = A,B,C, A,B, (1) A,B AB (2) (AB)C = A(BC) (3) 1 A 1A = A1 = A (4) A A 1 A 1 A = AA 1 = 1 AB = BA ( ) AB BA ( ) 3 SU(N),N 2 (Lie) A(θ 1,θ 2,

1 A A.1 G = A,B,C, A,B, (1) A,B AB (2) (AB)C = A(BC) (3) 1 A 1A = A1 = A (4) A A 1 A 1 A = AA 1 = 1 AB = BA ( ) AB BA ( ) 3 SU(N),N 2 (Lie) A(θ 1,θ 2,θ n ) = exp(i n i=1 θ i F i ) (A.1) F i 2 0 θ 2π 1

n ξ n,i, i = 1,, n S n ξ n,i n 0 R 1,.. σ 1 σ i .10.14.15 0 1 0 1 1 3.14 3.18 3.19 3.14 3.14,. ii 1 1 1.1..................................... 1 1............................... 3 1.3.........................

(e ) (µ ) (τ ) ( (ν e,e ) e- (ν µ,µ ) µ- (ν τ,τ ) τ- ) ( ) ( ) ( ) (SU(2) ) (W +,Z 0,W ) * 1) [ ] [ ] [ ] ν e ν µ ν τ e µ τ, e R,µ R,τ R (2.1a

1 2 2.1 (e ) (µ ) (τ ) ( (ν e,e ) e- (ν µ,µ ) µ- (ν τ,τ ) τ- ) ( ) ( ) ( ) (SU(2) ) (W +,Z 0,W ) * 1) [ ] [ ] [ ] ν e ν µ ν τ e µ τ, e R,µ R,τ R (2.1a) L ( ) ) * 2) W Z 1/2 ( - ) d u + e + ν e 1 1 0 0

反D中間子と核子のエキゾチックな 束縛状態と散乱状態の解析

.... D 1 in collaboration with 1, 2, 1 RCNP 1, KEK 2 . Exotic hadron qqq q q Θ + Λ(1405) etc. uudd s? KN quasi-bound state? . D(B)-N bound state { { D D0 ( cu) B = D ( cd), B = + ( bu) B 0 ( bd) D(B)-N

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

TOP URL 1

TOP URL http://amonphys.web.fc.com/ 3.............................. 3.............................. 4.3 4................... 5.4........................ 6.5........................ 8.6...........................7

0406_total.pdf

59 7 7.1 σ-ω σ-ω σ ω σ = σ(r), ω µ = δ µ,0 ω(r) (6-4) (iγ µ µ m U(r) γ 0 V (r))ψ(x) = 0 (7-1) U(r) = g σ σ(r), V (r) = g ω ω(r) σ(r) ω(r) (6-3) ( 2 + m 2 σ)σ(r) = g σ ψψ (7-2) ( 2 + m 2 ω)ω(r) = g ω ψγ

.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

NHK 204 2 0 203 2 24 ( ) 7 00 7 50 203 2 25 ( ) 7 00 7 50 203 2 26 ( ) 7 00 7 50 203 2 27 ( ) 7 00 7 50 I. ( ν R n 2 ) m 2 n m, R = e 2 8πε 0 hca B =.09737 0 7 m ( ν = ) λ a B = 4πε 0ħ 2 m e e 2 = 5.2977

: 2005 ( ρ t +dv j =0 r m m r = e E( r +e r B( r T 208 T = d E j 207 ρ t = = = e t δ( r r (t e r r δ( r r (t e r ( r δ( r r (t dv j =

72 Maxwell. Maxwell e r ( =,,N Maxwell rot E + B t = 0 rot H D t = j dv D = ρ dv B = 0 D = ɛ 0 E H = μ 0 B ρ( r = j( r = N e δ( r r = N e r δ( r r = : 2005 ( 2006.8.22 73 207 ρ t +dv j =0 r m m r = e E(

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

199 1 1 199 1 1. Vx) m e V cos x π x π Vx) = x < π, x > π V i) x = Vx) V 1 x /)) n n d f dξ ξ d f dξ + n f = H n ξ) ii) H n ξ) = 1) n expξ ) dn dξ n exp ξ )) H n ξ)h m ξ) exp ξ )dξ = π n n!δ n,m x = Vx)

多体問題

Many Body Problem 997 4, 00 4, 004 4............................................................................. 7...................................... 7.............................................

