Transcript L20 - Particle Physics and Particle Astrophysics

Lecture 20
Spherical Harmonics – not examined
last lecture !!!!!!
• There will be 1 revision lecture next
• Come to see me before the end of term
• I’ve put more sample questions and answers in Phils Problems
• Past exam papers
• Have a look at homework 2 (due in on 15/12/08)
Remember Phils Problems and your notes = everything
http://www.hep.shef.ac.uk/Phil/PHY226.htm
Introduction
We can all imagine the ground state of a particle in an infinite quantum well in 1D
ψ
X
Or the 2D representation of 2 harmonics of a wave distribution in x and y interacting
on a plate
ψ
Y
X
www.falstad.com/mathphysics.html
Visualisation of the spherical harmonics in a 3D spherical potential well is more tricky !!!!!
Introduction
Let’s think about the Laplace equation in 3D
In 3D Cartesian coordinates we write:
 2 ( x, y, z )  0
In spherical polar coordinates last lecture we stated that:
1  2 
1
 
 
1
2
  2 r

 sin 

r r  r  r 2 sin   
  r 2 sin 2   2
2
and so the Laplace equation in spherical polar coordinates is:
 2  (r ,  ,  )  0
Comparing the Cartesian case with the spherical polar case, it is not difficult to believe
that the solution will be made up of three separate functions, each comprising an integer
variable to define the specific harmonic solution.
e.g.
 ( x, y, z )  X ( x)Y ( y ) Z ( z )
 (r ,  ,  )  R(r ) P( ) F ( )
 ( x, y, z )  A sin k x x sin k y y sin k z z
(r , ,  )  Are r 2 a0 sin  e  i
Let’s look at electron orbitals for the Hydrogen atom
This topic underlies the whole of atomic and nuclear physics. Next semester in atomic
physics you will cover in more detail the radial spherical polar solutions of the
Schrödinger equation for the hydrogen atom.
Bohr and Schrodinger predicted the energy levels of the H atom to be: E n  
13.6eV
n2
This means that the energy of an electron in any excited orbital depends purely on the
energy level in which it resides. From your knowledge of chemistry, you will know that
each energy level can contain more than one electron. These electrons must therefore
have the same energy.
We say that there exists more than one quantum state corresponding to each energy
level of the H atom. (Actually there are 2n2 different quantum states for the nth level).
For the 1D case it was sufficient to define a quantum state fully using just one
quantum number, e.g. n = 2 because our well extended only along the x axis. In 3D
we have to consider multiple axes within a 3D potential well, and since the probability
density functions corresponding to the EPCs are mostly not radially symmetric, we
must represent wavefunctions with the same energy but different eigenfunctions,
using a unique set of quantum numbers.
The quantum numbers for polar coordinates corresponding to (r ,  ,  )
are (n, l , m)
Let’s look at electron orbitals for the Hydrogen atom
An electron probability cloud (EPC) is a schematic representation of the likely position
of an electron at any time.
This figure shows the EPCs
corresponding to the ground state
and some excited states of the
hydrogen atom.
For each energy level there are
several different EPC
distributions corresponding to
the different 3D harmonic
solutions for that energy level.
The quantum numbers for polar
coordinates corresponding to
(r ,  ,  ) are (n, l , m)
Let’s look at electron orbitals for the Hydrogen atom
n is defined as the principal quantum number (and sets the value of the energy level
of the wave).
For each wave with quantum number n, there exist quantum states of l from l = 0 to
l = (n - 1) where l is defined as the orbital quantum number.
So for example an electron in the 3rd excited state can be in (n=3, l=0), or (n=3, l=1)
or (n=3, l=2) quantum states.
Each one of these states has further states represented by quantum number m defined
as the magnetic quantum number, a positive or negative integer where | m | l .
Let’s look at electron orbitals for the Hydrogen atom
The full solution  (r , ,  )  R(r ) P( ) F ( ) , for the ground state and first few excited
states corresponding to each specific combination of quantum numbers is shown below.
a0 is the first Bohr radius corresponding to the ground state of the H atom …..
Let’s look at electron orbitals for the Hydrogen atom
Once we have the solution to the wave equation in 3D spherical polar coordinates we
can deduce the probability function.
For example the probability density function in 3D for ground state (1,0,0) is …..
1
1
r a

e
32

a
2
0
2  r a0 1
 (r ,  ,  )  R(r ) P( ) F ( )  3 2 e
a0
2
0
The radial probability density for the hydrogen ground state is obtained by multiplying
the square of the wavefunction by a spherical shell volume element.
2
 1

