# Physics 214 Lecture 11 - PowerPoint

Document Sample

```					“It was almost as incredible as if
you fired a 15-inch shell at a piece
of tissue paper, and it came back
to hit you!”
--E. Rutherford
(on the „discovery‟
of the nucleus)
Overview of the Rest of the Course

Up to now:
   General properties and equations of quantum mechanics
   Time-independent Schrodinger’s Equation (SEQ) and eigenstates.
   Time dependent SEQ, superposition of eigenstates, time dependence.
   Collapse of the wave function, Schrodinger’s cat
   Tunneling

This week:
 3 dimensions, angular momentum, electron spin, H atom
 Exclusion principle, periodic table of atoms
Next week:
 Molecules and solids, consequences of Q. M.
 Metals, insulators, semiconductors, superconductors, lasers, . .

Final Exam: Monday, March 8
Homework 6: Due next Saturday (March 6), 8 am
Lecture 11:
3D Potentials and the Hydrogen Atom

1            r / ao
 ( x, y, z )   ( x) ( y) ( z )                                  (r )                    e
ao
3

z
11   1

r = a0

L                                                 P(r)x) 0.5
L                     g(

x

0
L
0
0
0   0            r2              4a0 4
0             x                      4

                       
2
h               2       2       2                                13.6 eV
En                              nx  n y  nz                              En 
x n y nz               2                                                              2
8mL                                                                n
Today

3-Dimensional Potential Well:
 Product Wavefunctions
 Concept of degeneracy

Schrödinger’s Equation for the Hydrogen Atom:
 Semi-quantitative picture from uncertainty principle
 Ground state solution*
 Spherically-symmetric excited states (“s-states”)*

*contains details beyond what we expect you to learn here
Quantum Particles in 3D Potentials
A real (2D) “quantum dot”

So far, we have considered quantum
particles bound in one-dimensional
potentials. This situation can be
applicable to certain physical systems
but it lacks some of the features of
many real 3D quantum systems, such
as atoms and artificial structures.
http://pages.unibas.ch/phys-meso/Pictures/pictures.html

One consequence of confining a quantum particle in
two or three dimensions is “degeneracy” -- the existence
of several quantum states at the same energy level.

To illustrate this important point in a simple system, we extend our
favorite potential -- the infinite square well -- to three dimensions.
Particle in a 3D Box (1)
The extension of the Schrödinger Equation (SEQ) to 3D is
straightforward in Cartesian (x,y,z) coordinates:

  2  2  2 
2

     2            2 
 U ( x, y , z )  E             where    ( x, y , z )
2m  x   y      z 
2

1
Kinetic energy term
2m
p   2
x
 py  pz
2    2


Let’s solve this SEQ for the particle in a 3D box:
z
    outside box, x or y or z < 0
U(x,y,z) =    0    inside box
L                                     outside box, x or y or z > L
x
y                L                       This simple U(x,y,z) can be “separated”:
L
U(x,y,z) = U(x) + U(y) + U(z)
U =  if any of the three terms = .
Particle in a 3D Box (2)

Whenever U(x,y,z) can be written as the sum of functions of the individual
coordinates, we can write some wave functions as products of functions of
the individual coordinates: (discussion is in the supplementary slides)

 ( x, y , z )  f ( x )g ( y )h( z )                                 2D wave functions:
n         ny  
For the 3D square well, each function is                                                      sin  x x  sin      y
 L         L    
simply the solution to the 1D square well
(nx,ny) = (1,1)
problem:                                                                                 L
(nx,ny) = (1,2)
L
2                          y
n                                                                                         y
2
h  n 
fn ( x )  N sin  x x    Enx       x 
0
 L              2m  2L                                                              0
x                            x
Similarly for y and z.                                                                             L                           L

(nx,ny) = (2,1)               (nx,ny) = (2,2)
Each function contributes to the energy.                                                L                             L
The total energy is the sum:                                                      y                             y

Ex + Ey + Ez = E                                                           0                            0

x                              x
Particle in a 3D Box (3)

The energy eigenstates and energy values in a 3D cubical box are:

 nx         ny         nz         z
  N sin      x  sin      y  sin     z
 L           L           L   
2
h
En
x ny nz

8mL
2   
 nx  ny  nz
2    2       2
                  L
x
where nx,ny, and nz can each have values 1,2,3,….        y         L
L

This problem illustrates 2 important points.