Z: Q: R: C: sin 6 5 ζ a, b

Z: Q: R: C: 3 3 7 4 sin 6 5 ζ 9 6 6............................... 6............................... 6.3......................... 4 7 6 8 8 9 3 33 a, b a bc c b a a b 5 3 5 3 5 5 3 a a a a p > p p p, 3,

2019 1 5 0 3 1 4 1.1.................... 4 1.1.1......................... 4 1.1.2........................ 5 1.1.3................... 5 1.1.4........................ 6 1.1.5......................... 6 1.2..........................

Λ(1405) supported by Global Center of Excellence Program Nanoscience and Quantum Physics 2009, Aug. 5th 1

Λ(1405) supported by Global Center of Excellence Program Nanoscience and Quantum Physics 2009, Aug. 5th 1 S KN Λ(1405) Λ(1405) CDD Nc 2 L = q(i/ m)q P L = 1 2 (1 γ 5), P R = 1 2 (1 + γ 5), q L P L q, q

chap9.dvi

9 AR (i) (ii) MA (iii) (iv) (v) 9.1 2 1 AR 1 9.1.1 S S y j = (α i + β i j) D ij + η j, η j = ρ S η j S + ε j (j =1,,T) (1) i=1 {ε j } i.i.d(,σ 2 ) η j (j ) D ij j i S 1 S =1 D ij =1 S>1 S =4 (1) y j =

Einstein 1905 Lorentz Maxwell c E p E 2 (pc) 2 = m 2 c 4 (7.1) m E ( ) E p µ =(p 0,p 1,p 2,p 3 )=(p 0, p )= c, p (7.2) x µ =(x 0,x 1,x 2,x

7 7.1 7.1.1 Einstein 1905 Lorentz Maxwell c E p E 2 (pc) 2 = m 2 c 4 (7.1) m E ( ) E p µ =(p 0,p 1,p 2,p 3 )=(p 0, p )= c, p (7.2) x µ =(x 0,x 1,x 2,x 3 )=(x 0, x )=(ct, x ) (7.3) E/c ct K = E mc 2 (7.4)

³ÎΨÏÀ

2017 12 12 Makoto Nakashima 2017 12 12 1 / 22 2.1. C, D π- C, D. A 1, A 2 C A 1 A 2 C A 3, A 4 D A 1 A 2 D Makoto Nakashima 2017 12 12 2 / 22 . (,, L p - ). Makoto Nakashima 2017 12 12 3 / 22 . (,, L p

006 11 8 0 3 1 5 1.1..................... 5 1......................... 6 1.3.................... 6 1.4.................. 8 1.5................... 8 1.6................... 10 1.6.1......................

‚åŁÎ“·„´Šš‡ðŠp‡¢‡½‹âfi`fiI…A…‰…S…−…Y…•‡ÌMarkovŸA“½fiI›ð’Í

Markov 2009 10 2 Markov 2009 10 2 1 / 25 1 (GA) 2 GA 3 4 Markov 2009 10 2 2 / 25 (GA) (GA) L ( 1) I := {0, 1} L f : I (0, ) M( 2) S := I M GA (GA) f (i) i I Markov 2009 10 2 3 / 25 (GA) ρ(i, j), i, j I

dynamics-solution2.dvi

1 1. (1) a + b = i +3i + k () a b =5i 5j +3k (3) a b =1 (4) a b = 7i j +1k. a = 14 l =/ 14, m=1/ 14, n=3/ 14 3. 4. 5. df (t) d [a(t)e(t)] =ti +9t j +4k, = d a(t) d[a(t)e(t)] e(t)+ da(t) d f (t) =i +18tj

25 7 18 1 1 1.1 v.s............................. 1 1.1.1.................................. 1 1.1.2................................. 1 1.1.3.................................. 3 1.2................... 3

Microsoft Word - 信号処理3.doc

Junji OHTSUBO 2012 FFT FFT SN sin cos x v ψ(x,t) = f (x vt) (1.1) t=0 (1.1) ψ(x,t) = A 0 cos{k(x vt) + φ} = A 0 cos(kx ωt + φ) (1.2) A 0 v=ω/k φ ω k 1.3 (1.2) (1.2) (1.2) (1.1) 1.1 c c = a + ib, a = Re[c],

