Quantum Theory of Hydrogen
electron probability density
radiative transitions
Zeeman effect
“Every Hun has value—if only to serve as a bad example.”—Attila the Hun
From the last lecture...
Caution: here again is how to calculate the probability:
P(r1  r  r2 ) =

r2
r1
P(r) dr =

r2
r1
R(r) R(r) r 2 dr
comes from volume
element dV
Here is how to calculate the expectation value:
r =


0
R(r) r R(r) r 2 dr
comes from definition of <G>
Raise your hand if you plan on forgetting the r2.
Raise your hand if you plan on forgetting the r in <r>.
You’re joking, right?
Let’s use the convention that P(r) stands for the probability density function
R2r2, and P(r1rr2) stands for the probability of finding the electron
somewhere between r1 and r2 (i.e., the integral).
We have been manipulating the radial part of the wave
function, because that’s where all the “action” is.
But hydrogen is a 3D object.
As a start towards 3D visualization, let’s “spin” the 1s radial
wave function about r=0.
This is supposed to help you “see” how the wave function
drops off in all directions. Imagine a 3D object with this shape
(kind of like a certain chocolate candy).
But the probability density is the meaningful “thing.”
Here’s the 1s probability density, “spun” about r=0.
Actually, there is a small but finite probability of finding the 1s electron “inside” the nucleus.
I suppose you want to see 3D models of orbitals.
You have to take the radial probability density, “fold in” the
angular probability densities (which for the 1s are uniform) and
somehow display in 3D.
Because the 1s (or 2s, 3s, …) probability density is spherically
symmetric, you usually see it represented as a sphere. The
figure on the previous slide clearly shows that the sphere is
“hollow” and the probability density varies with r.
People who represent this with “3D” drawings try to use
shading to represent relative probabilities. It is difficult to
show the “hollowness” of the 1s probability density.
Here are a couple of
representations of
the 1s probability
density.
http://lompado.uah.edu/
HAtomPDFs.htm
http://www.phy.davidson.edu/
StuHome/cabell_f/Density.html
Here are more images illustrating how probability densities are
displayed. They are from Houghton Mifflin Online
(http://hmchemdemo.clt.binghamton.edu/zumdahl/docs/chemis
try/07atomstructure/library/0707.htm).
According to this source, the figure on the left illustrates the
electron probability distribution, and the figure on the right
illustrates “the surface that contains 90% of the total electron
probability (the size of the orbital, by definition).”
The images on the previous slide still imply that the 1s electron
probability distribution is something like a peach. A better
analogy would be a tennis ball with a thick but “fuzzy” skin.
Anyway, now that you have been exposed to some of the
subtleties of displaying probability densities, you should be
responsible enough to view the pages listed here.
http://lompado.uah.edu/HAtomPDFs.htm (broken, March 2005)
http://www.phy.davidson.edu/StuHome/cabell_f/Density.html
http://www.dauger.com/orbitals/
http://www.chemistry.mcmaster.ca/esam/Chapter_3/section_2.html
http://library.wolfram.com/webMathematica/MSP/Explore/Physics/Hydrogen
http://hmchemdemo.clt.binghamton.edu/zumdahl/docs/chemistry/07atomst
ructure/library/0707.htm
http://www.orbitals.com/orb/
Section 6.7 contains much of the “testable” material of chapter
6. The earlier sections are important (especially quantum
numbers and angular momentum) but many of the problems
come from 6.7, so be sure to study it well.
Important ideas (quantum mechanics works very well for
describing the hydrogen atom, but we need to modify our
classical thinking in several ways):
 We can only give probabilities for finding an
electron at some set of coordinates.
 The electron doesn't orbit the nucleus in any
conventional sense. You should think of the pictures
shown here as representing fuzzy clouds. The clouds
aren't electrons, but they show where electrons are
likely to be found. The denser the "cloud," the more
likely is the electron to be found.
6.8 Radiative Transitions
We are well on our way to understanding the hydrogen atom,
as well as why Bohr’s model had some real physics behind it.
We have found that electrons don’t orbit in the classical sense
(like a planet).
In fact, our electron wave functions are “stationary states.”
That helps us understand why the electrons don’t radiate
continuously, although we haven’t yet done the mathematics
to show this.
Now it’s time to do “that” mathematics.
First consider an electron in hydrogen in a state with principal
quantum number n.
Its energy is En, its wave function has a frequency fn=En/h,
and its wave function is
 jE t 
 n = n exp  - n  .


This wave function now includes time dependence.
The expectation value of the electron's position is
x =
*