1  2 r a0
 r a0


 (r , ,  )  
e
 3e
32

  a0
 a0
2
So

1  2 r a0
dP  3 e
4r 2 dr
a0

2
1 2r a0
4
r
4r 2 dr   3 e 2r a0 dr
If we integrate over all space P   3 e
a0
a0
0
0
we can show that the total probability is 1.
4  2 r
P  3 e
a0 

2
a02 r a03 
a0   a 0 r


   1
2
4  0
 2
Let’s look at electron orbitals for the Hydrogen atom
Probability density function in 3D for ground state (1,0,0) is
dP 
1  2 r a0
2
e
4

r
dr
3
a0
Let’s look at electron orbitals for the Hydrogen atom
It would be very interesting to plot the full 3D probability density distributions for each
combination of quantum states. Unfortunately, distributions for non spherically symmetric
solutions (i.e. p and d quantum states) would be a function of θ and φ as well as of radius
r making them exceedingly difficult to plot.
Let’s look at electron orbitals for the Hydrogen atom
If we were to plot only the probability density functions for spherically symmetric
solutions (i.e. s quantum states) for each quantum state n we would find the following
distributions corresponding to the EPCs shown earlier for hydrogen.
Let’s look at electron orbitals for the Hydrogen atom
Spherical Harmonics
The solution of a PDE in spherical polar coordinates is
 (r , ,  )  R(r ) P( ) F ( )
We can say that the solution is comprised of a radially dependent function R (r ) and
two angular dependent terms P ( ) F ( ) which can be grouped together to form
specific spherical harmonic solutions Yl m ( ,  ) .
Formally the spherical harmonics Yl m ( ,  ) are the angular portion of the solution to
Laplace's equation in spherical coordinates derived in the notes.
The spherical harmonics Yl m ( ,  ) can be directly compared with the P ( ) and F ( )
solutions for the wave function describing the electron orbitals of the hydrogen atom.
Let’s look at electron orbitals for the Hydrogen atom
Spherical Harmonics
Spherical harmonics are useful in an
enormous range of applications, not
just the solving of PDEs.
They allow complicated functions of
θ and φ to be parameterised in terms
of a set of solutions.
For example a summed series of
specific harmonics as a Fourier
series can be used to describe the
earth (nearly but not exactly
spherical).
Summing harmonics can produce
some really pretty shapes
http://www.lifesmith.com/spharmin.html
Oil droplets or soap bubbles oscillating
Spherical Harmonics Yl m ( ,  ) also describe the wobbling deformations of an
oscillating, elastic sphere.
What sine and cosine are for a one-dimensional,
linear string, the spherical harmonics are for the
surface of a sphere.
A tiny oil droplet is placed on an oil bath which is
set into vertical vibrations to prevent coalescence
of the droplet with the bath. The droplet, which at
rest would have spherical form due to surface
tension, bounces periodically on the bath.
A movie shows the oscillations of the drop and
the corresponding calculations using spherical
harmonics Yl m ( ,  ) with ℓ = 2, 3, 4 and m = 0.
The magnetic quantum number m determines
rotational symmetry of the wobbling around the
vertical axis. For m ≠ 0, deformations are not
symmetric with respect to the vertical, and in this
case, the droplet starts to move around on the oil
bath. This can be seen in a second movie.
Oil droplets or soap bubbles oscillating
Oil droplets or soap bubbles oscillating