(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

.5 z = a + b + c n.6 = a sin t y = b cos t dy d a e e b e + e c e e e + e 3 s36 3 a + y = a, b > b 3 s363.7 y = + 3 y = + 3 s364.8 cos a 3 s365.9 y =,

[ ] IC. r, θ r, θ π, y y = 3 3 = r cos θ r sin θ D D = {, y ; y }, y D r, θ ep y yddy D D 9 s96. d y dt + 3dy + y = cos t dt t = y = e π + e π +. t = π y =.9 s6.3 d y d + dy d + y = y =, dy d = 3 a, b

Note.tex 2008/09/19( )

1 20 9 19 2 1 5 1.1........................ 5 1.2............................. 8 2 9 2.1............................. 9 2.2.............................. 10 3 13 3.1.............................. 13 3.2..................................

http://www.ike-dyn.ritsumei.ac.jp/ hyoo/wave.html 1 1, 5 3 1.1 1..................................... 3 1.2 5.1................................... 4 1.3.......................... 5 1.4 5.2, 5.3....................

6 6.1 L r p hl = r p (6.1) 1, 2, 3 r =(x, y, z )=(r 1,r 2,r 3 ), p =(p x,p y,p z )=(p 1,p 2,p 3 ) (6.2) hl i = jk ɛ ijk r j p k (6.3) ɛ ijk Levi Civit

6 6.1 L r p hl = r p (6.1) 1, 2, 3 r =(x, y, z )=(r 1,r 2,r 3 ), p =(p x,p y,p z )=(p 1,p 2,p 3 ) (6.2) hl i = jk ɛ ijk r j p k (6.3) ɛ ijk Levi Civita ɛ 123 =1 0 r p = 2 2 = (6.4) Planck h L p = h ( h

untitled

0. =. =. (999). 3(983). (980). (985). (966). 3. := :=. A A. A A. := := 4 5 A B A B A B. A = B A B A B B A. A B A B, A B, B. AP { A, P } = { : A, P } = { A P }. A = {0, }, A, {0, }, {0}, {}, A {0}, {}.

4 2 Rutherford 89 Rydberg λ = R ( n 2 ) n 2 n = n +,n +2, n = Lyman n =2 Balmer n =3 Paschen R Rydberg R = cm 896 Zeeman Zeeman Zeeman Lorentz

2 Rutherford 2. Rutherford N. Bohr Rutherford 859 Kirchhoff Bunsen 86 Maxwell Maxwell 885 Balmer λ Balmer λ = 364.56 n 2 n 2 4 Lyman, Paschen 3 nm, n =3, 4, 5, 4 2 Rutherford 89 Rydberg λ = R ( n 2 ) n

量子力学Ａ

c 1 1 1.1....................................... 1 1............................................ 4 1.3.............................. 6 10.1.................................. 10......................................

高知工科大学電子 光システム工学科

卒業研究報告 題 目 量子力学に基づいた水素分子の分子軌道法的取り扱いと Hamiltonian 近似法 指導教員 山本哲也 報告者 山中昭徳 平成 14 年 月 5 日 高知工科大学電子 光システム工学科. 3. 4.1 4. 4.3 4.5 6.6 8.7 10.8 11.9 1.10 1 3. 13 3.113 3. 13 3.3 13 3.4 14 3.5 15 3.6 15 3.7 17

橋本研

-Goldstone ( ) 2011 4-2012 3 -Goldstone UV IR SSB+ : Chiral Lagrangian :,... -Goldstone Nambu( 60), Goldstone(61), Nambu, Jona-Lasinio( 61), Goldstone, Salam, Weinberg( 62) Lorentz ( ) = NG 2002~2004?

