Physics 249 Lecture 18, Oct 17th 2012
Reading: Chapter 7.1, 7.2, 7.3
HW due Friday Oct 19th. Available on web site.
1) Bound states of an electrostatic potential: The Hydrogen atom
The potential energy:
𝑉(𝑟) = −
𝑘𝑒 2
𝑟
The time independent Schrodinger equation in spherical coordinates is:
ℏ2 1 𝜕 2 𝜕𝜓
1 1 𝜕
𝜕𝜓
1 𝜕 2𝜓
𝐸𝜓 = − { 2 (𝑟
)− 2[
(𝑠𝑖𝑛𝜃 ) +
]} + 𝑉(𝑟)𝜓
2𝜇 𝑟 𝜕𝑟
𝜕𝑟
𝑟 𝑠𝑖𝑛𝜃 𝜕𝜃
𝜕𝜃
𝑠𝑖𝑛2 𝜃 𝜕𝜙 2
Separating variables:
𝜓(𝑟, 𝜃, 𝜙) = 𝑅(𝑟)𝑓(𝜃)𝑔(𝜙)
𝑔(𝜙) = 𝑒 𝑖𝑚𝜙
where 𝐶𝜙 = 𝑚2 , 𝑚 = 0, ±1, ±2 …
Lets consider m=0. The simplest solution for 𝑓(𝜃) is:
𝑓(𝜃) = 𝐶
Then
C𝑟 = 0
1 𝜕 2 𝜕𝑅
2𝜇
(𝑟
) + 2 𝑟 2 (𝐸 − 𝑉(𝑟)) = 0
𝑅(𝑟) 𝜕𝑟
𝜕𝑟
ℏ
1 𝜕 2 𝜕𝑅
2𝜇
𝑘𝑒 2
(𝑟
) + 2 𝑟 2 (𝐸 +
)=0
𝑅(𝑟) 𝜕𝑟
𝜕𝑟
ℏ
𝑟
𝜕 2 𝜕𝑅
2𝜇
𝑘𝑒 2
(𝑟
) + 2 𝑟 2 (𝐸 +
) 𝑅(𝑟) = 0
𝜕𝑟
𝜕𝑟
ℏ
𝑟
Again not an easy differential equation to solve: Let’s try a negative exponential.
Negative so that the probably goes down with large r.
𝑅(𝑟) = 𝐶𝑒 −𝑟/𝑎0
𝜕 2 −1
2𝜇 2
𝑘𝑒 2
(𝑟 ( ) 𝑅(𝑟)) + 2 𝑟 (𝐸 +
) 𝑅(𝑟) = 0
𝜕𝑟
𝑎0
ℏ
𝑟
1
1
2𝜇
𝑘𝑒 2
(−2𝑟 ( ) + 𝑟 2 2 ) 𝑅(𝑟) + 2 𝑟 2 (𝐸 +
) 𝑅(𝑟) = 0
𝑎0
ℏ
𝑟
𝑎0
2
1
2𝜇
2𝜇𝑘𝑒 2
(− ( ) 𝑟 + 2 𝑟 2 ) 𝑅(𝑟) + ( 2 𝐸𝑟 2 +
𝑟) 𝑅(𝑟) = 0
𝑎0
ℏ
ℏ2
𝑎0
𝑎0 =
ℏ2
𝜇𝑘𝑒 2
2
ℏ2
𝜇 𝑘𝑒 2
𝐸1 = −
=
−
(
) = 13.6𝑒𝑉
2 ℏ
2𝜇𝑎02
𝜓(𝑟, 𝜃, 𝜙) = 𝐶𝑒 −𝑟/𝑎0
where the constant will come from normalizing the wave function.
1 2 −𝑟/𝑎
0
𝜓(𝑟, 𝜃, 𝜙) = √
𝑒
4𝜋 √𝑎03
The first wave function is spherically symmetric with a distribution that falls off
exponentially characterized by a radius 𝑎0 , which is identical to the Bohr radius!
Looking at the wave functions above it is clear that the full set of wave functions can be
more complex than spherically symmetric though they will have phi symmetry. The
solutions are called orbitals where the distributions are analogs of orbits with a
characteristic vector z about which there is phi orbital symmetry but otherwise complex
functional dependences in theta.
The radial wave function, though independent of phi and theta will have complex
functional dependences as well.
