Transcript H-atom, spin

PH 401
Dr. Cecilia Vogel
Review
Commutators and Uncertainty
Angular Momentum
Radial momentum
Outline
Spherically Symmetric Hamiltonian
H-atom for example
Eigenstates of H, Lz, L2
Degeneracy
Spherically Symmetric Problem
Suppose the potential energy
depends only on r,
not q or f.
such as for hydrogen atom
ke2
V (r )  
r
then the Hamiltonian looks like.
 Hˆ = 
2
2m
 2  V (r )
Spherically Symmetric Problem
The Hamiltonian in spherical
coordinates, if V is symmetric

Hˆ = 
2
2m
 2  V (r )
2





1


1

2
ˆ
H =
r

sin q
 2
 V (r )
2 
2 

2mr  r r  sin q q
q sin q f  
Written in terms of pr, L2 and Lz:
2
2
2
p
L
 Hˆ = r 
 V (r )
2
2m 2mr
Commutation
Components of angular momentum
commute with r, pr, and L2.
Therefore Lz and L2 commute with H
We can (and will) find set of
simultaneous eigenstates of
L2, Lz, and H
Let Y(r,q,f) = R(r)f(q)g(f)
Let quatum numbers be n, ℓ, mℓ
Separation of variables
In the TISE, Hy=Ey, where
2





1


1

2
ˆ
H =
r

sin q
 2
 V (r )
2 
2 
2mr  r r  sin q q
q sin q f  
2
there is only one term with f or f
derivatives
So this part separates

2
y
 constant *y
2
f
Eigenstate of Lz
In fact ˆ

Lz = i
f
So eigenvalue eqn for Lz is
Lz|nlm> = m|nlm>
 means
 2y
2


m
y
2
f
Eigenstate of L2
Also, angular derivatives only show
up in L2 term, which also separates:
2


1


1

 ˆ2y = 2
L  
sin q
 2
2 
q sin q f 
 sin q q
y
 FYI solution is Spherical Harmonics
Note that L2=Lx2+Ly2+Lz2
 n m | Lˆ2 | n m  n m | Lˆ2x  Lˆ2y  Lˆ2z | n m 
so
(  1)
2
2
2
2
ˆ
ˆ
 Lx  Ly   m
2
(  1)  m 2 therefore | m |
Eigenstate of H
Radial part of eqn
ˆpr 2
(  1)
 2m R  2mr 2
2
R  V ( r ) R  En R
As r goes to infinity, V(r) goes to zero
For hydrogen atom
So solution is (polynomial*e-r/na.)
lowest order in polynomial is ,
highest order in polynomial is n-1,
so  is a non-negative integer,
n is a positive integer, &  n
Degeneracy of Eigenstates
Consider n=5
4th excited state of H-atom
What are possible values of ?
For each , what are possible values of
m?
for each n & , how many different states
are there? “subshell”
for each n, how many different states are
there? “shell”
what is the degeneracy of 4th excited
state?
Spin Quantum Number
Actually there turns out to be twice as many
H-atom states as we just described.
Introduce another quantum number that can
have two values
spin can be up or down (+½ or -½)
 It is called spin, but experimentally the
matter of the electron is not spinning.
It is like a spinning charged object, though, in
the sense that
it acts like a magnet, affected by B-fields
it contributes to the angular momentum,
when determining conservation thereof.
Quantized Lz
g (f )
i
 m g (f )
f
Solution
 g (f )  eim f
Impose boundary condition
g(2p)=g(0)
requires that ml=integer
Lz is quantized
or Lx or Ly, but not all
Eigenstate of Lx, Ly, and Lz
Recall
[Ly,Lz]=iLx is not zero
so there’s not complete set of
simultaneous eigenstates of Ly and Lz
What if Lx=0?
OK, but then also simultaneous
eigenstate of Lx, and
[Lx,Lz]=-iLy is not zero
unless Ly=0
Eigenstate of Lx, Ly, and Lz
Similarly
[Lx,Ly]=iLz is not zero
unless Lz=0
So we can have a simultaneous
eigenstate of Lx, Ly, Lz, and L2
if the eigenvalues are all zero
|n 0 0> is an eigenstate of all
components of L
Component and Magnitude
The fact that
 | m |
is a consequence of
the fact that
a component of a
vector ( m )
can’t be bigger
than
magnitude of the
vector ( (  1) )