from last time:

1. show that

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

22 )1()(x

m

exm

Ax

2. Sketch the probability distribution for the SHO wavefunctions, AND the probability distribution for a classical oscillator on the same axes.

U

x0

quantum

P=2

Classical

Pv -1

quantum physics 5:

more solutions to The SchroedingerEquation:

finite potentials

Solving the Schroedinger equation - a recipe

1. 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.

2. 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.

3. Apply boundary conditions – these will often limit your values of energy.

4. Evaluate any undetermined constants (like amplitudes), e.g. by using boundary conditions, applying normalisation.

5. 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:

xLxU

LxxU

,0

00)(

In this case our potential as finite outside the “well” so the electron has a finite probability of escaping!

x0 L

U

We’re working with the time independent form again:

)()()()(

2 2

22

xExxUx

x

m

and our wave function will be of the form:

xikxik BeAex 11)(

inside the well:

0)()( 2

12

2

xk

x

x

but now we can’t assume B equal to zero, because the well is finite!

)()(

2 2

22

xEx

x

m

or

mE

k2

1

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:

outside the well:

0)()(2)(22

2

xEU

m

x

x

)()()(

2 2

22

xExUx

x

m

or

xikxik DeCex 22)0(

xikxik GeFeLx 22)( )(2

2

EUmk

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:

BADC therefore

but as x -, the solution outside the well unless D = 0 if the wave function is to be normalisable.

so

0000 1122)0( ikikikik BeAeDeCex

xikeBAx 2)()0(

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

x0 L

U

tunelling!

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.

x0 L

U

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.

x0 L

U

xkCex 2)( xikxik BeAex 11)( xikxik FeDex 11)(

real!!

complex!!complex!!

probability of tunelling

Within the barrier the wavefunction is of the form:

xkCex 2)( where:

)(22

EUmk

Recall that the probability is:xkCeCxx 22*)()(*

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 C17Light as particlemechanics

Maxwell C19Light as wave

Greek Atomic theory,C5 BCDemocritus etc

Rutherford then Bohr,orbitals

Thompson and others,Pudding models

Einstein’s solution C20Light as particle

de Broglie

particles as waves

Quantum physics

Davisson and Germer’sconfirmation

Heisenberg and Pauli

uncertainty, exclusion

Technology

Problems – spectra

and Schrodinger equation

Technology

Problems – PE effect, Black bodies

Historical

development

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.

4. 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.

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.

r

qqU 21

041

r

mv

r

qq 2

221

041 F

r

qqmvKE

241

221

0

2

So the KE is ½ the PE, and the total energy is:

r

qqKEUE

241 21

0

quantising the energy by quantising L:

Only certain orbits are allowed, because angular momentum is quantised:

nmvrL hencemv

nr

which means the energy is:

mv

qqr

221

041

22

222

vm

nr

so

from our force equation we can write

divide one by the other to get r without v:mqq

nr

21

22

04

222

221

2

0

21

0

1

2

)(41

241

nk

n

qqm

r

qqE

2

1

nE

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:

22

11

mnkhfEphoton

20

21

n

E

n

EE nn and

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