Cơ học lượng tử 1.1

Quantum Mechanics I
Sally Seidel
Primary textbook: “Quantum Mechanics” by Amit Goswami

Please read Chapter 1, Sections 4-9

Outline
What you should recall from previous courses
Motivation for the Schroedinger Equation
The relationship between wavefunction ψ and probability
Normalization
Expectation values
Phases in the wavefunction
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10 facts to recall from previous courses

Fundamental particles (for example electrons, quarks, and photons) have all the usual classical properties (for example mass and charge) + a new one: probability of location.
Because their location is never definite, we assign fundamental particles a wavelength.
Peak of wave – most probable location
length of wave – amount of indefiniteness of location
Wavelength λ is related to the object’s momentum p




The object itself is not “wavy”...it does not oscillate as it travels. What is wavy is its probability of location.





Planck’s constant
4.13 x 10-15 eV-sec
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Example of an object with wavy location probability distribution

Consider a set of 5 large toy train cars joined end to end. Each car has a lid and a door leading to the next car.









Put a mouse into one box and close the lid. The mouse is free to wander among boxes.
At any time one could lift a lid and have a 20% chance to find the mouse in that particular car.

Now equip Boxes 2 and 4 with mouse repellent
Equip Boxes 1, 3, and 5 with cheese
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A diagram on the outside of the boxes shows how likely it is that the mouse is in any of the boxes. Now the probability of finding the mouse is not uniform in space: maxima are near the cheese, minima are near the poison.
very likely
sometimes
not likely
sometimes
very likely
Probability
Position
Conclude:
the mouse does not look like a wave---it looks like a mouse
the mouse does not oscillate like a wave---it moves like a mouse
but the map of probable locations for the mouse is shaped like a wave
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The situation for the electron or photon is almost the same, except
for the mouse example, Probability = Amplitude
for the electron (or any quantum mechanical object,

Probability = (Amplitude)*(Amplitude)

As with all waves, wavelength λ is related to frequency υ:

λυ = velocity of wave
6. QM says that the λ ( or υ) is also related to the energy:

E=hυ

Special relativity says that total E and momentum p are also related by




rest mass of the object
c = 3 x 108 m/s
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QM says that every object in the universe is associated with a mathematical expression that encodes in it every property that it is possible to know about the object.



This math expression is called the object’s wavefunction ψ.
As the object moves through space and time, some of its properties (for example location and energy) change to respond to its external environment.

So ψ has to track these
Conclude: ψ has to include information about the environment of the particle (for example location x, time t, sources of potential V)

10. So if you know the ψ of the object, you can find out everything possible about it.

The goal of all QM problems is: given an object (mass m, charge Q, etc.) in a particular environment (potential V), find its ψ. The way to do this (in 1-dimension) is to solve the equation
its charge, mass, location, energy...
The Schroedinger Equation
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II. Motivation for the Schroedinger Equation

We can develop the Schroedinger Equation by combining 6 facts:

FACT 1: The λ and p of the ψ produced by this equation must satisfy λ=h/p.
FACT 2: The E and υ of the ψ must satisfy E=hυ.
FACT 3: Total energy = kinetic energy + potential energy
Etota l= KE + PE

Restricting ourselves to non-relativistic problems, we can rewrite this as
Etotal = p2/2m + V.

(For relativistic problems, we would need ).
FACT 4: Because a particle’s energy, velocity, etc, depend on any force F it experiences, the equation must involve F. Insert this as a V-dependence through



To simplify initially, consider only cases where V = constant = V0. Later we will generalize to V=V(x,y,z,t).
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FACT 5: The only kind of wave that is present in the region of a constant potential is an infinite wave train of constant λ everywhere.
Example:
An ocean wave over the flat ocean floor extends in all directions with constant amplitude and λ.
When the wave reaches a change in floor level (i.e. a beach) then its structure changes.
Conclude: if V = constant,





Recall that the definition of a wave is an oscillation that maintains its shape as it propagates. For constant velocity v, “x-vt” ensures that as t increases, x must increase to maintain the arg=(x-vt)= constant. This is a rightward-traveling wave.
cos or sin indicate the wave shape.
k has units 1/length to make the argument of the cos dimensionless
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Again

Rewrite this as



Then

FACT 6: ψ represents a particle and wave simultaneously. Waves interfere. This means if we combine the amplitudes of 2 waves (A(ψ1) and A(ψ2)), we get A(ψTotal) = A(ψ1) + A(ψ2).

