1. The Hydrogen Atom
Schrödinger Equation
Angular Momentum
Solution of the Schrödinger Equation
for Hydrogen
Werner Heisenberg
(1901-1976)
The atom of modern physics can be symbolized only through a partial differential
equation in an abstract space of many dimensions. All its qualities are inferential; no
material properties can be directly attributed to it. An understanding of the atomic world
in that primary sensuous fashion…is impossible.
- Werner Heisenberg

2. 1. Schrödinger Equation
The time-independent Schrodinger equation for atom hydrogen:
Use the three-dimensional time-independent Schrödinger Equation.
For Hydrogen-like atoms (He+ or Li++)
Replace e2 with Ze2 (Z is the atomic number).
Replace m with the reduced mass, m..
Our task is to find the wave functions ª(r; µ; Á) that satisfy this
equation, and hence to find the allowed quantized energies E.

3. 2. Angular Momentum
Classical definition of angular momentum:
For circular orbits this simplifies to L = mvr, and in Bohr’s model, L
was quantized in integer units of h.
However, the full quantum treatment is more complicated, and requires
the introduction of two other quantum numbers l and ml, as we shall now
see.
The components of L are given by

4. Angular Momentum …
In quantum mechanics we know that the linear momentum by
differential operators:
Therefore, the quantum mechanical operators for the angular
momentum are given by:

5. Angular Momentum ...
The magnitude of the angular momentum is given by:
Now we therefore have the quantum mechanical operator
It should be noted that this operators should be understood in terms
of repeated operations:

6. Angular Momentum …
It can be shown that the components of the angular momentum
operator do not commute, that is
However, in fact we can show that:
where the “commutator bracket” [ˆLx; ˆLy] is defined by
Remember that: if two quantum mechanical operators do not
commute, then it is not possible to know their values simultaneously.

7. Angular Momentum …
For example, the operators for position and momentum in a one-
dimensional system:
Thus we have:

8. 3. Solution of the Schrödinger Equation for H
satisfies the azimuthal equation for any value of mℓ.
The solution must be single valued to be a valid solution for any f:
mℓ must be an integer (positive or negative) for this to be true.
Now set the left side equal to −mℓ2 and rearrange it [divide by sin2(q)].
Now, the left side depends only on r, and the right side depends only
on θ. We can use the same trick again!

9. Solution of the Schrödinger Equation for H
Set each side equal to the constant ℓ(ℓ + 1).
Radial equation
Angular equation
We’ve separated the Schrödinger equation into three ordinary second-
order differential equations, each containing only one variable.

10. Solution of the Radial Equation for H
The radial equation is called the associated Laguerre equation and the
solutions R are called associated Laguerre functions. There are
infinitely many of them, for values of n = 1, 2, 3, …
Assume that the ground state has n = 1 and ℓ = 0. Let’s find this solution.
The radial equation becomes:
The derivative of yields two terms.
This yields:

11. Solution of the Radial
Try a solution
 A is a normalization constant.
Equation for H
 a0 is a constant with the dimension of length.
 Take derivatives of R and insert them into the radial equation.

Tosatisfy this equation for any r, both expressions in parentheses
must be zero.
Set the second expression
equal to zero and solve for a0:
Set the first expression equal
to zero and solve for E:
Both are equal to the Bohr results!

12. Principal
Quantum
Number n
There are many solutions to the radial wave equation, one for
each positive integer, n.
The result for the quantized energy is:
A negative energy means that the electron and proton are bound
together.

13. Quantum Numbers
The three quantum numbers:
n: Principal quantum number
ℓ: Orbital angular momentum quantum number
mℓ: Magnetic (azimuthal) quantum number
The restrictions for the quantum numbers:
n = 1, 2, 3, 4, . . .
ℓ = 0, 1, 2, 3, . . . , n − 1
mℓ = −ℓ, −ℓ + 1, . . . , 0, 1, . . . , ℓ − 1, ℓ
Equivalently:
n>0
The energy levels are:
ℓ<n
|mℓ| ≤ ℓ

14. Hydrogen Atom Radial Wave Functions
First few
radial
wave
functions
Rnℓ
Sub-
scripts
on R
specify
the
values of
n and ℓ.

15. Solution of the Angular and Azimuthal
Equations
The solutions to the azimuthal equation are:
Solutions to the angular and azimuthal equations are linked
because both have mℓ.
Physicists usually group these solutions together into
functions called Spherical Harmonics:
spherical harmonics

16. Normalized
Spherical
Harmonics

17. Solution of the Angular and Azimuthal
Equations
The radial wave function R and the spherical harmonics Y determine
the probability density for the various quantum states. The total wave
function depends on n, ℓ, and mℓ. The wave function
becomes

18. Orbital Angular Momentum Quantum
Number ℓ
Energy levels are degenerate with respect to ℓ (the energy is
independent of ℓ).
Physicists use letter names for the various ℓ values:
ℓ= 0 1 2 3 4 5...
Letter = s p d f g h...
Atomic states are usualy referred to by their values of n and ℓ.
A state with n = 2 and ℓ = 1 is called a 2p state.

