7/17/2019 lec_3
http://slidepdf.com/reader/full/lec3563db9d0550346aa9aa02b1e 1/46
Lecture 3
The Schrödinger Equation
The Particle in a Box (part 1)
OrthogonalityPostulates of Quantum Mechanics
NC State University
Chemistry 431
7/17/2019 lec_3
http://slidepdf.com/reader/full/lec3563db9d0550346aa9aa02b1e 2/46
Derivation of the Schrödinger EquationThe 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 awave as a sine or cosine function. These functions areoscillating functions. We will derive the wave equation infree space starting with one of its solutions: sin(x).
Before we begin it is important to realize that bound statesmay provide different solutions of the wave equation than
those we find for free space. Bound states includerotational and vibrational states as well as atomic wavefunctions. These are important cases that will be treatedonce we have fundamental understanding of the origin of
the wave equation or Schrödinger equation.
7/17/2019 lec_3
http://slidepdf.com/reader/full/lec3563db9d0550346aa9aa02b1e 3/46
The derivativeThe 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.
7/17/2019 lec_3
http://slidepdf.com/reader/full/lec3563db9d0550346aa9aa02b1e 4/46
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.
7/17/2019 lec_3
http://slidepdf.com/reader/full/lec3563db9d0550346aa9aa02b1e 5/46
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.
7/17/2019 lec_3
http://slidepdf.com/reader/full/lec3563db9d0550346aa9aa02b1e 6/46
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.
7/17/2019 lec_3
http://slidepdf.com/reader/full/lec3563db9d0550346aa9aa02b1e 7/46
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.
7/17/2019 lec_3
http://slidepdf.com/reader/full/lec3563db9d0550346aa9aa02b1e 8/46
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.
7/17/2019 lec_3
http://slidepdf.com/reader/full/lec3563db9d0550346aa9aa02b1e 9/46
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.
7/17/2019 lec_3
http://slidepdf.com/reader/full/lec3563db9d0550346aa9aa02b1e 10/46
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.
7/17/2019 lec_3
http://slidepdf.com/reader/full/lec3563db9d0550346aa9aa02b1e 11/46
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.
7/17/2019 lec_3
http://slidepdf.com/reader/full/lec3563db9d0550346aa9aa02b1e 12/46
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 thederivative of sin(x) is cos(x).
7/17/2019 lec_3
http://slidepdf.com/reader/full/lec3563db9d0550346aa9aa02b1e 13/46
The derivative of sin(x)
sin(x) = cos(x)d
dx
7/17/2019 lec_3
http://slidepdf.com/reader/full/lec3563db9d0550346aa9aa02b1e 14/46
The derivative of cos(x)
cos(x) = -sin(x)d
dx
7/17/2019 lec_3
http://slidepdf.com/reader/full/lec3563db9d0550346aa9aa02b1e 15/46
The second derivative of sin(x)
sin(x) = -sin(x)d
dx
d
dx
7/17/2019 lec_3
http://slidepdf.com/reader/full/lec3563db9d0550346aa9aa02b1e 16/46
The second derivative of sin(x)
sin(x) = -sin(x)d2
dx2
7/17/2019 lec_3
http://slidepdf.com/reader/full/lec3563db9d0550346aa9aa02b1e 17/46
Sin(x) is an eigenfunction
sin(x) = sin(x)d2
dx2
If we define as an operator G then we have:
-
d2
dx2-
G sin(x) = sin(x)
which can be written as:
This is a simple example of an operator equation thatis closely related to the Schrödinger equation.
7/17/2019 lec_3
http://slidepdf.com/reader/full/lec3563db9d0550346aa9aa02b1e 18/46
Sin(kx) is also an eigenfunction
sin(k x) = -k cos(k x)ddx
We can make the problem more general by includinga constant k. This constant is called a wavevector.
It determines the period of the sin function. Now we musttake the derivative of the sin function and also thefunction kx inside the parentheses (chain rule).
-
Here we call the value k 2 the eigenvalue.
sin(k x) = k 2sin(k x)d2
dx2-
7/17/2019 lec_3
http://slidepdf.com/reader/full/lec3563db9d0550346aa9aa02b1e 19/46
Sin(kx) is an eigenfunctionof the Schrödinger equation
The example we are using here can easily be expressedas the Schrodinger equation for wave in space. We onlyhave to add a constant.
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 inthis equation is actually the energy.
sin(k x) = sin(k x)d2
dx2- h
2
2m
-h
2
k 2
2m
-
-
7/17/2019 lec_3
http://slidepdf.com/reader/full/lec3563db9d0550346aa9aa02b1e 20/46
The Schrödinger equationBased on these considerations we can write a compactform for the Schrödinger equation.
