Transcript QM lecture - The Evergreen State College

QM in 3D
Quantum Ch.4, Physical Systems, 24.Feb.2003 EJZ
Schrödinger eqn in spherical coordinates
Separation of variables (Prob.4.2 p.124)
Angular equation (Prob.4.3 p.128 or 4.23 p.153)
Hydrogen atom (Prob 4.10 p.140)
Angular Momentum (Prob 4.20 p.150)
Spin
Schrödinger eqn. in spherical coords.
The time-dependent SE in 3D
 2 2
(r, t )
  (r, t )  V (r, t )  i
2m
t

has solutions of form
where n(r,t) solves
 (r, t )   cn n (r) e
 i Ent
n 1
 2 2
  n (r)  V n (r)  En n (r)
2m
Recall how to solve this using separation of variables…
Separation of variables
To solve
Let
 2 2
  n (r)  V n (r)  En n (r)
2m
 (r)   (r,  )  R(r ) Y (  )
Then the 3D diffeq becomes two diffeqs (one 1D, one 2D)
Radial equation
Angular equation
Solving the Angular equation
To solve
Let Y() = Q() F() and separate variables:
The  equation has solutions F() = eim (by inspection)
and the  equation has solutions Q() = C Plm(cos) where
Plm = associated Legendre functions of argument (cos).
The angular solution = spherical harmonics:
Y()= C Plm(cos) eim where C = normalization constant
Quantization of l and m
In solving the angular equation, we use the Rodrigues
formula to generate the Legendre functions:
“Notice that l must be a non-negative integer for [this] to
make any sense; moreover, if |m|>l, then this says that
Plm=0. For any given l, then there are (2l+1) possible
values of m:”
(Griffiths p.127)
Solving the Radial equation…
…finish solving the Radial equation
Solutions to 3D spherical Schrödinger eqn
Radial equation solutions for V= Coulomb potential
depend on n and l (L=Laguerre polynomials, a = Bohr radius)
Rnl(r)=
Angular solutions = Spherical harmonics
As we showed earlier, Energy = Bohr energy with n’=n+l.
Hydrogen atom: a few wave functions
Radial wavefunctions
depend on n and l,
where l = 0, 1, 2, …, n-1
Angular wavefunctions
depend on l and m, where
m= -l, …, 0, …, +l
Angular momentum L:
review from Modern physics
Quantization of angular momentum
direction for l=2
Magnetic field splits l level in (2l+1)
values of ml = 0, ±1, ± 2, … ± l
L  l (l  1) where l  0,1, 2,..., n  1
Lz  ml  L cos
E
E1
(n  l )
2
where E1  Bohr ground state
Angular momentum L:
from Classical physics to QM
L=rxp
Calculate Lx, Ly, Lz and their commutators:  Lx , Ly   i Lz
Uncertainty relations:  L  L  Lz
x
y
2
Each component does commute with
L2:
 L2 , L   0
Eigenvalues:
H nlm  En nlm , L2 nlm 
l (l 1) nlm , Lz nlm  m nlm
2
Spin - review
• Hydrogen atom so far: 3D spherical solution to Schrödinger
equation yields 3 new quantum numbers:
l = orbital quantum number
L  l (l  1)
ml = magnetic quantum number = 0, ±1, ±2, …, ±l
ms = spin = ±1/2
• Next step toward refining the H-atom model:
Spin
with
1
1 1
s

m


s  2 ( 2  1)
z
s
2
Total angular momentum J=L+s
J
j ( j  1)
with j=l+s, l+s-1, …, |l-s|
Spin - new
Commutation relations are just like those for L:
Can measure S and Sz simultaneously, but not Sx and Sy.
Spinors = spin eigenvectors
  s m 
1 1
2 2
 
  s m 
1 1
2 2
 
An electron (for example) can have spin up or spin down
  a    b   a   b 
Next time, operate on these with Pauli spin matrices…
Total angular momentum:
Multi-electron atoms have total J = S+L where
S = vector sum of spins,
L = vector sum of angular momenta
Allowed transitions (emitting or absorbing a photon of spin 1)
ΔJ = 0, ±1 (not J=0 to J=0)
ΔL = 0, ±1 ΔS = 0
Δmj =0, ±1 (not 0 to 0 if ΔJ=0)
Δl = ±1 because transition emits or absorbs a photon of spin=1
Δml = 0, ±1 derived from wavefunctions and raising/lowering ops
Review applications of Spin
Bohr magneton m  e  9.27 x1024 Joule  5.79 x109 eV
B
2me
Tesla
Gauss
Stern Gerlach measures me = 2 m B:
Dirac’s QM prediction = 2*Bohr’s semi-classical prediction
Zeeman effect is due to an external magnetic field.
Fine-structure splitting is due to spin-orbit coupling (and a small
relativistic correction).
Hyperfine splitting is due to interaction of melectron with mproton.
Very strong external B, or “normal” Zeeman effect, decouples L
and S, so geff=mL+2mS.