Quantum Mechanics Formulas by R.L. Griffith pdf - Web Education

Quantum Mechanics Formula Sheet Ephoton = h= 1_{2 mv}2 + e   1_{2 mv}2 +   E = h= –h _n1 _ 22 – 1 n12  eix _{= cos x + i sin x} ei + e-i 2 = cos  e i – e-i

2i = sin  eikx k = pħ =

(2mE)½

ħ E= n_2ma2ħ222= n

2_h2

8ma2 =2_a½sinnx_{a (x,y)=}_ab4_½sinnx_{a sin}x ny_{b E=} y h 2 8m   n12 a2+ n22 b2 xx^ pħ i     x tt^ E(t) E^ = iħ     t Ek– ħ 2 2m  2 x2 [O1,O2] = O^1O^2 – O^2 O^1 (x) = A

e

–½ a2_x2 2 ₌ mk ħ A =  _ 2  ¼ (x) =       mo ħ ¼

e

– mox2 2ħ

x = r sincos y = r sinsin z = r cos d = r2 _{sindr dd}

r2 _{= x}2 _{+ y}2 _{+ z}2 _{: 0    : 0   2r: 0  r  } L^x = –iħ       y _{z – z}  _y  L^y = –iħ       z _{x – x}  _z  L^z = –iħ       x _{y – y}  _x  2 ₌ 1 r    2 r2 r +     1 r2 22 = 1 sin2_{ } _ 2 2 +     1 sin _  sin _   |L|= ħ l(l+1) Lz = ħml = _ _ 1 2 1_/2

e

iml _{– ħ}2 _2_Y l,ml l (l + 1)h–2 Yl,ml E = ħ2 2I l ( l + 1) – ħ2 2m   1 r d2 dr2 r R – Z e2 4or RR En = –     Z2 _e4 _m 322_o2 _ħ2 1 n2 ao = 4o ħ 2 me2 En = – _2maZ2 ħ_o22 _n12 En = –13.6 eV Z 2 n2 = –109,678 cm-1 Z 2 n2 = –1312 kJ mol-1 Z 2 n2 = – H₂ Z 2 n2 1s = 1      Z ao 3/2

e

-Zr/ao _{h = 6.626 x 10}-34 _{J s ħ = 1.055 x 10}-34 _{J s} 2s = 1 4 2     Z ao 3/2  2 – Zrao 

e

-Zr/2a o _{e = 1.602 x 10}-19 _C 2pz = 1 4 2     Z ao 3/2

e

-Zr/2ao Zr ao cos o = 8.85 x 10 -12 _J-1_c2_m-1 211 = 1 4 2     Z ao 3/2

e

-Zr/2ao Zr ao sine i _{ao = 0.0529 nm = 0.529Å} 21-1 = 1 4 2     Z ao 3/2

e

-Zr/2aoZr ao sine-i   1H = 2625.5 kJ mol-1 2px = 211 + ₂ 21-1 = 1 4 2     Z ao 3/2

e

-Zr/2aoZr ao sincos  me = 9.11x10 -31 _kg 2py = 211 – _2i 21-1 = 1 4 2     Z ao 3/2

e

-Zr/2aoZr ao sinsin  h = 109,678 cm -1

l

ml Yl,m_l 0 0 (1/4π)½ 1 0 (3/4π)½ _cos ±1 _±(3/8π)½ _sine±i 2 0 (5/16π)½ _{(3 cos}2_{ – 1)} ±1 _±(15/8π)½_{cossine}±i

±2 _±(15/32π)½ _sin2_e±i2

y =  x (y) = H

e

–½ y 2 d 2_H dy2 – 2y dH dy + 2 H = 0  H(y) H(x) 0 1 1 1 2y 2x 2 4y2 _{– 2 4}2_x2 _{– 2} 3 8y3_{– 12y 8}3_x3_{– 12x} H+1 = 2y H - 2 H-1   -∞ ∞ H'

e

–½y2 H

e

–½y2 dy = 0 if '  = π½_{2! if ' } 

sin2_{(x) dx = – ½ sin(x) cos(x) + ½ x}



cos2_{(x) dx = ½ sin(x) cos(x) + ½ x}



sin(x) cos(x) dx = ½ sin2_(x)

  0 /2 sin2_{(x) dx = } 0 /2 cos2_{(x) dx =}  4   0 /2 sin3_{(x) dx = } 0 /2 cos3_{(x) dx =} 2 3   0  sin2_{(ax) dx =} 0  cos2_{(ax) dx =}  2   0 /a cos(ax) sin(ax) dx =  0  cos(ax) sin(ax) dx = 0   0  sin(ax) sin(bx) dx =  0 

cos(ax) cos(bx) dx = 0 (a  b; a,b integers)

 

0 

sin(ax) cos(bx) dx = _a22a_-b2 if a-b is odd, or zero if a-b is even

  0  cos(ax) sin(ax) dx = 0 _ 0 n x2 _sin2_{(x) dx =} n33 6 – n 4