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Transcript
Chemistry 431
Lecture 3
The Schrödinger Equation
The Particle in a Box (part 1)
Orthogonality
Postulates of Quantum Mechanics
NC State University
Derivation of the Schrödinger Equation
The Schrödinger equation is a wave equation.
Just as you might imagine the solution of such an equation
in free space is a wave. Mathematically we can express a
wave as a sine or cosine function. These functions are
oscillating functions. We will derive the wave equation in
free space starting with one of its solutions: sin(x).
Before we begin it is important to realize that bound states
may provide different solutions of the wave equation than
those we find for free space. Bound states include
rotational and vibrational states as well as atomic wave
functions. These are important cases that will be treated
once we have fundamental understanding of the origin of
the wave equation or Schrödinger equation.
The derivative
The derivative of a function is the instantaneous rate of change.
The derivative of a function is the slope.
We can demonstrate the derivative graphically.
We consider the function f(x) = sin(x) shown below.
The derivative of sin(x)
The derivative of a function is the instantaneous rate of change.
The derivative of a function is the slope.
At sin(0) the slope is 1 as shown by the blue line.
The derivative of sin(x)
The derivative of a function is the instantaneous rate of change.
The derivative of a function is the slope.
At sin(π/4) the slope is 1/√2 as shown by the blue line.
The derivative of sin(x)
The derivative of a function is the instantaneous rate of change.
The derivative of a function is the slope.
At sin(π/2) the slope is 0 as shown by the blue line.
The derivative of sin(x)
The derivative of a function is the instantaneous rate of change.
The derivative of a function is the slope.
At sin(3π/4) the slope is -1/√2 as shown by the blue line.
The derivative of sin(x)
The derivative of a function is the instantaneous rate of change.
The derivative of a function is the slope.
At sin(3π/4) the slope is -1/√2 as shown by the blue line.
The slopes of all lines thus far are plotted as black squares.
The derivative of sin(x)
The derivative of a function is the instantaneous rate of change.
The derivative of a function is the slope.
At sin(π) the slope is -1 as shown by the blue line.
The slopes of all lines thus far are plotted as black squares.
The derivative of sin(x)
The derivative of a function is the instantaneous rate of change.
The derivative of a function is the slope.
At sin(5π/4) the slope is -1/√2 as shown by the blue line.
The slopes of all lines thus far are plotted as black squares.
The derivative of sin(x)
The derivative of a function is the instantaneous rate of change.
The derivative of a function is the slope.
At sin(5π/4) the slope is -1/√2 as shown by the blue line.
The slopes of all lines thus far are plotted as black squares.
The derivative of sin(x)
The derivative of a function is the instantaneous rate of change.
The derivative of a function is the slope.
We see from of the black squares (slopes) that the
derivative of sin(x) is cos(x).
The derivative of sin(x)
d
sin(x) = cos(x)
dx
The derivative of cos(x)
d
cos(x) = -sin(x)
dx
The second derivative of sin(x)
d
dx
d
sin(x) = -sin(x)
dx
The second derivative of sin(x)
d2
sin(x) = -sin(x)
dx2
Sin(x) is an eigenfunction
d2
If we define dx2
as an operator G then we have:
d2
sin(x) = sin(x)
dx2
which can be written as:
G sin(x) = sin(x)
This is a simple example of an operator equation that
is closely related to the Schrödinger equation.
Sin(kx) is also an eigenfunction
We can make the problem more general by including
a constant k. This constant is called a wavevector.
It determines the period of the sin function. Now we must
take the derivative of the sin function and also the
function kx inside the parentheses (chain rule).
d
sin(kx) = -k cos(kx)
dx
d2
sin(kx) = k2sin(kx)
dx2
Here we call the value k2 the eigenvalue.
Sin(kx) is an eigenfunction
of the Schrödinger equation
The example we are using here can easily be expressed
as the Schrodinger equation for wave in space. We only
have to add a constant.
-h2 d2
-h2k2
sin(kx) =
sin(kx)
2m dx2
2m
In this equation -h is Planck’s constant divided by 2π and
m is the mass of the particle that is traveling through space.
The eigenfunction is still sin(kx), but the eigenvalue in
this equation is actually the energy.
