Euler Appendix cos, sin 2π t = 0 kx = 0, 2π x = 0 (wavelength)λ kλ = 2π, k = 2π/λ k (wavenumber) x = 0 ωt = 0, 2π t = 0 (period)t T = 2π/ω ω = 2πν (fr

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This manuscript is modified on March 26, 2012 3 : 53 pm [1] 1 ( ) Figure 1: (longitudinal wave)

2) ( ) δt t u(x, δt) = f(x vδt) (1) u(x, δt) = f(x + vδt) (2) u(x, t) = f(x vt), propagate + x direction

1 This manuscript is modified on March 26, : 53 pm [1] 1 ( ) Figure 1: (longitudinal wave) (transverse wave). P 7km S 4km P S P S x t x u(x, t) t = t 0 = 0 f(x) f(x) = u(x, 0) v +x (Fig.2) ( ) δt t u(x, δt) = f(x vδt) (1) u(x, δt) = f(x + vδt) (2) u(x, t) = f(x vt), propagate + x direction (3) x v v u(x, t) = f(x + vt), propagate x direction (4) (Fig.3) Fig.2,3 f Fig.4 Ae i(kx ωt) = A cos(kx ωt) + ia sin(kx ωt) (5) 1

2 Euler Appendix cos, sin 2π t = 0 kx = 0, 2π x = 0 (wavelength)λ kλ = 2π, k = 2π/λ k (wavenumber) x = 0 ωt = 0, 2π t = 0 (period)t T = 2π/ω ω = 2πν (frequency)ν ν = 1/T ω (angular frequency) exp(iθ) θ kx ωt phase cos kx ωt = 0 x = (ω/k)t +x phase velocity v = ω/k x Ae i(kx+ωt) (Fig. 5, 6, 7) t 0 + δt vδt t = t 0 vδt t 0 δt x Figure 2: +x v t 0 + δt vδt t = t 0 vδt t 0 δt x Figure 3: x v 2 u x t X f f(x) X x vt, f(x vt) = f(x) (6) x = X df x dx = df dx (7) x 2 = df x dx = X df x X dx = d2 f dx 2 (8) = X df df = v t t dx dx (9) t 2 = v df X df = v t dx t X dx = v2 d2 f dx 2 (10) t 2 = v 2 2 u x 2 (11) 2

3 !" )./01$/2/-%-&-&34561*/3 ))7)') """"" ))7)') """" ))7)') """ "!# "!" $"!# &'#()*' ()+*,' -&- """" $%" $ " " " %" x Figure 4: f(x) = A exp( x 2 /2/ 2 ) exp(ikx) Gaussian wavepacket f(x) = A 2π dk exp[ 2 2 (k k ) 2 + ik x] [ ] Ref(x) = A exp( x 2 /2/ 2 ) cos(kx) = A + 2π dk exp 2 2 (k k ) 2 cos(k x) Ref(x) = A [ ] 2π k (δk ) exp 2 2 k 2 cos[(k k )x], k = k k = ndk, (n = N, N + 1,..., 2, 1, 0, 1, 2,..., N 1, N) cos[(k k )x] 2π exp [ 2 2 k 2 ] 2π π π kx 2π Figure 5: cos(kx) δ > 0 cos(kx δ), cos(kx+δ) kλ = 2π +x cos(kx δ), x cos(kx + δ) 3

4 Figure 6: 波の位相 ei(kx+ωt) Figure 7: +x 方向への進行波の頂点に乗ってサーフィンする頂点は波の位相がゼロであるこの条件から位相速度 v = ω/k が得られる波の最高到達地点の間隔は波長 λ である cos(ωt) ωt = 0, ±2π,... T = 2π/ω T = 2π/ω = 1/ν v = ω/k = 2πν/(2π/λ) = λν c = ω/k = λν Figure 8: ある場所における波の振動周期振動数角振動数 4

5 f(x + vt) f 1 (x vt) + f 2 (x + vt) ( f(x + vt) f 1 (x vt) + f 2 (x + vt) ) (11) (wave equation) 2 (11) f(x vt), f(x + vt), f 1 (x vt) + f 2 (x + vt) / t 2 / x 2 u X, Y X x vt, Y x + vt (12) x = X + Y, t = X + Y 2 2v (13) = X t t X + Y = v t Y X + v Y (14) = v t t t X + v t Y = v( X t X + Y t Y ) X + v( X t X + Y t Y ) Y (15) = v 2 2 u X 2 2v2 X Y + v2 2 u Y 2 (16) x = X x X + Y x Y = X + Y (17) x x = x X + x Y = ( X x X + Y x Y ) X + ( X x X + Y x Y ) Y (18) = 2 u X u X Y + 2 u Y 2 (19) X Y = 0 (20) u = f 1 (X) + f 2 (Y ) u(x, t) = f 1 (x vt) + f 2 (x + vt) Y X = f 1(X) (21) f 1(X) Y X X u = dxf 1(X) + f 2 (Y ) = f 1 (X) + f 2 (Y ) (22) f 2 (Y ) X Y u(x, t) u(x, t) = f 1 (x vt) + f 2 (x + vt) u (plane wave) 2π/λ k cos(kx ωt) ±x ±k r k r = 0 (23) k x x + k y y + k z z = 0 (24) u(r, t) = A exp[i(k r ωt)] (25) 5

