Download Precursors to Modern Physics

Hydrogen Atom and QM in 3-D
1.
2.
3.
HW 8, problem 6.32 and A review of the
hydrogen atom
Quiz 10.30
Topics in this chapter:


Today

The hydrogen atom
The Schrödinger equation in 3-D
The Schrödinger equation for central force
HW 8, problem 6.32
Following discussion in the textbook, page
1   inc  refl  1eikx  Beikx
196 – 199, to the left of the step ( x < 0 ):
To the left of the step ( x < 0 ):
 2   trans  Ceik' x
Apply smoothness condition:  1 x0   2
x 0
d 1
dx
Solve for B and C: B 
x 0
d 2

dx
1  ikx
4 ik' x


e


e
So: refl
trans
3
3
R
 refl
 inc
2
1 3


12
2
k1  B   k' C
x 0
k  k'
with k  2mE
k  k'
B  1 3 and C  4 3
2
 1 B  C
1

9
and k'  2m  E  34 E 
HW 9, problem 6.32
Conditions given:
Rectangular barrier : L  6 1011 m U 0 m  4 108 J/kg
The particle (you): m  65 kg v  4 m/s
 E
1 2
mv  520 J U 0  4 108 J/kg  m  2.6 1010 J
2
Because E  U 0 you will have to “tunnel” to Jupiter.
E 
E  2 L
The probability that you end up there is: T  16
1 
e
U0  U0 
Which is: T  e210  0
52
2 m E U 0 
The electrical potential:
The hydrogen atom
1 e2
U r   
4 0 r
E photon  Eatom,i  Eatom, f
To solve the Schrödinger equation
in a 3-D polar system is trivial.
Let’s start from one of its
solutions, the energy level:
En  

me 4
2  4 0 
13.6 eV
n2
2
2
1
, n  1, 2,3,...
n2
 1
1 
 13.6 eV  2  2 
 n f ni 


1


E photon
hc
13.6 eV  1
1 
 2  2 
hc  n f ni 
 1
1 
 1.097 107 m 1  2  2 
 n f ni 



Example 7.2
The Schrödinger equation, from 1-D to 3-D
2
d 2  x 
 U  x   x   E  x 
1-D:

3-D:
If choose Cartesian coordinates:
2m
dx
2
 2
2
2 




  x, y,z   U  x, y,z   x, y,z   E  x, y,z 
2m  x 2 y 2 z 2 
2
Or from:
2
2
2
  2 2 2
x y z
2
 
2
2m
 2  x, y,z   U  x, y,z   x, y,z   E  x, y,z 
In a more general case, the coordinate is represented by a vector
The Schrödinger equation in 3-D: 
Now the normalization condition is
2
2m
r
2  r   U  r   r   E  r 

  r  dV  1
2
all space
For bound states, the standing wave is a 3-D standing wave, with
energy quantized by 3 quantum numbers, each for one dimension.
The 3-D infinite well, just as an example
 0 0  x  Lx ,0  x  Ly ,0  x  Lz
U r  
otherwise

   x, y,z   0
Here Cartesian coordinates are natural choice, so:
2
2
2
  2 2 2
x y z
2

 2
2
2 




  x, y,z   U  x, y,z   x, y,z   E  x, y,z 
2m  x 2 y 2 z 2 
2
Inside the box, if we express the wave function like this:
  x, y,z   F  x  G  y  H  z  and because U  r   0
The Schrödinger equation becomes:
This is only possible if
2
2
2
1  F  x
1  F  y
1  H z
2mE




2
F  x  x 2
G  y  y 2
H  z  z 2
2
1 d F  x
 Cx
F  x  dx 2
and
2
1 d F  y
 Cy
G  y  dy 2
and
1
H  z
d 2H  z
dz 2
 Cz
n  
 nx 
n 
x  G  y   Ay sin  y y  H  z   Az sin  z z 
 Ly 
 Lz 
 Lx 


This leads to three solutions: F  x   Axsin 
And the energy quantization:
2 2
nx2 2 ny
nz2 2
2mE
 2  2  2  2
Lx
Ly
Lz

Enx ,ny ,nz
 nx2 n y2 nz2   2 2
 2  2  2 
 Lx Ly Lz  2m


Discussion about the energy levels and their wave functions
Energy levels:
Enx ,ny ,nz
 nx2 n y2 nz2   2 2
 2  2  2 
 Lx Ly Lz  2m


Now take
x
,n y ,nz
  n y    nz 
x  sin 
y sin
 Ly   Lz
 
 
 n x
 Lx
 x, y,z   Asin 

z

There are three quantum number nx ,ny ,nz
that define an energy state and its wave
function.
The ground state: nx ,ny ,nz  1,1,1 Can anyone be 0?
2
2
2
 2mL
2
E2 ,1,1  E1,2 ,1  E1,1,2  6
While the wave functions are
not the same:
 2
 L
 2mL and
2
E1,1,1  3

 
x  sin 
 L
  
y  sin  z 
 L 
    2    
x  sin 
y  sin  z 
L
L

 
 L 
Now a special symmetric case, a cube: Lx  Ly  Lz  L
2 2
2mL2
 1,2 ,1  Asin 
     
 1,1,1  x, y,z   Asin  x  sin  y  sin  z 


 Lx   Ly   Lz 

2
z
and nx ,n y ,nz  1,1, 2
The energy levels are the
same:
2 2
 2 ,1,1  Asin 
 1
1
1 
E1,1,1   2  2  2 
 Lx Ly Lz  2m


