Advanced Physical Chemistry
NOTES 1
Review of Developments leading to Quantum Mechanics
Nature of Light
Black Body Radiation
Experimental Results led to:
Stefan-Boltzman Law
M = sT4
s = 56.7 nW/(m2K4) independent of the material
Wien's Displacement Law
lmzx T = constant
Could Electromagnetic Theory explain these?
Rayleigh Jeans
Energy in a region divided by the volume of the region is the Energy density, E(l)
AND dE(l) = rR(l)dl
dE(l) is the contribution from radiation in the wavelength range l to l + dl
rR(l) is the spectral density of states and can be thought of as the distribution of energy emitted a particular wavelength, (the relative amount of the total energy or fraction of the total energy emitted at a particular wavelength l ).
FROM Electromagnetic Theory, Rayleigh and Jeans derived a theoretical value of the spectral density of states. The formula is:
rR(l) = 8pkT/l4
where k is
Note what happens as l approaches zero.
This did not match the experimentally derived values and it became known as the Ultraviolet Catastrophe
Planck however derived the spectral density distribution of a black-body radiator which was essentially based on the idea that an oscillation of the electromagnetic field of frequency n, could only be excited in steps of energy of magnitude hn, due to the fact that the radiator was an oscillator with oscillations of energy hv, h being a new constant of nature, Planck's constant.
With this he derived
rR(l) = (8phc/l5) (e-hc/ λkT/ (1-e-hc/ λkT))
h = 6.626x10-34Js
This fit the entire distribution well.
Heat Capacities
The idea that there were oscillations in a material that gave rise to radiation also led to a correct value for the temperature dependence of Cv
Dulong and Petit said that the atoms of simple bodies all have the same heat capacity of 25 J/(K mol), no temp. dependence. Found to be incorrect.
Cv,m = 3RfD(T) where fD(T) = 3(T/qD)3 0 ∫ q/T x4ex 1 / (ex-1)2 dx
qD is the Debye Temperature and is related to the maximum frequency of oscillation that can be supported by the solid.
This complemented Planck's ideas, and lent credence to the idea that matter was quantized.
Photoelectric Effect
Einstein studied the emission of electrons when exposed to electromagnetic radiation.
Found that the emission was instantaneous when the radiation was applied even at low radiation intensities, but there was no emission, whatever the intensity of the radiation when the frequency of the radiation was below some threshold value. The threshold value was different for different metals.
The experiment was performed as below
And a plot of the Kinetic Energy as a function of the frequency of light gave the following plot
Thus there is a linear relationship between the kinetic energy of the emitted electron and the frequency of the radiation, and a cutoff frequency, below which the electrons are not emitted.
An energy balance says:
Energy of light = Kinetic Energy of e- + work function of metal (ionization energy)
or
The slope of the line in the plot was equal to
So photons have particle character. If so they should possess a momentum and should be able to transfer their momentum.
Compton Scattering
The relativistic expression for particle's energy
E2 = m2c4 + p2c2
c is the speed of light, and based on photoelectric effect E=hn and m = 0 so
(hn)2 = p2c2 and c = ln
so p = h/l
Partial transfer of the momentum of a photon during a collision, should appear as a cange in the wavelength of the photon.
dl = lf - li
when the radiation was scattered through an angle q and assuming photon as momentum h/l:
dl = 2lc sin2(½q)
where lc = h/(mec) is called the Compton wavelength of the electron (lc = 2.426 pm).
Hydrogen Spectra and Bohr's Theory
Scientists were interested in explaining why H emitted discrete spectral lines, and not a broad spectrum of visible light.
Balmer experimentally determined that visible lines of H followed
n' = RH (1/22 - 1/n2) n = 3, 4, 5, ...
