can be evaluated as the determinant of a matrix, the first row of which consists of the unit vectorsthe second row of the components of the position operator, and the third row of the components of the momentum operator. We expand along the top row.

All the coefficients are by convention written as positive. We can do this by taking the negative sign belonging to the first and last terms inside the brackets, obtaining

This is the most general form of the angular momentum operator. We can separate this into it's components:

The operators can also be written with respect to and angle. This is common foroften written

]]>Perfect knowledge of a particles momentum means its position cannot be known with precision.

In fact we can say the product of the uncertainty in the position and the uncertainty in the momentum of a particle must be greater than or equal to

Particles created from the vacuum must have a certain nonzero energy. We can treat this energy as an uncertainty because it is energy created almost from nothing. Knowledge of this energy puts an upper limit on the lifetime of the particle.

In fact we can say the product of the reciprocal of uncertainty of the energy of the particle and the reciprocal in the lifetime of the particle must be greater than or equal to

We cannot know simultaneously the angular momentum of a particle around two different axes. The product of the uncertainties must be at least

Each of these three examples gives an example of two quantities that cannot be simultaneously known to arbitrarily high precision. For each pair, this is the case because their quantum mechanical operators do not commute.

For exampleand

Dropping thegives the result

Similar relationships hold for every pair of quantities that cannot be written down simultaneously to arbitrarily high precision.

If two quantum mechanical operators hat x and hat b do commute thenand the two physical quantities corresponding to these two quantum mechanical operators can simultaneously be know to arbitrarily high position,

]]>where {jatex options:inline};^2 =-1{/jatex}.

To prove anticommutativity we must show {jatex options:inline}AB = -BA{/jatex} for each pair of matrices.

{jatex options:inline}\sigma_x \sigma_y =\left( \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right) \left( \begin{array}{cc} 0 & -i \\ i & 0 \end{array} \right) = \left( \begin{array}{cc} i & 0 \\ 0 & -i \end{array} \right){/jatex}

{jatex options:inline}\sigma_y \sigma_ =\left( \begin{array}{cc} 0 & -i \\ i & 0 \end{array} \right) \left( \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right) = \left( \begin{array}{cc} -i & 0 \\ 0 & i \end{array} \right)=- \sigma_y \sigma_x{/jatex}

{jatex options:inline}\sigma_x \sigma_z -=\left( \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right) \left( \begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array} \right) = \left( \begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array} \right){/jatex}

{jatex options:inline}\sigma_z \sigma_x -=\left( \begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array} \right) \left( \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right) = \left( \begin{array}{cc} 0 & 1 \\ -1 & 0 \end{array} \right)= - \sigma_x \sigma_z{/jatex}

{jatex options:inline}\sigma_y \sigma_z -=\left( \begin{array}{cc} 0 & -i \\ i & 0 \end{array} \right) \left( \begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array} \right) = \left( \begin{array}{cc} 0 & i \\ i & 0 \end{array} \right){/jatex}

{jatex options:inline}\sigma_z \sigma_y -=\left( \begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array} \right) \left( \begin{array}{cc} 0 & -i \\ i & 0 \end{array} \right) = \left( \begin{array}{cc} 0 & -i \\ -i & 0 \end{array} \right)= - \sigma_y \sigma_z{/jatex}

Hence the set of Pauli Spin matrices are anticommutative.]]>

We can use the ordinary rules of classical Newtonian mechanics to derive the equation giving the differences in the energy levels of the electrons in the Hydrogen atom.

We can equate the Coulomb force to the force required to keep the electron in orbit around the nucleus.

From this we can obtain

(1)

and(2)

The total energy of the electron in it's orbit is the sum of the kinetic and potential energies: Use of (2) gives

This means an energy change ofwhereis the frequency andis the wavelength of the associated emitted or absorbed photon. Substitute (1) to give

Use this and divide byto get

is called the Rydberg constant and is given the symbolIt has the value

]]>The Compton effect is governed by the formula

Proof: If the electron is initially at rest then it's energy is it's rest mass energy, and afterwards it's energy is determined by the relativistic equation

Conservation of energy impliesrearrangement of which gives(1)

Conservation of momentum impliesor

The dot product property gives

for a photon hence(2)

PutAfter some simplification we obtain

Divide both sides byto obtainand becausewe have

]]>Avagadro's Constant

Planck's Constant

Mass of Electron

Mass of Proton

Mass of neutron

Elementary Charge

Bohr Magneton

Bohr Radius

Compton Wavelength of Electron

Coulomb's Law Constant

Electron Charge to Mass Ratio

Rydberg Constant for Hydrogen

Speed of Light in Vacuum

Boltzmann's Constant

]]>In general the wavepacket is made of of many differentfrequencies or wavelengths. They interfere constructively in the region of the particle and destructively outside it. Different parts of thewavepacket will in general travel at different velocities, with the velocity dependant on frequency or wavelength. This means that the wavepacket will spread out as it travels. We can derive the'disperson relation' from the wavefunction obeyed by the particle.

The basic, non – dispersive wave equation isor(1)in one dimension. This equation is obeyed only by ideal waves –which have no weight, for example. Electromagnetic waves travelling through a vacuum obey this form of wave equation.

If the conditions of ideality do not apply, for example if the wave is carried by a string which is not perfectly flexible, then we will need to introduce a term proportional to displacement with a negative coefficient. The wave equation becomes (2)

or the wave is moving through a resistive medium, so experiences damping, we need to introduce a term proprtional to with a negative coefficient. The wave equation becomes(3)

The dispersion relation is found by substituting the solution for the undamped wave equation (1) (4) into (2) or (3). Dividing by the exponential factor leaves the dispersion relation.

For the first example (2)

Differentiating (4) gives

Subsitution into (1)

Dividing by the non zerogivesor

The phase velocityand the group velocity

For the second example (3)

Dividing through by asbefore givesThecomplex factor here is a problem. It means that eitheroris complex. Because we included a damping or resistance ter mdifferentiated with respect to time, I will assume it isthat is complex, that

Cancelling the exponential term gives

Equating the real parts gives

Equating imaginary parts gives

]]>In quantum mechanics we have something similar. Remember that the probability density of a particle is given by

whereis the operator of the observable we want to find the expectation value of.

For example ifthe expectation value of the position is

We use the identity,rearranging to obtainThe integral becomes

We integrate by parts:

This is, as we would expect in the centre of the well, sinceis symmetric about the centre of the well.

]]>The number of radial nodes increases by one with each increase in energy level.

The wavefinction decreases to zero outside the quantum well since it is a bound wavefunction.

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