2
Enx ,ny ,nz  nx2  ny2  nz2
2
y
for nx ,n y ,nz  2,1,1 nx ,n y ,nz  1, 2,1
Eave function:
n

Enx ,ny ,nz  n  n  n
2
x
2
2
2mL2
  
x  sin 
L  L
 1,1,2  Asin 
  2
y  sin 
  L
This is called degeneracy.
The energy levels are called
degenerated.

z

Energy levels degenerated and splitting (symmetry broken)
E
The symmetric case: Lx  Ly  Lz  L
leads to energy levels degeneration.
For example, for:
2,1,1
1, 2,1
1,1, 2
nx ,n y ,nz  2,1,1 nx ,n y ,nz  1, 2,1 nx ,n y ,nz  1,1, 2
E2 ,1,1  E1,2 ,1  E1,1,2  6
2
2
1,1,1
2
2mL
Plot the energy levels
The wave functions are different, but symmetric.
    2    
2      
x  sin 
y  sin  z   2 ,1,1  Asin 
x  sin  y  sin 
L   L  L 
 L  L  L
 1,2 ,1  Asin 
x y


z   1,1,2  Asin 

L
 
x  sin 
 L
  2
y  sin 
  L

z

yz
When the symmetry is broken: for example: Lx  Ly  L, Lz  0.9L  energy level splits
Energy levels degenerated and splitting (symmetry broken)
Symmetry broken: Lx  Ly  L, Lz  0.9L
Symmetric: Lx  Ly  Lz  L
E1,1,1  3
2
1  2 2
2 2

E111
3
, ,  2

0.92  2mL2
2mL2

2
2mL2
E2 ,1,1  E1,2 ,1  E1,1,2  6
2
1  2 2
2 2

E2 ,11,  E1,2 ,1   5 
6

0.92  2mL2
2mL2

2
2mL2
E1,1,2

22   2 2
2 2
 2 2 
6
2
0
.
9
2
mL
2mL2


E11, ,2  E2 ,11,  E1,2 ,1
degenerated
Reveals more
details
splitting
E
1,1, 2
2,1,1
2,1,1
1, 2,1
1,1, 2
1,1,1
1,1,1
1, 2,1
The Schrödinger equation for central force
Central force:
U  U r 
For example:
1 e2
U r   
4 0 r
Polar coordinates are a natural choice:
x  rsin cos
y  rsin sin
z  rcos
dV  r 2sin drd d
1 e2
ˆr
the force: F  
2
4 0 r

2
2m
with:
2  r, ,   U  r   r, ,   E  r, , 
2 
1   2  
 
 
2 
2
r

csc

sin


csc

 




r 2  r  r 
 
 
 2 
Solve Schrödinger equation for central force
Separate variables:   r, ,   R  r  Θ   Φ  

2
2m
2  r, ,   U  r   r, ,   E  r, , 
Becomes three equations:
d  2 d 
2mr 2
 r
 R  r   2  E  U  r  R  r   CR  r 
dr  dr 
radial equation:
azimuthal equation:
d 2 Φ  
polar equation:
sin
d
2
  DΦ  
Both C and
are constants.
D
dΘ   
d 
2
sin


  Csin  Θ    DΘ  
d 
d 
The solution to the azimuthal equation is (the simplest):
Φ    eiml
ml  0,1,2,3,...
ml2  D
The z component Lz of the particle’s angular moment
Lz  ml
L
is quantized:
ml  0, 1, 2, 3,... is called the magnetic quantum number.
The angular momentum and its quantum numbers
The solution to the polar equation:
sin
dΘ   
d 
2
2
sin


  Csin  Θ    ml Θ  
d 
d 
Leads to the quantize the magnitude
L  l l  1
of the particle’s momentum:
Because Lz  L
we have ml  0,1,2,3,..., l
l
0
1
2
L
0
0
0
2
0 , 1
6
0 ,  1,  2
0,
0,  ,  2
ml
Lz
Where l  0,1,2,3,... is
called the orbital quantum
number, and C  l l  1
Example 7.3, 7.4 on black board.
The shape of an atom of central force
The angular probability density for a central force:
The radial equation of a central force
radial equation:
d  2 d 
2mr 2
 r
 R  r   2  E  U  r  R  r   CR  r 
dr  dr 
rearrange:
 2 d  2 d 
r
 R r  
2mr 2 dr  dr 
KE radial
2
l  l  1
2mr 2
R  r   U  r  R  r   ER  r 
PE
KE rotation
Here one needs to know the potential
explicitly. Assume hydrogen atom:
This leads to the solution for energy
levels, and the principal quantum
number. Degenerate (only depends on
n, not l and ml )
The relationship of the three quantum
numbers (magnetic, orbital and principal):
1 e2
U r   
4 0 r
En  

me 4
2  4 0 
2
2
1
, n  1, 2,3,...
n2
13.6 eV
n2
n  1, 2,3,...
l  0,1, 2,3,...,n 1
ml  0, 1, 2, 3,..., l
The electron “cloud” in the hydrogen atom
Electron probability density.
Surfaces of constant probability
density.
l 0
l 1
l 1
l 2
Radial probability: the “size” of the atom
Radial probability:
P  r   r 2 R2  r 
The Bohr radius:
a0
4 0 


me
2
Example 7.6
2
 0.0529 nm
Review questions


What are the steps in working out the
Schrödinger equation for hydrogen atom?
How do you connect the quantum numbers
introduced in the solutions with those learned
from a chemistry class?
Preview for the next class (10/28)

Text to be read:


8.1, 8.2 and 8.3
Questions:

What had the Stern-Gerlach experiment been designed to
prove? What it actually proved?
Homework 11, due by 11/6
Problems 7.37, 7.38 and 7.45 on page 281.