Transcript Hydrogen 1

Quantum
Physics
2002
The Hydrogen Atom
Recommended Reading:
Harris Chapter 6, Sections 3,4
• Spherical coordinate system
•The Coulomb Potential
•Angular Momentum
• Normalised Wavefunctions
•Energy Levels
• Degeneracy
The Hydrogen Atom
The hydrogen atom is just an electron and a proton moving together
under a mutual electrostatic (Coulomb) attraction.
Bound system total energy, E, is less than zero
Mass of proton Mp = 1836me so to a first approximation we can
assume that the proton is fixed and the electron moves around it in a
circular orbit.
Potential energy of electron is given by
1 e2
Ur   
4 πε 0 r
r
(1)
Where r is the distance between the electron and the proton.
We now have a three-dimensional problem and we need three
coordinates to specify the position of the electron relative to the
proton.
The Coulomb Potential
We want to find the
wave functions and
allowed energies for
the electron when it
is confined to this
potential.
Note that :
U(r)
U(r)  0 as r  
and
U(r)  -  as r  0
r
1 e2
Ur   
4 πε 0 r
(2)
Spherical Coordinates (r,,)
Instead of the three
coordinates (x,y,z), we can
specify the position of the
electron by three Spherical
coordinates (r,,).
z
r

range 0  
r = radius,
 = polar angle range 0  
 = azimuthal angle
range 0  2
y

y = r sin  sin 
x
x = r sin  cos 
Schrodinger Equation in Spherical Coordinates
The time independent Schrodinger equation in three-dimensions is
2 2

 ψr, θ, φ   Uψr, θ, φ   Eψr, θ, φ 
2m
(3)
We must also express the Laplacian operator 2 in polar coordinates
(see additional notes) the result is that
1   2  
1  
 
1 2 
2
 

sinθ  
 r
 (4)


2
2
2
θ  sin θ φ
r r  r  sinθ θ 


therefore the Schrodinger equation in 3D polar coordinates is
1   2  
1  
 
1  2  2m
 r
  2 E  Uψ  0

 sinθ  
2
2
2
θ  sin θ φ  
r  r  r  sinθ θ 
(5)
this looks terrible but it in not much different from a particle in a 3d
box or the SHO equation.
Schrodinger Equation in Spherical Coordinates
We note that the potential energy of the electron only depends on its
radial distance from the proton and does not depend on the angles 
and , U = U(r). Because the force asociated with such a potential is
always directed radially inward or outward, it is called a central force
We try the same approach as before and use Separation of Variables,
that is we assume we can write the wavefunction as a product of three
one-dimensional wavefunctions
ψr, θ, φ   R(r)  θ    φ 
(6)
We now look at the partial derivatives we need in the Schrodinger
equation
ψ
r, θ, φ      R
r
r
ψ
r, θ, φ   R   
θ
r
2ψ
φ
2
r, θ, φ   R  
 2
r 2
Now substitute these into the Schrodinger equation (just a lot of
algebra) and we find
   2 R 
R  
 
r

sin
θ

 2


2
r  r sin θ θ 
θ 
r r 
2
R
2
 
2
r sin θ φ
2

2m

2
E  UR  0
(7)
Multiply across by r2sin2 and divide across by R.. and rearrange:
sin2 θ   2 R  sinθ  
  2m 2
1  2
2

r

 sinθ
  2 r sin θE  U 
R r 
r 
 θ 
θ  
 φ2
(8)
On the left hand side we have only functions of r and , while the
right hand side is only a function of , (Separation of Variables)
This can only hold for all values of r,  and  if both sides are equal
to a constant. We call this constant mL2. Then taking the R.H.S. we
have
1  2
 2
2
2  0
 mL 
 mL
(9)
2
 φ 2
φ
The the azimuthal equation describes how the wavefunction varies with
angle  and  are the azimuthal wave functions. This is the same
equation as that for a Particle on a Ring and Sphere which we solved
previously.
substituting equation (9) into equation (8) we get
  2m 2
sin2 θ   2 R  sinθ  
2
2
θ
sin

r

  2 r sin θE  U   mL



θ  
 θ 
r 
R r 
divide across by sin2 and rearrange
2
mL
1   2 R  2m 2
1
 
 

sin θ
r
  2 r E  U 


2
R r 
r  
θ 
sin θ  sin θ θ 
(10)
again we have managed to separate the variables, the LHS only
depends on the variable r while the RHS only depends on the
variable  .
So each side must be equal to a constant, this time we call the
constant l(l +1) . Setting the LHS of (10) equal to l(l + 1) gives
2