 Three quantum numbers (nx,ny,nz) are needed to
identify the state of this three-dimensional system.
That is true for every 3D system.

 More than one state can have the same energy: “Degeneracy”.
Degeneracy reflects an underlying symmetry in the problem.
3 equivalent directions, because it’s a cube, not a rectangle.
Act 1
Consider a particle in a 2D well, with Lx = Ly = L.
1. Compare the energies of the (2,2), (1,3), and (3,1) states?
a. E(2,2) > E(1,3) = E(3,1)
b. E(2,2) = E(1,3) = E(3,1)
c. E(2,2) < E(1,3) = E(3,1)

2. If we squeeze the box in the x-direction (i.e., Lx < Ly)
compare E(1,3) with E(3,1).
a. E(1,3) < E(3,1)
b. E(1,3) = E(3,1)
c. E(1,3) > E(3,1)
Act 1: Solution
Consider a particle in a 2D well, with Lx = Ly = L.
1. Compare the energies of the (2,2), (1,3), and (3,1) states?
a. E(2,2) > E(1,3) = E(3,1)
b. E(2,2) = E(1,3) = E(3,1)     E(1,3) = E(3,1) = E0 (12 + 32) = 10 E0

c. E(2,2) < E(1,3) = E(3,1)     E(2,2)       = E0 (22 + 22) = 8 E0
2
h
E0            2
8mL
2. If we squeeze the box in the x-direction (i.e., Lx < Ly)
compare E(1,3) with E(3,1).
a. E(1,3) < E(3,1)
b. E(1,3) = E(3,1)
c. E(1,3) > E(3,1)
Act 1: Solution
Consider a particle in a 2D well, with Lx = Ly = L.
1. Compare the energies of the (2,2), (1,3), and (3,1) states?
a. E(2,2) > E(1,3) = E(3,1)
b. E(2,2) = E(1,3) = E(3,1)     E(1,3) = E(3,1) = E0 (12 + 32) = 10 E0

c. E(2,2) < E(1,3) = E(3,1)     E(2,2)        = E0 (22 + 22) = 8 E0
2
h
E0            2
8mL
2. If we squeeze the box in the x-direction (i.e., Lx < Ly)
compare E(1,3) with E(3,1).
a. E(1,3) < E(3,1)      Because Lx < Ly, for a given n, the x
b. E(1,3) = E(3,1)      motion contributes more to E than the y
motion. This effect is larger for larger nx.
c. E(1,3) > E(3,1)
Therefore, E(3,1) > E(1,3).

Example: Lx = ½ , Ly = 1. E(1,3)  124 + 321 = 13                 We say, “The degeneracy
E(3,1)  324 + 121 = 37                 has been lifted.”
Energy Level Exercise (1)
z
Now back to the 3D cubic box:
Show energies and label (nx,ny,nz) for the first 11
states of the particle in the 3D box, and write the
L
degeneracy D for each allowed energy.
x
Use Eo= h2/8mL2.
y                                L
L

E          (nx,nynz)

n                               
2
h                    2       2           2
En                                                n y  nz
x n y nz                    2        x
8 mL
nx,ny,nz = 1,2,3,...

6Eo          (2,1,1) (1,2,1) (1,1,2)   D=3
The first four are given.
