Since we cannot say exactly where an electron is, the Bohr picture of the atom, with electrons in neat orbits, cannot be correct.
Quantum theory describes electron probability distributions:

Hydrogen Atom:
Schrödinger Equation and Quantum Numbers
Potential energy for the hydrogen atom:

Hydrogen Atom:
Schrödinger Equation and Quantum Numbers
The time-independent Schrödinger equation in three dimensions is then:
where
Note, this 13.6 ev is the Rydberg constant, that is also found via QM
R

 e
2
4

0
2 m e
2  2
Rydberg constant = 13.6 eV

Intermezzo
Where does the quantisation in QM come from ?
The atomic problem is spherical so rewrite the equation in ( r , q
,

) x
 r sin q cos
 y
 r sin q sin
 z
 r cos q
Rewrite all derivatives in ( r , q
,

), gives Schrödinger equation;
 2

2 m

 r r
2

 r
 2
 
2 m sin
1 q

 q sin q

 q

1 sin
2 q
 2
  2
 
V ( r )
 
E

This is a partial differential equation, with 3 coordinates (derivatives);
Use again the method of separation of variables:
  r , q
,
  
R
   
Bring r -dependence to left and angular dependence to right (divide by

):
1
R


 d dr r
2 dR
 dr
2 mr
2

E
 2

V
  
R



 
O
QM q
Y
Y
,

 

Intermezzo
Where does the quantisation in QM come from ?
Radial equation
Angular equation
1
R


 d dr r
2 dR
 dr
2 mr
2
 2

E

V
  
R



 

O
QM q
Y
Y ,

,

 sin
1 q

 q sin q
 q
Y

 sin
 
1
2 q
 2
  2

Y ,

 
Once more separation of variables:

 2
Y
  2
 sin q

 q sin q

Y
 q
  sin
2 q
Y
Y
 
 
   
Derive:

1

 2 
  2

1
 sin q

 q sin q
 
 q
  sin
2 q   m
2
(again arbitrary constant)
Simplest of the three: the azimuthal angle;
 2 
 

  2 m
2 
 

0

Intermezzo
Where does the quantisation in QM come from ?
A differential equation with a boundary condition
 2 
 

  2 m
2 
 

0 and

  
2
 
 
(

)
Solutions:
    e im

Boundary condition;

  
2
 
 e im
  
2
 
 
 
 e im
 e
2
 im 
1 m is a positive or negative integer this is a quantisation condition
General: differential equation plus a boundary condition gives a quantisation

Intermezzo
Where does the quantisation in QM come from ?
First coordinate
    e im
 with integer m
(positive and negative)
Second coordinate
Results in
Third coordinate
1 sin q

 q sin q
 
 q
   m
2 sin
2 q
 
0


 

 
1
 and with
 
0 , 1 , 2 ,  m
   ,
  
1 ,  ,  
1 , 
1
R


 d dr r
2 dR
 dr
2 mr
2
 2

E

V
  
R



 

 
1

Differential equation
Results in quantisation of energy
E n
 
Z
2 n
2
R 
 
Z
2 n
2 e
2
4

0
2 m e
2 
2 with integer n (n>0) angular part angular momentum radial part

Intermezzo
Angular wave functions
Operators: L
2 
 i



1 sin q

 q sin q

 q
L z

 i

 

1 sin
2 q
 2
  2



Angular momentum

L

( L x
, L y
, L z
)
There is a class of functions that are simultaneous eigenfunctions
L
2
Y lm
  
 
1

 2
Y lm
  with
 
0 , 1 , 2 , 
L z
Y lm and
 m  Y lm m
   ,
  
1 ,  ,  
1 , 
Spherical harmonics (Bolfuncties)
Y
00

1
4
 Y
11
 
Y
10
 
3
8
 sin q e i

3
4
 cos q
Y
1 ,

1

3
8
 sin q e
 i

Y lm
Vector space of solutions


Y lm
2 d



Y
* lm
Y l ' m ' d
 

1
 ll '
 mm '
Parity
P op
 lm
 
 
   q
,
        lm
 

Intermezzo
The radial part: finding the ground state
1
R


 d dr r
2 dR
 dr
2 mr
2
 2

E

V
  
R



 
Find a solution for  
0 , m

0
 2

2 m
"
2
R ' r

Ze
2
4

0 r
R

ER
Physical intuition; no density for r
  trial:
R
  
Ae
 r / a
R '
R "



A a
A a
2 e
 r / e
 r a
/ a



R
R a a
2
 2

2 m
1 a
2

2 ar

Ze
2
4

0 r

E
 2 ma

Ze
2
4

0
1

 r
 2
 ma
2

E must hold for all values of r !!
Prefactor for
1/r
:  2 ma

Ze
2
4

0

0
Solution for the length scale paramater a
0

4

0
 2
Ze
2 m
Bohr radius
Solutions for the energy
E


2

2 ma
2
 
Z
2 e
2
4

0
2 m e
2 
2
E
 
 m e
Z
2
E
0
Ground state in the
Bohr model (n=1)
Isotope effect

Hydrogen Atom:
Schrödinger Equation and Quantum Numbers
 2

2 m

 r r
2

 r
 2
 
2 m sin
1 q

 q sin q

 q

1 sin
2 q
 2
  2
 
V ( r )
 
