Applications of the Schrodinger Wave Equation

The free particle Chapter 4.1

No boundary conditions

The free particle has V = 0. Assume it moves along a straight line in the x-direction

Solutions are exponentials of the form:

The two wavefunctions are degenerate

not quantized

The problem in this case is that the wavefunction is not normalizable, so it’s not a true wavefunction . It can be used to make true wavefunctions by superposition.

The probability of finding the particle =

Independent of position so the particle is equally likely to be anywhere.

In the case of the free particle the separable solutions do not represent physically realizable states.

A free particle cannot exist in a stationary state or in other words there is no such thing as a free particle with a definite energy.

We can find the linear momentum:

We know the momentum exactly, but nothing about the position.

4

Free Particle and a Step Potential (Engel 5.5+)

Let us now examine a problem which exhibits distinctly non-classical results.

Consider a particle of mass m and energy E, coming from the left that approaches the following step potential.

Note it is equally valid to interpret this problem as a single particle incident on the step potential or as a beam of non-interacting particles incident on the potential. Sometimes the latter is useful for interpretative proposes.

x V(x)=0

V(x)=Vo

x=0

Infinitely wide barrier

V

5

Classical Picture

• All particles with E < Vo will be reflected back.

• All particles with E > Vo will be pass through into zone II.

Quantum Mechanical Highlights

• Particles with E < Vo can penetrate the barrier and make it into zone II.

• Particles with E > Vo there is a finite chance that they will be reflected.

x=0 V(x)=0

V(x)=Vo

x

Zone II Zone I

6

The Schrödinger equation for the problem can be divided into two parts, one for each zone.

Here, since there is a zero potential in this region, the Hamiltonian is given by:

and the Schrödinger equation is:

ZONE I

x=0 V(x)=0

V(x)=Vo

x

Zone II Zone I

notice we have subscripted our wave function

7

Again the general solution is given by:

This is the same as the free particle, so again we make the argument that:

The amount of reflection can be given by the relative magnitudes of B versus A. (the amplitudes)

In this case, we cannot set A or B equal to zero because there is the possibility of reflection of the particle at the barrier.

8

rearranging we have:

Here, the potential is constant and equal to Vo

ZONE II

x=0 V(x)=0

V(x)=Vo

x

Zone II Zone I

the Schrödinger equation is then:

9

Again the general solution is given by:

Since particles that make it into the barrier will only be traveling in the positive x-direction, D = 0.

0

Summary of Zone I and II Wave Functions

Now we have the general form for our wave function in the two regions:

ZONE II

ZONE I

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Due to the restrictions that the wave function be continuous, single valued and smooth, we have some continuity relationships that connect the wave functions in the two regions.

at the boundary, x=0

We have 2 equations and three unknowns. We can only solve for B and C in terms of A.

If the particle was confined to some region of space, then a normalization condition would be our third condition. But since our particle can be anywhere, we can not normalize the wave function.

Solve for B and C in terms of A using the continuity relationships.

In other words the wave functions have to match up at the boundary.

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The constant ‘A’ can be specified from the initial conditions. e.g. how we shoot our particle at the potential barrier.

where:

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Consider Case of when E < Vo

When the energy of the incoming particle is less than the potential, classically we expect all particles to reflect back. Lets look at what happens in the quantum mechanical description.

Let us examine wave function in the barrier or Zone II:

If E < Vo, then E-Vo is negative and we have a negative root in KII.

What Happens to the Beam as it Encounters the Step Potential?

This largely depends on if the energy of the particles relative to Vo.

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If we let

k is a positive and real valued.

Since k is positive and real valued, inside the barrier we have a exponentially decaying wave function.

This is an example of quantum mechanical penetration into a classically forbidden barrier.

This gives us:

where

Zone II Zone I

representation of the real component of the wave function

x

x=0

14

Notice that the larger k, the steeper the decay. Thus, the larger the mass, the less penetration one will have.

For macroscopic particles, the penetration is negligible and for all practical purposes is zero no matter what the barrier is.

One can define a penetration depth as:

This is the depth at which the amplitude of the wave function will diminish to a factor of e-1 of its value at the edge of the barrier.

The penetration depth is inversely proportional to k and therefore the mass, m. So again, the penetration will be negligible for large m.

15

Consider Case of when E > Vo

Non-Classical Reflection

Classical picture.

• All particles with E > Vo will be pass through into zone II.

Zone I

x=0 V(x)=0

V(x)=Vo

Zone II

• Classically,100% of the particles are transmitted into the barrier region. 0% of the particles are reflected.

In quantum mechanics we will see that there can be a finite chance of reflected particles even if E>Vo.

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Here it can be useful to interpret our problem as a beam of non-interacting particles incident on the potential.

The magnitudes of the constants A, B and C can be interpreted as amplitudes of the beam of particles. For example, the larger |A| the larger the amplitude the beam of particles. (The more particles).

x=0 V(x)=0

V(x)=Vo

Zone II Zone I

Recall we could not specify A, B and C without the ‘initial’ conditions, but we could specify their ratios.