1   2  R  2m 
l l  1 
r

E

U

R0




2
2
2
r   
2m r

r r 


(11)
equation 11 is called the Radial Equation, it describes how the
wavefunction varies as the variable r changes R(r) are the Radial
wavefunctions.
Setting the RHS of (10) equal to l(l + 1) gives
2 

m
1  
  
L   0
 sin θ
   l l  1 2 
sin θ θ 
θ  
sin
θ

(12)
Equation (12) is the Angular Equation we encountered previously, and it
describes how the wave function varies with the polar angle . The
wavefunction () are called the Angular or Polar wavefunctions.
We now have three differential equations (9, 11 and 12) that provide
solutions for the r,  and  components of the wavefunction.
That is, we must solve these differential equations to find the
wavefunctions R(r), () and (). Once we have these solutions we
can then obtain the total, three-dimensional wavefunction from
ψr, θ, φ   R(r)  θ    φ 
We now look at each of these wavefunctions in turn starting with
equation 9
1  2
 2
2
 m l 
 m2
  0 (9)
l
2
2
 φ
φ
This is the same equation we found for the particle confined to
move on a ring and Sphere and we found that the solutions are:
 φ   A exp iml φ 
with m l  0,1,  2,  3,
(13)
Recall our definition of the z-component of the angular momentum
operator
d
d2
2
2
(14)
L̂ z  i
 L̂ z  
dφ
d2 φ
and using this definition equation (8) becomes
2  φ 
L̂2z  φ   m2

l
(15)
Equation (15) is an eigenvalue equation  the eigenvalues of the
Lz2 operator are
2
m2

l
Thus the wavefunction () represents states for which the zcomponent of the angular momentum Lz are quantised and can
only take on the values
Lz  m l 
where m l  0,  1,  2,
(16)
Visualisation of Azimuthal wavefunctions 1()
sin(1.)
cos(1.)
1
Imaginary
part
Real part
0
Angle 
2
sin(1.)
cos(1.)
1
Imaginary
part
Real part
+
-
+
2 ()
sin(2.)
cos(2.)
Imaginary
part
Real part
0
Angle 
2
2
Imaginary
part
Real part
+
-
-
+
+
+
-
Solution to the  - Equation, ()
We now look at the solutions to the -equation (eqn 16)
2 

mL
1  
  
  0
(16)
 sin θ
   l l  1 2 
sin θ θ 
θ  
sin
θ

Again this is a difficult equation to solve. It was first solved by the
mathematician Adrien Marie Legendre (1752 - 1833) and is named after
him: The Associated Legendre Equation.
Note that this equation depends on two quantum numbers l and mL, so
we have found another quantization. Applying appropriate boundary
conditions leads to the following restrictions on the quantum numbers l
and mL.
l  0, 1, 2, 3, 4, 
(17)
m l   l , (-l  1), - 2,-1,0,1,2,...(l - 1), l
i.e. the quantum number l must be zero or a positive integer, and for a
given value of l, mL can take on all integer values ranging from - l to + l.
For example, if l = 3 then mL can only have the values -3, -2, -1, 0, 1, 2, 3
What are the solutions of the The Associated Legendre Equation?
The solutions are a called the Associated Legendre Functions,
Pl,mL(cos ) they are polynomials that depend on the angle  and the
two quantum numbers l and ml. For each allowed set of quantum
numbers (l, mL) there is a solution l,mL(). The first few are given in
the following table
l
0
1
1
2
2
2
mL
Pl,mL(cos )
0
1
0
cos 
1
sin 
0 (3cos2 - 1)/2
 1 3cos sin 
2
3sin2
l
3
3
3
3
mL
0
1
2
3
Pl,mL(cos )
(5cos3 - 3cos)/2
3sin(5cos2 -1)/2
15cossin2
15sin3
the solutions to the angular equation are given by
l,mL() = Pl,mL(cos )
What do these wavefunctions look like? Exactly the same as a
particle on a sphere. Let us plot a few of them
Polar Plots to visualise the () wavefunctions
P1,0
l = 1, mL = 0
z

x-y plane
P1,1
l = 1, mL =  1
P2,0
l = 2, mL = 0
P2,1
l = 2, mL =  2
P2,1
l = 2, mL =  1
P3,0
P3, 2
l = 3, mL = 0
l = 3, mL =  2
P3,1
P3, 3
l = 3, mL =  1
l = 3, mL =  3
Solution to the Radial - Equation, R(r)
We now look for solutions to the radial-equation (eqn 11) rearranged as
2
2 1 

l l  1
(11)
 2 








r
R
r

R
r

U
(
r
)
R
r

ER
r


2
2m r r  r 
2m r 2
this equation is just expresses conservation of energy, the first two
terms are the kinetic energies due to radial and rotational motion
respectively, while the third term is the Potential energy. For the
hydrogen atom we just put in the Coulomb potential to get
2 1 
  2 l l  1
e2 
 2 