3Eo          (1,1,1)                   D=1
Exercise (1) Solution
z
Now back to the 3D cubic box:
Show energies and label (nx,ny,nz) for the first 11
states of the particle in the 3D box, and write the
L
degeneracy D for each allowed energy.
x
Use Eo= h2/8mL2.
y                                L
L

E          (nx,nynz)

12Eo          (2,2,2)                   D=1

n                               
2
11Eo          (3,1,1) (1,3,1) (1,1,3)   D=3                                 h                    2       2           2
En                                                n y  nz
x n y nz                    2        x
8 mL
9Eo          (2,2,1) (2,1,2) (1,2,2)   D=3                        nx,ny,nz = 1,2,3,...

6Eo           (2,1,1) (1,2,1) (1,1,2)   D=3

3Eo           (1,1,1)                   D=1
Act 2
For a cubical box, we just saw that the 5th energy level is at 12 E0, with
a degeneracy of 1 and quantum numbers (2,2,2).

1. What is the energy of the next energy level?
a. 13E0     b. 14E0      c. 15E0

2. What is the degeneracy of this energy level?
a. 2        b. 4         c. 6
Act 2: Solution
For a cubical box, we just saw that the 5th energy level is at 12 E0, with
a degeneracy of 1 and quantum numbers (2,2,2).

1. What is the energy of the next energy level?
a. 13E0     b. 14E0      c. 15E0
E1,2,3 = E0 (12 + 22 + 32) = 14 E0

2. What is the degeneracy of this energy level?
a. 2        b. 4         c. 6
Act 2: Solution
For a cubical box, we just saw that the 5th energy level is at 12 E0, with
a degeneracy of 1 and quantum numbers (2,2,2).

1. What is the energy of the next energy level?
a. 13E0     b. 14E0      c. 15E0
E1,2,3 = E0 (12 + 22 + 32) = 14 E0

2. What is the degeneracy of this energy level?
a. 2        b. 4         c. 6

Any ordering of the three numbers will give the
same energy. Because they are all different
(distinguishable), the answer is 3! = 6.
Exercise (2)
z
Now consider a non-cubic box:
The box is stretched along the y-direction.        L1
What will happen to the energy levels?                       x

L2 > L1
y
E                                            L1

11Eo

9Eo

6Eo

3Eo
Exercise (2) Solution
z
Now consider a non-cubic box:
The box is stretched along the y-direction.                                                                  L1
What will happen to the energy levels?                                                                                    x

2                                       2
h                                       h
En
x ny nz

8mL
2   n   x
2
 nz
2
  8mL  n 
2   y
2

y
L2 > L1
1                                       2
E                                                                                                    L1

11Eo
The others are left for you.
9Eo                                                                                1: The symmetry is “broken” for y, so
the 3-fold degeneracy is lowered. A
2-fold” degeneracy remains,
6Eo                    (2,1,1) (1,1,2) D=2                                            because x and z are still symmetric.

(1,2,1)         D=1                                         2: There is an overall lowering of
3Eo                                                                                   energies due to decreased
(1,1,1)                                 D=1                    confinement along y.
Another 3D System:            The Atom
-electrons confined in Coulomb field of a nucleus
Early hints of the quantum nature of atoms:
   Discrete Emission and Absorption spectra                        Atomic hydrogen
   When excited in an electrical discharge, atoms
emitted radiation only at discrete wavelengths
   Different emission spectra for different atoms                  (nm)

   Geiger-Marsden (Rutherford) Experiment (1911):
Au
    Measured angular dependence of a particles                          (He ions)
scattered from gold foil.                                              
   Mostly scattering at small angles                           supported the
v
“plum pudding” model. But…
   Occasional scatterings at large angles Something massive in there!

   Conclusion: Most of atomic mass is
concentrated in a small region of the atom             a nucleus!
Rutherford Experiment
Hydrogen atom qualitatively
Why doesn’t the electron collapse into the nucleus,
where its potential energy is lowest?