2) Full electrostatic potential (Hydrogen atom) solution:
We explored the simplest solution to the problem, which corresponds to the ground state
of all three quantum numbers. Both the theta and radial wave equations clearly have
more complex solutions. Also, since each separate wave equation is related by constants,
which are different depending on the quantum numbers there will be relationships
between the three quantum numbers of the system.
𝜓𝑛𝑙𝑚 (𝑟, 𝜃, 𝜙) = 𝐶𝑛𝑙𝑚 𝑅𝑛𝑙 (𝑟)𝑓𝑙𝑚 (𝜃)𝑔𝑚 (𝜙)
𝑔𝑚 (𝜙) = 𝑒 𝑖𝑚𝜙
𝑙+|𝑚|
(𝑠𝑖𝑛𝜃)|𝑚|
𝑑
(𝑐𝑜𝑠 2 𝜃 − 1)𝑙
𝑓𝑙𝑚 (𝜃) =
[
]
2𝑙 𝑙!
𝑑(𝑐𝑜𝑠𝜃)
𝑚 = −𝑙 … 0 … + 𝑙
Where the f functions are known as the Legendre functions
The combination of f and g functions are often expressed as spherical harmonics
𝑌𝑙𝑚 (𝜃, 𝜙) = 𝐶𝑙𝑚 𝑓𝑙𝑚 (𝜃)𝑔𝑚 (𝜙)
l=0, m=0
𝑌00 (𝜃, 𝜙) = √
1
4𝜋
l=1, m=1,0,-1
3
𝑌11 (𝜃, 𝜙) = −√ 𝑠𝑖𝑛𝜃𝑒 𝑖𝜙
8𝜋
3
𝑌10 (𝜃, 𝜙) = √ 𝑐𝑜𝑠𝜃
4𝜋
𝑌1−1 (𝜃, 𝜙) = √
3
𝑠𝑖𝑛𝜃𝑒 −𝑖𝜙
8𝜋
Where we have normalized the wave functions over the angular space. The l=0, m=0
state is the one we already solved.
C𝑟 = 𝑙(𝑙 + 1) connecting the radial and angular equations and quantum numbers.
Note that this is a constant factor with no dependence on the angular or radial
coordinates. However, it is a different constant for each value of l each of which is a
separate solution the angular portion of the Schrodinger equation. For each value of l,
and thus different constants, the solutions to the radial equation will be different.
𝑅𝑛𝑙 (𝑟) = 𝐶𝑛𝑙 𝑒 −𝑟/𝑛𝑎0 𝑟 𝑙 ℒ𝑛𝑙 (𝑟/𝑎0 )
n= 1,2,3 …
l= 0, …, n-1
Where the L functions are known as the Laguerre polynomials.
n=1, l=0
𝑅10 (𝑟) =
2
√𝑎03
𝑒 −𝑟/𝑎0
n=2, l=0,1
𝑅20 (𝑟) =
𝑅21 (𝑟) =
1
√2𝑎03
1
(1 −
𝑟
2√6𝑎03 𝑎0
𝑟
) 𝑒 −𝑟/2𝑎0
2𝑎0
𝑒 −𝑟/2𝑎0
and
2
ℏ2
𝜇 𝑘𝑒 2
𝐸1
𝐸𝑛 = −
=
−
(
) =− 2
2
2
2
2𝑛
ℏ
𝑛
2𝜇𝑛 𝑎0
3) Exploring this solutions
There are three quantum numbers.
n: principle quantum number. The energy is set by this quantum number. For a given n
all the states of l and m will be degenerate in energy. As expected we find an energy
degeneracy.
l: the orbital quantum numbers: We will find that this quantum number is associated with
the angular momentum of the orbit.
m: the magnetic quantum number: We will find that the projection of the angular
momentum on the z axis is associated with this quantum number.
Radial distributions.
The radial distributions are easiest to understand in case where l=m=0. In those cases the
probability distribution is spherically symmetric. These cases are the so-called s orbitals.
Though since the angular portion is separated we can also look at the radial distributions
in the cases where l>0. The probability distributions as a function of the radial coordinate
are an exponential or an exponential times a polynomial distribution. A distance scale set
by the Bohr radius characterizes them. They can be understood by graphing the
probability distribution and quantified by calculating the maxima, minima and
expectation values. For higher n these distributions will be pushed out to higher values.