That is...add the first powers of the ψ1 and ψ2 amplitudes, not functions that are more complicated.

Conclude: if we want the Schroedinger Equation to produce a wavelike ψ, then it too must include only first powers of ψ...that is, ψ, dψ/dx, dψ/dt, etc., but NOT, for example, ψ2.
units are

So call kv = ω.
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Now use all 6 facts to construct the Schroedinger Equation:
Notice we are already using FACT 4 (i.e. V is included. Consider the simplified case V = constant = V0. This implies


Recall this produces an infinite, single-λ wave.
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III. The connection between ψ and probability

Max Born proposed (1926) that the probability of finding a particle at a specific location x at time t,

Prob(x,t) = ψ*ψ.

Justification:

If the particle that ψ describes is assumed to last forever [this must later be revised by Quantum Field Theory] then the probability associated with finding it somewhere must always be 1. So probability must have an associated continuity equation like the one that applies to electric charge.

In electricity and magnetism:



electric current density
electric charge density
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We need an analogous expression to describe
probability density ρProb and
probability current JProb which can flow in space but remain conserved.

Assume ρProb and Jprob involve ψ somehow, but in an unspecified function.

Plan:
Use the only equation we have for ψ: the Schroedinger Equation
Manipulate it to get the form
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IV. Normalizing a wavefunction

Recall that when we were deriving the Schroedinger Eq. for a free particle, we got to this step:
We guessed ψ=δcos(kx-ωt)+γsin(kx-ωt)
We found that γ=±iδ
So ψ = δcos(kx-ωt) ± iδsin(kx-ωt)
=δ[cos(kx-ωt) ± isin(kx-ωt)]






Although this function corresponds to ψfree, all ψ’s have a “δ”.

Next goal: find a general technique for obtaining δ. This is called normalizing the wavefunction.
2 options correspond to waves traveling right and left. We can choose either one.
As-yet unspecified overall amplitude
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To find δ, recall
P(x,t) = ψ*ψ
The sum of probabilities of all possible locations of the particle must be 1.





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V. Expectation values

Although particle is never in a definite location, it is more likely to be in one location than others, if any potential V is active.

Recall the definition of a weighted average position:

By convention, place x between ψ’s
If ψ has been normalized, this denominator is 1.
This is the “expectation value of x”
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Please read Goswami Chapter 2.

Outline
Normalizing a free particle wavefunction
Acceptable mathematical forms of ψ
The phase of the wavefunction
The effect of a potential on a wave
Wave packets
The Uncertainty Principle
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II Acceptable mathematical forms of wavefunctions

ψ must be normalizable, so must be a convergent integral-

i.e., the at minimum, require



A ψ that satisfies this is called “square integrable.”
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III The phase of the wavefunction

FACT 1: We cannot observe ψ itself we only observe ψ*ψ. So overall phase is physically irrelevant.

FACT 2: The relative phase of two ψ’s in the same region affects the probability distribution, which is measurement, through superposition:

Suppose ψ1=Aeiα and ψ2=Beiβ, where A and B are real.
ψtot=Aeiα+Beiβ=eiα[A+Bei(β-α)], so
Prob=ψ*ψ=[A+Bei(β-α)][A+Be-i(β-α)]=A2+B2+AB[ei(β-α)+e-i(β-α)]= A2 + B2 + 2ABcos(β-α)

FACT 3: The flow of probability depends on both the amplitude and the phase:
Consider ψ=Aeiα where A can be complex.
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phase dependence
amplitude dependence
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IV The effect of a potential upon a wave

If everywhere in the universe, V were constant, all particle/waves would be free and described by ψfree=e i(kx-ωt), an infinite train of constant wavelength λ.
If somewhere V≠constant, then in that region ψ will be modulated.
schematic potential
schematic wavefunction response
A modulated wave is composed of multiple frequencies (i.e., Fourier components) that create beats or packets.
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V. Wave packets






The more Fourier component frequencies there are constituting a wave packet, the more clearly separated the packet is from others. Specific requirements on a packet:

To achieve a semi-infinite gap on each side of the packet (i.e. a truly isolated packet/particle), we need an infinite number of waves of different frequencies.
Each component is a plane wave


To center the packet at x = x0, modify


so at x≅x0, all the k’s (ν’s) superpose constructively.