 Rr   ERr 
r
Rr   
2
2
2m r r  r 
4 πε 0r 
 2mr
(18)
Again equation 18 is very difficult to solve, it is known as the Associated
Laguerre equation after Edmund Nicolas Laguerre (1834 - 1886) who
was the first to solve it. The solutions are called the Associated Laguerre
Functions Rn,l(r), and they depend on two quantum numbers n and l.
Let us look at the solutions in the special case where l = 0 (no angular
momentum). In this case equation (18) becomes
d2R
2 dR 2m 
e2 


E 
R0
2
2
r dr  
4 πε 0r 
dr
(19)
we guess that the solutions to this equation have the form
 r 
Rr   A exp 

 a0 
(20)
where A is a normalisation constant and a0 is some constant, to be
determined that has dimensions of length. Substitute (20) into (19)
using the fact that
d2R R
 r 
dR
1
R


A exp 
  
and
2
dr
a0
a0
dr
a2
 a0 
0
 1 2m   2me2
2  1




E 

 0 (21)
 2

2
 4 πε  2 a0  r
a


0
 0
 
this equation can only be satisfied for all values of r if each of the terms
in brackets are equal to zero,
2
 2me
2 


0
 4 πε  2 a0 


0
(22)
rearrange equation 22
a0 
4 πε 0 2
me2
(23)
substuting equation 23 into the first term in equation (21) gives
E
2
2ma2
0

me 4
24 πε0 2  2
 E0  13.6eV (24)
Amazingly, equation (27) is the same result we obtained for the ground
state energy level of the hydrogen atom in the Bohr Model! While
equation (23) is the Bohr radius of the atom! We have now obtained
these parameters from the Schrodinger equation!
What do the full solutions of the radial equation look like, (the
assiciated Laguerre functions). Again they are polynomials in the
variable r, and depend on two quantum numbers n and l. The allowed
energies are given by
En  
2
2
2ma2
n
0

me 4
1
24 πε0 2  2 n2
where n is a positive integer, n = 1, 2, 3, 4, ...

13.6
n2
eV (25)
for a given value of n, the quantum numbers l and mL can take on
the values
l  0, 1, 2, (n  1)
m L   l , (-l  1), - 2,-1,0,1,2,...(l - 1), l
n
1
2
four degenerate states
since they all have the
same values of n and
therefore the same
energy En
l
0
0
1
3
0
1
2
mL
0
0
-1
0
+1
0
-1
1
+1
-2
-1
0
+1
+2
only one nondegenerate state
nine degenerate
states
Radial wave functions R(r)
Solutions to equation (18), the Associated Laguerre Polynomials
n
1
2
2
3
3
3
l Rn,l(r)
0
 r 
2

exp 
 a0 
a0 3 2
0


2
r 
r 
 1 
 exp 

2a0 3 2  2a0   2a0 
1
 r 

1
r 

 exp 

2a0 3 2  3a0   2a0 
0


1
4r
4r 2 
r 
2 



exp 


3
2
2
3a0   3a0 27a0   3a0 
1
 4 2 r 

1
r 
r 

 1 
 exp 

3
2
3a0   9a0  6a0   3a0 
2
 2 2r 
1

 exp  r 
 3a 
 27 5a2 
3
2

0
3a0  
0
Radial wave functions R(r), n = 1 and 2
R2,0
1s wavefunction
n = 2, l = 0
2s wavefunction
R1,0
n = 1, l = 0
R2,1
n = 2, l = 1
2p wavefunction
Radial wave functions R(r) for n = 3
3s wavefunction
3p wavefunction
R3,1
R3,0
n = 3, l = 0
n = 3, l = 1
3d wavefunction
R3,2
n = 3, l = 2
Since there is no upper limit to
the value n can have, there are
an infinite number of radial
wavefunctions, but they all have
finite energies En