19. Orbital Angular Momentum Quantum
Number ℓ
It’s associated with the R(r) and f(θ) parts of the wave function.
Classically, the orbital angular momentum with L = mvorbitalr.
L is related to ℓ by
In an ℓ = 0 state,
This disagrees with Bohr’s
semi-classical “planetary”
model of electrons orbiting Classical orbits—which do not
a nucleus L = nħ. exist in quantum mechanics

20. The angle f is the angle from the Magnetic Quantum
z axis.
Number mℓ
The solution for g(f) specifies that
mℓ is an integer and is related to
the z component of L:
Example: ℓ = 2:
Only certain orientations of are
possible. This is called space
quantization.
And (except when ℓ = 0) we just
don’t know Lx and Ly!

21. Rough derivation of ‹L2› = ℓ(ℓ+1)ħ2
We expect the average of the angular momentum components
squared to be the same due to spherical symmetry:
But
Averaging over all mℓ values (assuming each is equally likely):

because: 
n  
m 2  (  1)(2  1) / 3

22. 4. Magnetic Effects on Atomic
Spectra—The Zeeman Effect
In 1896, the Dutch physicist Pieter Zeeman
showed that spectral lines emitted by atoms
in a magnetic field split into multiple energy
levels. It is called the Zeeman effect.
Consider the atom to behave like a small magnet.
Think of an electron as an orbiting circular current loop of I = dq / dt
around the nucleus. If the period is T = 2p r / v,
then I = -e/T = -e/(2p r / v) = -e v /(2p r).
The current loop has a magnetic moment m = IA = [-e v /(2p r)] p r2
=
 e 
[-e/2m] mrv: m  L
2m
where L = mvr is the magnitude of the orbital angular momentum.

23.  e 
The Zeeman Effect m  L
2m
The potential energy due to the
magnetic field is:
If
the magnetic field is in the z-direction, we only care about the z-
component of m:
e e
mz   Lz   (m )   mB m
2m 2m
where mB = eħ / 2m is called the Bohr magneton.

24. The Zeeman Effect
A magnetic field splits the mℓ levels. The potential energy is quantized
and now also depends on the magnetic quantum number mℓ.
When a magnetic field is applied, the 2p level of atomic hydrogen is
split into three different energy states with energy difference of ΔE =
mBB Δmℓ.
mℓ Energy
1 E0 + μBB
0 E0
−1 E0 − μBB

25. The
Zeeman
Effect
The transition
from 2p to 1s,
split by a
magnetic field.

26. The Zeeman Effect
An atomic beam of particles in the ℓ = 1 state pass through a
magnetic field along the z direction.
  m B m (dB / dz )
The mℓ = +1 state will be deflected down, the mℓ = −1 state up, and
the mℓ = 0 state will be undeflected.

27. 5. Energy Levels
and Electron
Probabilities
For hydrogen, the energy
level depends on the prin-
cipal quantum number n.
In the ground state, an atom
cannot emit radiation. It can
absorb electromagnetic
radiation, or gain energy
through inelastic
bombardment by particles.

28. We can use the wave functions
Selection Rules to calculate transition probabilities
for the electron to change from
one state to another.
The probability is proportional to
the mag square of the


dipole moment: 
d   er  *
Allowed transitions:
Electrons absorbing or emitting photons can change states
when Δℓ = 1 and Δmℓ = 0, 1.
Forbidden transitions:
Other transitions are possible
but occur with much smaller
probabilities.

29. Probability Distribution Functions
We use the wave functions to calculate the probability
distributions of the electrons.
The “position” of the electron is spread over space and is not
well defined.
We may use the radial wave function R(r) to calculate radial
probability distributions of the electron.
The probability of finding the electron in a differential volume
element dτ is .

30. Probability Distribution Functions
The differential volume element in spherical polar coordinates is
Therefore,
At the moment, we’re only interested in the radial dependence.
The radial probability density is P(r) = r2|R(r)|2 and it depends only on n
and ℓ.

31. Probability
Distribution
Functions
R(r) and P(r)
for the lowest-
lying states of
the hydrogen
atom.

32. Probability Distribution Functions
The probability density for the hydrogen atom for three different
electron states.

