The Quantum OscillatorQuantum Physics Lecture 18

7/28/2019 The Quantum OscillatorQuantum Physics Lecture 18

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PHYS 2040

Quantum Physics

Step 1: Potential

Step 2: Schrdinger equation

What do we know about the solutions

So what will the wavefunction look like?

Where does the particle spend most of its t ime?

The wavefunction solutions are

Getting the energy eigenvalues

Significance in the solutions

Binding atoms together to make molecules

When the quantum oscillator comes in handy

Molecular vibrations

PHYS2040 Lecture 18 The Quantum Oscillator

Updated: 14/5/2008 10:59 AM

Lecture 18: The Quantum Oscil latorLecture 18: The Quantum Oscil lator


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PHYS 2040

Quantum Physics

After dealing with free particles in Newtonian mechanics, usually the next problem isa partic le subject to a restoring force. The classic example of this is the partic le on a

spring.

The quantum oscillatorThe quantum oscillator

The particle is subject to a force proportional to its displacement from some

equilibrium position, with the force directed oppositely to the displacement.

In other words, F = k(x x0) where x x0 is the displacement from the equilibriumposition x0 and k is a constant (Hookes constant) that determines the strength of the

restoring force.

Classical mechanics predicts that the particle will oscillate about the equilibrium

position with frequency f= (k/m)/2, with some amplitude (and energy) that can takeany value and depends on the initial displacement of the system.


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PHYS 2040

Quantum Physics

At first this might seem completely pointless. After all, howcan you attach a particle to a spring and still have it all smallenough to behave in a quantum manner?

Why this is a useful exerciseWhy this is a useful exercise

Despite the fact that you actually can do this (see later on),

you dont really need a spring to have simple harmonic

motion, just a restoring force thats proportional to a

displacement. One really important place where this occurs is

in solids!

In a solid, the atoms are neatly arranged in a lattice. Suppose

one atom suffers a small displacement. The neighbours that it

becomes closer to will repel it, and the neighbours that

become further away will attract it, thereby providing a

restoring force that tr ies to move it back into position.

In effect here, the chemical bonds act like little springs.

Because of this, there is so much about materials that we can

understand by solving the quantum oscillator problem.


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PHYS 2040

Quantum Physics

So lets work through this problem, see what answers come out and then try to makesome sense of those answers.

Step 1: PotentialStep 1: Potential

The first step is getting the potential V(x). We will make our lives easy here and set x0= 0. We know that the force F(x) = kx, and that F(x) = dV(x)/dx, so

Work out the potential for yourselves, is it:

a) V(x) = k (i.e., just a constant)b) V(x) = kx2 + Cc) V(x) = exp(kx)

d) V(x) = kx2 + C

The correct answer is d), you need to take the integral of F(x) and then reverse the

sign. We can safely set the constant C to zero here. If you plot V(x), you can see that

F(x) is proportional to x (it gets bigger as you increase x) and is the right potential to

use.


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PHYS 2040Quantum Physics

Were now about to put our potential into the Schrdinger equation and go from there,but fi rst, a quick stop to see how much you learned in first year

Revision: Simple harmonic motionRevision: Simple harmonic motion

Is the energy of a particle undergoing simple harmonic motion constant true or

false? Why?

True - The total energy is alwaysconstant unless theres

dissipation, its just converted

from KE to PE and back again.

At the limit of motion, theKE = 0 and all the particles

energy is potential.

At the center of motion, thePE = 0 and all the part icles

energy is kinetic.


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OK, so: Kinetic Energy + Potential Energy = Total Energy, which is just theSchrdinger equation

Step 2: The Schrdinger equationStep 2: The Schrdinger equation

We can simpli fy this using the classical result f= (k/m)/2, which gives k = m2, and:

(18.1))()()()(

2 2

22

xExxVdx

xd

m

=+

h

(18.2))()(2

1)(

2

2

2

22

xExkxdx

xd

m

=+

h

(18.3))()(2

1)(

2

22

2

22

xExxmdx

xd

m

=+

h

(18.4))(2

12)( 2222

2

xExm

m

dx

xd

= h

Yikes, now this is a tough one to solve because of the x2 term. So lets start by

thinking about requirements for the solution and what the wavefunctions should look

like.


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So lets go back to the potential. What can we guess are some of the properties of thewavefunction solutions.