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can be defined as the reflection coefficient. In other words, it gives the fraction of particles that are reflected by the barrier.

The ratio:

Recall that for this problem we derived:

so we have

18

Using our definitions for KI and KII,

It is a matter of algebra to derive an expression for the reflectance coefficient.

R depends on the relative size of the Vo, the barrier, and the energy of the particles.

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0

0.2

0.4

0.6

0.8

1

1.2

0 1 2 3 4E/Vo

Ref

lect

ion

Co

effi

cien

t, R

E=Vo E<Vo

E>Vo

Thus, even when E>Vo there is a finite chance of reflection. This is a non-classical result and it is sometimes called non-classical reflection.

Amount of reflection decreases rapidly as E increases over Vo.

Consider reflection coefficient changes as we alter the ratio E/Vo.

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The amount of transmission through the barrier is given by:

Notice that if E<Vo, there is zero chance of transmission. This is different from ‘penetration’.

Barrier penetration is NOT transmission.

0

0.2

0.4

0.6

0.8

1

1.2

0 1 2 3 4

E/Vo

Ref

lect

ion

Co

effi

cien

t, R

E=Vo E<Vo

E>Vo

21

Below the probability density is plotted for increasing energy of the particle, but where E is less than Vo.

E about half Vo

The probability oscillates in zone I due to interference with the reflected particles.

E close to Vo

Notice that the penetration of the particle into the barrier increases as the energy increases.

Visualization of the Probabilities, ψ*ψ

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E slightly higher than Vo

The probability oscillates in zone I due to interference with the reflected particles. In zone II, they are only moving in one direction so the probability is constant. (remember the free particle with no barrier)

E much greater than Vo

Notice that the amplitude in zone I diminishes as the energy increases. This is because there is less non-classical reflection as E increases.

E greater than Vo

23

Plotted below are the real part of the stationary state wave functions of the particle incident on the barrier of infinite width at various energies.

E about half Vo

E close to Vo

E less than Vo

Visualization of the Wave Functions

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E greater than Vo E slightly higher than Vo

E much greater than Vo

Although the probability is constant in zone II, the real component of the wave function in the zone II still oscillates. Again consider the free particle wave function with no barrier.

Further notice that there is a change in the wave length of the oscillations.

Why does the wavelength in zone II decrease as the energy increases? (Ignore the wave length in zone I)

25

Finite Width Barrier and Quantum Mechanical Tunneling

x=0 V(x)=0

V(x)=Vo

x

Zone II Zone I Zone III

V(x)=0 x=L

Classical picture.

• All particles with E < Vo will be reflected back.

• All particles with E > Vo will be pass through the barrier into Zone III.

Consider a particle of mass m and energy E, coming from the left.

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x=0 V(x)=0

V(x)=Vo

x

Zone II Zone I Zone III

V(x)=0 x=L

The quantum mechanical treatment of this problem will show that particles with energies less than the barrier can tunnel through to the other side!

Quantum mechanical tunneling is important in many areas of chemistry (proton tunneling; STM).

In other words when E<Vo there will be a finite transmission coefficient through to zone III.

Finite Width Barrier and Quantum Mechanical Tunneling

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x

Zone II

x=0 V(x)=0

V(x)=Vo

Zone I Zone III

V(x)=0 x=L

We can again set up our Hamiltonian and Schrödinger equation in each of the regions. Using this, we have wave functions for each zone given by:

at the boundary, x=0

at the boundary, x=L

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Again, by applying the continuity relationships, it is a matter of algebra to derive analytic expressions for the transmission and reflection coefficients. Let us not worry about these derivations, but rather focus on interpreting the results.

x=0 V(x)=0

V(x)=Vo

x

Zone II Zone I Zone III

V(x)=0 x=L

Non-classical reflection

The results are similar to that of the step potential problem. But now, even if E>Vo, there is a finite chance of reflection at both boundaries.

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Quantum Mechanical Tunneling

For this system, an expression for the transmission coefficient can be derived that shows that tunneling can occur.

When E<Vo, there can be a finite chance that particles will make it all the way through the barrier, into zone III. This is called quantum mechanical tunneling because particles are ‘seen’ to tunnel through barriers.

Again it depends on Vo and E, but now also on the length of the barrier L.

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Plot of the Tunneling probability as a Function of the kinetic energy of the incoming particle relative to the barrier height, E/Vo for E<Vo.

As the energy E approaches the barrier height, or E/Vo approaches 1, the more tunneling we have.

The different plots show that:

• tunneling decreases as barrier length, L, increases

• tunneling decreases as mass of the particle, m, increases

• tunneling decreases as the barrier height Vo increases.

T

E/V

0.2 0.4 0 0.8 0.6

0.1

0.3

0.2

0.5

0.4

10 8

6

4

3

2

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Plot of the Tunneling or transmission probability as a function of the kinetic energy of the incoming particle relative to the barrier height, E/Vo for E>Vo.