33. 6. Intrinsic Spin
In 1925, grad students, Samuel
Goudsmit and George Uhlenbeck,
in Holland proposed that the
electron must have an
intrinsic angular momentum
and therefore a magnetic moment.
Paul Ehrenfest showed that, if so, the surface of the spinning
electron should be moving faster than the speed of light!
In order to explain experimental data, Goudsmit and Uhlenbeck
proposed that the electron must have an intrinsic spin
quantum number s = ½.

34. Intrinsic Spin
The spinning electron reacts similarly to
the orbiting electron in a magnetic field.
The magnetic spin quantum number ms
has only two values, ms = ½.
The electron’s spin will be either “up” or
“down” and can never be spinning with its
magnetic moment μs exactly along the z
axis.

35.  e 
Intrinsic Spin Recall: m L   L
2m
The magnetic moment is .
The coefficient of is −2μB and is a
consequence of relativistic quantum mechanics.
Writing in terms of the gyromagnetic ratio, g: gℓ = 1 and gs = 2:
and
The z component of .
In an ℓ = 0 state: no splitting due to .
there is space quantization due to the intrinsic spin.
Apply ms and the potential energy becomes:

36. Multi-electron atoms
When more than one electron is involved, the potential and the
wave function are functions of more than one position:
     
V  V (r1 , r2 ,..., rN )    (r1 , r2 ,..., rN , t )
Solving the Schrodinger Equation in this case can be very hard.
But we can approximate the solution as the product of single-
particle wave functions:
     
 (r1 , r2 ,..., rN , t )  1 (r1 , t )  2 (r2 , t )   N (rN , t )
And it turns out that, for electrons (and other spin ½ particles), all
the i’s must be different. This is the Pauli Exclusion Principle.

37. Molecules have many energy levels.
A typical molecule’s energy levels:
E = Eelectonic + Evibrational + Erotational
2ndexcited
electronic state
Lowest vibrational and
rotational level of this
electronic “manifold”
Energy
1st excited
Excited vibrational and
electronic state
rotational level
Transition There are many other
complications, such as
Ground spin-orbit coupling,
electronic state nuclear spin, etc.,
which split levels.
As a result, molecules generally have very complex spectra.

38. Generalized Uncertainty Principle
Define the Commutator of two operators, A and B:
 A, B  AB  BA
Then the uncertainty relation between the two corresponding
observables will be:
A B  1
2  *  A, B  
So if A and B commute, the two observables can be measured
simultaneously. If not, they can’t.
     
Example:  p, x     px  xp     i   x   x  i  
 x   x 
 x     
  i   ix   x  i   i
 x x   x 
So:  p, x  i and p x   / 2

39. Two Types of Uncertainty in Quantum
Mechanics
We’ve seen that some quantities (e.g., energy levels) can be
computed precisely, and some not (Lx).
Whatever the case, the accuracy of their measured values is
limited by the Uncertainty Principle. For example, energies can
only be measured to an accuracy of ħ /t, where t is how long
we spent doing the measurement.
And there is another type of uncertainty: we often simply don’t
know which state an atom is in.
For example, suppose we have a batch of, say, 100 atoms,
which we excite with just one photon. Only one atom is excited,
but which one? We might say that each atom has a 1% chance
of being in an excited state and a 99% chance of being in the
ground state. This is called a superposition state.

40. Superpositions of states
Stationary states are stationary. But an atom can be in a
superposition of two stationary states, and this state moves.
  
 (r , t )  a1 1 (r ) exp(iE1t / )  a2 2 (r ) exp( iE2t / )
where |ai|2 is the probability that the atom is in state i.
Interestingly, this lack of knowledge means that the
atom is vibrating:
 2  2  2
(r , t )  a1 1 (r )  a2 2 (r ) 
 * 
2 Re a1 1 (r )a2 2 (r ) exp[i( E2  E1 )t / ]

41. Superpositions of states
Vibrations occur at the frequency difference between the two levels.
 2  2  2
(r , t )  a1 1 (r )  a2 2 (r ) 
 * 
2 Re a1 1 (r )a2 2 (r ) exp[i( E2  E1 )t / ]
Excited level, E2
 Energy
E = hn
Ground level, E1
The atom is vibrating The atom is at least partially in
at frequency, n. an excited state.

42. Calculations in Physics: Semi-classical
physics
The most precise computations are performed fully quantum-
mechanically by calculating the potential precisely and solving
Schrodinger’s Equation. But they can be very difficult.
The least precise calculations are performed classically,
neglecting quantization and using Newton’s Laws.
An intermediate case is semi-classical computations, in
which an atom’s energy levels are computed quantum-
mechanically, but additional effects, such as light waves, are
treated classically. Our Thomson Scattering calculation of a and
n was an example of this.