4. To tune the shape of the packet, adjust the amplitude of each component separately---so
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A(k) is called the Fourier Transform of ψ(x)
infinite number of ν’s (k’s)
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VI. The Uncertainty Principle

The shape of a packet depends upon the spectrum of amplitudes A(k) of its constituent Fourier components.

Examples of possible spectra:















Note this is the A(k) not the ψ(x).
k
A(k)
k
A(k)
A(k)=δ(k-k0) Dirac delta
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Each A(k) spectrum produces a different wavepacket shape, for example
.


versus






Qualitatively it turns out that
large number of constituent k’s in the A spectrum (=large “Δk”) produces a short packet (small “Δx”).

So



So ΔpΔx cannot be arbitrarily small for any wave packet.
We begin to see that the Uncertainty Principle is a property of all waves, not just a Quantum Mechanical phenomenon.
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The proportionality in is qualitative at this point.

To derive the Uncertainty Principle from this, we need to know:
a precise definition of Δx
a precise definition of Δp
what is the smallest combined choice of ΔpΔx (or ΔkΔx) that is geometrically possible for a wave.

To answer these, use the Gaussian wave packet in k-space to answer the questions above, in the reverse order.
k
A(k)
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Find this amplitude by normalization later.
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Answer to (2)--- “What is Δx?” :

For all A(k) spectra, the precise definition of Δx is



For simplicity, choose center of the packet at x0 = 0. Then
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Please read Goswami Chapter 3.

Outline
I. Phase velocity and group velocity
II. Wave packets spread in time
III. A longer look at Fourier transforms, momentum conservation, and packet dispersion.
Operators
Commutators
VI. Probing the meaning of the Schroedinger Equation
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Phase velocity and group velocity

A classical particle has an unambigous velocity Δx/Δt or dx/dt because its “x” is always perfectly well known.
A wave packet has several kinds of velocity:












In general vphase ≠ vgroup. Which velocity is related to the velocity of the particle that this wave represents?
vgroup, the rate of travel of the peak of the envelope.
vphase, the rate of travel of the component ripples
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Recall a traveling wave packet is described by





Bear in mind the definitions
k = 2π/λ “inverse wavelength” and
ω = 2πυ “angular frequency”

Recall ω = ω(k).
If the packet changes shape as it travels, the function may be complicated.
If the packet changes shape rapidly and drastically, the notion of a packet with well-defined velocity becomes vague.

For clarity, consider only those packets that do not change shape “much” as they travel.
For them, ω(k)=constant + small terms proportional to some function of k.
Taylor expand ω about some k=k0.




Plug this into ψ:
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This is an overall phase which has no meaning in ψ*ψ, so forget it.
This is identical to ψ(x,0) except the position of the packet is shifted by
so that must be the velocity of the packet:
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Conclude: vgroup = dω/dk is the velocity of the packet envelope AND of the associated particle.

vphase = ω/k is usually different from dω/dk.

Notice
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II. Wave packets spread in time

The lecture plan:
Recall ψGeneral A’s(x,t =0)
Specialize to ψGaussian A’s(x,t =0)
Extrapolate from x to x-vt, so eikxei(kx-ωt)
Find P(x,t)=ψ*(x,t)ψ(x,t).

We will find that |ψ(x,t)|2 is proportional to exp(-x2/(stuff)2).
Since the width of ψ is defined as the distance in x over which ψ decreases by e, this “stuff” is the width.
We will see that the “stuff” is a function of time.

Carry out the plan...
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call this T, the characteristic spreading time.
Notice this is Δx(t=0).
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Conclusions:

The width Δx of the probability distribution increases with t, i.e., the packet spreads.

This only works because the amplitudes A are time-independent, i.e., the A(k) found for ψ(t=0) can be used for ψ(all t). The A(k) distribution is a permanent characteristic of the wave.

Notice the “new x”:


The group velocity naturally appears because this Δx describes a property of the packet as a whole.

(4) Recall the A(k) are not functions of t, so
Prob(k,t) = A*A does not have time-dependence, so Δp does not spread as Δx does.
This is momentum conservation.
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A longer look at Fourier transforms, momentum conservation, and packet dispersion

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These ω’s are the frequencies of the Fourier components. The components are plane waves---the ψ’s of free particles.
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Operators

Recall earlier we wanted

but we needed to represent p as a function of x. How to find this representation:

Recall the wavefunction for a free particle is ψ = ei(kx-ωt).
Notice


















This says: if ψ represents a free particle, any time we have “pψ”, we can replace it with
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Q. What if ψ is NOT a free particle?...what if ψ is influenced by a potential V so is a packet?