RH is Rydberg constant for hydrogen know as RH = 1.097 x 105 cm-1
Suggested that the energies of atoms were confined to discrete values
Bohr showed that this could be explained by assuming that the angular momentum of an electron around a central nucleus was only allowed to possess certain values corresponding to circular orbits at certain discrete radii away from the nucleus
Based on this deduced
En = -me4/(8h2eo2) (1/n2) n = 1, 2, 3, ……
m is the reduced mass of the atom and eo is the vacuum permittivity
Thus this marriage of the model of mechanics and radiation was able to quantitatively account for the know spectrum of hydrogen including the Lymann, Balmer, and Paschen lines.
deBroglie's Relation
Since light which was thought to be strictly a wave, can act as both a particle and wave, deBroglie suggested that fundamental quantities that were thought to be just particles, could also possess wave properties. The wavelength of these "particles" should be given by:
l = h/p (just as for photons)
Davisson and Germer used a crystal of nickel to show that electrons shot at the crystal yielded a diffraction pattern consistent with the electrons having a wavelength as proposed by deBroglie.
If all particles have wave-like character, then there should be observational consequences.
Just as a wave of definite wavelength cannot be localized at a point, one should expect that one cannot expect a definite linear momentum to be defined at a single point.
Is this important for all particles, large and small?
Compare wavelength to the size of the particle.
Heisenberg Uncertainty Principle
Dp Dx ≥ h/4p
Complementarity of pairs of observables
The mutual exclusion of the specification of one property by the specification of another
A days few later, around Christmas, Schrodinger left Zurich for vacation in the Swiss Alps. There he developed an equation to take into account this wave particle duality.
Impossible to ‘rigorously derive’ the equation (the Quantum Mechanical Eqn.) that described microscopic particles which show both wave and particle duality, but from his notes we can determine that it was derived similar to the following:
Starting point:
Well known Equation of Classical Wave Motion (interrelates the space and time dependence of waves)
Separate into to further eqns.: one dealing only with spatial variations of the waves, the other with the time. For waves oscillating in three dimensions, the spatial wave eqn takes the form.
...... 2y = -k2 y
where 2 is known as the Laplacian operator and is given by d2/dx2 + d2/dy2 + d2/dz2 in Cartesian coordinates, and k, is the waver vector, and is equal to 2p/l, where l is the wavelength. There is a whole range of functions y (called wavefunctions) that satisfy the eqn., ranging from simple sine and cosine functions to more complicated functions.
But according to de Broglie, l = h/p and if p = mv then
...... k = 2p/l = 2pp/h = 2pmv/h
and hence: ...... 2y = (4p2m2v2/h2)* y ...... (A)
Also the total energy of a particle E is the sum of its kinetic and potential energies:
...... E = ½ mv2 + V
rearranging: ...... mv2 = 2(E-V)
and inserting this into Expression (A) gives:
...... 2y = -
or
...... -2/(2m) 2y + Vy = Ey ...... where = h/2p
What is E?
Known as three dimensional Schrodinger Equation and is one of the fundamental relationships in quantum mechanics, first described by Heisenberg and Schrodinger in 1926.
Schrodinger was able to apply his eqn to the hydrogen atoms and obtained the results consistent with Bohr's results.
An essential characteristic of Schrodinger's Eqn was that an observable like the energy of the system can be represented as an operator and the possible results of that observation are the eigenvalues of that operator in an eigenvalue equation:
In general if for the operator A, an eigenvalue eqn of the form:
AUn = an Un
can be formed for the operator, then:
Un are said to be the eigenfunctions or eigenstates of the operator A and
an are the allowed observable values of the operator operating on each of the eigenstates
In the case of Schrodinger's eqn this is usually written as
Hy = Ey or H yn = En yn
where yn are the energy eigenstates for the Hamiltonian which is the Energy operator and En are the energy eigenvalues or the possible energy values that can be observed.
Example:
Determine if the wavefunction y2 = (2/L)1/2 sin (2px/L) is an eigenfunction of the operator d2/dx2 and if it is, what is the eigenvalue corresponding to this eigenstate?
An important concept to note is that in general any function can be expanded in terms of a linear combination of all of the eigenfunctions (a complete set) of a particular operator. These are called the basis functions.