[ ] 0.1 lim x 0 e 3x 1 x IC ( 11) ( s114901) 0.2 (1) y = e 2x (x 2 + 1) (2) y = x/(x 2 + 1) 0.3 dx (1) 1 4x 2 (2) e x sin 2xdx (3) sin 2 xdx ( 11) ( s

[ ]. lim e 3 IC ) s49). y = e + ) ) y = / + ).3 d 4 ) e sin d 3) sin d ) s49) s493).4 z = y z z y s494).5 + y = 4 =.6 s495) dy = 3e ) d dy d = y s496).7 lim ) lim e s49).8 y = e sin ) y = sin e 3) y =

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

1 I 1.1 ± e = = - =1.602 10 19 C C MKA [m], [Kg] [s] [A] 1C 1A 1 MKA 1C 1C +q q +q q 1 1.1 r 1,2 q 1, q 2 r 12 2 q 1, q 2 2 F 12 = k q 1q 2 r 12 2 (1.1) k 2 k 2 ( r 1 r 2 ) ( r 2 r 1 ) q 1 q 2 (q 1 q 2

ohpmain.dvi

fujisawa@ism.ac.jp 1 Contents 1. 2. 3. 4. γ- 2 1. 3 10 5.6, 5.7, 5.4, 5.5, 5.8, 5.5, 5.3, 5.6, 5.4, 5.2. 5.5 5.6 +5.7 +5.4 +5.5 +5.8 +5.5 +5.3 +5.6 +5.4 +5.2 =5.5. 10 outlier 5 5.6, 5.7, 5.4, 5.5, 5.8,

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

all.dvi

5,, Euclid.,..,... Euclid,.,.,, e i (i =,, ). 6 x a x e e e x.:,,. a,,. a a = a e + a e + a e = {e, e, e } a (.) = a i e i = a i e i (.) i= {a,a,a } T ( T ),.,,,,. (.),.,...,,. a 0 0 a = a 0 + a + a 0

201711grade1ouyou.pdf

2017 11 26 1 2 52 3 12 13 22 23 32 33 42 3 5 3 4 90 5 6 A 1 2 Web Web 3 4 1 2... 5 6 7 7 44 8 9 1 2 3 1 p p >2 2 A 1 2 0.6 0.4 0.52... (a) 0.6 0.4...... B 1 2 0.8-0.2 0.52..... (b) 0.6 0.52.... 1 A B 2

量子力学 問題

3 : 203 : 0. H = 0 0 2 6 0 () = 6, 2 = 2, 3 = 3 3 H 6 2 3 ϵ,2,3 (2) ψ = (, 2, 3 ) ψ Hψ H (3) P i = i i P P 2 = P 2 P 3 = P 3 P = O, P 2 i = P i (4) P + P 2 + P 3 = E 3 (5) i ϵ ip i H 0 0 (6) R = 0 0 [H,

TOP URL 1

TOP URL http://amonphys.web.fc2.com/ 1 6 3 6.1................................ 3 6.2.............................. 4 6.3................................ 5 6.4.......................... 6 6.5......................

0. Intro ( K CohFT etc CohFT 5.IKKT 6.

E-mail: sako@math.keio.ac.jp 0. Intro ( K 1. 2. CohFT etc 3. 4. CohFT 5.IKKT 6. 1 µ, ν : d (x 0,x 1,,x d 1 ) t = x 0 ( t τ ) x i i, j, :, α, β, SO(D) ( x µ g µν x µ µ g µν x ν (1) g µν g µν vector x µ,y

III 1 (X, d) d U d X (X, d). 1. (X, d).. (i) d(x, y) d(z, y) d(x, z) (ii) d(x, y) d(z, w) d(x, z) + d(y, w) 2. (X, d). F X.. (1), X F, (2) F 1, F 2 F

III 1 (X, d) d U d X (X, d). 1. (X, d).. (i) d(x, y) d(z, y) d(x, z) (ii) d(x, y) d(z, w) d(x, z) + d(y, w) 2. (X, d). F X.. (1), X F, (2) F 1, F 2 F F 1 F 2 F, (3) F λ F λ F λ F. 3., A λ λ A λ. B λ λ

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

2017 1 2 1.1...................................... 2 1.2......................................... 4 1.3........................................... 10 1.4................................. 14 1.5..........................................

chap10.dvi

. q {y j } I( ( L y j =Δy j = u j = C l ε j l = C(L ε j, {ε j } i.i.d.(,i q ( l= y O p ( {u j } q {C l } A l C l