Ans. The packet is a superposition of free particle states, so the replacement is still valid.

Use a similar method to find the operator for energy E:
Begin with is ψfree = ei(kx-ωt).
Notice
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Facts about this procedure:
The order of the symbols is important here: so far this applies to “pψ” not “ψp”.
when a mathematical expression (like p or E) precedes a function (like ψ) and has the possibility of changing the function (for example multiplying it, taking its derivative), call the expression an operator.
So p is an operator applied to ψ.
Operators can be expressed in coordinate space or momentum space:

Operator Coordinate Space Representation Momentum Space Representation
p p


Pick the representation that matches the space in which the function is expressed.
ex.—is p acting on ψ(x) or on A(k)?
Any measureable attribute of a particle has an associated operator Examples....
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6. If you need to apply 2 or more operators successively to a function, the order in which you apply them affects the answer.
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Why do we care about commutators?

An operator represents a measurement.

A commutator’s value indicates whether the order of doing 2 measurements matters.

It does not matter in all cases----this depends upon the particular pair of measurements.

If the order does matter, this means that the first measurement disrupts the system in a way that influences the result of the second.

This is the Uncertainty Principle.

In QM we care mostly about operators that represent measureable quantities.

This kind of operator
always produces real, not complex, expectation values
is called “Hermitian.”
“observables”
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V. How to calculate a commutator

Act with it on a dummy wavefunction








Then remove the dummy wavefunction and see what is left over:



(2) If the operators in the commutator are functions of simpler operators whose commutators you know, expand the commutator with the functions expressed explicitly and look for simpler commutators whose value you know.

Example: Find [x,p2].
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The student will show all of the following commutator identities in homework. Once they have been proven explicitly one time, they can afterward be used whenever convenient without re-proof:

[A,B]=-[B,A]

[A,B+C]=[A,B]+[A,C]

[AB,C]=[A.C]B+A[B,C]

[A,BC]=[A,B]C+B[A,C]
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A longer look at the Schroedinger Equation

Recall the equation







These operator relationships were the basis of Schroedinger’s thought when he discovered the equation. He thought:
Assume ψ is built of plane waves ei(kx-ωt)
Find the operators
Guarantee non-relativistic energy conservation, p2/2m + V = E.

The operator

represents the total energy and is also called the Hamilton operator H. Applying H to ψ evolves ψ in time. Time itself is not an operator.
Notice this is poperator2/2m
Notice this is Eoperator
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Please read Goswami Chapter 4

Outline
How to solve the Schroedinger Equation
Why do we concentrate on ψ’s that are separable?
Why E is real if ψ=u(x)T(t)
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How to solve the Schroedinger Equation

Recall the equation:




Why we want to solve it:
Solving it gives us Ψ(x,t), which includes EVERYTHING that can be known about a particle, including its mass, energy, location, response to V, etc.
Once we have Ψ(x,t), we extract these properties by using Ψ to compute expectation values.

How to solve the Schroedinger Equation
Consider the case where V=V(x only, not t). [Later we will consider more general V.
GUESS that Ψ(x,t)=u(x)T(t) and plug this guess into the equation:
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Recall that e - i(stuff) can be written as cos(stuff) – i sin(stuff), so




The argument of these trigonometric functions must be dimensionless, so

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Facts about the Time-independent Schroedinger Equation:

The ψ’s are called the eigenfunctions of the Hamiltonian

Sometimes, depending on the form of V, more than one value of ψ can solve it.

For example: if V is the Coulomb potential of a nucleus, then each ψ includes information about the properties an electron would have in one of the energy shells around the nucleus.

3. Method: Plug in a V, solve for u(x) (which is ψ), then solve for E. [We will work examples of this.] Then insert u(x) into the full solution
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Why do we concentrate on Ψ’s that are separable?

Note, some systems have so separable Ψ’s are not the most general kind.
Reasons why they are interesting:

The energy E associated with them is mathematically real; i.e. measureable in the lab.
Note one can’t take this for granted since Ψ itself is complex. The reality of E leads to

The probability density is not a function of time, so the states do not change their properties with time.

These states have a definite energy; i.e. uncertainty ΔE=0.