So if Un are the eigenfunctions corresponding to the operator A, then a general function, f can be written as a linear combination of the Un as:
f = S cn Un
cn are the coefficients and the sum is over the complete set of the eigenfunctions of the A operator. The sum is over the defined space.
The advantage of this type of expansion is that it allows one to determine the effect of an operator on a function that is not simply an eigenfunction of that operator.
The operators for the physical observables such as the position, the momentum, and the energy must reflect the ideas laid out by the Heisenberg uncertainty principle. There are two ways that this can be checked.
1. Commutation
Based on the Heisenberg uncertainty principle, it would seem to matter whether the position is measured first, and then the momentum, or if the momentum is measure before the position. Thus if x is the position operator and px is the momentum operator in the x direction, in one dimension if the Heisenberg Uncertainty principle is valid:
(x px - px x)g ¹ 0 g where g is the function describing the system. Thus since
(x px - px x) ¹ 0 , then x and px are said to "not-commute"
if on two operators say A and B are two operators that commute, then AB-BA = 0
The commutator of x and px is defined as [x, px] = (x px - px x) or more generally, the commutator of two operators A and B is [A,B] = AB-BA
Thus by the Heisenberg uncertainty principle, the commutator of the operators of two observables that cannot be specified simultaneously, should not commute, and if they do commute then they can be specified simultaneously.
Example:
if the operator for the position is multiplication by x or x = x. and the operator for the momentum, px, is px is ħ/i d/dx. ħ = h/(2p) determine whether or not these operators commute.
There are different representations of quantum mechanics which define the form of the operators corresponding to the observables. Two common ones are the
Position Representation
Momentum Representation
The energy operator, the Hamiltonian (H) is constructed from the sum of the operators corresponding to the kinetic energy (T) and to the potential energy.(V). Thus
H = T + V
In one dimension, in the x representation:
T = px2/ (2m) = 1/(2m) (ħ/i)2 (d/dx)2 or -ħ2/(2m) d2/dx2
And in 3 dimensions
T = -ħ2/(2m) (δ2/δx2 + δ2/δy2 + δ2/δz2) = -ħ2/(2m) V2 where V2 is the Laplacian and is the sum of the three second partial derivatives, at least in the Cartesian Coordinate system
The potential energy typically depends only on the position coordinates and so it is basically just multiplication by the potential function.
So
H = T + V = -ħ2/(2m) V2 + V(x)
So to construct operators in the position rpresentation, one should
1. Write the classical expression of the observable in terms of the position coordinates and the linear momentum.
2. Replace x by multiplication by x and replace px by ħ/i d/dx and similarly for higher dimensions or other coordinates.
Using the operator to get the value of the observable
The relation between the outcome of an experiment and a calculation done using the operator is most generally given by:
I = ∫g* O g dt / ∫ g* g dt (gives the average or expectation value of the observable corresponding to the operator O)
where g* is the complex conjugate of the function g, and g is the function describes the state of the system, which may or may not be an eigenfunction of the Operator, O, dt is the volume element corresponding to the defined space. If one dimensional it is dx or if three dimensional then it is dx dy dz in Cartesian coordinates. Remember in spherical
polar coordinates it is r2sinq dr dq df. Also the denominator is related to the overlap integral, S, but in this case the functions can be completely different functions. If they are the same and if g is a normalized function, then the denominator is equal to one. It may be useful to think of the overlap integral.
Example:
The first excited state of a particle in a box is y2 = N sin(2px/L). In one dimension the defined space is along the x axis from x=0 to x=L. Determine the expression for the normalization constant.
As you might expect, there are many instances where integrals similar to the ones shown above for the expectation value of an observable will be needed. These can be simplified to using the following notation which is related to the matrix formulation of Quantum Mechanics.
Called Dirac Bracket Notation
Integrals are written as follows:
<m|O|n> = ∫ ym* O yn dt often abbreviated as Omn