Download Basic Physical Chemistry Lecture 1

Basic Physical Chemistry
Lecture 1
Keisuke Goda
Summer Semester 2015
Lecture schedule
Since we only have three lectures, let’s focus on a few important
topics of quantum chemistry and structural chemistry
Lecture 1
Hydrogen atom
– The most basic unit of chemistry
Lecture 2
Angular momentum (spin)
– This is what makes chemistry interesting
Lecture 3
Molecular structure
– The origin of various unique properties of molecules
Chapter 1: The Wave Function
1.1: The Schrodinger Equation
1.2: The Statistical Interpretation
1.3: Probability
1.4: Normalization
1.5: Momentum
1.6: The Uncertainty Principle
Classical mechanics
In classical mechanics, we use Newton’s second law [F = ma]
to find the position [x(t)]
F  ma
V
d x

m 2
x
dt
2
Classical mechanics
In classical mechanics, if we know the initial conditions of the particle,
we can determine the final state of the particle (position & momentum)
by using Newton’s second law [F = ma]
Why not classical mechanics for small objects?
Unfortunately, we cannot use classical mechanics to
describe microscopic particles (atoms and molecules)
Incorrect
Correct
Because microscopic particles exist probabilistically
(we do not know their states until we measure them)
Quantum mechanics
The quantum-mechanical analog of Newton’s F = ma is the
Schrodinger equation.
Schrodinger
Equation
ᴪ = wave function
1. Quantum mechanics is all about the Schrodinger equation.
2. Our goal is to solve this equation for the wave function
under various conditions.
Erwin Schrodinger
Schrodinger’s grave
So, what is this Schrodinger equation??
Schrodinger
Equation
ᴪ = wave function
2
p
E
V
2m

E  i
t
 p

E  
 V 
 2m

2

p  i
x
The Schrodinger equation looks similar to the equation of
energy conservation in classical mechanics, but not quite!
Derivation of the Schrodinger equation (1)
Let’s assume that is a wave function with wavenumber k and angular frequency .
  0 ei kx t 
Then, let’s differentiate this wave function with respect to x.

 ik0 e i kx t   ik
x
Differentiating this function one more time yields
 2
2
i  kx t 
2



ik

e


k

0
2
x
Using the de Broglie equation k 
2

p

p2
 2
 2 
2
x


On the other hand, the derivative of the wave function with respect to t is given by

 i0 e i kx t   i
t
Using the Planck relation  
E


iE
 
t

Derivation of the Schrodinger equation (2)
The derivatives of the wave function can be rewritten as
p2
 2
 2 
2

x

iE
 
t

2


2
2
p   
x 2

E  i
t
Now, let’s consider the energy equation which is the sum of the kinetic energy and the
potential energy.
p2
E
2m
V
Putting the wave function on both sides of the equation gives
2
p
E 
  V
2m
Substituting E and p2 into this equation gives the Schrodinger equation.
 2  2

i

 V
2
2m x
t
Operators
Comparing the following two equations gives birth to the concept of “operators.”
2
p
E
V
2m

E  i
t
 2  2


 V
i
2
2m x
t

p  i
x
• Never mind why the energy and momentum become operators.
• We can do quantum mechanics if we accept this assumption.
So, what is the wave function??
Born’s statistical interpretation of the wave function
Collapse of the wave function
Collapse of the wave function
after a measurement
Probability of continuous variables
We have dealt with discrete variables, but it is simple enough
to generalize the theory to continuous distributions.
Probability density
(probability of getting x)
Probability that x lies
between a and b
Obvious relations
The integration of the probability
density over the entire range is 1
(of course).
Average value of x
Average value of f(x)
Variance of x
Need for normalization
Remember Born’s statistical interpretation of the wave function
Probability density for finding
the particle at point x at time t
The integration of the probability density over the entire range is 1
(of course again).
Requirement for
normalization
Expectation value of position x
Expectation value of x
It does not mean that if you measure the position of one particle
over and over again, <x> is the average of the results you will get.
Interpretation of the expectation value
The expectation value is the average of repeated measurements on
an ensemble of identically prepared systems, not the average of
repeated measurements on one and the same system.
Expectation value of momentum p
Take the derivative with respect to time
Use integration by parts
Use integration by parts again
Let’s postulate that the expectation value of the velocity is the time derivative of the
expectation value of the position (we will prove this in Chapter 3)
More suggestive expressions of x and p
•
The operator
•
The operator
x
 
