3.2 Finite well and harmonic oscillator

Slides: Video 3.2.1 Particles in potential wells – introduction

Particles in potential wells

David MillerQuantum mechanics for scientists and engineers

3.2 Finite well and harmonic oscillator

Slides: Video 3.2.2 The finite potential well

Text reference: Quantum Mechanics for Scientists and Engineers

Section 2.9

Particles in potential wells

The finite potential well

Quantum mechanics for scientists and engineers David Miller

Finite potential wellInsert video here (split

screen)

Lesson 7Particles in potential

wells

Insert number 2

Particle in a finite potential well

We will choose the height of the potential barriers as Vowith 0 potential energy at the

bottom of the wellThe thickness of the well is Lz

Now we will choose the position origin in the center of the well

oV

00/ 2zL / 2zL z

Particle in a finite potential well

If there is an eigenenergy E for which there is a solutionthen we already know what

form the solution has to takesinusoidal in the middleexponentially decaying on either side

oV

00/ 2zL / 2zL z

Particle in a finite potential well

For some eigenenergy Ewith and

for

for

for

with constants A, B, F, and G

oV

00/ 2zL / 2zL z

/ 2zz L expz G z

/ 2 / 2z zL z L sin cosz A kz B kz

/ 2zz L expz F z

22 /k mE 22 /om V E

Particle in a finite potential well

Now we need to apply the boundary conditions to solve for the unknown coefficientsconstants A, B, F, and G

