Precursors to Modern Physics
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Transcript 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 Beikx
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
2
x 0
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
k1 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 U0 m 4 108 J/kg
The particle (you): m 65 kg v 4 m/s
E
1 2
mv 520 J U0 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 e210 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
me4
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
2 2 2 x, y,z U x, y,z x, y,z E x, y,z
2m x y z
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
1
F x
d 2F x
dx 2
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
Cx
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
2 2
nx2 2 ny
nz2 2
quantization: 2 2 2 2mE
2
Lx
Ly
Lz
Enx ,ny ,nz
nx2 ny2 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 ny2 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
nx
Lx
x, y,z Asin
z
There are three quantum number nx ,ny ,nz
that define an energy state and its wave
function.
, , Can anyone be 0?
The ground state: nx ,ny ,nz 111
2
z
2
2
2mL
2
Lx
1,1,1 x, y,z Asin
2
x sin
y sin
Ly Lz
Enx ,ny ,nz nx2 ny2 nz2
E2 ,1,1 E1,2 ,1 E1,1,2 6
2mL and
2
2
While the wave functions are
not the same:
2
L
x sin
L
y sin z
L
2
x sin
y sin z
L
L
L
z
E1,1,1 3
2mL2
1,2 ,1 Asin
Now a special symmetric case, a cube: Lx Ly Lz L
2
and nx ,ny ,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
2
y
for nx ,ny ,nz 2,1,1 nx ,ny ,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 ,ny ,nz 2,1,1 nx ,ny ,nz 1, 2,1 nx ,ny ,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
L L L
L L
1,2 ,1 Asin
x y
y sin z 1,1,2 Asin
L
L
x sin
L
2
y sin
L
z
yz
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
6
2
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
Polar coordinates are a natural choice:
1 e2
For example: U r
4 0 r
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
r2
2
2
2
r
csc
sin
csc
2
r r
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 2Φ
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
ml
0
2
0 , 1
6
0, 1, 2
Lz
0
0,
0, , 2
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
l l 1
2 d 2 d
r
R
r
R r U r R r ER r
2mr 2 dr dr
2mr 2
KEradial
PE
KErotation
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
me4
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.
l0
l 1
l 1
l2
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.