E

This partial differential equation can be separated into three equations;
General solution:

 r , q
,
       
1) Radial equation:
1
R

 d dr r
2 dR dr

2 mr
2
 2

E

V
  
R


 

 
1
 Eigenfunctions:
Eigenvalues: n
R
Eigenfunctions:
Eigenvalues: 

  and E n
 
Z
2 n
2
R

2) Equation for q
:
1 sin q

 q sin q
 
 q


 

 
1

 m 2 sin 2 q


 
0
3) Equation for

:
 2   
 m
2 
 
  2

0
Eigenfunctions:
Eigenvalues:
 m
NB: connections between n , l and m derive from solutions of Schrödinger Equation

Hydrogen Atom:
Schrödinger Equation and Quantum Numbers
There are three different quantum numbers needed to specify the state of an electron in an atom ( plus one ).
1. The principal quantum number n gives the total energy.
2. The orbital quantum number gives the angular momentum; can take on integer values from 0 to n - 1.
3. The magnetic quantum number, m , gives the direction of the electron’s angular momentum, and can take on integer values from – to + .
NB: connections between n , l and m derive from solutions of Schrödinger Equation

Hydrogen Atom Wave Functions
The wave function of the ground state of hydrogen has the form:
 r
0
 a
0

The probability of finding the electron in a volume dV around a given point is then | ψ | 2 dV .
 2 dV

4
 r
2
 2 dr

P r dr with
P r

4
 r
2
 2
Probability to find electron at position r

The ground state is spherically symmetric; the probability of finding the electron at a distance between r and r + dr from the nucleus is:

This figure shows the three probability distributions for n = 2 and
= 1 (the distributions for m = +1 and m = -1 are the same), as well as the radial distribution for the n = 2 states.
most probable

Atomic Hydrogen Radial part
Analysis of radial equation yields:
E nlm
 
Z
2 n
2
R  with R

 m e e
4
8

0 h
3 c
Wave functions:
 nlm
 

R nl
    lm
Traditional nomenclature for orbitals
 
0
; s-orbitals
 
1
; p-orbitals




2
3
; d-orbitals
; f-orbitals

Intermezzo
Quantum analog of electromagnetic radiation
Classical electric dipole radiation
Classical oscillator
I rad
 e r
 2
Transition dipole moment
Quantum jump
I rad
  
1
* e r


2 d

2
The atom does not radiate when it is in a stationary state !
The atom has no dipole moment
 ii
   *
1 r


1 d
 
0
Intensity of spectral lines linked to Einstein coefficient for absorption:
B if

 fi
2
6

0

2

Intermezzo
Selection rules
Wave functions
 have well defined parity
      

 
  
  even odd
Mathematical background: function of odd parity gives 0 when integrated over space
In one dimension:  f x
 i





 f
* x
 i dx



  f ( x ) dx with f ( x )
  f
* x
 i



 f ( x ) dx

0

  f ( x ) dx



0 f ( x ) dx



0 f (
 x ) d



0 f ( x ) dx



0 f (
 x ) dx



0 f ( x ) dx

2


0 f ( x ) dx

0 if f (
 x )
 f ( x )
 i and
 f opposite parity because: f (
 x )
  * f
(
 x )(
 x )
 i
(
 x )

0 if f (
 x )
  f ( x )
 i and
 f same parity
Electric dipole radiation connects states of opposite parity !

Intermezzo
Selection rules depend on angular behavior of the wave functions
Parity operator r

 q
 r

P r

  r

( r x
,
, y
,
, z )

 q   
(
 r ,
 x ,

 y ,
 q
,
 z

)
 
All quantum mechanical wave functions have a definite parity

 
  
 
 f r

 i

0
 
Rule about the
Y
 m functions
PY
 m
( q
,

)

 

Y
 m
( q
,

)

Hydrogen Atom:
Schrödinger Equation and Quantum Numbers
“Allowed” transitions between energy levels occur between states whose value of differ by one:
Other, “forbidden,” transitions also may occur but with much lower probability.
“selection rules, related to symmetry (parity)”
 
1
From s-orbitals to p-orbitals

Selection rules in Hydrogen atom
|
Intensity of spectral lines given by
 fi
|
2 
|
  * f
  i
|
2 
|
 f
 e r

 i
|
2
1) Quantum number n no restrictions
2) Parity rule for
   odd

3) Laporte rule for

Angular momentum rule:

 f


 i


1 so
  
1
From 2. and 3.
   
1
Balmer series
Lyman series