Physical Chemistry 20130520 week 8 Monday May 20, 2013 page 1
Exam on Friday, just chapter 18, not perturbation theory.
perturbation theory
Variation theory is for the ground state only. Purturbation theory works for the ground state or excited
states. Sometimes Schrӧdinger’s equation can’t be solved exactly because V (the potential energy) gets
too complicated when we have many particles.
Ĥψn = Enψn
Ĥ°ψn = En°ψn
Ĥ = Ĥ° + Ĥ’
exact
the Ĥ’ is the perturbation
(1)
°
En = ∫ ψ°*
n H°'ψn dτ
°
ψ°*
n H°'ψn is like an average, gives average energy
En=En°+En1+En2+…
En1 is the first order correction (the largest correction)
En2 is second order correction (smaller than the first order correction)
Each successive term is smaller than the previous.
We’re skipping Hermition operators.
Ch 19 builds on ch 18.
We’re no longer talking about a particle in a box.
Read section 19.1 (units) and 19.2 (historical background).
The H atom (1 electron, 1 proton, no neutron) is a two particle system.
H-like “atoms” or actually ions (He+, Li2+…) also have one electron.
-(ze')(e' )
PE=
r
z is atomic number
e' is charge of electron in cgs units
We’ll use cgs units to avoid extra unit conversion constants.
(ze’) is the total charge of the nucleus, the nuclear charge.
-e’ is the charge of the electron in stat C (stat coulomb) units, same as Gaussian unit
e’ = 4.80 * 10-10 esu
e = 1.60 * 10-19 C C is coulombs
Etotal = Etrans + Einternal motion
Etrans is the same as with the particle in a box.
For H atom, Ĥ for internal motion is:
̂=
H
-ћ2 ∂2
∂2
∂2
ze'2
( 2 + 2 + 2) 2µ ∂x
∂y
∂z
r
µ=
m1 m2
m1 +m2
m2 =me =mass of electron
µ=
Ĥψ = Eψ
1⁄
2
r=(x 2 +y 2 +z 2 )
µ is reduced mass
m1 =mnucleus =1836.15me
1836m2e
=.99946me =me
1837me
(time-independent)
Sperical coordinates:
x = rsinθcosφ
y = rsinθsinφ
z = rcosθ
0≤r≤∞
0≤θ≤π
0≤φ≤2π
r2=x2+y2+z2
complete wave function
ψ=Rn,l(r)Θl,m(θ)Φm(ϕ)
l,m,n are quantum numbers
Yl,m = Θl,m(θ)Φm(ϕ) = spherical harmonics
(n-l-1)!
1⁄
2
R n,l (r)=- (
)
3
2n((n+l)!)
3
2 l+2 l na-r 2l+1 2r
(
) r e 0 Ln+l (
)
na0
na0
2r
L2l+1
) is associated Laguerre function
n+l (
na0
a0 is Bohr radius.
n = 1,2,3,…∞
l=0,1,2,3,…,n-1
so that function is well behaved
We have 3 independent variables (r,θ,φ) so we have 3 quantum numbers (n,l,m).
m = ml = -l, -l+1,…,0,…,l-1,l
So if l = 1 then m can be -1, 0 or +1.
E=
-z 2 e'2
( )
n2 2a
a0 = .5292Å
a= .5295Å
If we approximate µ = me then use a0 (rounding).
If we calculate µ then use a (no rounding).
ћ2
a= '2
µe
a has dimension of length
Eα
-1
n2
Notice it doesn' t depend on l or m so we get degeneracy.
orbitals (wave functions)
E1
E2
2s
ψ200
2p0
ψ210
2p1
ψ21+1
1s
ψ100
2p-1
ψ21-1
These all have the same energy.
p0 is pz
In H atom, 2s, 2px, 2py, 2pz have the same energy level, so it’s 4 fold degenerate.