We must balance two effects:
 As the electron moves closer to the nucleus,             e
2

U  
its potential energy decreases:                              r

2
 However, as it becomes more and more           p                      KE 
confined, its kinetic energy increases:              r                          2m r
2

e
2                   2

Therefore, the total energy is:                  E  KE  PE                         2

2m r                r
2
This function has a minimum at:                  r                     a 0  0 .0 5 3 n m
m e
2

At this radius, E   m  e   1 3 .6 e V
2   4
The “Bohr radius” of the H atom.
2
2
This is the ground state energy of the hydrogen atom.
Heisenberg’s uncertainty principle prevents the atom’s collapse.
Act 3
Consider an electron around a nucleus that has two protons,
like an ionized Helium atom.
1. Compare the “effective Bohr radius” a0,He with the usual
a. a0,He > a0
Hint:
b. a0,He = a0                    Look at the expression for a0 on
the previous slide. What changes
c. a0,He < a0                    when we change H to He?

2. What is the ratio of ground state energies E0,He/E0,H?
a. E0,He/E0,H = 1
b. E0,He/E0,H = 2
c. E0,He/E0,H = 4
Act 3 - Solution
Consider an electron around a nucleus that has two protons,
like an ionized Helium atom.
1. Compare the “effective Bohr radius” a0,He with the usual
Look at how a0 depends on the charge:
a. a0,He > a0                             2                         2
a0
a0                a0, He               
m e                    m (2e)e
2
b. a0,He = a0                                                                  2
c. a0,He < a0                  This should make sense:
more charge  stronger attraction
 electron “sits” closer to the nucleus
2. What is the ratio of ground state energies E0,He/E0,H?
a. E0,He/E0,H = 1
b. E0,He/E0,H = 2
c. E0,He/E0,H = 4
Act 3 - Solution
Consider an electron around a nucleus that has two protons,
(an ionized Helium atom).
1. Compare the “effective Bohr radius” a0,He with the usual
Look at how a0 depends on the charge:
a. a0,He > a0                                2                            2
a0
a0                   a0, He                   
m e                       m (2e)e
2
b. a0,He = a0                                                                          2
c. a0,He < a0                   This should make sense:
more charge  stronger attraction
 electron “sits” closer to the nucleus
2. What is the ratio of ground state energies E0,He/E0,H?
a. E0,He/E0,H = 1               Clearly the electron will be more tightly
b. E0,He/E0,H = 2               bound, so |E0,He| > |E0,H| . How much
c. E0,He/E0,H = 4               more tightly? Look at E0:
 e                         (2e)e
2

E0, H                    E0, He          ao
 4 E0, H
2ao                           2   2

In general, for a “hydrogenic” atom with Z protons: E0, Z  Z 2 E0, H
Potential Energy for the Hydrogen Atom

To solve this quantum mechanics problem, we                       U(r)
must specify the potential energy of the electron.
We assume that the Coulomb force binds the                   0                                       r
electron to the nucleus.

 e2
This problem does not separate in Cartesian                                    U (r )  
coordinates, because we cannot write                                                         r
U(x,y,z) = Ux(x)+Uy(y)+Uz(z).
1
             9  10 Nm / C
9         2       2

However, we can separate the potential in                              4 0
spherical coordinates (r,,f), because:

U(r,,f) = Ur(r) + U() + Uf(f)        Therefore, we will be able to write:
e
  r , , f   R  r        f 
2

            0    0
r

Question: How many quantum numbers will be needed
to describe the hydrogen wave function?
Product Wave Function in Spherical Coordinates
We saw that because U depends only on the radius,
the problem is separable. The hydrogen SEQ can                                      
be solved (but not by us). We will show you the
solutions and discuss their physical significance.
r           z
We can write:                                                                           y
 nlm  r , , f   R nl  r  Y lm   , f                x       f
There are three quantum numbers:
 n “principal” (n  1)
 l “orbital”    (0  l < n)                             This is what we called
 m “magnetic” (-l  m  +l)                                   f  before.

The Ylm are called “spherical harmonics.”
Today, we will only consider l = 0 and m = 0. These are called “s-states”.