What do we know about the solutions?What do we know about the solutions?

a) The wavefunction should be wavelike inside the well, and decay exponentially

outside it (the walls arent infini te, so no nodes at the walls).

b) The wavefunction needs to be continuous at the potential wall, but exactly

where the wall is changes with the energy E of the particle, which is why thisproblem is so hard to solve.

c) The potential is symmetric about x = 0, so the observables, in particular*,must be symmetric about x = 0 also. Note this doesnt quite make symmetric.


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Where does the particle spend most of its time?Where does the particle spend most of its time?

Where does the particle spendmore of its time, the outer half ,

or the inner half ?

The most important factor in sketching the wavefunction is where the particle spendsmost of its time


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The most important factor in sketching the wavefunction is where the particle spendsmost of its t ime.

Where does the particle spend most of its time?Where does the particle spend most of its time?

Where does the particle spendmore of its time, the outer half ,

or the inner half ?


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On the basis of that, you should be able to sketch up the wavefunct ion.

So what will the wavefunction look like?So what will the wavefunction look like?

(a)

(b)

(c)

(d)

The correct answeris (b). At the walls

is larger becausethe particle travelsmore slowly and is larger becausethe probability of

finding the particlenear the wall isconsequently

higher.


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The full solution to Eqn 18.4 can be found in Appendix I of Eisberg and Resnick, butvery briefly...

For the mathematically inclined (non-examinable)For the mathematically inclined (non-examinable)

(18.4))(2

)(2

12)(2

2

2

2222

22

2

xmE

xm

xExmm

dx

xd

=

=

hhh

When a differential equation (DE) gets tough, generalise it as much as possible,

classify it, and look for an approach that works with that class of DE. If you make it

dimensionless by substi tuting in u:

(18.5)

where = m/h and = 2mE/h2 (n.b., E&R write C instead ofk). Then you arrive at:

( ) xkmxu21

41

h==

(18.6)

= 2

2

2

udu

d


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Now for any finite value ofE, then / wil l be neglig ible as |u| , so we get:

For the mathematically inclined (non-examinable)For the mathematically inclined (non-examinable)

In this case the general solution is:

(18.8)

where we set B = 0 to keep fini te as |u| .

for |u| (18.7) 22

2

udu

d=

222 222 uuuAeBeAe

=+=


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Now for any finite value ofE, then / wil l be neglig ible as |u| , so we get:

The wavefunction solutions areThe wavefunction solutions are

In this case the general solution is:

(18.8)

where we set B = 0 to keep fini te as |u| . Thats the easy part. For small u, werestil l stuck with Eqn. 18.6, the solutions turn out to be of the form:

for |u| (18.7) 22

2

udu

d=

222 222 uuuAeBeAe

=+=

The term Hn(u) representssomething called a Hermite

polynomial, a special class ofpolynomial functions that are

solutions to the DE I-15 in E&R.

22

)( unnn

euHA=


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So lets start with the first eigenfunction 0, we can get E0 by substituting i t back intothe Schrodinger equation (Eqn 18.3). But first, we should get A0 by normalising 0(n.b., in this case we dont need A0 to get E0, but in some cases you do, so well do it

here for practice).

Getting the energy eigenvaluesGetting the energy eigenvalues

where (18.7)

2

00

2u

eA

=( )

x

km

xu 21

41

h==

=a

dxeax 2

Note that:


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Hint: LeaveA0 asA0 because it wi ll cancel out.

Substitute the wavefunction 0 (Eqn. 18.7) into the Schrdinger equation (Eqn. 18.3)to obtain the corresponding energy E0.


Note that if you dont know

or remember here, you canobtain it by knowing that Ecant be x-dependent (as i tsa constant) and solv ing for itby making the sum of the x2

terms equal to zero.


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If we do this again with 1 to obtain the corresponding energy E1:


So now it will become apparent that En = (n + ) h, so lets recap before looking atthe signif icance of these results a litt le more.


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Summary of eigenfunctions and eigenvaluesSummary of eigenfunctions and eigenvalues

The most vital point here is thatthe energy levels are all equally

spaced in energy by h.


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Significance in the solutionsSignificance in the solutions

The most vital point here is that theenergy levels are all equally spaced

in energy by h, hence photons ofsome frequency wil l be absorbed

or emitted in moving up or down

through the energy levels.

The energy of the oscil latorcannot be zero, asthis violates the Heisenberg uncertainty

principle, because p, p, x and x are all zero.

The minimum allowed energy his called the zero point energy, it

explains a lot of things in physics,including why helium doesnt

freeze even at T = 0.


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Significance in the solutionsSignificance in the solutions

In the classical case, the particle cantake any total energy it likes, this

corresponds to different amplitudes

of the oscil lator. In all cases isfixed by k and m, the partic le justtravels faster to make a complete

rotation in the required time.