i x
represents position
represents momentum
To obtain the expectation value of position x and momentum p,
we sandwich each operator between ᴪ* and ᴪ and then integrate.
Generalized expectation value
Q = Q(x, p) = any function of x and p
To obtain the expectation value of Q, we simply replace every p
with the operator, insert the resulting operator between ᴪ* and ᴪ,
and integrate.
For example, the expectation value of the kinetic energy is given by
T is a function of p
Replace p with the
operator in the integral
Chapter 4: Quantum Mechanics in Three
Dimensions
4.1: Schrodinger Equation in Spherical
Coordinates
4.2: The Hydrogen Atom
4.3: Angular Momentum
4.4: Spin
Generalization to 3D
So far we have considered Schrodinger’s equation only in 1D, but it can
be extended to 3D
The momentum operator in each dimension is given by
Schrodinger’s equation hence becomes
where
Schrodinger’s equation in 3D
(This is called the Laplacian)
Schrodinger’s equation in spherical coordinates
In most cases in chemistry, the potential V is a function only of the
distance from the origin, which means that V = V(r) = V(r, , )
Schrodinger’s equation in spherical coordinates
Schrodinger’s equation in spherical coordinates
Cartesian coordinates
Spherical coordinates
Schrodinger’s equation in spherical coordinates (in terms of r, , )
Separation of variables in spherical coordinates
In order to solve the Schrodinger equation, we separate the wave
function into the radial part, R(r), and the angular part, Y(, )
Putting this into the Schrodinger equation yields
Simplifying this equation gives
The first tem is only a function of r whereas the second term only
depends on  and , such that each term must be constant
Separation of variables in spherical coordinates
Constant
Constant
The constant is written in this form for some reason (Never mind this
for now, but this substitution will turn out to be very convenient and
meaningful and be explained in Chapter 4.3)
Radial
part
Angular
part
Our next step is to solve each equation for R(r) and Y(, )
Angular part
Again, the angular part is given by
As always, we separate variables
Putting this into the angular equation, we find
The first tem is only a function of  whereas the second term is only a
function of , such that just like before, each term is written as a
constant
Never mind why we use
the constant m2 for now
(this will be explained in
Chapter 4.3)
Solving the  equation
The  equation is easy to solve
Since  is the azimuthal angle, the solution also allows
This means m must be an integer
Solving the  equation
The  equation is not so easy to solve
The solution is given by
where
is called the associated Legendre function defined by
and
is called the Legendre polynomial defined by the Rodrigues
formula
Legendre polynomial
The Legendre polynomial is even or odd, depending on the value of l
Associated Legendre functions
If m is odd, P is a function of sin
If m is even, P is a function of cos
Requirement for l and m
Notice that l must be a non-negative integer for the Rodrigues formula
to make any sense
If
For any given l, then there are
, then
possible values of m
For example, if l = 0, m = 0; if l = 1, m = -1, 0, 1; if l = 2, m = -2, -1, 0,
1, 2; and so forth
Normalization and solution
The volume element in spherical coordinates is given by
So the normalization condition becomes
It is convenient to normalize R and Y separately
The normalized angular wave functions are called spherical harmonics
Orthogonality and quantum numbers
Spherical
harmonics
The spherical harmonics are orthogonal (I will leave the proof to you)
l is called the azimuthal quantum number and m is called the magnetic
quantum number
Radial part
Notice that the angular part of the wave function, Y(, ), is the same
for all spherically symmetric potentials whereas the actual shape of the
potential, V(r), affects only the radial part of the wave function, R(r),
which is given by
This equation can be simplified if we use the substitution
Radial
equation
This radial equation is identical in form to the 1D Schrodinger equation
except for the effective potential
Normalization of the radial wave function
The normalization condition becomes
since
This looks like the normalization condition in 1D, but remember that
because of this substitution, we have the extra term in the effective
potential
Effective potential
So, the strategy for solving the equation is first to find u(r) and then
obtain R(r) = u(r) / r
Short summary of the 3D Schrodinger equation
Schrodinger’s equation in spherical coordinates (in terms of r, , )
The function is separated into the radial part, R(r), and the angular part,
Y(, )
The angular part is found to be
The radial part can be solved for
when V is provided
Hydrogen atom
Finally, we are getting closer to quantum “chemistry” based on our
knowledge of quantum mechanics, but be warned that even the wave
function of the simplest model (hydrogen atom) is not so simple
Hydrogen atom
Based on the Schrodinger equation in 3D spherical coordinates, we are
ready to solve the radial part of the Schrodinger equation for hydrogen
  me
In theory, we should use the reduced mass, but since mp is about 1000
