Download Applications of the Schrodinger Wave Equation The free particle

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 xdirection
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.
Free Particle and a Step Potential (Engel 5.5+)
Let us now examine a problem which exhibits distinctly nonclassical results.
Consider a particle of mass m and energy E, coming from the left
that approaches the following step potential.
V(x)=Vo
Infinitely wide
barrier
V
V(x)=0
x=0
x
Note it is equally valid to interpret this problem as a single
particle incident on the step potential or as a beam of noninteracting particles incident on the potential. Sometimes the
latter is useful for interpretative proposes.
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V(x)=Vo
Zone I
V(x)=0
Zone II
x=0
x
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.
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V(x)=Vo
Zone I
V(x)=0
Zone II
x=0
x
The Schrödinger equation for the problem can be divided into two
parts, one for each zone.
ZONE I
Here, since there is a zero potential in this region, the
Hamiltonian is given by:
and the Schrödinger equation is:
notice we have subscripted
our wave function 6
Again the general solution is given by:
This is the same as the free particle, so again we make the
argument that:
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.
The amount of reflection can be given by the relative magnitudes
of B versus A. (the amplitudes)
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V(x)=Vo
Zone I
V(x)=0
Zone II
x=0
x
ZONE II
Here, the potential is constant and equal to Vo
the Schrödinger equation is then:
rearranging we have:
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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 I
ZONE II
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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
In other words the wave functions have to match up at the boundary.
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.
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where:
The constant ‘A’ can be specified from the initial conditions. e.g.
how we shoot our particle at the potential barrier.
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What Happens to the Beam as it Encounters the Step Potential?
This largely depends on if the energy of the particles relative to Vo.
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.
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If we let
where
k is a positive and real valued. This gives us:
Since k is positive and real valued, inside the barrier we have a
exponentially decaying wave function.
representation of
the real
component of the
wave function
x
Zone I
x=0
Zone II
This is an example of quantum mechanical penetration into a
classically forbidden barrier.
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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
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for large m.
Non-Classical Reflection
Consider Case of when E > Vo
Classical picture.
• All particles with E > Vo will be pass through into zone II.
V(x)=Vo
Zone I
V(x)=0
Zone II
x=0
• 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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V(x)=Vo
Zone I
Zone II
V(x)=0
x=0
Here it can be useful to interpret our problem as a beam of noninteracting 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).
Recall we could not specify A, B and C without the ‘initial’
conditions, but we could specify their ratios.
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The ratio:
can be defined as the reflection coefficient. In other words, it
gives the fraction of particles that are reflected by the barrier.
Recall that for this problem we derived:
so we have
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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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Consider reflection coefficient changes as we alter the ratio
E/Vo.
1.2
Reflection Coefficient, R
E<Vo
E=Vo
1
0.8
0.6
0.4
E>Vo
0.2
0
0
1
2
3
4
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.
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1.2
E<Vo
Reflection Coefficient, R
1
E=Vo
0.8
0.6
0.4
E>Vo
0.2
0
0
1
2
3
4
E/Vo
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.
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Visualization of the Probabilities, ψ*ψ
Below the probability density is plotted for increasing energy of the
particle, but where E is less than Vo.
E about half Vo
E close to Vo
The probability oscillates in zone I due to interference with the
reflected particles.
Notice that the penetration of the particle into the barrier increases
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as the energy increases.
E greater than Vo
E slightly higher than Vo
E much greater
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)
Notice that the amplitude in zone I diminishes as the energy
increases. This is because there is less non-classical reflection as E
increases.
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Visualization of the Wave Functions
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 less than Vo
E about half Vo
E close to Vo
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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)
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Finite Width Barrier and Quantum Mechanical Tunneling
V(x)=Vo
Zone I
V(x)=0
Zone II
x=0
x
Zone III
x=L
V(x)=0
Consider a particle of mass m and energy E, coming from the left.
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.
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Finite Width Barrier and Quantum Mechanical Tunneling
V(x)=Vo
Zone I
V(x)=0
Zone II
x=0
x
x=L
Zone III
V(x)=0
The quantum mechanical treatment of this problem will show that
particles with energies less than the barrier can tunnel through to
the other side!
In other words when E<Vo there will be a finite transmission
coefficient through to zone III.
Quantum mechanical tunneling is important in many areas of
chemistry (proton tunneling; STM).
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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:
V(x)=Vo
Zone I
V(x)=0
Zone II
x=0
x
x=L
Zone III
V(x)=0
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.
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.
V(x)=Vo
Zone I
V(x)=0
Zone II
x=0
x
x=L
Zone III
V(x)=0
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Quantum Mechanical Tunneling
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.
For this system, an expression for the transmission coefficient
can be derived that shows that tunneling can occur.
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.
0.5
As the energy E approaches the
barrier height, or E/Vo approaches
1, the more tunneling we have.
0.4
0.3
The different plots show that:
2
T
• tunneling decreases as barrier
length, L, increases
0.2
3
4
0.1
10
0
0.2
0.4
E/V
0.6
8
6
0.8
• tunneling decreases as mass of
the particle, m, increases
• tunneling decreases as
barrier height Vo increases.
the
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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
1
3
10
0.8
2
T
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.
0.6
0.4
The wild oscillations are a quantum
mechanical effect.
0.2
0
1
2
3
E/V
4
The peaks are called scattering
resonances.
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More on potential boxes, barriers and unbound states
System in a BOUND particle state
All particles making up the system are
localized or “bound” to certain regions
of coordinate space.
V=
L
x=0 V = 0
V=
x=L
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.
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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:
V(x) = 0
+
energy of the system will correspond to a continuum of energy
levels. The energy is NOT quantized.
For at least one particle:
as
We can’t normalize the wave function in the normal sense.
(can be handled but not in this course)
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).
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Discrete, Continuous and Mixed Spectra of the Energy Levels
discrete energy
levels
continuous energy
levels
Mixed energy
levels
Occurs when
V=
V=
V(x) = 0
L
V=0
Let us examine qualitatively, why the mixed energy level occurs.
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Consider a Square Well Potential
V(x) = Vo
energy
V(x) = Vo
 ( x)
V(x) = 0
When E < Vo, we will have bound states.
 ( x)
There will be penetration into the
non-classical region.
The wave function of these states is
normalizable, notice that:
energy
The energy levels will be quantized.
2
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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.
E
E
e
a
f
b
g
c
d
x
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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
Questions:
How many nodes in a wavefunction?
How many wavefunctions?
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)
Parity is + 1 for n odd and -1 for n even
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)
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
Let P= parity operator
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)
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 timeindependent 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