Download Hydrogen 2

Quantum
Physics
2002
The Hydrogen Atom
continued..
Recommended Reading:
Harris Chapter 6, Sections 3,4
• Spherical coordinate system
•The Coulomb Potential
•Angular Momentum
• Normalised Wavefunctions
•Energy Levels
• Degeneracy
Put it all together
We now have all the components necessary to write down the solutions
to the Schrodinger Equation for the Hydrogen Atom. We applied the
method of Separation of Variables to wavefunction
ψr, θ, φ   R(r)  θ    φ 
and have found the following solutions for the radial and angular parts.
R(r)  Rn,l r   AssociatedLaguerrePolynomials
θ   Pl,mL θ   AssociatedLegendrePolynomials
 φ   A exp imL φ   Azimuthal wavefunction
And so the total wavefunction can be written as:
ψ n,l ,m r, θ, φ   Nn,l ,ml Rn,l r Pl ,ml θ exp iml φ   Nn,l ,ml Rn,l r Yl ,ml (, )
l
where Nn,l,mL is a normalisation factor which we still have to determine.
Note that the wavefunction depends on three quantum numbers, n, the
principal quantum number, l = the total angular momentum and ml the
z-component of the angular momentum.
Example:
the wavefunction corresponding to the state n = 1, l = 0, ml = 0, what is
the explicit form of this wavefunction
ψ 1,0,0 r, θ, φ   N1,0,0  R1,0 r   P0,0 θ   exp i0φ 
writing the explicit forms of the Laguerre and Legendre polynomials
gives




2
r
ψ 1,0,0 r, θ, φ   N1,0,0  
exp 
 
 a3 2
 a0  
 0

since P0,0() = 1 and exp(i.0.) = 1.
Similary for n = 2, l = 2, mL = 1 we find
ψ 2,1,1 r, θ, φ   N2,1,1  R2,1r   P1,1θ   exp iφ 

 r 

1
r 
ψ 2,1,1 r, θ, φ   N2,1,1  

 exp 
  sinθ  exp iφ 
3
2
 3a0 
 2a0  
 2a0 
The angular part of the wavefunction
Lets examine the angular part in more detail. We can write the angular
part as
ψθ, φ   θ   φ   Pl,mL θ exp imL φ 
as before we can combine these two terms together into a single
function Yl,mL(,) the Spherical Harmonics, this function combines the
 and  dependent part of the solution. The Spherical represent the
solutions to the Schrodinger equation for a particle confined to move on
the surface s a sphere of unit radius. The first few are tabulated on the
next page.
Note that the Spherical harmonics depends on two quantum numbers l
and ml.
The Spherical Harmonics Yl,mL(, )
l
0
mL
0
1
0
1
1
2
0
2
1
2
2
Yl,mL(,)
1
2 π
1 3
cos θ
2 π
1 3

sinθe  iφ
2 2π
1 5
3 cos2 θ  1
4 π
1 15

sinθ cos θe  iφ
2 2π
1 15
sin2 θe  2iφ
4 2π


The Spherical Harmonics Yl,mL(, )
continued...
l
3
mL
0
3
1
3
2
3
3

Yl,mL(,)

1 7
5 cos3 θ  3 cos θ
4 π
1 21

sinθ 5 cos2 θ  1 e  iφ
8 π
1 105
sin2 θ cos θe  2iφ
4 2π
1 35

sin3 θe  3iφ
8 π


to see what these wavefunctions look like see the following websites
http://www.quantum-physics.polytechnique.fr/en
http://www.uniovi.es/~quimica.fisica/qcg/harmonics/harmonics.html
http://wwwvis.informatik.uni-stuttgart.de/~kraus/LiveGraphics3D/
java_script/SphericalHarmonics.html
Y0,0
Y1,0
Y1,+1
Y1,-1
Y2,+1
Y2,+2
Y2,0
Y2,-1
Y2,-2
z-Component of Angular momentum
What happens if we operate on the angular part of the wavefunction
with the Lz operator ?