This simplifies the problem, because Y00(,f) is a constant and the wave
function has no angular dependence:

 n 00 (r , ,f )  Rn 0 (r )
Note:
Some of this nomenclature
dates back to the 19th century,
and has no physical significance.
Here are graphs of the s-state wave functions, Rno(r) , for the electron in
the Coulomb potential of the proton. The zeros in the subscripts are a
reminder that these are states with l = 0 (zero angular momentum!).

1                                                  1
1                                                 1                                                 3

2
R10                                                  R20
f( x) 0.5
0.5
d4( x)
R30
h( x)

0                                                0
0   00                                            .2
0                                             0
.5

00                2                  4
00                5          10a0
10            00      5
r    10   15a0
15
rx                 4a0                              r                                  0        x          15
0                                4                  0                x           10
            r    r / 3 a0
2

 r / a0                                          r   r / 2a0                     2r
R1,0 (r )  e                                   R2,0 (r )   1      e           R3,0 (r )   3      2      e
2a0                              a0     3a0  
                                                      

13.6 eV
2

a0                         0.053 nm                                En 
me e
2                                                                  2
n
Probability Density of Electrons
2 = Probability density = Probability per unit volume  Rn02 for s-states.
The density of dots plotted below is proportional to Rn02.
1s state                               2s state

1                                        1
1                                       1

R10                             0.5       R20
f( x) 0.5                               h( x)

0

0   00                                  .2
0
00                                           0          10a0
4a0                                 rx
0    5     10
rx
2     4
0          4                              0         10
Summary of wave functions and radial probability densities
for some s-states.
1
1
1                                     1
P10                      The radial probability density has an extra
R10
factor of r2 because there is more volume
f( x) 0.5                                  g( x) 0.5
at large r. That is, P(r)  r2|(r)|2.

0
00 0                   4a0
0
00
0
This means that:
00          r
2             4
0
2
r
4
4a0
1
1
x        4                        0           x                  4
The most likely r is not 0 !!!
R20                                       .5

0.4
P20             Even though that’s where
h( x)
0.5                                                                                                           |(r)|2 is largest.
h2( x)
0.2

00
.2

00
0                    5
10a0
10                         00 0
0

10a0
r   x            10
0
0   r
5
x
10
10
3

P30
2               R30
( x)

0
.5
00      0           5
r   x
10
15a0
15
15                            0             r              20a0

Example: Optical Transitions in Hydrogen
An electron, initially excited to the n = 3 energy
E
level of the hydrogen atom, falls to the n = 2 level,
0
emitting a photon in the process. What are the                   n=3
energy and wavelength of the photon emitted?                     n=2

n=1

-15 eV
Solution
An electron, initially excited to the n = 3 energy
E
level of the hydrogen atom, falls to the n = 2 level,
0
emitting a photon in the process. What are the                                      n=3
energy and wavelength of the photon emitted?                                        n=2

13.6 eV
En            2
n
n=1
 1   1        
Eni n f    13.6  2  2         eV                  -15 eV
n   nf        
 i            
1 1
E photon  E32  13.6    eV  1.9eV                         Atomic hydrogen
9 4
hc            1240eV  nm
                                  656nm
E photon          1.9eV
 (nm)

Question:
Which transition is this?
You will measure several transitions in Lab.             = 486 nm)
Next week: Laboratory 4
Cross Focusing       Light Collimating
Eyepiece hairs   lens Grating shield   lens Slit
Light source

Deflected
beam

Undeflected
beam

13.6eV
En          2
n
Example Problem 1

Consider the three lowest energy states of the hydrogen atom.
What wavelengths of light will be emitted when the electron
jumps from one state to another?

Solution:
E21 = 10.2 eV   E = -13.6 eV/n2, so E1 = -13.6 eV, E2 = -3.4 eV,
E31 = 12.1 eV   and E3 = -1.5 eV. There are three jumps to
E32 = 1.9 eV    consider, 2-to-1, 3-to-1, and 3-to-2. The photon
carries away the energy that the electron loses.