The next lecture topics demonstrate these 3 points.
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Outline

Stationary States
If Ψ is a stationary state, its ΔE=0.
Degeneracy
Required properties of eigenfunctions
Solving the time-independent Schroedinger Equation when V=0
Solving the time-independent Schroedinger Equation for a Barrier Potential
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If Ψ is a stationary state, its ΔΕ=0

Recall when Ψ=u(x)T(t), this leads to
the Time-dependent Schroedinger Eq:

the Time-independent Schroedinger Eq:
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III. Degeneracy

If 2 states of the same system have the same energy, we call them degenerate.
Example: the rightward and leftward traveling components of the free particle Ψ:
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Required properties of eigenfunctions

-in addition to being square integrable-

Recall Probability density(x,t)=Ψ*Ψ. This means Ψ is related to the probability of locating an object at (x,t).
By definition of probability, no probability can be > 100%, and especially, no probability can be infinite.
So Ψ must be finite in amplitude.

(2) Recall

Momentum in the physical world cannot be infinite, so
dΨ/dx must be finite.

(3) Recall the Schroedinger Equation





In the physical world, V, E, and m cannot be infinite, so
d2Ψ/dx2 must be finite.
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Recall the continuity relations:

If a variable (Ψ) is discontinuous,
then its first derivative is infinite at the discontinuity.

So conversely, if dΨ/dx is finite, then Ψ is continuous.

We already showed this. This is not really new information but it deserves emphasis.

(5) Similarly, if d2Ψ/dx2 is finite, then dΨ/dx is continuous.

(6) If the amplitude of Ψ represents information about a physical object, Ψ must be single-valued (i.e, one value at each x, not necessarily same value at all x).

(7) If Ψ is single-valued, dΨ/dx is single-valued.
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Solving the Time-independent Schroedinger Equation for a Barrier Potential


Consider:
Region 1 Region 2 Region 3
The procedure to solve this:
Solve the Time-indep Schroedinger Eq separately in each of the 3 regions.
Incorporate initial conditions.
Make sure that 3 solutions join smoothly at boundaries between regions (this is the continuity requirement on ψ and dψ/dx).
Normalize the final ψ over the full range .
particle with E x=-a x=+a
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Focus on the meaning of

Notice we are considering the case where E





This means that in Region 2, the ψ components are real decaying exponentials, e.g.
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Now apply initial conditions:

Assume incident wave comes from the left: this is .
It can reflect at the x=-a boundary, so in Region 1 there can also be a leftward-going wave: this is .

The wave can be transmitted through the x=-a boundary ( ),
then reflected at the x=+a boundary ( ). So in Region 2 there are both leftward and rightward going waves.

Transmit rightward through the x=a boundary into Region 3 ( ).

In Region 3 there is no way to develop a leftward-traveling wave since there is no boundary to cause reflection there. So D = 0.

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We have 4 equations and 5 unknowns (A, B, C, F, and G).
Solve for B, C, F, and G in terms of A.
Then get A by normalizing.
This produces the complete wavefunction. What it looks like:
wave with E
wave with E>V0 is amplified above the barrier
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Outline

Reflection and Transmission Coefficients
The response of a particle to being trapped in a square well potential
Energy of a particle trapped in a square well
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I. Reflection and transmission coefficients

Consider 2 questions:

What is the probability that the original particle is transmitted past any particular boundary?

What is the probability that it is reflected at any particular boundary?

Recall probability density ρ(x,t) = ψ*ψ.
This is the probability per location x and time t; i.e., the probability that the particle is located AT point x if one tries to observe it at time t.
Now suppose the particle has velocity v, and you want to know its

“probability flux”: probability density PER SECOND that the particle crosses location x, heading in a particular direction.
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Consider a volume:
Area dA
v dt
particle somewhere inside has velocity v

Choose the orientation of the volume so the particle is moving parallel to the edge “v dt”.

If the particle is somewhere in the box at t = 0, and has velocity v, then by the time dt, the particle is guaranteed to have crossed through dA. The probability that this will happen is (Probability per unit volune that particle is in box) ✕ (Volume of box)

(ψ*ψ ) ✕ (v dt dA)

So the probability of crossing dA during dt per unit volume per unit area is

Probability Flux = ψ*ψ v.
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Now consider a specific particle of velocity =v, approaching step potential V. It passes through dA on its way there:
 v
dA
Define the probability of an initial pass through dA as P(I), “probability of incidence upon the step”.