L̂ z ψθ, φ   L̂ z Pl ,ml θ exp iml φ   L̂ z Yl ,m θ, φ 
l
 i


d
 d

Pl ,ml θ exp iml φ   iPl ,ml θ 
exp iml φ 
dφ
 dφ



 m l  Pl ,ml θ exp iml φ   m l Yl ,ml θ, φ 
Thus we see that the Spherical Harmonics are eigenfunctions of the Lzoperator with eigenvalues ml 
L̂ z Yl ,m θ, φ   ml Yl ,m θ, φ 
l
l
this shows that the z-component of the angular momentum is quantised
in units of 
What else can we learn from the angular part of the wavefunction ?
Total Angular Momentum
Since the angular momentum is a vector we can write the total angular
momentum as
L2  L2x  L2y  L2z
From this (after some math) we find that
2 

1


1



L̂2   2 

 sinθ  
2
2
sin
θ

θ

θ

 sin θ φ 

Let us operate on the angular part of the wavefunction and see what
result it gives us, that is
2 

1


1



L̂2 Yl ,ml θ, φ    2 
 Yl ,ml θ, φ 
 sinθ  
2
2
θ  sin θ φ 
 sinθ θ 
  2 l l  1Yl ,ml θ, φ 
this shows that the the angular part of the wave function is an
eigenfunction of the L2 operator, with eigenvalue l l  1 2
Total Angular Momentum
Since the square of a vector is equal to the square of its magnitude, this
means that the magnitude of the angular momentum can take on only
the values
L  L2  ll  1 2 where l  0, 1, 2, 3... 
and recall that no physical solution exists unless ml   l. So with this
restriction we have
L Z  mL 
where mL
 0,  1,  2, ...  l

The information we now have is the magnitude of the quantised angular
momentum Land the magnitude of the z-component of the angular
momentum.
For example, if l = 2 then
L  22  1   6 
and the z-components of the allowed angular momenta are
L Z  2,1 , 0, 1, 2
we can represent this graphically as follows
Space Quantisation
The angular momentum vector can only point in certain directions in
space  Space Quantisation
z
LZ  mL 
2

6
0
θ
L  ll  1 
-
- 2
L
cosθ  z 
L
 mL 
mL 

 θ  cos1
l l  1 
 ll  1 
Example
What is the minimum angle that the angular momentum vector may make
with the z-axis in the case where (a) l = 3 and (b) for l = 1?
When l = 3 the magnitude of the angular momentum is
L  33  1   12 
with 7 possible z-components, mL = -3, -2, -1, 0, 1, 2, 3. The angular
momentum vector will make the minimum angle with the z-axis when
the z-component is as large as possible, 3
Lz
3
 3 

1

cosθ 

 θ  cos 
  30
L
12
 12 
for l = 1:
L  11 1   2 
and then
Lz
1
 1 

1

cosθ 

 θ  cos 

45

L
2
2


L Z  3
θ
L  12
Notation
We can specify the angular momentum states of a single particle as
follows
l=
state
Number of states
0
s
1
1
p
3
2
d
5
3
f
7
4
g
9
5
h
11
6
i
13
so for example, if the particle has angular momentum l = 2, then it is
called a d-state. In this state the magnitude of the angular momentum is
L  22  1   6 
and its z-component can have the following values,
L Z  2,1 , 0, 1, 2
Note that each angular momentum state has a degeneracy of 2l+1 . Thus
for example, a d-state has a degeneracy of 5. this means that there are 5
states with l = 2 which all have the same energy.
Degeneracy
We have found that the energy levels are given by
me 4
1
me 4 1
13.6
En  