変 位 変位とは 物体中のある点が変形後に 別の点に異動したときの位置の変化で あり ベクトル量である 変位には 物体の変形の他に剛体運動 剛体変位 が含まれている 剛体変位 P(x, y, z) 平行移動と回転 P! (x + u, y + v, z + w) Q(x + d x, y + dy,

変 位 変位とは 物体中のある点が変形後に 別の点に異動したときの位置の変化で あり ベクトル量である 変位には 物体の変形の他に剛体運動 剛体変位 が含まれている 剛体変位 P(x, y, z) 平行移動と回転 P! (x + u, y + v, z + w) Q(x + d x, y + dy, z + dz) Q! (x + d x + u + du, y + dy + v + dv, z +

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

C : q i (t) C : q i (t) q i (t) q i(t) q i(t) q i (t)+δq i (t) (2) δq i (t) δq i (t) C, C δq i (t 0 )0, δq i (t 1 ) 0 (3) δs S[C ] S[C] t1 t 0 t1 t 0

1 2003 4 24 ( ) 1 1.1 q i (i 1,,N) N [ ] t t 0 q i (t 0 )q 0 i t 1 q i (t 1 )q 1 i t 0 t t 1 t t 0 q 0 i t 1 q 1 i S[q(t)] t1 t 0 L(q(t), q(t),t)dt (1) S[q(t)] L(q(t), q(t),t) q 1.,q N q 1,, q N t C :

42 3 u = (37) MeV/c 2 (3.4) [1] u amu m p m n [1] m H [2] m p = (4) MeV/c 2 = (13) u m n = (4) MeV/c 2 =

3 3.1 3.1.1 kg m s J = kg m 2 s 2 MeV MeV [1] 1MeV=1 6 ev = 1.62 176 462 (63) 1 13 J (3.1) [1] 1MeV/c 2 =1.782 661 731 (7) 1 3 kg (3.2) c =1 MeV (atomic mass unit) 12 C u = 1 12 M(12 C) (3.3) 41 42 3 u

X G P G (X) G BG [X, BG] S 2 2 2 S 2 2 S 2 = { (x 1, x 2, x 3 ) R 3 x 2 1 + x 2 2 + x 2 3 = 1 } R 3 S 2 S 2 v x S 2 x x v(x) T x S 2 T x S 2 S 2 x T x S 2 = { ξ R 3 x ξ } R 3 T x S 2 S 2 x x T x S 2

QMII_10.dvi

65 1 1.1 1.1.1 1.1 H H () = E (), (1.1) H ν () = E ν () ν (). (1.) () () = δ, (1.3) μ () ν () = δ(μ ν). (1.4) E E ν () E () H 1.1: H α(t) = c (t) () + dνc ν (t) ν (), (1.5) H () () + dν ν () ν () = 1 (1.6)

77

O r r r, F F r,r r = r r F = F (. ) r = r r 76 77 d r = F d r = F (. ) F + F = 0 d ( ) r + r = 0 (. 3) M = + MR = r + r (. 4) P G P MX = + MY = + MZ = z + z PG / PG = / M d R = 0 (. 5) 78 79 d r = F d

2 G(k) e ikx = (ik) n x n n! n=0 (k ) ( ) X n = ( i) n n k n G(k) k=0 F (k) ln G(k) = ln e ikx n κ n F (k) = F (k) (ik) n n= n! κ n κ n = ( i) n n k n

. X {x, x 2, x 3,... x n } X X {, 2, 3, 4, 5, 6} X x i P i. 0 P i 2. n P i = 3. P (i ω) = i ω P i P 3 {x, x 2, x 3,... x n } ω P i = 6 X f(x) f(x) X n n f(x i )P i n x n i P i X n 2 G(k) e ikx = (ik) n

I ( ) 2019

I ( ) 2019 i 1 I,, III,, 1,,,, III,,,, (1 ) (,,, ), :...,, : NHK... NHK, (YouTube ),!!, manaba http://pen.envr.tsukuba.ac.jp/lec/physics/,, Richard Feynman Lectures on Physics Addison-Wesley,,,, x χ,

1 (Contents) (1) Beginning of the Universe, Dark Energy and Dark Matter Noboru NAKANISHI 2 2. Problem of Heat Exchanger (1) Kenji