6 Figure 9: X u = f(x) X = k x x + k y y + k z z ωt (26) = X f f = ω t t X X (27) = ω X f t t t X X = ω2 2 f X 2 (28) x = X f x X = k f x X (29) x x = k X f x x X X = 2 f k2 x X 2 (30) y 2 = ky 2 2 f X 2, z 2 = 2 k2 z X 2 (31) 2 u = k 2 2 f X 2, 2 = ( 2 / x 2 ) + ( 2 / y 2 ) + ( 2 / z 2 ) (32) kx 2 + ky 2 + kz 2 = k 2 (33) 2 f X 2 = 1 ω 2 t 2 = 1 k 2 2 u (34) ( ω ) 2 t 2 = 2 = v 2 2 u (35) k 3 v = ω/k X Laplacian 2 qu = 2 r u θ φ ru(r, t) = U(r, t) 2 qu = 2 r 1 2 U r t 2 = v 2 ( 2 r r + 2 u r r 2 θ 2 + cos θ r 2 sin θ θ + 1 r 2 sin 2 θ φ 2 (36) r + 2 u r r 2 (37) U r + 2 U r 2 r ) = 1 2 U v2 r r 2, 2 U t 2 = v2 2 U r 2 (38) 6

7 U(r, t) = f 1 (r vt) + f 2 (r + vt) u(r, t) = U(r, t) r = f 1(r vt) + f 2 (r + vt) r (39) (spherical wave) OUT IN Figure 10: IN f 2 (r + vt) OUT f 1 (r vt) 4 Appendix e iθ = exp(iθ) = cos θ + i sin θ (40) Taylor If the function f(x) can be approximated by the polynominals around x = a f(x) a 0 + a 1 (x a) + a 2 (x a) 2 + a 3 (x a) 3 + a 4 (x a) a n (x a) n (41) If x a << 1, the above equations can be converged. We can get the 1st to n-th derivative such as Then we have f (x) = a 1 + 2a 2 (x a) + 3a 3 (x a) 2 + 4a 4 (x a) na n (x a) n 1 (42) f (x) = 2a a 3 (x a) + 4 3a 4 (x a) n(n 1)a n (x a) n 2 (43) f (x) = 3 2a a 4 (x a) n(n 1)(n 1)a n (x a) n 3 (44) f (x) = 4 3 2a n(n 1)(n 2)(n 3)a n (x a) n 4 (45) f (n) (x) = n!a n a 1 = f (a), a 2 = 1 2! f (a), a 3 = 1 3! f (a), a 4 = 1 4! f (a), a n = 1 n! f (n) (a) (47) f(x) f(a) + f (a)(x a) + 1 2! f (a)(x a) ! f (a)(x a) n! f (n) (a)(x a) n (48) This is the Taylor expansion around x = a. (46) 7

8 e x = 1 + x + 1 2! x ! x n! xn +... (49) e ix = 1 + ix 1 2! x2 1 3! ix ! x ! ix5... (50) cos(x) = 1 1 2! x ! x4..., sin(x) = x 1 3! x ! x5..., then we have e ix = cos x + i sin x (51) e ix x << 1 x = x 0 x 0 = x 1 + dx, dx << 1 e ix 0 = e i(x 1+dx) = e ix 1 e idx = e ix 1 [1 + idx 1 2! (dx)2...] = e ix 1 [cos(dx) + i sin(dx)] x 1 = x 2 + dx, x 2 = x 3 + dx,..., x n 1 = x n + dx, 0 = x n = x 0 ndx e ix 1 = e ix 2 [cos(dx) + i sin(dx)], e ix 0 = e ix 2 [cos(dx) + i sin(dx)] 2... e ixn 1 = e ixn [cos(dx) + i sin(dx)], e ix0 = e ixn [cos(dx) + i sin(dx)] n = [cos(dx) + i sin(dx)] n [cos(dx) + i sin(dx)] 2 = cos 2 (dx) sin 2 (dx) + 2i cos(dx) sin(dx) = cos(2dx) + i sin(2dx) [cos(dx) + i sin(dx)] 3 = [cos(2dx) + i sin(2dx)][cos(dx) + i sin(dx)] = cos(2dx) cos(dx) sin(2dx) sin(dx) + i[sin(2dx) cos(dx) + sin(dx) cos(2dx)] = cos(3dx) + i sin(3dx) (53)... [cos(dx) + i sin(dx)] n = cos(ndx) + i sin(ndx) = cos(x 0 ) + i sin(x 0 ) e ix0 = cos(x 0 ) + i sin(x 0 ) (54) x e ix = cos x + i sin x i = -1 虚数軸 r θ z = r e iθ 複素平面実数軸 (52) : 長さ r で実軸となす角 θ z = r e iθ = r (cosθ + i sinθ) z = Re(z) + i Im(z), Re(z) = r cosθ, Im(z) = r sinθ Figure 11: z z = re iθ ( r θ ), Rez = r cos θ Imz = r sin θ 8

9 e i(kx ωt) t = 0 x = x x = 0 t = + t x = + x 2π π π kx 2π Figure 12: +x e i(kx ωt). x = 0( ) t = 0, e 0 = 1 t = + t e iω t = cos(ω t) i sin(ω t) 1 ω +x e i(kx+ωt) t = + t t = 0 x = x x = 0 x = + x 2π π π kx 2π Figure 13: x e i(kx+ωt) x = 0( ) t = 0, e 0 = 1 t = + t e iω t = cos(ω t) + i sin(ω t) 1 ω x 9

10 References [1], ( (8)) 1986, 10

QMI_09.dvi

QMI_09.dvi 25 3 19 Erwin Schrödinger 1925 3.1 3.1.1 3.1.2 σ τ 2 2 ux, t) = ux, t) 3.1) 2 x2 ux, t) σ τ 2 u/ 2 m p E E = p2 3.2) E ν ω E = hν = hω. 3.3) k p k = p h. 3.4) 26 3 hω = E = p2 = h2 k 2 ψkx ωt) ψ 3.5) h