高知工科大学電子光システム工学科

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えのしんまつ
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1 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

2 高知工科大学電子光システム工学科

4 1. 3

5 .1. m v p = mv h E = hν, p = (-1) λ (3-1)- -34 E ν h J s h λ = p.3 u x φ = Asin πν t (-) u A x t x 4

6 φ 4π ν x = Asin πν t x u u 4πν = φ u -3 t φ x = 4πν Asinπν t t u = 4 πνφ -4 (-3)(-4) φ 1 φ = (-5) x u t (-) x t ψ x t φ( x, t) = ψ( x) T( t) (-6) x t φ () d ψ = Tt x dx φ dt = ψ (-7) t dt -5 1 () d ψ Tt = ψ d T dx u dt -8 d 1 d T ψ = ψ ( x) dx u T ( t) dt -9 k d ψ 1 d T = = k ψ ( x) dx u T( t) dt -10 d ψ = k ψ ( x) (-11) dx 5

7 dt kutt () = (-1) dt k z z = k (-13) + k = 0 (-14) (-10) z =± ik ( i = 1) (-15) ψ ( x) = Aexp( ikx) + Bexp( ikx) (-17) (-11) -6 Tt ( ) = Cexp( ikut) + Dexp( ikut) (-18) φ( xt, ) = ψ( xtt ) ( ) = A exp ik x ut + A ik x ut { ( )} exp ( ) ( ) exp { } { } { ( )} 1 + A exp ik x+ ut + A ik x+ ut 3 4 A 1 A 4 (-14) (-18) πν π k = = (-19) u λ kx ( ut) x kx ( ut) = πν t u x = π νt λ λ ν (-10) k π = λ (-0) 6

8 d ψ 4π = ( x) ψ (-1) dx λ d ψ 4π + ( x) 0 ψ = (-) dx λ.4 h p λ p = λ E T V p E = T + V = + V m (-3) p = mv p me V = ( ) (-4) 1 p m ( E V ) λ = h = h (-5) (-18) d ψ 8π m + ( E V ) ψ ( x ) = 0 (-6) dx h ψ 1 3 ψ ψ 8π m + + ( ) (, ) 0 E V ψ x y = (-7) x y h ψ ψ ψ 8π m ( ) (,, ) 0 E V ψ x y z = 3 (-8) x y z h 7

9 8π m E h ψ + ( V ) ψ = 0 (-9).5 r t ψ (,) rt t r dv ψ (r,t ) dv (-30) ψ r ψ = 0 ψ ψ ψ (,) rt dr= 1 (-31) 1 ψ 8

10 ψ ψ ψ ψ ψ ψ (-31) ψ 1 ( x, yz, ) q t ψ q q+ dq ψ ( qt, ) dq (-3) F ˆf F f ˆf ˆf ψ = fψ (-33) fˆ( Ψ+Φ ) = fˆψ+ fˆφ (-34) ˆf 9

11 Ψ1, Ψ,, Ψn Ψ Ψ Ψ= c1ψ+ 1 cψ + + cnψn n (-35) = c Ψ i= 1 i i n n i i i i i= 1 i= 1 (-36) fˆψ= fˆ( cψ ) = c ( fˆψ) (-5) x φ( x, t) = Aexpπi νt (-37) λ -1 (, ) exp i φ x t = A ( px Et) (-38) h = π (-38) x t x φ ip i = exp ( px Et) x 10

12 ip = φ (-39) t φ ie i = Aexp ( px Et) t ie = φ (-40) (-38),(-39) φ i = pφ x (-41) φ i = Eφ (-4) t p i E i x t i p i E x t E.7 1 T = mv p = mv p p v = (-43) m T p T = (-44) m p i x p T = m = + + m x y z 11

13 = m (-45) p 3 T m T V Ĥ Ĥ = T + V (-46) T (-45) ˆ H = + V (-47) m Ĥ (-4) Ĥ Ĥφ = Eφ (-48) ψ Ĥψ = Eψ (-49) (-4)(-49) ˆ ψ Hφ = + Vφ = i (-50) m t.8.9 1

14 , Hˆ ψ(1, ) = Eψ(1, ) (3-1)

15 3.3.1 Ĥ Hˆ m = + V ˆ H = + + M m ( A B) ( 1 ) e e 4πε r 4πε r 01A 01B e e 4πε r 4πε r 0 A 0 B e e + + 4πε r 4πε R 01 0 e (3-) (3-3) M m e A B F QQ 1 F = (3-4) 4πε0r Q 1, Q r ε 0 1 ε 0 =

16 kg kg 1840 (3-3) R (3-3) ˆ H = + m e ( 1 ) e e 4πε r 4πε r 01A 01B e e 4πε r 4πε r 0 A 0 B e + + 4πε r 01 (3-5) e 4πε r

17 3.6 1 φ(1) φ() 1 r1 V( r) = e dv (3-6) φ(1) r1 V( r ) = e dv (3-7) V( r 1 ) V( r ) r 1, r Ĥ 1 Hˆ = hˆ + hˆ 1 ˆ e e h ( ) 1 = 1 + V r1 me 4πε0ra1 4πε0rb 1 ˆ e e h ( ) = + V r me 4πε0ra 4πε0rb (3-8) E φ(1) φ() Hˆ φ(1) φ() dv dv = φ φ (1) () dv dv 1 1 (3-9) Ĥ = h ˆ ˆ 1+ h Ĥ 16

