Quantum Mechanics Formula Sheet
1
1
1
1
Ephoton = h= 2 mv2 + e   2 mv2 +  
E = h= –h n 2 – n 2
2
1 

ei + e-i
ei – e-i
p (2mE)½
ikx
eix = cos x + i sin x
=
cos

=
sin

e
k
=
2
2i
ħ=
ħ
2
2
2
2
2
2
2
nħ nh
h n1 n22
2 ½ nx
 4 ½ nxx nyy
E= 2ma2 = 8ma2 =a  sin a
(x,y)=ab sin a sin b
E= 8m  a2 + b2 
 
 


2
2
ħ  
–ħ 
^ = iħ  
^ 1O
^2 – O
^2 O
^1
xx^
p i  
tt^ E(t) E
Ek  2m 2
[O1,O2] = O
x
x
t
2
¼
mo¼
mox2
– 2ħ
mk

2 = ħ
A= 
(x) = 
 e

 ħ 
x = r sincos
y = r sinsin
z = r cos d = r2 sindr dd
r2 = x2 + y2 + z2
: 0   
: 0   2r: 0  r  





^ = – iħ y – z 
^ = –iħ z
^ = –iħ x  – y  
L
L
–x 
L
x
y
z
y
z
x
 z
 x
 y
2
2
1    1
1    1  

2 = r  2r + r22 2 = 2  2+ sin  sin 
|L|= ħ l(l+1)
sin    

 
r   
ħ2
 1 1/2
Lz = ħml
=   eiml
– ħ2 2Yl,ml l (l + 1)h–2 Yl,ml
E = 2I l ( l + 1)
2
–½ a2x2
(x) = A e
2 4
– ħ2 1 d2
4o ħ2
 Z e2
 Z e m  1
r
R
–
RR
E
a



n=–
o=
2
2
2
2
2
2m  r dr
me2
 4or
32 o ħ  n
2
2
2
2
2
Z ħ 1
Z
Z
Z
H Z2
En = – 2ma 2 n2 En = –13.6 eV n2 = –109,678 cm-1 n2 = –1312 kJ mol-1 n2 = – 2 n2 
o
1  Z 3/2 -Zr/ao
h = 6.626 x 10-34 J s
ħ = 1.055 x 10-34 J s
  e
 ao
1  Z 3/2 
Zr 
2s =
e = 1.602 x 10-19 C
a   2 – a  e-Zr/2ao
o
o




4 2
1  Z 3/2 -Zr/2ao Zr
2pz =
cos
o = 8.85 x 10-12 J-1c2m-1
a  e
a
o
o


4 2
1  Z 3/2 -Zr/2ao Zr
i
211 =
ao = 0.0529 nm = 0.529Å
  e
ao sine
4 2 ao
1  Z 3/2 -Zr/2ao Zr
-i 

1H = 2625.5 kJ mol-1
21-1 =
  e
ao sine
4 2 ao
211 + 21-1
1  Z 3/2 -Zr/2ao Zr
-31

2px =
=
  e
2
ao sincos  me = 9.11x10 kg
4 2 ao
211 – 21-1
1  Z 3/2 -Zr/2ao Zr
-1
2py =
=
  e
2i
ao sinsin  h = 109,678 cm
4 2 ao
1s =
l
ml
Yl,ml
0
1
0
0
±1
(1/4π)½
(3/4π)½ cos
2
0
±1
±2
±i
±(15/8π)½cossine
±i2
±(15/32π)½ sin2e
(y) = H e
y=x

0
1
2
3
±i
±(3/8π)½ sine
(5/16π)½ (3 cos2 – 1)
H(y)
1
2y
4y2 – 2
8y3 – 12y
–½ y2
H(x)
1
2x
42x2 – 2
83x3 – 12x
d2H
dH
–
2y
2
dy
dy + 2 H = 0
H+1 = 2y H - 2 H-1
∞
2
= π½2!


0
/2
/2

2
dx = 
cos (x) dx = 4
0
/2
2
3
3

sin (x) dx = 
cos (x) dx = 3
0

0

0
/a
0

sin2(ax) dx =
cos2(ax) dx = 2



cos(ax) sin(ax) dx = 
cos(ax) sin(ax) dx = 0
0

0

0

0

sin(ax) sin(bx) dx = 
cos(ax) cos(bx) dx = 0
2a

sin(ax) cos(bx) dx = a2-b2
(a  b; a,b integers)
if a-b is odd, or zero if a-b is even
0

n
0
0

cos(ax) sin(ax) dx = 0
if ' 
2(x) dx = ½ sin(x) cos(x) + ½ x
cos

sin(x) cos(x) dx = ½ sin2(x)

sin2(x)
if ' 
-∞
2(x) dx = – ½ sin(x) cos(x) + ½ x
sin

/2
2
–½ y
–½ y

H e
dy = 0
 H' e
n33 n
2 sin2(x) dx =
x


6 – 4