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6.1 The Schrödinger Wave Equation

6.2 Expectation Values

6.3 Infinite Square-Well Potential

6.4 Finite Square-Well Potential

6.5 Three-Dimensional Infinite-Potential Well

6.6 Simple Harmonic Oscillator

6.7 Barriers and Tunneling

CHAPTER 6

Quantum Mechanics II

I think it is safe to say that no one understands quantum mechanics.

Do not keep saying to yourself, if you can possibly avoid it, But how can it be like that? because you will get down the drain into a blind alley from which nobody has yet escaped. Nobody knows how it can be like that.

- Richard Feynman

(2)

6.2: Expectation Values

Useful to relate the wavefunction to measurable quantities.

In quantum mechanics the expectation value is:

the expected result of the average of many measurements of a given quantity.

The expectation value of x is denoted by <x>.

Conversely, for a single measurement the expectation value predicts the most probable outcome.

Any measurable quantity for which we can calculate the expectation

value is called a physical observable.

The expectation values of physical observables (for example,

position, linear momentum, angular momentum, and energy) must be

real, because the experimental results of measurements are real.

Example: The average value of x (statistics) is

(3)

Continuous Expectation Values

We can change from discrete to

continuous variables by using the probability P(x,t) of observing the particle at a particular x.

In quantum mechanics, the probability is P(x) = ||2 = by using the wave function and the expectation value is:

If the wave function is properly

normalized, the expectation value becomes:

The expectation value of any function g(x) for a normalized wave function:

(4)

Examples

A a particle is described by the wave function Ψ(x) = ax between x = 0 and x = 1, and zero elsewhere.

Find a:

Find the probability of finding the particle between x1 = 0.45 and x2 = 0.55.

Find <x>

(5)

Momentum Operator

Suppose we want to find the expectation value for the momentum, p.

But what do we insert for p?

we need a “momentum extractor” that represents p in such a way that we can evaluate <p>.

given the variable of integration is x, this “extractor” should be expressed in terms of x.

If we start with the wave function for a free particle, this wave function

can also be written as:

(6)

Momentum Operator:

So how do we extract p from the wave function?

If we take the derivative of the wave function of the free particle with respect to x:

Rearranging:

The quantity on the right has “extracted” the momentum of the wavefunction.

This extractor is defined to be the momentum operator:

Using the operator the expectation value for the momentum is

(7)

Energy operator is able to extract the energy of the particle.

The time derivative of the free-particle wave function is

Substituting ω = E / ħ yields

Therefore we can also define the energy operator:

The expectation value of the energy is

Energy Operator:

(8)

We developed expressions for the momentum and energy operators

using the wave function of the free particle, however the results

are valid for any wave function.

In general, an operator is a mathematical operation that transforms one function into another.

Each physical observable has an associated operator that is used to find that observable's value.

Given the form of some operators, the order of operation is important.

   the operator must be placed between Ψ* and Ψ to operate on Ψ.

Operators

(9)

For non-relativistic particle the total energy is the sum of the kinetic and potential energies:

Writing the equation in terms of operators and letting both sides of the equation operate on the wave function:

Therefore the Schrödinger equation can be considered an operator equation for the total energy.

Schrödinger equation as an operator equation

(10)

6.3: Infinite Square-Well Potential

Will now apply the formalism developed to several potentials.

In each case we will:

Determine the wave function and associated observables.

will see that some observables are quantized.

The simplest example is that of a particle trapped in a box with infinitely hard walls that the particle cannot penetrate. This

potential is called an infinite square well and is given by:

We have looked at this situation previously (particle confinement in section 5.6, particle in

a box in section 5.8), however we now revisit

the problem using Schrödinger wave mechanics.

Clearly the particle is restricted to reside in the region where the potential is zero

(the wave function must be zero outside).

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6.3: Infinite Square-Well Potential

Where the potential is zero inside the box, the Schrödinger wave

equation becomes where .

The general solution is

Boundary conditions dictate that the wave function must be zero

at x = 0 and x = L.

to ensure ψ=0 at x = 0  B = 0.

to ensure ψ=0 at x = L  k = nπ/L (where n= 1,2,3,...).

The wave function is now:

(12)

Quantization

We normalize the wave function:

The normalized wave function becomes

These functions are identical to those obtained

in classical physics for a vibrating string with fixed ends.

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Quantized Energy

The quantized wave number now becomes

Solving for the energy yields

Note that the energy depends on the integer values of n.

 Hence the energy is quantized and nonzero.

The special case of n = 1 is called the ground state energy 

Classically probability is uniform P(x) = 1/L

For large n classical and quantum results should agree.

(14)

Example 6.8

Determine expectation value for p and p

2

of a particle in an infinite square well in the first excited state.

The first excited state corresponds to n = 2 therefore:

The answer means particle is moving left as often as right.

 For <p

2

>:

Comparing with E2:

(15)

6.4: Finite Square-Well Potential

(16)

Finite Square-Well Solution

(17)

Finite Square-Well Solution

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Other differences

Wavelength is somewhat larger for finite well because the wave function extends beyond the well.