This is the same quantity as plotted in the previous slide but now the particles have E>Vo

The wild oscillations are a quantum mechanical effect.

The peaks are called scattering resonances.

As the energy becomes larger relative to the barrier, the more chance that the particles are transmitted and the less chance there is non-classical reflection.

10

0.2

0.4

0.6

0.8

1

2 1 3 4 E/V

T

0

2 3

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More on potential boxes, barriers and unbound states

All particles making up the system are localized or “bound” to certain regions of coordinate space.

System in a BOUND particle state

In all cases:

energy of the system will be quantized

The probability of finding ANY particle will approach zero as x approaches infinity.

The wave function for bound states are normalizable.

L V = V =

V = 0 x=0 x=L

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System in an UNBOUND particle state

At least one of the particles making up the system behaves like a free particle in the sense that it is not localized in any particular region of coordinate space.

In all cases:

NOTE the confusing terminology. Although we are talking about a UNBOUND state, the wave function is still a bound function of the coordinates (i.e. does go to infinity).

energy of the system will correspond to a continuum of energy levels. The energy is NOT quantized.

as

For at least one particle:

We can’t normalize the wave function in the normal sense. (can be handled but not in this course)

V(x) = 0 - +

34

Discrete, Continuous and Mixed Spectra of the Energy Levels

Occurs when

discrete energy levels

continuous energy levels

Mixed energy levels

Let us examine qualitatively, why the mixed energy level occurs.

V = V =

L

V = 0

V(x) = 0

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Consider a Square Well Potential

V(x) = 0

V(x) = Vo V(x) = Vo

When E < Vo, we will have bound states.

The energy levels will be quantized.

There will be penetration into the non-classical region.

The wave function of these states is normalizable, notice that:

ener

gy

( )x

ener

gy

2( )x

36

When E > Vo, We Will Have Unbound States.

The allowed energy levels will form a continuum.

There will be non-classical reflection at BOTH boundaries.

Notice the ‘speed’ up of the particles in the well.

x

E

E

a

b

c

d

e

f

g

The particle in a 1D box

Schrodinger equation:

Outside the box V = ∞

→ Ψ(x) =0 because the second derivative of Ψ(x) must exist.

As Ψ(x) must be continuous → Ψ(0) = 0 and Ψ(a) =0. These are the boundary conditions.

Inside the box where V(x) = 0

Solutions of the same form as before, but to make it easier to apply boundary conditions use:

At x =0

At x =L

A ≠ 0 because that would make Ψ(x) = 0 everywhere in the box. It would imply no particle at all!

Therefore sin(ka) = 0. This will happen when ka = nπ; n = 1, 2, 3. Note n = 0 no good because again this would make Ψ(x) = 0 everywhere in the box.

Thus:

Solve for E:

Quantization comes from the boundary conditions; that is, from confinement!

How to find A?

Normalization:

Change variables:

Limits:

Therefore:

Energy level spacings increase with increasing n

For a box of length “a”.

Plot Ψ

How many nodes in a wavefunction? How many wavefunctions?

Questions:

Examples:

1. Calculate the probability that the particle in a 1D box of length a is between 0 and a/2

Change variables:

for any n!

2. Is Ψ3 an eigenfunction of px2? What is the average of many measurements?

Yes, Ψ3 is an eigenfunction with eigenvalue =

= the average of many measurements.

Can determine expectation value formally:

Application of PIB is molecular conjugation of polyenes.

Additional points to make about PIB wavefunctions

1. 1st derivative is discontinuous at box edges. This is an allowed exception because V = ∞

2. Wavefunctions are orthonormal. • We did normalization when we found A.

Orthogonality:

As an exercise can show if n ≠ m the integral = 0.

3. They form a complete set → Fourier series.

4. They are alternatively even and odd about the center of the box.

A function f(x) is even if f(-x) = f(x) and odd if f(-x) = -f(x)

5. # nodes increases with n.

6. There is a zero point energy. In the lowest state (ground state = n =1) the energy (which is all kinetic energy ≠ 0 even if T → 0K)

Parity is + 1 for n odd and -1 for n even

7. Look at what happens as a changes: When a is very large:

Free particle, unquantized

8. Time dependence

Still normalized.

Is Ψn still an eigenfunction of H?

yes

5. # nodes increases with n.

6. There is a zero point energy. In the lowest state (ground state = n =1) the energy (which is all kinetic energy ≠ 0 even if T → 0K)

Let P = parity operator

A general solution of the time-dependent Schrodinger equation is a linear combination of stationary states

Check that this is a solution.

Suppose that the PIB is in a state:

Is this normalized?

If we measure E what do we find? Is it a solution of the time-independent Schrodinger equation?

9. Classical limit: as n → ∞, Ψn becomes close to smooth. See Fig. 4.4 (Engel): ΔE/E → 0 as n → ∞

Time dependence continued