The Schrödinger equation
Based on these considerations we can write a compact
form for the Schrödinger equation.
HΨ = EΨ
-h2 d2
H=2m dx2
Energy operator, Hamiltonian
-h2k2
E=
2m
Energy eigenvalue, Energy
Ψ= sin(kx)
Wavefunction
The momentum
The momentum is related to the kinetic energy. Classically
The kinetic energy is:
1 2
E = mv
The momentum is:
2
p = mv
So the classical relationship is:
p2
E=
2m
If we compare this to the quantum mechanical energy:
-h2k2
E=
2m
we see that: p = hk
The general solution to the
Schrödinger equation in free space
The preceding considerations are true in free space.
Since a cosine function has the same form as a sine
function, but is shifted in phase, the general solution
is a linear combination of cosine and sine functions.
Ψ= Asin(kx) + Bcos(kx)
Wavefunction
The coefficients A and B are arbitrary in free space.
However, if the wave equation is solved in the presence
of a potential then there will be boundary conditions.
The particle in a box problem
Imagine that a particle is confined to a region of space.
The only motion possible is translation. The particle has
only kinetic energy. While this problem seems artificial at
first glance it works very well to describe translational
motion in quantum mechanics.
0 Allowed Region
L
The solution to the Schrödinger
equation with boundary conditions
Suppose a particle is confined to a space of length L.
On either side there is a potential that is infinitely large.
The particle has zero probability of being found at the
boundary or outside the boundary.
0 Allowed Region
L
The solution to the Schrödinger
equation with boundary conditions
The boundary condition is that the wave function will
be zero at x = 0 and at x = L.
Ψ(0) = Asin(k0) + Bcos(k0) = 0
From this condition we see that B must be zero.
This condition does not specify A or k.
The second condition is:
Ψ(L) = Asin(kL) = 0 or kL = arcsin(0)
From this condition we see that kL = nπ. The conditions
so far do not say anything about A. Thus, the solution
for the bound state is:
Ψn(x) = Asin(nπx/L)
Note that n is a quantum number!
The probability interpretation
The wave function is related to the probability for finding
a particle in a given region of space. The relationship is
given by:
P = Ψ 2dV
If we integrate the square of the wave function over a
given volume we find the probability that the particle is
in that volume. In order for this to be true the integral
over all space must be one.
Ψ 2dV
1=
all space
If this equation holds then we say that the wave function
is normalized.
The normalized bound state
wave function
For the wave function we have been considering,
all space is from 0 to L. So the normalization constant A
can be determined from the integral:
L
1=
0
L
2
Ψ dx =
0
A sin nπx
L
2
2
dx = A
L
2
0
sin nπx
L
2
dx
The solution to the integral is available on the
downloadable MAPLE worksheet. The solution is just L/2.
Thus, we have:
2
2
2
1=A L ,A = 2 ,A=
L
L
2
As you can see the so-called normalization constant has
been determined.
Normalization
What is the normalization constant for the wave function
exp(-ax) over the range from 0 to infinity?
A. a
B. 2a
C. √a
D. √2a
Normalization
What is the normalization constant for the wave function
exp(-ax) over the range from 0 to infinity?
A. a
B. 2a
C. √a
D. √2a
L
1=
0
∞
Ψ 2dx =
0
∞
2
A 2exp – ax dx = A 2
2
= A 1 , A = 2a
2a
0
exp – 2ax dx
The appearance of the wave functions
The appearance of the wave functions
Note that the wave functions have
nodes (i.e. the locations where they
cross zero). The number of nodes is
n-1 where n is the quantum number
for the wave function. The appearance
of nodes is a general feature of
solutions of the wave equation in bound states. By bound
states we mean states that are in a potential such as the
particle trapped in a box with infinite potential walls. We
will see nodes in the vibrational and rotational wave functions
and in the solutions to the hydrogen atom (and all atoms).
Note that the wave functions are orthogonal to one another.
This means that the integrated product of any two of these
functions is zero.
