1. PHYS 274: Atomic Structure Part II
Question: What are the three integers shown here ?
Ans: n, l, ml
Today’s topics:
Copenhagen Interpretation of QM
Review 3D particle in a box
Understanding solutions of the Schrodinger equation for the hydrogen atom.

2. Recent photo of Copenhagen
Little Mermaid
Werner Heisenberg and Niels Bohr in Copenhagen in 1934
Recent photo of Copenhagen

3. The Copenhagen Interpretation of QM has three primary parts:
1. The wave function is a complete description of a wave/particle. Any information that cannot be derived from the wave function does not exist. For example, a wave is spread over a broad region, therefore does not have a specific location.
2. When a measurement of the wave/particle is made, its wave function collapses. In the case of momentum, a wave packet is made of many waves each with its own momentum value. Measurement reduced the wave packet to a single wave and a single momentum.
3. If two properties are related by an uncertainty relation, no measurement can simultaneously determine both properties to a precision greater than the uncertainty relation allows. So, if we measure a wave/particles position, its momentum becomes uncertain.
Credit: University of Oregon
Credit: University of Oregon

4. Particle in a 3-D box: Review
Boundary conditions; on the walls, X, Y, Z must be zero 
X(x)=0 at x=0 and x=L;
Y(y)=0 at y=0 and y=L;
Z(z)=0 at z=0 and z=L 4

5. Review: Particle in a three-dimensional box
For a particle enclosed in a cubical box with sides of length L (see Figure below), three quantum numbers nX, nY, and nZ label the stationary states (states of definite energy).
The three states shown here are degenerate: Although they have different values of nX, nY, and nZ, they have the same energy E. 5

6. Q29.3
Compare (1,1,1) to (2,1,1) or (1,2,1) or (1,1,2) gives 6 / 3 =2 A 6

7. Q29.3 Compare (1,1,1) to (2,1,1) or (1,2,1) or (1,1,2)
gives 6 / 3 =2 A
Compare (1,1,1) to (2,1,1) or (1,2,1) or (1,1,2) gives 6 / 3 =2 A 7

8. Q29.4
For a particle confined inside a 3-D box, the energies are given by
Which of the following sets of states are degenerate ?
(1,1,1),(2,2,2),(3,3,3)
(1,2,1),(2,1,2),(1,1,2)
(3,1,1),(1,1,3),(3,1,1)
(1,1,1),(0,0,0),(-1,1,0)
Hint: Degenerate means the energies are equal. C 8

9. Q29.4
For a particle confined inside a 3-D box, the energies are given by
Which of the following sets of states are degenerate ?
(1,1,1),(2,2,2),(3,3,3)
(1,2,1),(2,1,2),(1,1,2)
(3,1,1),(1,1,3),(3,1,1)
(1,1,1),(0,0,0),(-1,1,0)
Hint: Degenerate means the energies are equal. C 9

10. Review The hydrogen atom: Quantum numbers
The Schrödinger equation for the hydrogen atom is best solved using coordinates (r, θ, ϕ) rather than (x, y, z) (see Figure at right).
The stationary states are labeled by three quantum numbers: n (which describes the energy), l (which describes orbital angular momentum), and ml (which describes the z-component of orbital angular momentum). 10

11. The hydrogen atom: Results
This result agrees with the Bohr model !
Here l=0,1,2,….n-1
This result does not agree with the Bohr model.
Question: Why ? What happens for n =1 ?
Here m=0,±1, ±2,…. ±l
The Bohr model does not include this part at all. 11

12. The hydrogen atom: Results
This result agrees with the Bohr model !
Here l=0,1,2,….n-1
This result does not agree with the Bohr model.
Question: Why ? What happens for n =1 ?
Here m=0,±1, ±2,…. ±l
The Bohr model does not include this part at all. 12

13. For large quantum numbers, we recover the classical results, i. e
For large quantum numbers, we recover the classical results, i.e. the classical angular momentum formula for this case
For example, let’s compute the orbital quantum number for a macroscopic stone
A stone with a mass of 1.0 kg is spun around in a horizontal circle of radius 1.0 m with a period of 1.0 s
This is called “Bohr’s correspondence principle”

14. The hydrogen atom: Results from Schrodinger’s eqn.
Here z is the quantization axis. Note that Lx and Ly are unconstrained. Why ?
Ans: This is an application of Heisenberg’s uncertainty principle. Otherwise we would know both z and pz precisely !
Question: How do we pick the z-direction ? 14

15. The hydrogen atom: Quantum states
Table 41.1 (below) summarizes the quantum states of the hydrogen atom.
For each value of the quantum number n, there are n possible values of the quantum number l. For each value of l, there are 2l + 1 values of the quantum number ml. 15

16. What do these letters s, p, d, f mean ?
For atomic structure, we distinguish the orbital angular momentum states as follows:
s-wave: l=0
p-wave: l=1
d-wave: l=2
f-wave: l=3
For the principal quantum numbers, in x-ray spectroscopy we use the old labeling:
K-shell: n=1
L-shell: n=2
M-shell: n=3
N-shell: n=4 16

17. Function follows form
The way the electrons distribute determines atomic behavior
Question: What doesn’t the nucleus matter much for chemistry ? 17

18. What are the quantum numbers of the hydrogen atom in the figure on the lower right
l=2, ml=4, n=3
n=4, ml=3, l=2
n=4, l=3, ml=2
n=4, l=3, ml=1

19. Example of counting hydrogen states
How many distinct (n,l,ml) states of the hydrogen atom with n=3 are there ? What are their energies ?
Answer:
n=3 l=0,1,2 (s,p and d waves are possible)
For l=0, there is one state. For l=1, ml=-1,0,1 (3 states)
l=2, ml=-2,1,0,1,2 (5 states)
So all together there are 1+3+5= 9 states of the hydrogen atom in n=3
E=-13.6eV/9=-1.51 eV
All 9 states are degenerate. 19

20. The hydrogen atom: Degeneracy
Hydrogen atom states with the same value of n but different values of l and ml are degenerate (have the same energy).
The figure on the right shows the states with l = 2 and different values of ml. The orbital angular momentum has the same magnitude L for each these states, but has different values of the z-component Lz. 20

21. Clicker atom on 3-D hydrogen atom
This illustration shows the possible orientations of the angular momentum vector in a hydrogen atom state with l = 2. For a given value of Lz,
A. the angular momentum vector can point in any direction tangent to the cone for that value of Lz.
B. the electron orbits along the corresponding red circle, so the orbit may or may not have the nucleus at its center.
C. both A. and B. are true.
D. neither A. nor B. is true.
Answer: A 21

22. Clicker question on 3-D hydrogen atom
This illustration shows the possible orientations of the angular momentum vector in a hydrogen atom state with l = 2. For a given value of Lz,
A. the angular momentum vector can point in any direction tangent to the cone for that value of Lz.
B. the electron orbits along the corresponding red circle, so the orbit may or may not have the nucleus at its center.
C. both A. and B. are true.
D. neither A. nor B. is true. 22

23. Angular momentum in an excited state of hydrogen
Consider the n=4 states of hydrogen. (a) What is the maximum magnitude L of the orbital angular momentum ? (b)What is the minimum angle between the L vector and the z-axis
Answer:
n=4 l=0,1,2,3 (s,p,d and f waves are possible)
So l=3 is the maximum possible 23

24. For next time Atomic concepts
Read material in advance
Review spherical coordinates, angular momentum
> Study spin and the Pauli Exclusion Principle.