times larger than me and we can replace the reduced mass with me
Schrodinger’s equation for the hydrogen atom
From Coulomb’s law, the potential energy (which only depends on r) is
given by
The radial equation is given by
This equation allows both continuum states
(E > 0), describing electron-proton scattering,
as well as discrete bound states (E < 0),
representing the hydrogen atom
Solution to the radial equation
Unfortunately, the calculation of this radial equation is too lengthy to
cover in this lecture. Therefore, I will leave the calculation to you (it’s
covered over several pages in Griffiths), and let me get straight to the
answers directly.
Hint: If you want to really solve this differential equation, focus on the
first few values of l (l = 0, l = 1, etc.) and you can solve the equation.
Radial wave functions for hydrogen
Remember R(r) = u(r) / r
Here a is the characteristic
constant defined to be
This is called the famous
Bohr radius.
Radial wave functions for hydrogen
Normalized hydrogen wave functions
Combining the radial and angular wave functions
yields the normalized wave functions for the hydrogen atom
where
is called the associated Laguerre polynomial
is called the Laguerre polynomial
Normalized hydrogen wave functions
The wave functions look quite ugly, but the hydrogen atom is one of the
very few realistic systems that can be solved analytically (the
Schrodinger equation in all other chemical systems needs to be solved
numerically by using computer simulations)
Just like in the previous examples (infinite square well and harmonic
oscillator), wave functions with different (n, l, m) values are orthogonal
The wave functions are mutually orthogonal
Density plots for the hydrogen wave functions
Surfaces of constant probability densities
Probability densities for various orbitals
Bohr formula
Substitute the radial wave functions
into the radial part of the Schrodinger
equation
Bohr formula
Niels Bohr obtained this function in 1913 by using the Bohr model (a
premature mixture of classical physics and quantum physics). Note that
the Schrodinger equation did not come until 1924.
Hydrogen energy levels in more detail
The allowed energy levels for the hydrogen atom (without corrections)
are given by
A few major corrections (which can be obtained by using the
perturbation theory) are given by
Fine structure
(relativistic correction
+ spin-orbit coupling)
Hyperfine splitting
(spin-spin coupling)
Energies of the first few states
The ground state (the state of the lowest energy) is the case n = 1
(hence, l = 0), and therefore, the ground state energy is given by
If n = 2, the first excited state energy is given by
Here l is either l = 0 (with m = 0) or l = 1 (with m = -1, 0, 1)
Spectrum of hydrogen
• In principle, if you put a hydrogen atom into some stationary state, it
should stay there forever.
• However, if you tickle it slightly (by collision with another atom or
shining light on it), the electron may undergo a transition to some
other stationary state – either by absorbing energy (absorbing a
photon and moving up to a higher energy state) or by giving off
energy (emitting a photon and moving down to a lower energy state).
• This transition energy corresponds to the difference in energy
between the initial and final states.
Energy
difference
Initial
state
Final
state
Spectrum of hydrogen
According to the Planck formula, the energy of a photon is proportional
to its frequency
Meanwhile, the wavelength is given by
Combining these two equations, we find the so-called Rydberg formula
where R is called the Rydberg constant
Energy levels and transitions
Lyman series
 Transitions to the ground
state (n = 1)
 In the ultraviolet range
Balmer series
 Transitions to the first
excited state (n = 2)
 In the visible range
Paschen series
 Transitions to the second
excited state (n = 3)
 In the infrared range
Energy levels and transitions
Energy levels and transitions
Spectroscopy
Absorption spectra of various elements
Quiz
Answer the following questions about the hydrogen atom. The table of the radial wave
functions for hydrogen on the right may be useful for your calculations. Provide your
answers in terms of the Bohr radius ܽ.
(a) Find the position of the electron in the 1s orbital where it is most likely to be found.
(b) Find the position of the electron in the 2s orbital where it is most likely to be found.
(c) Find the position of the electron in the 2p orbital where it is most likely to be found.
(d) Find the position of the electron in the 3d orbital where it is most likely to be found.
Second report
• Read the paper “The Early History of Spectroscopy” by N. C.
Thomas in Journal of Chemical Education 68, 631 (1991) and write
a 1-page report about it in English.
• The theme of your report is about how spectroscopy has been
developed
• The report should be written in a A4-sized paper.
• Only single spacing is acceptable (no double spacing).
• Font size: 11 pt
• Font type: Arial, Times New Roman, or Calibri
• Write down your name and ID number at the top of your report.
• Do not forget to write the title for your report.
• Submission deadline: 4/21
• Put your report on the lecturer’s desk before the lecture begins.