eV
2
2
2
2
2
2
8ε 0h n
n
24 πε 0   n
note that the energy levels are degenerate
n l mL state
since they do not depend explicitly on l and
1 0 0
1s
ml . Recall that n = 1, 2, 3, 4, …
2 0 0
2s
l = 0, 1, 2, 3, … n-1
-1
2p
1 0
2p
ml = - l, -(l -1), ….(l -1), l
+1
2p
so each energy level has a degeneracy of
3 0 0
3s
-1
3p
n 1
2
 2 l  1  n
1 1
3p
l 0
+1
3p
Example: the n = 2 state has a degeneracy of
-2
3d
-1
3d
4:
2 0
3d
a single 2-s state with l = 0, ml = 0 and a 3+1
3d
fold degenerate 2-p state with l = 1,
+2
3d
ml = -1, 0, 1
Energy
-13.6 eV
-3.40 eV
-3.40 eV
-1.51 eV
-1.51 eV
-1.51 eV
Normalisation
Just as in the case of one dimensional problems, the total probability of
finding the electron somewhere in space must be unity. Using the volume
element in polar coordinates, the nomalisation condition becomes

all
space
2
Ψ n,l,m L dV  1

π
2π
2
2 2
imL φ  imL φ
R
r
dr
P
sin
θdθ
e
e
dφ  1
 n,l
 l,mL

0
0
0
The last integral is just 2 so we are left with two integrals.

2 2
 Rn,l r dr  1 and
0
π
2 π  Pl,mL
2
sinθdθ  1
0
Using these two integrals we can find the correctly normalised Radial
and Angular wavefunctions
Example
Find the angular normalisation constant for P1,+1().
For l = 1, mL = +1 we have P1,+1() = sin  (see table). then
π
π
2
2
2
2 πA1,1  sinθ sinθdθ  2 πA1,1  sin3 θdθ  1
0
0
Carry out the integral, to find
 43  1
2
2 π A1
,1
 A1,1 
and the correctly normalised wavefunction is
3
P1,1 θ  
sinθ
8π
3
8π
Probability Density
What is the probability of finding the electron within a small volume
element dV at position r,, ?
2
2
P  ψn,l,m dV  ψn,l,m r 2 sinθdrdθdφ
2 2
2 2
2
 Rn,lYl,m r sinθdrdθdφ  Rn,l r dr Yl,m sinθdrdθdφ
If we want to find the probability of finding the electron at a distance
between r and r + dr (i.e. in a spherical shell of radius r and thickness
dr), then we just integrate over  and .
π 2π
2 2
2
Pdr  Rn,l r dr   Yl,m sinθdθdφ
0 0
But, since the spherical harmonics are normalised (by definition), so the
integral is equal to 1 and we then have
2
Pr dr  Rn,l r 2dr
or
2
Pr  Rn,l r 2
is the probability of finding the electron at a distance r from the nucleus
at any angle . What do these distributions look like ?
1s State
n = 1, l = 0
R1,0
2
Pr  R1,0 r 2
2s State
n = 2, l = 0
R2,0
Pr  R2,0
2 2
r
2p State
n = 2, l = 1
R2,1
2
Pr  R2,1 r 2
3s State
n = 3, l = 0
R3,0
2
Pr  R3,0 r 2
3p State
n = 3, l = 1
R3,1
2
Pr  R3,1 r 2
3d State
n = 3, l = 2
R3,2
2
Pr  R3,2 r 2
Summary of Key Points
1) The wavefunctions of the hydrogen atom can be expressed as the
product of three one dimensional functions in the variables r,  and 
ψr, θ, φ   R(r)  θ    φ 
2) The mathematical form taken by the wavefunction depends upon
three quantum numbers n, l, ml.
3) These three quantum numbers can only take integral values subject
to the following restrictions:
n = 1, 2, 3, … l < n, ml = 0, 1, 2 …  l
orbitals with l = 0, 1 and 2 are known as s, p and d orbitals respectively.
The radial solutions Rn,l(r) are given by the Associated Laguerre
Polynomials
The angular solutions are given by the Spherical Harmonics Yl,ml( ,)
where Yl,ml( ,) =Pl,ml().exp(i ml ), and Pl,ml() are the Associated
Legendre Polynomials.