 = h/p = hc/E   hc = 1240 eV·nm
21 = 122 nm     Two wavelengths are in the ultraviolet.
31 = 102 nm
32 = 653 nm     The 3-to-2 transition gives a visible (red) photon.
Example Problem 2

What is the normalization constant N?                                                                     1
1

R10
 r / ao
 100 (r , ,f )  NR10 (r )  Ne                                                                f( x) 0.5

0

2
0
The probability density is                                     =   N2   exp(-2r/ao).                          00
0    r
2
x   4a0
4
4

In 3D, this means “probability per unit volume”.

Require that the total probability = 1:                                               2 dV = 1
dV = r2 sin dr d df
With spherical symmetry, the angular integrals give 4, so we are left with:

 2 r / a0                                          1
4 N       r                        dr  1                         
2        2                                                2
e                                        N                                     (integral table)
a
3
0                                                                    o

1           r / ao           Ground-state wave function
 100 (r )                               e
 ao
3                            for hydrogen
Example Problem 3

1
Estimate the probability of finding the                    1

electron within a small sphere of radius                                    r  R10 (r)
rs = 0.2 ao at the origin.                             f( x) 0.5
rs
If it says “estimate”, don’t “integrate”.                  0   00
00              2            4a0
4

The wave function is nearly constant near r = 0:                    0           x             4

 r / ao
 ( r )  Ne
1        0 / ao        1
 (0)            e             
 ao
3
 ao
3

Simply multiply 2 by the volume V = (4/3)rs3.

3
4  rs 2
Probability   (0) V         0.0107
3  ao 
Example Problem 4
At what radius are you most likely to                       1
1

 r / ao
find the electron?                                                                 ( r )  Ne
Looks like a no-brainer. r = 0, of course!               f( x) 0.5

Well, that’s not the answer.                                             r
0
0
You must find the probability P(r)r that the                     00             rmax      2                   4a0
4

electron is in a shell of thickness r at radius r.                  0                     x                   4

The volume, V, of the shell increases with                                         rmax
Radial Probability  P ( r )r   ( r ) V                1
1
P(r)r
 Cr e                r
2   -2 r / ao

 4r22
Set dP/dr = 0 to find rmax = a0                  g( x) 0.5

V = 4r2 r     r                                                  0
0
00                    r2              4a0
4

r                                                   0                 x               4
Supplement: Separation of Variables
In the 3D box, the SEQ is:

2
  2  2  2                                                    NOTE:
     2            2 
 U ( x )  U ( y )  U ( z )   E
2m  x   y      z                                                   Partial derivatives.
2

Let’s see if separation of variables works.
Substitute this expression for  into the SEQ:

 ( x, y , z )  f ( x )g ( y )h( z )

2
    2
d f
2
d g       d h
2
NOTE:
     gh 2  fh    2
 fg    2 
 U ( x )  U ( y )  U (z )  fgh  Efgh      Total derivatives.
2m    dx      dy        dz 

Divide by fgh:
2
 1 d 2f 1 d 2 g 1 d 2 h 
                         2 
 U ( x )  U ( y )  U ( z )   E
2m  f x
2        2
g dy     h dz 
Supplement: Separation of Variables
Regroup:

   2   2
1d f                 2   2
1d g                 2   2
1d h           
           U ( x )             U ( y )             U ( z )  E
 2m f x
2                        2                        2
  2m g dy                2m h dz              

A function of x                      A function of y     A function of z

We have three functions, each depending on a
different variable, that must sum to a constant.
Therefore, each function must be a constant:
2     2
1d f
                       U( x )  Ex
2m f x
2

2         2
1d g
                   2
 U ( y )  Ey           Each function, f(x), g(y), and h(z)
2m g dy
satisfies its own 1D SEQ.
2         2
1d h
                   2
 U ( z )  Ez
2m h dz

E x  E y  Ez  E

```
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