 n x n dx ,
which Beiser shows is independent of time.
There is nothing here which tells us the electron radiates
energy, and in fact, it doesn't (the electron doesn't oscillate
with time).
When the electron is excited to another quantum state m, it
doesn't radiate energy after it reaches its final state; that is, as
long as it is in the state m.
However, the electron does have to get from state n to m, and
in the process it does radiate (or absorb) energy.
We calculate the energy radiated (or absorbed) by considering
the wave function
 = a n +b m ,
which represents an electron which may exist in state m or n.
The magnitudes of a and b tell us the relative probabilities of
the two states. Kind of like Schrödinger’s cat.
It is straightforward, if slightly tedious, to calculate the
expectation value <x> for such an electron. You could
reproduce this calculation with some assistance on my part.
However, I have never asked this on an exam.
The only "trick" in the calculation is the fact that
 *n  m =  m*  n
and a*b = b* a.
The above two equations come from a more thorough
investigation of the mathematics of quantum mechanics.
Here, we just have to accept them.
With these “tricks” we can calculate <x>.
The result of the calculation is:
x = a2  *n x n dx + b2  *m x m dx
+ 2a*b cos  2f t   *n x m dx ,
Em - En
.
where f =
h
If the electron is in state n, a=1, b=0 and only the first
integral contributes to <x>, just as we would expect.
If the electron is in state m, a=0, b=1, and only the second
integral contributes to <x>, just as we would expect.
However, when the electron is in transition between states n
and m, both a and b are nonzero, and all three integrals,
including the third one, contribute to <x>.
x = a2  *n x n dx + b2  m* x m dx
+ 2a*b cos  2f t   n* x m dx
The third integral corresponds to an electron "jumping"
between quantum state n to quantum state m.
The cosine term gives the frequency of radiation emitted by
the electron oscillating ("jumping") between quantum states n
and m.
The important result here is that an electron radiates
electromagnetic energy when it jumps between quantum
states, and that the frequency of emitted radiation is
Em - En
f=
.
h
We do have a problem here--we've mixed classical
(accelerated electron emits radiation) and quantum theory.
Strictly speaking, according to this interpretation and
calculation, the electron should emit electric dipole radiation
instead of a single photon.
Beiser clears this up in the next section. Our ideas are OK; but
a complete and correct treatment requires quantum
electrodynamics, and there we find that single photons are,
indeed, emitted.
6.9 Selection Rules
One more time:
x = a2  *n x n dx + b2  *m x m dx
+ 2a*b cos  2f t   *n x  m dx .
For an electron to get from one quantum state to another, it
must radiate/absorb energy. For this to happen, the integral
must be nonzero.
*

 n x m dx
What if the integral is zero?
If the integral is zero, the electron can never get between the
two quantum states. There is no path connecting the states!
*

Thus, calculation of the integral  n x m dx tells us which
transitions are allowed (the integral is finite) or forbidden (the
integral is zero).
The wave functions for the hydrogen atom are known, and the
integral can be evaluated for all possible transitions. The
integral is nonzero when
Δ =±1 and Δm = 0, ±1.
These are selection rules for the allowed transitions. In other
words, the integral is nonzero only for these particular changes
in ℓ and mℓ.
Hey! Things are starting to fit together. When the integral is “big”, the
probability of transition is large (the transition happens “fast”). When the
integral is small, the probability of transition is small (the transition
happens “slow”—a metastable state!).
Note that ℓ = 1 implies a change in the angular momentum
of the atom during an allowed transition. (Why?) Angular
momentum must be conserved, so the photon must carry off
angular momentum.
r
p
Photon of linear momentum p carries off
angular momentum of r x p. Angular
momentum of electron participating in
event must change (angular momentum is
conserved).
When mℓ = 0, there is no change in the angular momentum
along the z axis. This means that the photon is emitted in the
x-y plane. When mℓ = 1, the photon is emitted along the z
axis.
Study figure 6.13 (see next slide). Be able to explain parts of
it to me on an exam or quiz!
Figure 6.13 shows allowed
transitions in hydrogen.
It is only concerned with the
ℓ = 1 requirement.
The arrows in the figure
indicate the allowed
transitions.
Why is the transition indicated
by the red arrow forbidden?
Notice the energy scaling;
“ground” is 0 eV instead of
-13.6 eV (perfectly “legal”).
The subsection on quantum electrodynamics is interesting
(virtual photons???) but is not testable.
6.10 Zeeman Effect
Good physics! Skip most! (sob) Why—not enough time!
Here’s the idea: electrons with different n differ in energy.
This section shows that if you put an atom in a magnetic field,
states with the same n but different mℓ are split slightly in
energy. (Aha—that’s why mℓ is called the magnetic quantum
number!)
2p means n=0 ℓ=1, mℓ=0,
1.
mℓ=-1
mℓ=0
mℓ=1
Spectral lines corresponding to a particular n (we studied those
in chapter 4) are “split” into separate lines corresponding to
different mℓ.
The splitting is small but measurable. It
is called the Zeeman effect, named after
the scientist who discovered it in 1896.
(Nobel Prize, 1902)
Explanation of the Zeeman effect is a dramatic confirmation of
the validity of quantum mechanics.
None of the material in this section (including these two
Powerpoint slides on it) will be tested or quizzed.
Homework problems for me to set up on the board:
What are the angles between L and the z-axis for ℓ=2?
Find the most probable value of r for a 3d electron in
hydrogen.
How much more likely is the electron in a ground-state
hydrogen atom to be at the distance a0 from the nucleus than
at the distance 2a0?
R1s  r  =
2
- r / a0
e
.
3/2
a0
There are “zillions” of web sites about the hydrogen atom. Every physics teacher
who does anything relating to modern physics and puts class material on the web
seems to do the hydrogen atom. A search using Google for “hydrogen atom
schrodinger equation” (but don’t put the words in quotes when you do the search)
comes up with about 4300 sites. Many of them are much nicer looking than mine,
but the basic concepts presented are (not surprisingly) virtually the same. The
sites below are repeats of the sites on slide 15.
Here are a few choice sites (search words in parentheses): (hydrogen atom
electron probability density)
http://lompado.uah.edu/HAtomPDFs.htm
http://www.phy.davidson.edu/StuHome/cabell_f/Density.html
http://www.dauger.com/orbitals/
http://www.chemistry.mcmaster.ca/esam/Chapter_3/section_2.html
http://library.wolfram.com/webMathematica/MSP/Explore/Physics/Hydrogen
http://hmchemdemo.clt.binghamton.edu/zumdahl/docs/chemistry/07atomstructure
/library/0707.htm
Slightly different search: (hydrogen atom electron orbital)
http://www.orbitals.com/orb/
All of these links were “live” as of fall 2003, but some may now be “dead.”