HΨ = EΨ
d2
dx2-h2
2m
-
E = h2
k 2
2m
-
H =
Ψ= sin(k x)
Energy operator, Hamiltonian
Energy eigenvalue, Energy
Wavefunction
7/17/2019 lec_3
http://slidepdf.com/reader/full/lec3563db9d0550346aa9aa02b1e 21/46
The momentum
The momentum is related to the kinetic energy. ClassicallyThe kinetic energy is:
E = mv2
The momentum is:p = mv
So the classical relationship is:p2
E =2m
If we compare this to the quantum mechanical energy:
we see that: p = hk
1
E =h2k 2
2m
-
2
7/17/2019 lec_3
http://slidepdf.com/reader/full/lec3563db9d0550346aa9aa02b1e 22/46
The general solution to theSchrödinger equation in free space
The preceding considerations are true in free space.Since a cosine function has the same form as a sinefunction, but is shifted in phase, the general solutionis a linear combination of cosine and sine functions.
The coefficients A and B are arbitrary in free space.
However, if the wave equation is solved in the presenceof a potential then there will be boundary conditions.
Ψ= A sin(k x) + Bcos(k x) Wavefunction
7/17/2019 lec_3
http://slidepdf.com/reader/full/lec3563db9d0550346aa9aa02b1e 23/46
The particle in a box problemImagine 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 atfirst glance it works very well to describe translationalmotion in quantum mechanics.
0 L Allowed Region
7/17/2019 lec_3
http://slidepdf.com/reader/full/lec3563db9d0550346aa9aa02b1e 24/46
The solution to the Schrödingerequation with boundary conditionsSuppose 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 theboundary or outside the boundary.
0 L Allowed Region
7/17/2019 lec_3
http://slidepdf.com/reader/full/lec3563db9d0550346aa9aa02b1e 25/46
The solution to the Schrödinger
equation with boundary conditionsThe boundary condition is that the wave function willbe zero at x = 0 and at x = L.
From this condition we see that B must be zero.This condition does not specify A or k.The second condition is:
From this condition we see that kL = nπ. The conditionsso far do not say anything about A. Thus, the solutionfor the bound state is:
Note that n is a quantum number!
Ψ(0) = A sin(k 0) + Bcos(k 0) = 0
Ψ(L) = A sin(k L) = 0 or k L = arcsin(0)
Ψn(x) = A sin(nπx/L)
7/17/2019 lec_3
http://slidepdf.com/reader/full/lec3563db9d0550346aa9aa02b1e 26/46
The probability interpretationThe wave function is related to the probability for findinga particle in a given region of space. The relationship is
given by:
If we integrate the square of the wave function over a
given volume we find the probability that the particle isin that volume. In order for this to be true the integralover all space must be one.
If this equation holds then we say that the wave functionis normalized.
P = Ψ2dV
1 = Ψ2
dV all space
7/17/2019 lec_3
http://slidepdf.com/reader/full/lec3563db9d0550346aa9aa02b1e 27/46
The normalized bound statewave functionFor 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:
The solution to the integral is available on thedownloadable MAPLE worksheet. The solution is just L/2.Thus, we have:
As you can see the so-called normalization constant has
been determined.
1 = Ψ2dx
0
L
= A 2sin n π x
L
2
dx 0
L
= A 2
sin n π x
L
2
dx 0
L
1 = A 2L
2, A
2= 2
L , A = 2
L
7/17/2019 lec_3
http://slidepdf.com/reader/full/lec3563db9d0550346aa9aa02b1e 28/46
NormalizationWhat is the normalization constant for the wave functionexp(-ax) over the range from 0 to infinity?
A. aB. 2aC. √a
D. √2a
7/17/2019 lec_3
http://slidepdf.com/reader/full/lec3563db9d0550346aa9aa02b1e 29/46
NormalizationWhat is the normalization constant for the wave functionexp(-ax) over the range from 0 to infinity?
A. aB. 2aC. √a
D. √2a
1 = Ψ2dx 0
L
= A 2exp – ax 2dx 0
∞
= A 2 exp – 2ax dx 0
∞
= A 2 12a
, A = 2a
7/17/2019 lec_3
http://slidepdf.com/reader/full/lec3563db9d0550346aa9aa02b1e 30/46
The appearance of the wave functions
7/17/2019 lec_3
http://slidepdf.com/reader/full/lec3563db9d0550346aa9aa02b1e 31/46
The appearance of the wave functions
Note that the wave functions havenodes (i.e. the locations where they
cross zero). The number of nodes isn-1 where n is the quantum numberfor the wave function. The appearanceof nodes is a general feature of
solutions of the wave equation in bound states. By boundstates we mean states that are in a potential such as theparticle trapped in a box with infinite potential walls. Wewill 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.
