4/15 No.

Similar documents
02-量子力学の復習

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

QMI_09.dvi

QMI_10.dvi

QMI_10.dvi

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 (

p = mv p x > h/4π λ = h p m v Ψ 2 Ψ

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

数学の基礎訓練I

I

untitled

輻射の量子論、選択則、禁制線、許容線


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

* 1 1 (i) (ii) Brückner-Hartree-Fock (iii) (HF, BCS, HFB) (iv) (TDHF,TDHFB) (RPA) (QRPA) (v) (vi) *

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.

1 filename=mathformula tex 1 ax 2 + bx + c = 0, x = b ± b 2 4ac, (1.1) 2a x 1 + x 2 = b a, x 1x 2 = c a, (1.2) ax 2 + 2b x + c = 0, x = b ± b 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 (fr

Gmech08.dvi

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

SiC SiC QMAS(Quantum MAterials Simulator) VASP(Vienna Ab-initio Simulation Package) SiC 3C, 4H, 6H-SiC EV VASP VASP 3C, 4H, 6H-SiC (0001) (11 20) (1 1

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

) ] [ 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

A = A x x + A y y + A, B = B x x + B y y + B, C = C x x + C y y + C..6 x y A B C = A x x + A y y + A B x B y B C x C y C { B = A x x + A y y + A y B B

Xray.dvi

Radiation from moving charges#1 Liénard-Wiechert potential Yuji Chinone 1 Maxwell Maxwell MKS E (x, t) + B (x, t) t = 0 (1) B (x, t) = 0 (2) B (x, t)


鉄鋼協会プレゼン

, 1.,,,.,., (Lin, 1955).,.,.,.,. f, 2,. main.tex 2011/08/13( )


note4.dvi

19 σ = P/A o σ B Maximum tensile strength σ % 0.2% proof stress σ EL Elastic limit Work hardening coefficient failure necking σ PL Proportional


量子力学 問題

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

Onsager SOLUTION OF THE EIGENWERT PROBLEM (O-29) V = e H A e H B λ max Z 2 Onsager (O-77) (O-82) (O-83) Kramers-Wannier 1 1 Ons

ssp2_fixed.dvi

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

kawa (Spin-Orbit Tomography: Kawahara and Fujii 21,Kawahara and Fujii 211,Fujii & Kawahara submitted) 2 van Cittert-Zernike Appendix A V 2

1 1.1,,,.. (, ),..,. (Fig. 1.1). Macro theory (e.g. Continuum mechanics) Consideration under the simple concept (e.g. ionic radius, bond valence) Stru


.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

液晶の物理1:連続体理論(弾性,粘性)

SFGÇÃÉXÉyÉNÉgÉãå`.pdf

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

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

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

positron 1930 Dirac 1933 Anderson m 22Na(hl=2.6years), 58Co(hl=71days), 64Cu(hl=12hour) 68Ge(hl=288days) MeV : thermalization m psec 100

A 99% MS-Free Presentation

i

( ; ) C. H. Scholz, The Mechanics of Earthquakes and Faulting : - ( ) σ = σ t sin 2π(r a) λ dσ d(r a) =

PDF

2012専門分科会_new_4.pptx

08-Note2-web

. 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

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

PowerPoint Presentation

ii p ϕ x, t = C ϕ xe i ħ E t +C ϕ xe i ħ E t ψ x,t ψ x,t p79 やは時間変化しないことに注意 振動 粒子はだいたい このあたりにいる 粒子はだいたい このあたりにいる p35 D.3 Aψ Cϕdx = aψ ψ C Aϕ dx


E 1/2 3/ () +3/2 +3/ () +1/2 +1/ / E [1] B (3.2) F E 4.1 y x E = (E x,, ) j y 4.1 E int = (, E y, ) j y = (Hall ef

°ÌÁê¿ô³ØII

Microsoft Word - 信号処理3.doc

²ÄÀÑʬΥ»¶ÈóÀþ·¿¥·¥å¥ì¡¼¥Ç¥£¥ó¥¬¡¼ÊýÄø¼°¤ÎÁ²¶á²òÀÏ Asymptotic analysis for the integrable discrete nonlinear Schrödinger equation

(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

II No.01 [n/2] [1]H n (x) H n (x) = ( 1) r n! r!(n 2r)! (2x)n 2r. r=0 [2]H n (x) n,, H n ( x) = ( 1) n H n (x). [3] H n (x) = ( 1) n dn x2 e dx n e x2