The energy levels
The energy levels are:
2 2
n
h
E=
2
8mL
The probability of finding the
particle in a given region of space
Using the normalized wave function
Ψx =
2 sin nπx
L
L
one can calculate the probability of finding the particle
in any region of space. Since the wave function is
normalized, the probability P is a number between 0 and 1.
For example: What is the probability that the particle is
between 0.2L and 0.4L. This is found by integrating over
this region using the normalized wave function (see MAPLE
worksheet). 0.4L
0.4L
2
2
nπx
2
P =
Ψ x dx =
sin
dx ≈ 0.25
L
L
0.2L
0.2L
The appearance of the probability Ψ2
ö
The probability of finding the
particle in a given region of space
Using the normalized wave function
Ψx =
2 sin nπx
L
L
one can calculate the probability of finding the particle
in any region of space. Since the wave function is
normalized, the probability P is a number between 0 and 1.
For example: What is the probability that the particle is
between 0.2L and 0.4L. This is found by integrating over
this region using the normalized wave function (see MAPLE
worksheet). 0.4L
0.4L
2
2
nπx
2
P =
Ψ x dx =
sin
dx ≈ 0.25
L
L
0.2L
0.2L
Solutions of the Schrodinger equation
are orthogonal
If the wavefunctions have different quantum numbers
Then their “overlap” is zero. We can call the integral
Of the product of two wavefunctions an overlap. We write:
L
ΨmΨndx = δ mn
0
Where δmn is called the Kronecker delta. It has the property:
δ mn =
0 if m ≠ n
1 if m = n
For the particle-in-a-box the orthogonality is written:
L
2
L
0
sin mπx sin nπx dx = δ mn
L
L
Postulates of quantum mechanics
are assumptions found to be
consistent with observation
The first postulate states that the state of a
system can be represented by a wavefunction
Ψ(q1, q2,.. q3n, t). The qi are coordinates of
the particles in the system and t is time.
The wavefunction can also be
time-independent or stationary, ψ(q1, q2,.. q3n).
Corollary: An acceptable
wavefunction must be continuous and
have a continuous first derivative
Since the wavefunction is a solution of the
Schrödinger equation it must be differentiable.
The wavefunction must also be single-valued.
Postulate 2. The probability of
finding a particle in a region of
space is given by
a
P(a) =
Ψ Ψdτ
∗
0
Assumptions for the Born interpretation
1. Ψ*Ψ is real (Ψ is Hermitian).
2. The wavefunction is normalized.
3. We integrate over all relevant space.
Normalization is needed so that
probabilities are meaningful
Normalization means that the integral of the
square of the wavefunction (probability density)
over all space is equal to one.
Ψ Ψdτ = 1
∗
all space
The significance of this equation is that the
probability of finding the particle somewhere
in the universe is one.
Postulate 3. Every physical
observable is associated with a
linear Hermitian operator
Observables are energy, momentum, position,
dipole moment, etc.
operator P → observable P
The fact that the operator is Hermitian ensures
that the observable will be real.
Postulate 4. The average
value of a physical property
can be calculated by
Ψ PΨdτ
∗
P =
Normalization
Ψ Ψdτ
∗
Postulate 4. The calculation of
a physical observable can be
written as an eigenvalue
equation
PΨ = PΨ
This is an operator equation that returns the
same wavefunction multiplied by the constant P.
P is an eigenvalue. An eigenvalue is a number.
The form of the operators is
Position
Momentum
q
P
Time
t
Energy
H
q
∂
ih
∂q
t
∂
ih
∂t
Postulate 5. It is impossible to specify
with arbitrary precision both the
position and momentum of a particle
This postulate is known as the Heisenberg
Uncertainty Principle. It applies not only to
the pair of variables position and momentum,
but also to energy and time or any two conjugate
variables.
Conjugate variables are Fourier transforms of
one another.
Conjugate variables do not commute.
Commutator in quantum mechanics
A commutator is an operation that compares the
order of operation for two operators:
[ p,x] = px – x p
Since the momentum, p involves a derivative,
the order of application affects the result. The way
to see the result of the commutator is to apply it to
a test function f(x).
pxf(x) = -ih(f(x) + xf’(x)) while xpf(x)= -ihxf’(x).
Therefore [p,x] = - ih.
We say that position and momentum do not
Commute.