Transcript L - Purdue Physics

CHAPTER 7
The Hydrogen Atom
Orbital Angular momentum
Application of the Schrödinger
Equation to the Hydrogen Atom
Solution of the Schrödinger Equation
for Hydrogen
Werner Heisenberg
(1901-1976)
The atom of modern physics can be symbolized only through a partial differential
equation in an abstract space of many dimensions. All its qualities are inferential; no
material properties can be directly attributed to it. An understanding of the atomic world
in that primary sensuous fashion…is impossible.
- Werner Heisenberg
Orbital Angular Momentum Quantum Number ℓ
Classically, angular momentum is L  r  p.
But quantum-mechanically, the total orbital angular momentum L is an
operator, e.g.,

 
 
 
 

Lˆz  xˆ pˆ y  yˆ pˆ x  x  i

y

i


i
x

y






y

x

y

x






The eigenvalues of L̂ are functions
of ℓ:
L  (  1)
So, in an ℓ = 0 state:
L  0(1)  0
This disagrees with Bohr’s semiclassical planetary model of
electrons orbiting a nucleus, for
which L = nħ, where n = 1, 2, …
Classical orbits—which don’t
exist in quantum mechanics
Orbital Magnetic Quantum Number mℓ
mℓ is an integer. It determines
the z component of L:
Lz  m 
Possible total angular momentum
directions when ℓ = 2
Lz
2
m 2
Example: ℓ = 2:
m 1
L  (  1)  6
Only certain orientations of L are
possible.
And (except when ℓ = 0) we just
don’t (and cannot!) know Lx and Ly!
m 0
0
L2x  L2y

2
m  1
m  2
Application of the Schrödinger Equation
to the Hydrogen Atom
The potential energy of the electron-proton system is electrostatic:
V r   
e2
4 0 r
For Hydrogen-like atoms (He+ or Li++),
replace e2 with Ze2 (where Z is the
atomic number).
The three-dimensional time-dependent Schrödinger Equation:
2
  2  2  2 

i

 2  2  V 

2
t
2m  x
y
z 
where:
   ( x, y , z , t )
which becomes the three-dimensional time-independent
Schrödinger Equation:
1   2  2  2

 2  2

2
2m   x
y
z
2

  E V

where:
    x, y , z 
Spherical Coordinates
The potential (central force) V(r) depends only on the distance r
between the proton and electron, where r 2  x 2  y 2  z 2 .
As a result of this radial symmetry, we must transform to spherical
coordinates:
x  r sin  cos 
y  r sin  sin 
z  r cos 
The inverse relations:
r 2  x2  y 2  z 2
cos   z / r
tan   y / x
Polar
angle
Azimuthal
angle
The
Schrödinger
Equation in
Spherical
Coordinates
Transformed into
spherical coordinates,
the Schrödinger
equation becomes:
Also, use the reduced mass 
instead of the electron mass me
1   2  
1
 
 
1
 2 2 
 2  E  V   0
r
 2
 sin 
 2 2
2
2
r r  r  r sin   
  r sin  
Separable Solution
The wave function  is a function of r, , . This is a potentially
complicated function.
Assume optimistically that  is separable, that is, a product of three
functions, each of one variable only:
 (r , ,  )  R(r ) f ( ) g ( )

dR
 fg
r
dr

df
 Rg

d
 2
d 2g
 Rf
2

d 2
This would make life much simpler—and it turns out to work!
Solution of the Schrödinger Equation
for Hydrogen
Start with Schrodinger’s Equation:
1   2  
1
 
 
1
 2 2 
 2  E  V   0
r
 2
 sin 
 2 2
2
2
r r  r  r sin   
  r sin  

dR
Substitute:
 fg
r
dr

df
 Rg

d
 2
d 2g
 Rf
2

d 2
fg   2 R 
Rg d 
df 
Rf d 2 g 2
 2  E  V  Rfg  0
r
 2
 sin 
 2 2
2
2
r r  r  r sin  d 
d  r sin  d
Multiply both sides by r2 sin2 / R f g:
sin 2    2 dR  2
sin  d 
df  1 d 2 g
2
2

r
  2  E  V  r sin  
 sin 

R r  dr 
f d 
d  g d  2
Solution of the Schrödinger Equation for Hydrogen
2
sin 2    2 dR  2
sin

d
df
1
d
g


2
2

r

E

V
r
sin


sin








R r  dr  2
f d 
d  g d  2
r and  appear only on the left side and  appears only on the right side.
The left side of the equation cannot change as  changes.
The right side cannot change with either r or .
Each side needs to be equal to a constant for the equation to be true.
Set the constant to be −mℓ2
d 2g
2


