Physical Chemistry 20130529 week 9 Wednesday May 29, 2013 page 1
For chapter 19, do problem 41c instead of 41d.
angular momentum
a. classical mechanics
Consider a 3-D graph with a point anywhere but the origin. The distance from the origin to the point is r.
From this point any direction is the momentum of the particle P. P is the linear momentum of the
particle. P⃑ = ma⃑ m = mass v⃑ = velocity a⃑ = acceleration L is perpendicular to both r and p. |L|
= rpsinβ For right hand rule: r then p then get L. L⃑ = r⃑ x P⃑ That’s cross product (or vector
product).
b. quantum mechanics has two kinds of angular momentum
1. orbital angular momentum: analogous to classical angular momentum
2. intrinsic (spin) angular momentum: no counterpart in classical mechanics
L̂ 2ψnlm = l(l+1)ħ2ψnlm
eigenvalue problem
ψnlm is eigenfunction
l(l+1)ħ2 is eigenvalue
|L|=orbital angular momentum= √l(l+1)ħ
Lz = mħ
m is quantum number
For s states: l = 0
|L|= 0 Lz = 0
For p states: l=1 |L|=√2ħ
Consider a 2-D graph with z on the vertical axis. For m = 1, the radius goes from the origin out diagonally
to r=ħ√2 such that the vertical height is ħ. For m=0, the radius goes straight to the right from the origin
to r=ħ√2. For m=-1, the radius goes from the origin out diagonally to r=ħ√2 such that the vertical
component is –ħ, making it the mirror image of the first radius. All three of these are actually cones that
go 360° around the z axis.
For d states:l=2 |L|= √6ħ
Consider a 2-D graph with z on the vertical axis. For m=2, the radius goes from the origin out diagonally
to ħ√6 such that the vertical height is 2ħ. For m=1, the radius goes from the origin up and right
diagonally a distance of ħ√6 such that the vertical height is ħ. For m=0, the radius goes straight right
from the origin to a distance of ħ√6. For m=-1, the radius goes from the origin down and right a
distance of ħ√6 such that the vertical height is –ħ. For m=-2, the radius goes from the origin diagonally
down and right a distance of ħ√6 such that the vertical height is -2ħ.
Ĥψnlm = Enψnlm
2
z 2 e'
En = - ( 2 ) ( )
n
2a
for H-like atom
e' means cgs system
L̂2ψnlm = l(l+1)ħ2ψnlm
L̂zψnlm = mħψnlm
L̂2 = L̂x2 + L̂y2 + L̂z2
L̂ x = -iħ (y
2
L̂ =
∂
∂
-z )
∂z ∂y
-h2
1 ∂
∂
1 ∂2
(sinθ
)
+
(
)
4π2 sinθ ∂θ
∂θ
sin2 θ ∂φ2
It’s easier in spherical coordinates than in Cartesian coordinates.
Wave functions of a degenerate level
Assume we have E1 with ψ1 and ψ2.
If Ĥψ1 = E1ψ1 and Ĥψ2 = E1ψ2 then Ĥ(c1ψ1+c2ψ2) = E1(c1ψ1+c2ψ2) linear combination
Any linear combination will work as long as the wave functions are at the same energy level and it’s
normalized so c12 + c22 =1. It’s not true that the save functions are at different energy levels so if we
have E2ψ6 and E1ψ5 and Ĥψ5 = E5ψ5 and Ĥψ6 = E6ψ6 but Ĥ(c1ψ5+c2ψ6) ≠ energy(c1ψ5+c2ψ6).
Real wave functions
2p+1 =be
-zr⁄
2a rsinθeiφ
b≡
+1 is m
z 5⁄2
( )
8√π a
1
2p-1 =be
b is constant
rsinθeiφ is spherical harmonics
-zr⁄
2a rsinθe-iφ
x
rsinθ=
cosφ
complex function
z 5⁄2
c≡
( )
√π 2a
1
-zr⁄
2a rcosθ
2pz =2p0 =ce
eiφ =cosφ+isinφ
2px =
1
√2
(2p-1 +2p+1 )
complex function
1
√2
real function
e-iφ =cosφ-isinφ
is normalization constant
real function
2py =
1
i√2
1
2
2
1
Notice ( ) + ( ) =1 so it ' s normalized.
√2
√2
(2p-1 -2p+1 )
See table 19.1
We want to show (wts): 2px =
1
√2
(2p-1 +2p+1 ) handout coming
See handout from May 22, 2013.
z 3⁄2 -zr⁄
ψ1s =
( ) e a
√π a
1
plot the square
|ψ1s|2 α |R(r)|2 From this it SEEMS that the most probable location of the electron is in the nucleus,
but that interpretation is wrong. Graphing just |ψ1s|2 against r is misleading because that’s only the
radial portion of the wave function. Instead, we need to multiply by 4πr2 so we’re doing a sphere
instead of a circle. Consider a sphere and we want to find the volume of a shell layer:
outer volume=
4
π(r+dr)3
3
inner volume=
4 3
πr
3
difference=4πr 2 dr
so that’s the factor to multiply by |ψ1s|2.
Consider a 2-D graph with 4πr2|ψ1s|2 on the vertical axis and r on the horizontal axis. The curve starts at
the origin (0,0) so there is no probability of finding the electron at the nucleus. The curve goes up and
right until peaking at r=0.529Å (for hydrogen) then curves back down asymptotic to the r axis. To find
the peak radius, take the derivative and set it equal to zero:
d
4πr 2 |ψ1s |2 =0
dr
r=
a
z
for H atom: z=1 so r=a=0.529Å
Average value (expectation value)
∫ ψ* M̂ψdτ
⟨M⟩=
all space
2π
⟨r⟩= ∫
0
π
∞
∫ ∫ ψ* r̂ψr 2 drsinθdθdφ
0
0
2π
π
∞
The ∫ is for φ
The ∫ is for θ
The ∫ is for r
0
0
0
In spherical coordinates dτ = r2drsinθdθdφ
Spherical harmonics drop out since they’re already normalized. So just:
∞
⟨r⟩= ∫ ψ* r̂ ψr 2 dr
0
r̂ is just r