18 E = { ˆ + ˆ } φ(1) φ() h h (1) φ() dv dv = 1 1 φ(1) φ() dv dv 1 φ(1) hˆφ(1) dv φ() dv + φ(1) dv φ() hˆφ() dv φ(1) φ() dv1dv (1) ˆ (1) () (1) () ˆ φ hφ dvφ dv φ dvφ hφ() dv = + φ(1) dv φ() dv φ(1) dv φ() dv (3-10) (3-11) ( 3-1) (1) ˆ (1) () ˆ φ hφ dv φ hφ() dv = + φ(1) dv φ() dv (3-13) = ε + ε (3-14) Ĥ ˆ hφ (1) = ε φ (1) (3-15) 1 1 h ˆ φ () = ε φ () (3-16) (3-14) ˆ φ( jh ) jφ( jdv ) j ε j =, j = 1, (3-17) φ( j) dv j

19 Ψ1, Ψ,, Ψn Ψ Ψ= cψ+ c Ψ + + c Ψ 1 1 n = c Ψ i= 1 i i n n (3-18) Ψ 3.8 ψ(1, ) = Cφ(1) + C φ() (3-19) 1 φ(1), φ() C 1, C 3.9 hφ = εφ (3-0) (3-17) ε φĥφdv = φ dv (3-1) ε (3-19) 18

20 = ˆ{ } { Cφ(1) + C φ() } dv Cφ(1) + C φ() h ( Cφ(1) + C φ() dv (3-) Ch + CCh + Ch = CS CS CS (3-3) h = φhˆ φdv h = φ hˆ φ dv h = φhˆ φ dv 1 1 S = S = φφ dv 1 1 (3-4) h11 h h 1 S 1 S (3-3) C ( h εs ) + CC ( h εs ) + C ( h εs ) = 0 (3-5) ε ε = = 0 (3-6) c c 1 C ( h S ε) + C ( h S ε) = 0 (3-7)

21 C ( h S ε) + C ( h S ε) = 0 (3-8) h εs h εs h εs h εs 1 1 = 0 (3-9) (3-9) ε ε ε = (3-30) ( h11 S11)( h S) ( h1 S1) 0 h h S S = h = α = β = S = 1 = S (3-31) (3-30) α ε β εs = (3-3) ( ) ( ) 0 αε± β ε (3-33) ε ε 1, ε α + β ε1 = (3-34) 1 + S α β ε = (3-35) 1 S α, β ε > ε 1 (3-34)(3-7) C1 = C(3-36) (3-35)(3-8) C1 = C(3-37) (-31) ψ (,) rt dr= 1 (3-38) dr = dτ (3-39) 0

22 (3-19) ψ (,) rt dτ = 1 (3-40) { } { } ψ ψ d τ = C φ 1 (1) + C φ () C φ 1 (1) + C φ () d τ = 1 (3-41) { } C φ 1 (1) + C φ () d τ = 1 (3-4) (3-43) ( C ) χ dτ + CC χ χ dτ + ( C ) χ dτ = φ(1) = χ1, φ() = χ 1 1 χ d τ = χ d τ = 1, χ χ d τ = S ( C1) + CC 1 S+ ( C) = 1 (3-44) (3-36)(3-44) C = C = 1 ε 1 ψ 1 χ1+ χ ψ1 = + S C 1 + S (3-45) (3-46) (3-37)(3-46) 1 = 1 S (3-47) 1 C = S (3-48) ε ψ χ1 χ ψ = S (3-49)

23 α + β ε1 = 1 + S α β ε = 1 S χ1+ χ ψ1 = + S χ1 χ ψ = S ψ 1 ψ p

25 R 4

26 4..1 { a( R R ) } V(R)=D 1 exp e (4-1) D A R R D 4.35 ev A 1.94 R =0.74 e e 4.3 5

27 F F = kr ( R e ) (4-) R e R k R e R V R R k V = FdR= k( R Re) dr ( R Re) R = (4-3) e Re ab, M a, M b a dr a a kr ( Re) = M (4-4) dt b dr b b kr ( Re) = M (4-5) dt R, R a b Ra + Rb = R, MaRa = MbRb R a R b M b Ra = R M + M a b (4-6) R b M a = M + M a b R (4-7) (5-6)(5-7)(5-4)(5-5) MM a b µ = M + M a b MM dr dr a b kr ( Re ) = =µ M + M dt dt a b (4-8) µ 6

28 Hˆ 1 V = µ + (4-9) (4-9) Hˆ 1 V = µ + (4-10) (5-1) { a( R R ) } V(R)=D 1 exp e (4-11) x = 0 e 3 n x x x x = 1+ x (4-1)! 3! n! x = ar ( R e ) V(R)=D 1 { exp a( R Re ) } D{ 1 1 ( a( R Re )) } + = D a ( R R ) e (4-13) 7

29 ev 1 ev ev

30 4.4.1 核間距離モース曲線二次曲線誤差核間距離モース曲線二次曲線誤差 E

31 E E E-06.4E E E (4-1) (4-13) (4-1)(4-13) D 4.35 ev A 1.94 R e =0.74 ev, ev ev

32 { } a( R R ) V(R)=D 1 exp e MM a b µ = M + M a b

33 6. M1 3

34 Donald A.McQuarrie John D.Simon 33

卒業研究報告題目 Hamiltonian 指導教員山本哲也教授報告者汐月康則平成 14 年 2 月 5 日 1

卒業研究報告題目 Hamiltonian 指導教員山本哲也教授報告者汐月康則平成 14 年 2 月 5 日 1 卒業研究報告題目 Hamiltonian 指導教員山本哲也教授報告者汐月康則平成 4 年月 5 日 .....4.....4......6.. 6.. 6....4. 8.5. 9.6....7... 3..... 3.... 3.... 3.3...4 3.4...5 3.5...5 3.5....6 3.5.... 3.5...... 3.5...... 3 3.5.3..4 3.5.4..5

高知工科大学電子 光システム工学科

高知工科大学電子光システム工学科