m
g Azimuthal-angle equation
2
d
Sines and cosines satisfy this equation, but it’s convenient to choose a
complex exponential:
g() = exp(imℓ)
Quantization of mℓ
g() = exp(imℓ)
exp(imℓ) satisfies the azimuthal equation for any value of mℓ.
But a type of boundary condition exists because  and   2 are
actually the same angle, so:
g    g   2 
So:
Canceling:
exp(im  )  exp[im (  2 )]
1  exp(2 im )
mℓ must be an integer (positive or negative) for this to be true.
Solution of the Schrödinger Equation for H
2
sin 2    2 dR  2
sin

d
df
1
d
g


2
2

r

E

V
r
sin


sin








R r  dr  2
f d 
d  g d  2
1 d 2g
2
Now replace
2 by −mℓ :
g d
sin 2    2 dR  2
sin  d 
df 
2
2
2

r

E

V
r
sin


sin



m






2
R r  dr 
f d 
d 
Divide by sin2 and rearrange:
m2
1   2 dR  2 r 2
1
d 
df 
r
  2  E V   2 
 sin 

R r  dr 
sin  f sin  d 
d 
Now, the left side depends only on r, and the right side depends only
on . We can use the same trick again!
Set each side equal to the constant ℓ(ℓ + 1).
Separation of the Schrödinger Equation for H
m2
1   2 dR  2 r 2
1
d 
df 
r

E

V


sin






2 
R r  dr 
sin 2  f sin  d 
d 
Set each side equal to the constant ℓ(ℓ + 1).
1 d  2 dR  2 
r
 2
2
r dr  dr 
2

E V 
2

1 d 
df  
 sin 

sin  d 
d  


The Associated
Laguerre equation
 1 
 R  0 Radial equation
2
r

m2 
 1  2  f  0
sin  
Polar-angle equation
We’ve separated the Schrödinger equation into three ordinary secondorder differential equations, each containing only one variable.
Solution of the Radial Equation for H
The radial equation is called the associated Laguerre equation and the
solutions R are called associated Laguerre functions. There are
infinitely many of them, for values of n = 1, 2, 3, …
Assume that the ground state has n = 1 and ℓ = 0. Let’s find this solution.
The radial equation becomes:
1 d  2 dR  2 
r
  2  E V  R  0
2
r dr  dr 
The derivative of r 2
dR
yields two terms:
dr
d 2 R 2 dR 2 
e2 
 R  0

 2  E 
2
dr
r dr  
40 r 
Solution of the Radial
Equation for H
r /a
Try a solution R(r )  Ae 0
A is a normalization constant.
a0 is a constant with the dimension of length.
Take derivatives of R and insert them into the radial equation.
d 2 R 2 dR 2 
e2 
 R  0

 2  E 
2
dr
r dr  
40 r 

 1 2    2 e 2
2 1
 2  2 E
   0
2

a


a0  r
 0
  40 
To satisfy this equation for any r, both expressions in parentheses
must be zero.
Set the second expression equal to zero
and solve for a0:
Set the first expression equal to zero and
solve for E:
Both are equal to the Bohr results!
40 2
a0 
e 2
2
E
  E0
2
2 a0
Hydrogen-Atom Radial Wave Functions
The solutions
R to the radial
equation are
called
Associated
Laguerre
functions.
There are
infinitely
many of
them, for
values of
n = 1, 2, 3, …
and ℓ < n.
The Polar-Angle Equation
2
m 
1 d 
f  
 sin      1  2  f  0
sin  d 
  
sin  
Polar-angle equation
Note the presence of mℓ. Recall that the solutions to the azimuthal
equation, exp(imℓ), also involve mℓ.
So solutions to the polar- and azimuthal-angle equations are linked.
We usually group these solutions together into functions called
Spherical Harmonics:
Y m ( ,  )  f m ( ) g m ( )
Spherical harmonics
Solutions to the polar-angle equation above involve sums and
products of sines and cosines.
Normalized
Spherical
Harmonics
Spherical
harmonics are
the solutions
to the
combination
polar- and
azimuthalangle
equations.
Complete Solution of the Radial,
Angular, and Azimuthal Equations
The total wave function is the product of the radial wave function Rnℓ
and the spherical harmonics Yℓmℓ and so depends on n, ℓ, and mℓ.
The wave function becomes:
 n m (r , ,  )  Rn (r ) Y m ( ,  )
where only certain values of n, ℓ, and mℓ are allowed.