Transcript from last time:

from last time:
m
m 2  2 x 2
1. show that  ( x)  A(1 
x )e

is also a solution of the SE for the SHO, and find the
energy for this state
2. Sketch the probability distribution for the SHO
wavefunctions, AND the probability distribution for a
classical oscillator on the same axes.
3. Ponder this: how can a particle in the n=2 state get
from one side of the well to the other?
2. Sketch the probability distribution for the
SHO wavefunctions, AND the probability
distribution for a classical oscillator on the
same axes.
quantum
P=2
U
Classical
Pv -1
x
0
quantum physics 5:
more solutions to
The
Schroedinger
Equation:
finite potentials
Solving the Schroedinger equation - a recipe
1.
2.
3.
4.
5.
Start by writing down the S.E. with the appropriate potential
energy, e.g. for the S.H. Oscillator U = ½ kx2. If U is not cts
you may need to write down the S.E. for each distinct
region where there is a different U.
Find a wave function which is a solution to the S.E. – this is
often done by educated guessing, and there may be more
than one solution.
Apply boundary conditions – these will often limit your
values of energy.
Evaluate any undetermined constants (like amplitudes), e.g.
by using boundary conditions, applying normalisation.
Check your solution, if it gives you something dodgy like
P=, check for errors or start again.
1D finite potential well
This time consider a particle, e.g. an electron, which is
confined to an finite square potential well, in 1D. This is
more realistic for most situations.
We can write the potential energy as:
 0 0 xL
U ( x)  
U 0  x, L  x
U
In this case our potential
as finite outside the “well”
so the electron has a finite
probability of escaping!
0
x
L
We’re working with the time independent form again:
 2  2 ( x)

 U ( x) ( x)  E ( x)
2
2m x
inside the well:

   ( x)
 E ( x)
2
2m x
2
2
or
 2 ( x)
2

k
1  ( x)  0
2
x
and our wave function will be of the form:
 ( x)  Ae
ik1 x
 Be
 ik1 x
k1 
2mE

but now we can’t assume B equal to zero, because the
well is finite!
outside the well:
 2  2 ( x)

 U ( x)  E ( x)
2
2m x
or
 2 ( x) 2m
 2 (U  E ) ( x)  0
2
x

We’re only going to consider the U > E case, because E>U
is not a confined particle anyway
We have two regions to worry about, to the left and the
right, so lets write the two possible solutions:
 ( x  0)  Ce
 ( x  L)  Fe
ik2 x
ik2 x
 De
 Ge
 ik2 x
 ik2 x
k2 
2m(U  E )

Now we use the requirements that the wave function is
well behaved – it must be continuous across boundaries,
and so must its derivative w.r.t x…
at x = 0:
 ( x  0)  Ce
ik2 0
 De
 ik2 0
 Ae
ik1 0
 Be
 ik1 0
therefore C  D  A  B
but as x -, the solution outside the well  unless D
= 0 if the wave function is to be normalisable.
so
 ( x  0)  ( A  B)e
ik2 x
on the RHS of the well, to prevent the wave function 
we require F = 0
then you do LOTS of algebra, using the two conditions of
continuity at the boundaries, to solve for A, B, D and then
to get the energy – homework problem, or next year!
The main results are:
-energy is again quantised inside the well
-outside the well the wavefunction has an exponentially
decreasing form BUT its not zero! The particle can escape
even though E < U!
-inside the well its standing waves again
Our wave functions and probability densities
look like this:

P=2
U
tunelling!
0
x
L
barriers and tunneling
Imagine what would happen if we had a barrier…
Classically, a particle with less energy than the
barrier height could not pass the barrier. But a
wave-like particle CAN, because its
wavefunction extends some distance into the
barrier, and beyond it.
U
0
L
x
To either side we have a “typical” wavefunction for a free
particle. Note that it is complex, and could be written in
terms of sin, cos instead. Inside the barrier the
wavefunction is exponentially decreasing, and real.
Also note that the wave vector is the same on each side.
real!!
 ( x)  Ce
complex!!
 ( x)  Aeik x  Be ik x
1
1
 k2 x
complex!!
 ( x)  De ik x  Feik x
1
U
0
L
x
1
probability of tunelling
Within the barrier the wavefunction is of the form:
 ( x)  Ce
 k2 x
where:
k2 
2m(U  E )