This leads to  a smaller momentum (p = h/λ)

slightly lower energy levels.

For a finite well,

Number of levels is limited by the potential height, V0.

Define the penetration depth x as the distance outside the potential well where the probability significantly decreases:

penetration depth deeper as E approaches V0.

It should not be surprising to find that the penetration distance

that violates classical physics is proportional to Planck’s constant.

Actually, detecting particle in classically forbidden region difficult.

∆x ~ δx (small) implying ∆p and thus ∆K ~ ћ/(2m(δx)2) (large)

no experiment could be certain the particle has negative K.

Infinite well finite well

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In 3-d the wave function is a function of all three spatial coordinates, ψ(x,y,z).

In 3-d the time independent Schrödinger wave equation takes on the form:

Consider a free particle in a 3-d box of dimensions L1, L2, L3.

Find the wave functions and energies.

6.5: Three-Dimensional Infinite-Potential Well

(20)

Given the fact that the particle is free inside the box the x, y and z dependent parts of the wave function must be independent or:

A reasonable solution 

Find k1, k2, k3 by applying boundary conditions in x, y, & z.

Therefore if particle is constrained in 3-d,

3 quantum numbers (n

1

, n

2

, n

3

) are required to describe the state.

Particle in 3-d box - Solution

( , , ) x y z X x Y y Z z ( ) ( ) ( )

 

1 2 3

( , , ) x y z A sin( k x ) sin( k y ) sin( k z )

 

(21)

To obtain the energies plug wave function into the 3-d wave equation.

Allowed energies are defined by 3 quantum numbers; n

1

, n

2

, n

3

.

Ground state corresponds to n

1

= n

2

= n

3

= 1.

If L

1

= L

2

= L

3

= L,

Particle in 3-d box – Solution (Continued)

1 2 3

( , , ) sin( ) sin( ) sin( ); n n

n

x y z A k x k y k z k n

L

    

 

 

1 2 3

2 2

2 2 2

1 2 3

2

2 n n n

E n n n

mL

    

(22)

3-D box (Degeneracy)

A quantum state is degenerate when there is

more than one wave function for a given energy.

(ex) E211 = E121 = E112

Degeneracy results from particular properties

of the potential energy function that describes the system.

 A perturbation of the potential energy can remove the degeneracy.

 

1 2 3

( , , ) sin( ) sin( ) sin( )

n n / n

x y z A k x k y k z

k n L

   

1 2 3

2 2 2

2 2 2 2 2 2

1 2 3 2 1 2 3

2 2

n n n

E k k k n n n

m mL

(23)

6.6: Simple Harmonic Oscillators

Simple harmonic motion (SHM) can be applied to describe many physical situations: springs, diatomic molecules and atomic lattices.

The most obvious example is the motion of a spring.

During oscillation the total energy E = K + V is conserved.

(24)

Simple Harmonic Oscillators

Potentials of many other physical systems can be approximated by SHM.

In general if we expand any potential in a Taylor expansion about the equilibrium position:

Redefining the zero of potential (V0= 0), and since dV/dx = 0 at equilibrium (V1=0), we may approximate the potential

Diatomic molecules

2 2

2 0 0

1 1

( ) ( ) ( ) ( 0)

2 2

V xV xxV x  x x

(25)

Want to look at the quantum treatment of simple harmonic motion (SHM).

The time independent form of the wave equation is:

The solutions are wave functions of the following form:

Hn(x) are Hermite polynomials of order n (n = 0, 1, 2, 3,..).

1 2

( ) 2 V x x

2/ 2

( ) ( )

x

n

x H x e

n

 

Quantum treatment of simple harmonic motion

(26)

Quantum treatment of simple harmonic motion

2/ 2

( ) ( ) x

n x H x en

n( )x 2

0( )x

1( )x

2( )x

3( )x

2 10( )x

Contrary to the classical case in ground state the largest probability at the center.

Classically, the probability of finding the particle is greatest at the ends of motion and smallest at the center (see dotted line at n =10)

this because particle slows down and spends more time at the turning points

in agreement with Bohr’s correspondence principle for large n quantum and classical predictions merge.

classical probability

(27)

SHM: Classical results

In the classical case the particle oscillates back and forth, the total energy of the system remains constant.

E0 = K + V = conserved

At the equilibrium position all energy is kinetic.

At the turning points (±a) all energy is potential i.e. velocity is zero, and a = amplitude of oscillation.

(28)

SHM: Quantum results

As was the case for the potential well, the allowed energies are quantized and given by:

Now n = 0 is allowed, E = 0 is not allowed since it would violate the uncertainty principle.

Zero energy would mean x and K (p) simultaneously zero.

The zero point energy (E0) is called the Heisenberg limit 

∆E or spacing between levels is constant  ∆E = ½ħω.

Wave function is non-zero beyond the classical turning points where it decays exponentially.

0

( ) x

12

 

12

En n n

m

0

1 E  2

(29)

Example

Find expectation value for x2 for the ground state of the harmonic oscillator.