7/17/2019 lec_3
http://slidepdf.com/reader/full/lec3563db9d0550346aa9aa02b1e 32/46
The energy levels
The energy levels are:
E = n2
h2
8mL
2
7/17/2019 lec_3
http://slidepdf.com/reader/full/lec3563db9d0550346aa9aa02b1e 33/46
The probability of finding theparticle in a given region of spaceUsing the normalized wave function
one can calculate the probability of finding the particle
in any region of space. Since the wave function isnormalized, the probability P is a number between 0 and 1.For example: What is the probability that the particle isbetween 0.2L and 0.4L. This is found by integrating overthis region using the normalized wave function (see MAPLEworksheet).
Ψ x = 2L
sin n π x
L
P = Ψ x 2dx
0.2 L
0.4 L
= 2L
sin n π x
L
2
dx 0.2 L
0.4 L
≈ 0.25
7/17/2019 lec_3
http://slidepdf.com/reader/full/lec3563db9d0550346aa9aa02b1e 34/46
The appearance of the probability Ψ2
ö
7/17/2019 lec_3
http://slidepdf.com/reader/full/lec3563db9d0550346aa9aa02b1e 35/46
The probability of finding theparticle in a given region of spaceUsing the normalized wave function
one can calculate the probability of finding the particle
in any region of space. Since the wave function isnormalized, the probability P is a number between 0 and 1.For example: What is the probability that the particle isbetween 0.2L and 0.4L. This is found by integrating overthis region using the normalized wave function (see MAPLEworksheet).
Ψ x = 2L
sin n π x
L
P = Ψ x 2dx
0.2 L
0.4 L
= 2L
sin n π x
L
2
dx 0.2 L
0.4 L
≈ 0.25
7/17/2019 lec_3
http://slidepdf.com/reader/full/lec3563db9d0550346aa9aa02b1e 36/46
Solutions of the Schrodinger equation
are orthogonal
If the wavefunctions have different quantum numbers
Then their “overlap” is zero. We can call the integralOf the product of two wavefunctions an overlap. We write:
Where δmn is called the Kronecker delta. It has the property:
For the particle-in-a-box the orthogonality is written:
2
L
sin m π x
L
sin n π x
L
dx = δmn
0
L
ΨmΨndx
0
L
= δmn
δmn =0 if m ≠ n
1 if m = n
7/17/2019 lec_3
http://slidepdf.com/reader/full/lec3563db9d0550346aa9aa02b1e 37/46
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 ofthe particles in the system and t is time.
The wavefunction can also be
time-independent or stationary, ψ(q1, q2,.. q3n).
7/17/2019 lec_3
http://slidepdf.com/reader/full/lec3563db9d0550346aa9aa02b1e 38/46
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.
7/17/2019 lec_3
http://slidepdf.com/reader/full/lec3563db9d0550346aa9aa02b1e 39/46
Postulate 2. The probability of
finding a particle in a region of
space is given by
P(a) = Ψ
∗
Ψd τ0
a
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.
7/17/2019 lec_3
http://slidepdf.com/reader/full/lec3563db9d0550346aa9aa02b1e 40/46
Normalization is needed so thatprobabilities are meaningful
Ψ ∗Ψd τall space
= 1
Normalization means that the integral of the
square of the wavefunction (probability density)
over all space is equal to one.
The significance of this equation is that theprobability of finding the particle somewhere
in the universe is one.
7/17/2019 lec_3
http://slidepdf.com/reader/full/lec3563db9d0550346aa9aa02b1e 41/46
Postulate 3. Every physical
observable is associated with a
linear Hermitian operator
operator P → observable P
Observables are energy, momentum, position,
dipole moment, etc.
The fact that the operator is Hermitian ensures
that the observable will be real.
7/17/2019 lec_3
http://slidepdf.com/reader/full/lec3563db9d0550346aa9aa02b1e 42/46
Postulate 4. The average
value of a physical property
can be calculated by
P = Ψ
∗
PΨd τ
Ψ∗
Ψd τNormalization
7/17/2019 lec_3
http://slidepdf.com/reader/full/lec3563db9d0550346aa9aa02b1e 43/46
Postulate 4. The calculation of
a physical observable can be
written as an eigenvalue
equation
PΨ = PΨThis is an operator equation that returns the
same wavefunction multiplied by the constant P.P is an eigenvalue. An eigenvalue is a number.
7/17/2019 lec_3
http://slidepdf.com/reader/full/lec3563db9d0550346aa9aa02b1e 44/46
The form of the operators is
Position q q
Momentum P ih
∂
∂qTime t t
Energy H ih ∂
∂t
7/17/2019 lec_3
http://slidepdf.com/reader/full/lec3563db9d0550346aa9aa02b1e 45/46
Postulate 5. It is impossible to specifywith arbitrary precision both the
position and momentum of a particleThis 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 ofone another.
Conjugate variables do not commute.
7/17/2019 lec_3
http://slidepdf.com/reader/full/lec3563db9d0550346aa9aa02b1e 46/46
Commutator in quantum mechanics
A commutator is an operation that compares the
order of operation for two operators:
Since the momentum, p involves a derivative,
the order of application affects the result. The wayto 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.
[ p, x] = px – x p