8 4 2018 6 2018 6 7 1 (Contents) 1. 2 2. (1) 22 3. 31 1. Beginning of the Universe, Dark Energy and Dark Matter Noboru NAKANISHI 2 2. Problem of Heat Exchanger (1) Kenji SETO 22 3. Editorial Comments Tadashi

all.dvi

72 9 Hooke,,,. Hooke. 9.1 Hooke 1 Hooke. 1, 1 Hooke. σ, ε, Young. σ ε (9.1), Young. τ γ G τ Gγ (9.2) X 1, X 2. Poisson, Poisson ν. ν ε 22 (9.) ε 11 F F X 2 X 1 9.1: Poisson 9.1. Hooke 7 Young Poisson G

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

構造と連続体の力学基礎

II 37 Wabash Avenue Bridge, Illinois 州 Winnipeg にある歩道橋 Esplanade Riel 橋6 6 斜張橋である必要は多分無いと思われる すぐ横に道路用桁橋有り しかも塔基部のレストランは 8 年には営業していなかった 9 9. 9.. () 97 [3] [5] k 9. m w(t) f (t) = f (t) + mg k w(t) Newton

9. 05 L x P(x) P(0) P(x) u(x) u(x) (0 < = x < = L) P(x) E(x) A(x) P(L) f ( d EA du ) = 0 (9.) dx dx u(0) = 0 (9.2) E(L)A(L) du (L) = f (9.3) dx (9.) P

9 (Finite Element Method; FEM) 9. 9. P(0) P(x) u(x) (a) P(L) f P(0) P(x) (b) 9. P(L) 9. 05 L x P(x) P(0) P(x) u(x) u(x) (0 < = x < = L) P(x) E(x) A(x) P(L) f ( d EA du ) = 0 (9.) dx dx u(0) = 0 (9.2) E(L)A(L)

H.Haken Synergetics 2nd (1978)

27 3 27 ) Ising Landau Synergetics Fokker-Planck F-P Landau F-P Gizburg-Landau G-L G-L Bénard/ Hopfield H.Haken Synergetics 2nd (1978) (1) Ising m T T C 1: m h Hamiltonian H = J ij S i S j h i S

y = x x R = 0. 9, R = σ $ = y x w = x y x x w = x y α ε = + β + x x x y α ε = + β + γ x + x x x x' = / x y' = y/ x y' =

y x = α + β + ε =,, ε V( ε) = E( ε ) = σ α $ $ β w ( 0) σ = w σ σ y α x ε = + β + w w w w ε / w ( w y x α β ) = α$ $ W = yw βwxw $β = W ( W) ( W)( W) w x x w x x y y = = x W y W x y x y xw = y W = w w

D = [a, b] [c, d] D ij P ij (ξ ij, η ij ) f S(f,, {P ij }) S(f,, {P ij }) = = k m i=1 j=1 m n f(ξ ij, η ij )(x i x i 1 )(y j y j 1 ) = i=1 j

6 6.. [, b] [, d] ij P ij ξ ij, η ij f Sf,, {P ij } Sf,, {P ij } k m i j m fξ ij, η ij i i j j i j i m i j k i i j j m i i j j k i i j j kb d {P ij } lim Sf,, {P ij} kb d f, k [, b] [, d] f, d kb d 6..

( ) ) AGD 2) 7) 1

( 9 5 6 ) ) AGD ) 7) S. ψ (r, t) ψ(r, t) (r, t) Ĥ ψ(r, t) = e iĥt/ħ ψ(r, )e iĥt/ħ ˆn(r, t) = ψ (r, t)ψ(r, t) () : ψ(r, t)ψ (r, t) ψ (r, t)ψ(r, t) = δ(r r ) () ψ(r, t)ψ(r, t) ψ(r, t)ψ(r, t) = (3) ψ (r,

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

3 3 5 5 5 3 3 7 5 33 5 33 9 5 8 > e > f U f U u u > u ue u e u ue u ue u e u e u u e u u e u N cos s s cos ψ e e e e 3 3 e e 3 e 3 e 3 > A A > A E A f A A f A [ ] f A A e > > A e[ ] > f A E A < < f ; >

