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Lecture 3

The Schrödinger Equation

The Particle in a Box (part 1)

OrthogonalityPostulates of Quantum Mechanics

NC State University

Chemistry 431

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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.

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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.

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The derivative of sin(x)The derivative of a function is the instantaneous rate of change.


At sin(0) the slope is 1 as shown by the blue line.

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At sin(π /4) the slope is 1/√2 as shown by the blue line.

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At sin(π /2) the slope is 0 as shown by the blue line.

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At sin(3π /4) the slope is -1/√2 as shown by the blue line.

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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.

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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.

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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.

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We see from of the black squares (slopes) that thederivative of sin(x) is cos(x).

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The derivative of sin(x)

sin(x) = cos(x)d

dx

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The derivative of cos(x)

cos(x) = -sin(x)d

dx

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The second derivative of sin(x)

sin(x) = -sin(x)d

dx

d

dx

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The second derivative of sin(x)

sin(x) = -sin(x)d2

dx2

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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.

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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-

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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

-

-

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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

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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

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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

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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

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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

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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)

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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

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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

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NormalizationWhat is the normalization constant for the wave functionexp(-ax) over the range from 0 to infinity?

A. aB. 2aC. √a

D. √2a

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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

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The appearance of the wave functions

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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.

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The energy levels

The energy levels are:

E = n2

h2

8mL

2

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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

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The appearance of the probability Ψ2

ö

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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

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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

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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).

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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.

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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.

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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.

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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.

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Postulate 4. The average

value of a physical property

can be calculated by

P = Ψ

∗

PΨd τ

Ψ∗

Ψd τNormalization

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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.

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The form of the operators is

Position q q

Momentum P ih

∂

∂qTime t t

Energy H ih ∂

∂t

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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.

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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

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