SO(3) 49 u = Ru (6.9), i u iv i = i u iv i (C ) π π : G Hom(V, V ) : g D(g). π : R 3 V : i 1. : u u = u 1 u 2 u 3 (6.10) 6.2 i R α (1) = 0 cos α

LCR e ix LC AM m k x m x x > 0 x < 0 F x > 0 x < 0 F = k x (k > 0) k x = x(t)

85 4

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

EGunGPU

q quark L left-handed lepton. λ Gell-Mann SU(3), a = 8 σ Pauli, i =, 2, 3 U() T a T i 2 Ỹ = 60 traceless tr Ỹ 2 = 2 notation. 2 off-diagonal matrices

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

振動と波動

chap1.dvi

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

0.1 I I : 0.2 I

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

QMII_10.dvi

磁性物理学 - 遷移金属化合物磁性のスピンゆらぎ理論

atomic line spectrum emission line absorption line atom proton neutron nuclei electron Z atomic number A mass number neutral atom ion energy

Mott散乱によるParity対称性の破れを検証

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

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

第3章

December 28, 2018

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 =

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


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

1 ( ) Einstein, Robertson-Walker metric, R µν R 2 g µν + Λg µν = 8πG c 4 T µν, (1) ( ds 2 = c 2 dt 2 + a(t) 2 dr 2 ) + 1 Kr 2 r2 dω 2, (2) (ȧ ) 2 H 2

untitled

I A A441 : April 15, 2013 Version : 1.1 I Kawahira, Tomoki TA (Shigehiro, Yoshida )

vol5-honma (LSR: Local Standard of Rest) 2.1 LSR R 0 LSR Θ 0 (Galactic Constant) 1985 (IAU: International Astronomical Union) R 0 =8.5

総研大恒星進化概要.dvi

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

画像工学特論

The Physics of Atmospheres CAPTER :

x A Aω ẋ ẋ 2 + ω 2 x 2 = ω 2 A 2. (ẋ, ωx) ζ ẋ + iωx ζ ζ dζ = ẍ + iωẋ = ẍ + iω(ζ iωx) dt dζ dt iωζ = ẍ + ω2 x (2.1) ζ ζ = Aωe iωt = Aω cos ωt + iaω sin

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

chap03.dvi

Transcription:

4/15 No. 1

4/15 No.

4/15 No. 3

Particle of mass m moving in a potential V(r) V(r) m i ψ t = m ψ(r,t)+v(r)ψ(r,t) ψ(r,t) = ϕ(r)e iωt ψ(r,t) Wave function steady state m ϕ(r)+v(r)ϕ(r) = εϕ(r) Eigenvalue problem ε = ω ϕ(r) : Energy eigenvalue : Eigen function 4/15 No. 4

Bare Coulomb potential from the nucleus V(r) = V(r) = Ze 4πε 0 r (Time-independent Schrödinger equation) m ϕ(r) Ze ϕ(r) = εϕ(r) 4πε 0 r cumbersome coefficients Introduction of atomic unit (a.u.) 1 ϕ(r) Z r ϕ(r) = εϕ(r) 4/15 No. 5

Electron Unit system in which = m = e = e =1 4πε 0 Length Energy a 0 = $ ' m& e ) % 4πε 0 ( = 4πε 0 me e 4πε 0 a 0 = 7.1 ev = 5.9 10 11 Bohr radius m (ionization potential of H) 1 ev =1.60 10 19 J Time Velocity 3 $ ' = a 0 αc m& e ) % 4πε 0 ( a 0 a 0 αc = αc = 0.04 fs α = fine structure constant e 4πε 0 c = 7.97 10 3 = 1 137.0 Atomic scale of length, energy, and time 4/15 No. 6

Atomic unit is closely related to Bohr hydrogen atom Dimension Expression Value Meaning length a 5.9 10-11 0 =4 0 /me m Bohr radius energy velocity time electric field laser intensity E h = me4 (4 0 ) = e 4 0a 0 v = e 4 0 = a 0 E h v F = e = c 4 0a 0 1 c 0F 7. ev.19 10 6 m/s 4. attoseconds 5.14 10 11 V/m Coulomb potential energy at the Bohr radius electron orbital velocity time during which the electron proceeds 1 radian field at the Bohr radius laser field = electric 3.51 10 16 W/cm field at the Bohr radius 4/15 No. 7