09kyoto

F [ ] F [ ] Nambu( 60), Goldstone(61), Nambu Jona-Lasinio( 61), Goldstone, Salam, Weinberg( 62). N NG = N BS NG : CC by-sa Aney CC by Zouavman Le Zouave CC by-sa Roger McLassus U(1) CC by-sa Aney -Goldstone

all.dvi

29 4 Green-Lagrange,,.,,,,,,.,,,,,,,,,, E, σ, ε σ = Eε,,.. 4.1? l, l 1 (l 1 l) ε ε = l 1 l l (4.1) F l l 1 F 30 4 Green-Lagrange Δz Δδ γ = Δδ (4.2) Δz π/2 φ γ = π 2 φ (4.3) γ tan γ γ,sin γ γ ( π ) γ tan

2.1: n = N/V ( ) k F = ( 3π 2 N ) 1/3 = ( 3π 2 n ) 1/3 V (2.5) [ ] a = h2 2m k2 F h2 2ma (1 27 ) (1 8 ) erg, (2.6) /k B 1 11 / K

2 2.1? [ ] L 1 ε(p) = 1 ( p 2 2m x + p 2 y + pz) 2 = h2 ( k 2 2m x + ky 2 + kz) 2 n x, n y, n z (2.1) (2.2) p = hk = h 2π L (n x, n y, n z ) (2.3) n k p 1 i (ε i ε i+1 )1 1 g = 2S + 1 2 1/2 g = 2 ( p F

,. Black-Scholes u t t, x c u 0 t, x x u t t, x c u t, x x u t t, x + σ x u t, x + rx ut, x rux, t 0 x x,,.,. Step 3, 7,,, Step 6., Step 4,. Step 5,,.

9 α ν β Ξ ξ Γ γ o δ Π π ε ρ ζ Σ σ η τ Θ θ Υ υ ι Φ φ κ χ Λ λ Ψ ψ µ Ω ω Def, Prop, Th, Lem, Note, Remark, Ex,, Proof, R, N, Q, C [a, b {x R : a x b} : a, b {x R : a < x < b} : [a, b {x R : a x < b} : a,

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

(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

Compton Scattering Beaming exp [i k x ωt] k λ k π/λ ω πν 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 O O v k α k α β, γ k γ k βk, k γ k + βk k γ k k, k γ k + βk 3 k k 4 k 3 k

v v = v 1 v 2 v 3 (1) R = (R ij ) (2) R (R 1 ) ij = R ji (3) 3 R ij R ik = δ jk (4) i=1 δ ij Kronecker δ ij = { 1 (i = j) 0 (i

1. 1 1.1 1.1.1 1.1.1.1 v v = v 1 v 2 v 3 (1) R = (R ij ) (2) R (R 1 ) ij = R ji (3) R ij R ik = δ jk (4) δ ij Kronecker δ ij = { 1 (i = j) 0 (i j) (5) 1 1.1. v1.1 2011/04/10 1. 1 2 v i = R ij v j (6) [

211 kotaro@math.titech.ac.jp 1 R *1 n n R n *2 R n = {(x 1,..., x n ) x 1,..., x n R}. R R 2 R 3 R n R n R n D D R n *3 ) (x 1,..., x n ) f(x 1,..., x n ) f D *4 n 2 n = 1 ( ) 1 f D R n f : D R 1.1. (x,

講義ノート 物性研究 電子版 Vol.3 No.1, (2013 年 T c µ T c Kammerlingh Onnes 77K ρ 5.8µΩcm 4.2K ρ 10 4 µωcm σ 77K ρ 4.2K σ σ = ne 2 τ/m τ 77K

2 2 T c µ T c 1 1.1 1911 Kammerlingh Onnes 77K ρ 5.8µΩcm 4.2K ρ 1 4 µωcm σ 77K ρ 4.2K σ σ = ne 2 τ/m τ 77K τ 4.2K σ 58 213 email:takada@issp.u-tokyo.ac.jp 1933 Meissner Ochsenfeld λ = 1 5 cm B = χ B =

医系の統計入門第 2 版 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, 第 2 版 1 刷発行時のものです.

医系の統計入門第 2 版 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. http://www.morikita.co.jp/books/mid/009192 このサンプルページの内容は, 第 2 版 1 刷発行時のものです. i 2 t 1. 2. 3 2 3. 6 4. 7 5. n 2 ν 6. 2 7. 2003 ii 2 2013 10 iii 1987