(30)

6.7: Barriers and Tunneling

Consider a particle of energy E approaching a potential barrier of height V0 and the potential everywhere else is zero.

Let’s consider two cases: E > V0 and E < V0

incident transmitted

reflected

incident transmitted

reflected

Case I: E > V

0

Case II: E < V

0

Classical: Reflection 0 Quantum: Reflection 0

Classical: Transmission 0 Quantum: Transmission 0

(Barrier) (Tunneling)

(31)

Case I: E > V 0 (Barriers)

Classically the particle would pass the barrier,

however in region II the kinetic energy would reduced by a factor V0 (energy, E conserved)

Because of wavelike properties, quantum mechanics predicts different behavior.

When waves are incident on a boundary between two media a portion of the wave is transmitted and a portion can be reflected.

In regions I and III the wave numbers are:

In the barrier region II we have

(32)

Reflection and Transmission (when E > V

0

)

The wave functions can be determined by solving the Schrödinger equation and applying boundary conditions.

Potentials and Schrödinger wave equations are:

The corresponding solutions are:

As the incident wave moves from left to right,

ik xI

Ae

ik xi

e

ik xI

Be

ik xI

Fe

ik xi

e

(propagate to +x direction) (propagate to -x direction)

(G = 0)

(33)

Probability of Reflection and Transmission

The reflection probability R and transmission probability T are:

Because the particles must be either reflected or transmitted, R + T = 1

By applying the boundary conditions at x = 0, L:

we arrive at the transmission probability 

 Notice that T = 1 only when: i) V0 = 0, or ii) kIIL = nπ.

(34)

Now we consider the situation where classically the particle does not have enough energy to surmount the potential barrier, E < V0.

The quantum mechanical result, however, is one of the most remarkable features of modern physics, and there is ample experimental proof of its existence.

There is a small, but finite, probability that the particle can penetrate the barrier and even emerge on the other side.

The wave function in region II becomes:

The transmission probability that

describes the phenomenon of tunneling is

Case II: E < V 0 (Tunneling)

Classically expect total reflection (K in the barrier would be negative)

ik

II

 

2 ( 0 ) ( ) ik xII ik xII x x;

II

m V E

x Ce De Ce De

      

2 ( 0)

II

m E V

k

(35)

Uncertainty Explanation of Tunneling

So quantum result is that there is a non-zero probability

 the particle can penetrate the barrier and emerge on the other side

the incident particle can tunnel the barrier.

Consider when κL >> 1 then the transmission probability becomes:

The particle is allowed in the barrier region by quantum mechanics and the uncertainty principle lets it penetrate into a classically forbidden region.

The minimum such kinetic energy is:

This violation allowed by the uncertainty principle is equal to the negative kinetic energy required!

incident transmitted

reflected

(Case II: E < V

0

)

Tunneling

1

sinh(kL)kL 2ekL

I I

ik x ik x

I Ae Be

ik xI

III Fe

x x

II Ce De

Exponential decay

(36)

Example 6.14 L 0.8 nm, E 5 eV V ,

0

10 V

In a particular semiconducting device, electrons that are accelerated through a potential difference of 5 V attempt to tunnel through a barrier of width 0.8 nm and height 10 V. What fraction of electrons are able to tunnel through the barrier?

Approximately 

(37)

Example 6.16

a) Find the penetration distance (penetration depth) (distance where the probability drops by 1/e).

b) Calculate the depth for a 5 eV electron approaching a step barrier of 10 eV.

L  

( ) ikx ikx; 2

I

x Ae Be k mK

  

2 ( 0 )

( ) x x;

II

m V E x Ce De

    

Boundary condition for x  : C  0 II( )xDex

Penetration distance l :

1

l 2

 

(EK 5 eV V, 0 10 V) 2

0

0.044 nm

4 2 ( )

l hc

mc V E

  

(38)

Analogy of tunneling with Wave Optics

If light passing through a glass prism reflects from an internal surface with an angle greater than the critical angle, total internal reflection occurs.

However, the electromagnetic field is not exactly zero just outside the prism. If we bring another prism very

close to the first one, experiments show that the

electromagnetic wave (light) appears in the second prism.

The situation is analogous to the tunneling described here. This effect was observed by Newton and can be demonstrated with two prisms and a laser.

 The intensity of the second light beam decreases

exponentially as the distance between the two prisms increases.

Te

2 L

L

(39)

Alpha-Particle Decay

The phenomenon of tunneling explains the alpha-particle decay of heavy, radioactive nuclei.

Inside the nucleus, an alpha particle feels the strong, short-range attractive nuclear force as well as the repulsive Coulomb force.

The nuclear force dominates inside the nuclear radius where the potential is approximately a square well.

The Coulomb force dominates outside the nuclear radius.

The potential barrier at the nuclear radius is several times greater than the energy of an alpha particle.

According to quantum mechanics, however, the alpha particle can

“tunnel” through the barrier.

Hence this is observed as radioactive decay.

(40)

Scanning Tunneling Microscopy (STM)

(41)

Direct Measurement of IψI 2 using STM

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