Recall that the probability is:
 * ( x) ( x)  C * Ce2 k x
2
So the probability of transmission:
- decreases with U, m
- decreases with x, i.e. barrier width
- increases with C, E
barriers, wells, atoms and conductivity…
We can consider a solid to be a collection of potential wells
with barriers between them.
Depending on the depth of the wells, and the
wavefunctions for the electrons bound to the atoms, the
material will have more or less conductivity.
Quantum mechanics is the basis of solid state physics,
which is the basis of our understanding of semiconductors,
which is the basis of all electronics and computers!
so lets look at atoms now…
Newton C17
Light as particle
mechanics
Maxwell C19
Light as
Problems –
wave
PE effect,
Black bodies Einstein’s solution C20
Light as particle
Historical
de Broglie
particles as waves
development
Thompson
and others,
Pudding
models
Greek Atomic
theory,C5 BC
Democritus
etc
Rutherford
then Bohr,
orbitals
Problems –
spectra
Davisson and
Germer’s
confirmation
Heisenberg
and Pauli
uncertainty,
Technology
exclusion
and Schrodinger
equation
Quantum physics
Technology
So far we’ve followed the wave particle and photon
side, mostly.
Now lets have a look at atoms, now that we have the
SE to help us understand them as systems with a
potential energy well which holds electrons (Fermions)
to a confined space…
Atoms
What we knew ca. 1920:
1. Atoms are stable – they don’t (usually) fall apart.
2. Atoms are very small.
3. Atoms have electrons in them, but are electrically
neutral.
4. Atoms emit and absorb radiation of discrete
wavelengths.
The big problem with most models was number 4 –
how to account for discrete energy changes. We’re
going to start with Bohr’s model because it was the
first to really explain this. Sort of.
Bohr’s model of the atom
Bohr took what was known about atoms, and made some
postulates, and came up with a model.
The postulates:
1.
an electron in an atom moves in a circular orbit about the
nucleus under the influence of the Coulomb attraction.
2.
It is only possible for an electron to move in an
orbit for which its orbital angular momentum is
quantised as L =nħ.
3.
In spite of its constant acceleration E remains constant and
the electron does not radiate EM radiation and collapse.
EM radiation is emitted if an electron discontinuously
changes from one orbital to another. The frequency of EM
radiation emitted is given by Eelectron=hf.
4.
The model:
electrons move in orbits in which the centripetal force due
to the coulomb attraction keeps them in a stable circular
orbital, like gravity keeping planets in orbit. We only
consider a one electron atom so we don’t have to worry
about interactions of electrons.
2
q
q
mv
F  1 1 22 
40 r
r
2
q
q
mv
q1q2
1 2
1
1
U 
KE 

40 r
40 2r
2
So the KE is ½ the PE, and the total energy is:
q1q2
1
E  U  KE  
40 2r
quantising the energy by quantising L:
Only certain orbits are allowed, because angular
momentum is quantised:
L  mvr  n
n

hence r 
so
mv
from our force equation we can write r 
2 2
n

2
r  2 2
mv
1 q1q2
40 v 2 m
n 2 2
divide one by the other to get r without v: r  40
q1q2 m
which means the energy is:
2
2
q
q
m
(
q
q
)
1
 1 
1 2
1 2
1
E

k 2

2 2
40 2r  40  2 n
n
1
E 2
n
and
En 1 E0
En  2  2
n
n
pops out of Bohr’s model as a result of quantisation of
angular momentum. This is good, because it matches the
experimental observation that spectral lines from hydrogen
could be fitted by the equation:
1 
 1
E photon  hf  k  2  2 
m 
n
this is good stuff, and the model
predicted the spectrum of hydrogen, but
there wasn’t any a priori justification for
the postulates…
for next time:
Readings:
T4: 37.3 – 37.4
T&M5: 36.3 – 36.4
PLUS look at hyperphysics on SE in 3D and atoms
Examples to do:
T4: 37.1, 37.2
T&M5: 36.1, 36.2
Homework problem:
Calculate the first Bohr radius for:
a. a hydrogen atom
b. a doubly ionised Lithium atom