At the moment when it reaches x=-a, there is
some probability that it reflects P(R) and some probability that it is transmitted P(T).
If it reflects, then some time later it will pass through dA at x=-vdt, traveling leftward.
If it is transmitted, then some time later it will pass through dA at x=+vt, traveling rightward.

-v dt +v dt
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leftward-going reflected particle = ψreflected
rightward-going incident particle = ψincident
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neglect
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The response of a particle to being trapped in a square well potential

Consider:

x=-a/2 x=+a/2
a particle with energy EHow did the particle get in there in the first place, if E
This is an approximation to the situation of a proton bound in a nucleus or an electron bound in an atom. (They are not really square.)

Perhaps before it was bound (i.e., trapped down in the well), it was in the vicinity of –a/2V0, then lost some energy (“gave up binding energy”)/
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Goal for this section:

Find the ψ of the particle. Remember: once we know ψ, we know everything it is possible to know about the particle in this situation. Everything is encoded in ψ.
Learn that if a potential has a shape that can bind a particle, then the particle can NOT have arbitrary energy.

The particle must have an energy selected from a limited set of allowed energies.

Examples of potentials that can bind particles:






Examples of potentials that do NOT bind particles:
restrict it to a limited region
The method to find ψ...









...is almost identical to the method for the Barrier Potential:

Write the Schroedinger Equation for each region
Solve the Schroedinger Equation for each region
Match ψ’s at boundaries to find A, B, C, D, etc.

The discovery we will make is that energy E of the particle cannot be arbitrary. It has to be treated like A, B, C etc. That is: only particular values of E will allow ψ to satisfy the boundary conditions.
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x=-a/2 x=+a/2
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Now apply boundary conditions

Initial conditions. Whereas for the barrier potential we might say “particle is travelling rightward”, here the particle is bound: not traveling.
ψ must be finite everywhere. This was true automatically for previous shapes of V so we did not explicitly consider it. Her we have to enforce it.
To guarantee ψ finite as x  -∞, D must =0.
To guarantee ψ finite as x  +∞, F must =0.
(3) ψ continous at x=-a/2
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For a given k1, a, and K2, it is mathematically impossible to have both relations simultaneously true. So we continue to treat them separately:
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}
{
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Energy of a particle trapped in a square well

The message of this section is:

A particle in a well cannot have any arbitrary energy. To satisfy all the boundary conditions, only certain energies are allowed.

How to find the allowed energies:

Recall that there are 2 classes of wavefunctions ψ supported by the square well. Each has its own relationship between k1 and K2:
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ε
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What is important about this energy result:
1.) A particle in a well cannot have arbitrary energy. It has a limited set of options.

This well is a rough approximation to the nuclear potential that binds protons in a nucleus or the Coulomb potential that binds electrons in an atom. So these permitted energies correspond to the allowed energy shells in which electrons. (Protons in the nucleus are restricted to shells too.)
2.) Consider the radius of the quarter-circle that is the graph of q(ε) versus ε.
The larger that radius is, the more intersections there will be: that is, the more allowed energy solutions that exist.

The radius length is given by



so larger a (well width) or larger V0 (well depth) both produce more allowed energies.
intersection #1
intersection #2
intersection #3
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3.) The ability of the potential to limit the allowed energies of the particle is not unique to the square well. It is a property of any potential that binds a particle (i.e. limits it to a specific region of space). So we would get a limited set of allowed energies for potentials shaped like:







4.) The fact that the set of allowed energies is limited and that a particle cannot ramp up or down its energy in transitioning from one energy level to another is called “energy quantization.”

5.) There is a one-to-one association between allowed ψ’s and allowed E’s. Consider Class 1 (cosine) solutions only:


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lowest energy “Class 1 E1” is ½ cycle of cosine.
next higher energy “Class 2 E1” is 1 cycle
Class 1 E2, 3/2 cycle, etc.
The Class 1 E’s come from here, where
ε
Notice higher energy goes with more cycles. This is reasonable as more cycles means higher frequency, and we know that E=hν.
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Recall each ψ is called an eigenfunction of the Hamiltonian. The E that goes with it is called its eigenvalue or “eigenenergy.”

6.) Recall we only plotted Class 1. We must not forget Class 2 solutions---these are defined by



The E’s for Class 2 solutions are different from the E’s for Class 1 solutions and interspersed with them:
Class 1 cosine




Class 2 sine

Class 1 cosine