In the quantum case, the Schrdingerequation means that only certain energies

are allowed, this corresponds to aninteger number of wavelengths along the

oscillation path. is still fixed by k and

m, but the quantum theory adds morerestrictions beyond that.


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Binding atoms together to make moleculesBinding atoms together to make molecules

If you take two neutral atoms andbring them relatively close together,then there are two forces actingbetween them. An attractive forcecalled van der Waals force and a

repulsive force due to the Pauliexclusion principle. If the atoms arecharged, then there are also Coulombforces involved.

If you add these together, you get theLennard-Jones potential, which has aminimum energy at some r = r0, whichis the equilibrium separation of thetwo atoms forming the molecule.


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Where the quantum oscillator comes in handyWhere the quantum oscillator comes in handy

A parabola is an excellent approximation to many more complicated potentials inphysics, hence you can often use the quantum osci llator instead unless you need thebetter accuracy provided by an exact model of the actual potential. An importantexample of this is stretching of chemical bonds.

Remember, you can express any functionas a Taylor polynomial, and one of the

first terms will be quadratic!


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Molecular vibrationsMolecular vibrations

A simple diatomic molecule can vibrate such that the bond length r oscillates aboutthe equil ibrium separation r0 given by the minima in the Lennard-Jones potential. The

vibration can only have discrete allowed energies En = (n + ) h.

Photons of frequency areabsorbed/emitted in transitions between

allowed vibrational energy levels.


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PhononsPhonons

You can extend that concept into materials, where a deformation wave or a latticevibration carries an energy h proportional to its frequency. These waves can betreated simi larly to photons, and are called phonons. They play an important role in

the physics of solids, youll learn about these properly in solid state physics

courses.

Phonons obey the same principles as electrons and photons (and other particles),

exhibiting wave-particle duality, obeying the Heisenberg uncertainty principle,

conservation rules, statistics, etc.


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Nanoelectromechanical systems (NEMS)Nanoelectromechanical systems (NEMS)

Very recently its become possible to build oscillators small enough to study the

quantum properties of mechanical systems, including quantization of thermal

conductivity (due to phonons being particles), uncertainty principle, etc.

If youre curious, see the artic le Putting Mechanics into

Quantum Mechanics by Schwaband Roukes on the course

website.


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Simple harmonic motion can also be considered from a quantum mechanicalperspective, a problem known as the quantum oscillator problem. It is an incrediblyuseful problem to solve, because it can be used to model and understand a lot of thevibrational behaviour of objects ranging from simple diatomic molecules through tocrystals and solids.

The eigenfunctions of the quantum osci llator have a wave-like nature inside the wellwith an increasing wavelength and amplitude towards the edges of the potentialbecause the particle moves more slowly and spends more time in the regions near theedge of the potential. The wavefunct ion decays exponentially outside the well.

The energy eigenvalues are equally spaced with En = (n + ) h and a constant levelseparation of h. A particle can thus move between vibrational states by eitherabsorbing (to move up) or emitting (to move down) a photon of frequency .

Applying these ideas to simple molecules such as water, allows you to use photonsto excite molecular vibrations increasing the molecules energy. Using microwavephotons, you can excite water molecules and warm/cook food.

Vibrations in a crystal lattice can behave like wave-particles (just like electrons andphotons do) which are called phonons. They have similar properties to other particles

in physics and carry an energy h based on their frequency of vibration .

SummarySummary

= Brooks/Cole - Thomson


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How your microwave works.How your microwave works.

Your microwave exploits this principle to heat food. A microwave oven has a devicecalled a magnetron which produces photons with frequency ~2.45 GHz ( ~ 12.24 cm)that are tuned to the rotational/vibrational spectra of the water molecules in yourfood.

The water molecules absorb the photons one by one and move up the energy levels,

increasing the internal energy of these molecules, and thus heating your food.


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How your microwave works.How your microwave works.

You can also see this process classically, as the fluctuating electromagnetic fieldflipping the molecule faster and faster, but you need quantum mechanics to explain

why ~ 2.45 GHz is the best frequency to use to excite the vibrations.

This also explains why petrol wont heat up in the microwave (its a non-polar

molecule, so the EM waves cant couple to the vibrational behaviour), and why putt ing

metals in a microwave is bad (the metal acts as an antenna, absorbing the energy andturning it in to large voltages/currents that result in arcing.)

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The Quantum OscillatorQuantum Physics Lecture 18