Bare Coulomb potential from the nucleus V(r) = V(r) = Ze 4πε 0 r = Z r (Time-independent Schrödinger equation) m ϕ(r) Ze ϕ(r) = εϕ(r) 4πε 0 r Polar coordinate 1 ϕ(r) Z r r = (r,θ,φ) ϕ(r) = εϕ(r) Bound state ε < 0 Energy eigenvalue ε n = Z me 4 4πε 0 Eigen function ϕ(r) = R nl (r)y lm (θ,φ) ( ) 1 n = Z n n =1,,3 0 l n 1 l n l Radial wave function Spherical harmonics 4/15 No. 8

Energy eigenvalue ε = Z me 4 n 4πε 0 ( ) 1 n = Z n =1,,3 n Energy (ev) Coulomb potential r in a 0 (Bohr radius ) a 0 = 4πε 0 me = 5.3 10 11 m = 0.053 nm ϕ(r) = R nl (r)y lm (θ,φ) 0 l n 1 l n l Ground state me 4 ε 1 = 4πε 0 ( ) = 13.6 ev 4/15 No. 9

Energy eigenvalue ε n = Z n =1,,3 n Balmer series Lyman series 4/15 No. 10

Z = 1 3/ " R 1s = 1 % $ ' e r / a 0 # a 0 & R s = 1 3/ " % $ ' # a 0 & 1 e r /a 0 " 1 r % $ ' # a 0 & R p = 1 3/ " % 1 $ ' # a 0 & 6 e r /a 0 r a 0 R 3s = 1 3/ " % ) $ ' # a 0 & 3 3 e r / 3a 0 1 r + " r % + $ ' * + 3 a 0 7 # a 0 & Orthonormality (r)r n # l (r) r dr = δ n n # R 0 nl ϕ nlm ϕ n % l % %,. -. Y 1,0 = Y 00 = 3 4π cosθ Y,0 = 1 4π Y 1,±1 = 5 16π 3cos θ 1 ( ) Y,±1 = 15 sinθ cosθ e±iφ 8π 15 Y,± = 3π sin θ e ±iφ Orthonormality Y lm (θ,φ) m r sinθdrdθdφ = δ n % Y l & & n δ l l % δ m m % 3 sinθ e±iφ 8π m (θ,φ)sinθdθdφ = δ l & l δ m m & 4/15 No. 11

Radial wave function Probability density r (atomic unit) 4/15 No. 1

ε > 0 Necessary when ionization is considered ε > 0 Arbitrary positive number ϕ(r) = R εl (r)y lm (θ,φ) l 0 l n l Radial wave function Coulomb wave function R εl (r) = l Z s + % 1 e π n % s=1 wave number k = me / = E n " = Z k ε ε $ R εl (r)r ε $ l (r) r dr = 0 n 0 (kr) l (l +1)! e ikr F(i n % +l +1,l +,ikr) confluent hypergeometric function R εl (r) r dr > 0 Density of states 0 4/15 No. 13

Bound states Continuum states r a 0 ( ) 4/15 No. 14

V(r) = 0 In a free space R El (r) = 1 ϕ(r) = εϕ(r) 1 # d % dr + $ r k π j (kr) l % % r πk d dr l(l +1) & r ( R(r) = εr(r) ' 1 r cos ' kr π * (l +1) () +, Spherical Bessel function 1 Coulomb wave function R El (r) $ $ r πk r cos ( kr + Z k logkr π (l +1) σ + l )*,- rr El (r) l =1 (p-wave) E = 13.6 ev 0.5 0.5 Phase shift σ l = argγ(l +1+iZ /k) 10 0 30 40 50 r Couomb V(r)=0 4/15 No. 15

V(r) = 0 1 ϕ(r) = εϕ(r) 1 # d % dr + r > r $ 0 r R El (r) = k π ( c j (kr)+ c y (kr) j l y l ) d dr Spherical Bessel function l(l +1) & r ( R(r) = εr(r) ' Figure 10.48.1: j n (x),n= 0(1)4, 0 x 1. Figure 10.48.: y n (x),n= 0(1)4, 0 <x 1. j l (kr) y l (kr) """ (kr)l """ 1 r 0 r (l +1)!! kr cos % kr π & ' (l 1)!! """ """ 1 r 0 (kr) l+1 r kr sin % kr π & ' ( (l +1) ) * ( (l +1) ) * Phase shift 4/15 No. 16

i ψ t = 1 ψ(r,t) Z r ψ(r,t)+v I (r,t)ψ(r,t) Interaction i ψ t = (H 0 + H I )ψ(r,t) H 0 = 1 Z r H I = V I (r,t) Without the external field H 0 ϕ n (r) = ε n ϕ n (r) ψ n (r,t) = ϕ n (r)e iω n t With the external field ω n = ε n Eigen state ψ(r,t) = c n ϕ n (r)e iω n t c n = e iω n t ϕ * n (r)ψ(r,t)dv = e iω n t n ψ n H 0 n = ω n n (atomic unit) 4/15 No. 17

i t ψ = (H 0 + H I ) ψ n ψ = c n e iω nt i t n ψ = n H 0 + H I ψ = n H 0 ψ + n H I ψ = ω n n ψ + n H I ψ i c n = n H I ψ e iω nt i c n = m m = I Identity operator can be inserted anywhere n H I m m ψ e iω n t = n H I m c m e i(ω n ω m )t m m m i c n = n H I m c m e i(ω n ω m )t m n H I m Image Transition matrix element Transition from m to n due to the interaction H I m H I n The interaction H I couples m to n. 4/15 No. 18

Resonance frequency ω 0 = ε ε 1 Two-level atom C ε If the laser frequency ω is close to ω 0, only the two levels are relevant. ω 0 C 1 ω 0 ε 1 ψ(r,t) = C 1 (t)ψ 1 (r,t)+c (t)ψ (r,t) 4/15 No. 19

i ψ t = 1 ψ(r,t) Z r ψ(r,t)+v I (r,t)ψ(r,t) C ε ψ(r,t) = C 1 (t)ψ 1 (r,t)+c (t)ψ (r,t) ω 0 ψ(r,t) d 3 r = C 1 (t) + C (t) =1 $ V I ( C 1 ψ 1 +C ψ ) = i C 1 t ψ 1 + C ' & ψ % ) t ( multiply with ψ 1 from the left and take a volume integral ψ 1 C 1 ε 1 i C 1 t = C 1 V 11 +C V 1 e iω 0t V ij = i V I j = ϕ i V I ϕ j d 3 r Similarly i C t = C 1 e iω 0 t V 1 +C V 4/15 No. 0

Dipole approximation is often sufficient. Wave number z k = π E 0 cos(kx ωt) λ Wavelength r x << λ kx <<1 E 0 cos(kx ωt) E 0 cosωt V I = ze 0 cosωt Dipole approximation " x Ze H 0 cos(kx ωt) k y 4/15 No. 1

i C 1 t i C t = C 1 V 11 +C V 1 e iω 0 t = C 1 e iω 0 t V 1 +C V V I = ze 0 cosωt V ij = i V I j = ϕ i V I ϕ j d 3 r = cosωt ze 0 ϕ i ϕ j d 3 r = X ij cosωt X 11 = X = 0 How V I couples the two levels. V I j i V ij = i V I j = ϕ i V I ϕ j d 3 r X 1 = X 1 = γ Real i C 1 t = γc e iω 0 t cosωt i C t = γc 1 e iω 0 t cosωt i C 1 t ( )t [ ] i C = γc e i ( ω ω 0 )t + e i ω+ω 0 t ( )t [ ] = γc 1 e i ( ω+ω 0 )t + e i ω ω 0 4/15 No.

i C 1 t ( )t [ ] i C = γc e i ( ω ω 0 )t + e i ω+ω 0 Rotating wave approximation t ( )t [ ] = γc 1 e i ( ω+ω 0 )t + e i ω ω 0 i C 1 t = γe i ( ω ω 0 )t C i C t Initial condition C 1 =1, C = 0 = γe i ( ω ω 0 )t C 1 ( ) % C 1 (t) = cosωt i ω ω 0 ' & Ω sinωt ( + * exp i ( ) ω ω. 0)t,- / 0 C ε C (t) = iγ Ω sinωt exp & i ( ω ω ) 0)t '( Ω = γ *+ + (ω ω 0) 4 C 1 ω 0 ε 1 4/15 No. 3

ω ω Population 0 = 3.5γ C 1 (t) =1 C (t) C (t) = γ Ω sin Ωt C 1 (t) Ω = γ + (ω ω 0 ) 4 Absorption-emission cycle γ t ω = ω 0 C (t) γ t γ t ω ω 0 = 0.9γ 4/15 No. 4

Dipole interaction can be expressed in either the length or velocity gauge Length gauge velocity gauge i L t = p + V (r)+r E(t) L i V t = (p + A(t)) + V (r) V gauge transformation L = e ir A(t) V vector potential All physical observables are gauge invariant. probability density L = V projection on eigenstate i (or population of eigenstate i) depends on gauge! i Ldr 3 = i V dr 3 Level population (such as C 1 and C ) is meaningful only if or 4/15 No. 5