or at least three of themthe fourth could be found by normalization

oV

00/ 2zL / 2zL z

/ 2zz L expz G z

/ 2 / 2z zL z L sin cosz A kz B kz / 2zz L expz F z

Particle in a finite potential well

From continuity of the wavefunction at

Writing

gives

oV

00/ 2zL / 2zL z

/ 2 exp / 2z zL F L

/ 2zz L

exp / 2L zX L

sin / 2L zS kL cos / 2L zC kL

L L LFX AS BC

sin / 2 cos / 2z zA kL B kL

Particle in a finite potential well

Similarly at

Continuity of the derivative givesat

at

oV

00/ 2zL / 2zL z

/ 2zz L

L L LGX AS BC

L L LGX AC BSk

/ 2zz L

/ 2zz L

L L LFX AC BSk

Particle in a finite potential well

So we have four relations

Now we need to find what solutions are compatible with these

oV

00/ 2zL / 2zL z

L L LGX AS BC

L L LGX AC BSk

L L LFX AC BSk

L L LFX AS BC

Particle in a finite potential well

Adding

gives

Subtracting

from

gives

oV

00/ 2zL / 2zL z

L L LGX AS BC

L L LFX AS BC

2 L LBC F G X

L L LFX AC BSk

L L LGX AC BSk

2 L LBS F G Xk

Particle in a finite potential well

As long aswe can divide

by

to obtain

This relation is effectively a condition for eigenvalues

oV

00/ 2zL / 2zL z

2 L LBC F G X

2 L LBS F G Xk

F G

tan / 2 /zkL k

Particle in a finite potential well

Subtractingfrom gives

Adding

and

gives

oV

00/ 2zL / 2zL z

L L LGX AS BC

L L LFX AS BC

L L LFX AC BSk

L L LGX AC BSk

2 L LAS F G X

2 L LAC F G Xk

Particle in a finite potential well

Similarly, as long aswe can divide

by

to obtain

This relation is also effectively a condition for eigenvalues

oV

00/ 2zL / 2zL z

F G

2 L LAS F G X

2 L LAC F G Xk

cot / 2 /zkL k

Particle in a finite potential well

For any case other thanwhich leaves

but notor

which leavesbut not

then the solutions and are contradictory

oV

00/ 2zL / 2zL z

F G

cot / 2 /zkL k tan / 2 /zkL k

F G cot / 2 /zkL k

tan / 2 /zkL k

tan / 2 /zkL k cot / 2 /zkL k

Particle in a finite potential well

So the only possibilities are

1 -and

2 –and

oV

00/ 2zL / 2zL z

F G cot / 2 /zkL k

F G tan / 2 /zkL k

Particle in a finite potential well

1 -and

Note from

and

SL and CL cannot both be 0so

Hence in the well we have

which is an even function

oV

00/ 2zL / 2zL z

F G tan / 2 /zkL k

2 L LAS F G X

2 L LAC F G Xk

0A

cosz kz

Particle in a finite potential well

1 -and

Note from

and

SL and CL cannot both be 0so

Hence in the well we have

which is an odd function

oV

00/ 2zL / 2zL z

0B

sinz kz

F G cot / 2 /zkL k

2 L LBS F G Xk

2 L LBC F G X

Particle in a finite potential well

Though we have found the nature of the solutionswe have not yet formally

solved for the eigenenergiesEand hence for k and

We do this by solving

and

oV

00/ 2zL / 2zL z

tan / 2 /zkL k

cot / 2 /zkL k

Solving for the eigenenergies

Change to “dimensionless” unitsUse the energy of the first level in the

“infinite” potential well width Lzleading to a dimensionless

eigenenergyand a dimensionless barrier height

Also

22

1 2 z

Em L

1/E E

1/o ov V E

212 / / / /z zk mE L E E L

212 / / / /o z o z om V E L V E E L

Consequently

and

So becomes

or

and becomes

or cot / 2 ov

tan / 2 ov

Solving for the eigenenergies

o oV E vk E

12 2 2zkL E

E

12 2 2oz

oV EL v

E

tan / 2 /zkL k tan / 2 /ov

cot / 2 /zkL k cot / 2 /ov

Graphical solution

Choose a specific well depth and plot the curve

ov

2 4 6 8

-2

0

2

4

6

2o

Graphical solution

Choose a specific well depth and plot the curve

2 4 6 8

-2

0

2

4

6

5o

ov

Graphical solution

Choose a specific well depth and plot the curve

2 4 6 8

-2

0

2

4

6

8o

ov

Graphical solution

Choose a specific well depth and plot the curve

Now add the curves

2 4 6 8

-2

0

2

4

6

8o

ov

Graphical solution

Choose a specific well depth and plot the curve

Now add the curves

2 4 6 8

-2

0

2

4

6

8o

tan2

ov

Graphical solution

Choose a specific well depth and plot the curve

Now add the curves

2 4 6 8

-2

0

2

4

6

8o

cot2

tan

2

ov

Graphical solution

For a specific the solutions are the values

of at the intersections of

and

or

ov

2 4 6 8

-2

0

2

4

6

8o

0.663 2.603 5.609

tan2

cot2

Solutions

These are the solutions for a well depth Vo ofNote that

they are all lower energies

than the corresponding solutions for the infinitely deep well of the same width

18E

1n

2n

3n

18oV E

00.663

2.603

5.609

3.2 Finite well and harmonic oscillator

Slides: Video 3.2.4 The harmonic oscillator

Text reference: Quantum Mechanics for Scientists and Engineers

Section 2.10

Particles in potential wells

The harmonic oscillator

Quantum mechanics for scientists and engineers David Miller

Mass on a spring

A simple spring will have a restoring force F acting on the mass M

S

Mass M

S

Mass on a spring

A simple spring will have a restoring force F acting on the mass Mproportional to the amount y by which

it is stretchedFor some “spring constant” K

The minus sign is because this is “restoring”it is trying to pull y back towards zero

This gives a “simple harmonic oscillator”

F Ky

y

Mass MForce F

Mass on a spring

From Newton’s second law

i.e.,

where we definewe have oscillatory solutions of angular frequencye.g., S

2

2

d yF Ma M Kydt

2

22

d y K y ydt M

2 /K M

/K M

siny t

S

Mass on a spring

From Newton’s second law

i.e.,

where we definewe have oscillatory solutions of angular frequencye.g.,

2

2

d yF Ma M Kydt

2

22

d y K y ydt M

2 /K M

/K M

siny t

S

Mass on a spring

From Newton’s second law

i.e.,

where we definewe have oscillatory solutions of angular frequencye.g.,

2

2

d yF Ma M Kydt

2

22

d y K y ydt M

2 /K M

/K M

siny t

S

Mass on a spring

From Newton’s second law

i.e.,

where we definewe have oscillatory solutions of angular frequencye.g.,

2

2

d yF Ma M Kydt

2

22

d y K y ydt M

2 /K M

/K M

siny t

S

Mass on a spring

From Newton’s second law

i.e.,

where we definewe have oscillatory solutions of angular frequencye.g.,

2

2

d yF Ma M Kydt

2

22

d y K y ydt M

2 /K M

/K M

siny t

S

Mass on a spring

From Newton’s second law

i.e.,

where we definewe have oscillatory solutions of angular frequencye.g.,

2

2

d yF Ma M Kydt

2

22

d y K y ydt M

2 /K M

/K M

siny t

S

Mass on a spring

From Newton’s second law

i.e.,

where we definewe have oscillatory solutions of angular frequencye.g.,

2

2

d yF Ma M Kydt

2

22

d y K y ydt M

2 /K M

/K M

siny t

S

Mass on a spring

From Newton’s second law

i.e.,

where we definewe have oscillatory solutions of angular frequencye.g.,

2

2

d yF Ma M Kydt

2

22

d y K y ydt M

2 /K M

/K M

siny t

S

Mass on a spring

From Newton’s second law

i.e.,

where we definewe have oscillatory solutions of angular frequencye.g.,

2

2

d yF Ma M Kydt

2

22

d y K y ydt M

2 /K M

/K M

siny t

S

Mass on a spring

From Newton’s second law

i.e.,

where we definewe have oscillatory solutions of angular frequencye.g.,

2

2

d yF Ma M Kydt

2

22

d y K y ydt M

2 /K M

/K M

siny t

Mass on a spring

From Newton’s second law

i.e.,

where we definewe have oscillatory solutions of angular frequencye.g., S

2

2

d yF Ma M Kydt

2

22

d y K y ydt M

2 /K M

/K M

siny t

S

Mass on a spring

From Newton’s second law

i.e.,

where we definewe have oscillatory solutions of angular frequencye.g.,

2

2

d yF Ma M Kydt

2

22

d y K y ydt M

2 /K M

/K M

siny t

S

Mass on a spring

From Newton’s second law

i.e.,

where we definewe have oscillatory solutions of angular frequencye.g.,

2

2

d yF Ma M Kydt

2

22

d y K y ydt M

2 /K M

/K M

siny t

S

Mass on a spring

From Newton’s second law

i.e.,

where we definewe have oscillatory solutions of angular frequencye.g.,

2

2

d yF Ma M Kydt

2

22

d y K y ydt M

2 /K M

/K M

siny t

S

Mass on a spring

From Newton’s second law

i.e.,

where we definewe have oscillatory solutions of angular frequencye.g.,

2

2

d yF Ma M Kydt

2

22

d y K y ydt M

2 /K M

/K M

siny t

S

Mass on a spring

From Newton’s second law

i.e.,

where we definewe have oscillatory solutions of angular frequencye.g.,

2

2

d yF Ma M Kydt

2

22

d y K y ydt M

2 /K M

/K M

siny t

S

Mass on a spring

From Newton’s second law

i.e.,

where we definewe have oscillatory solutions of angular frequencye.g.,

2

2

d yF Ma M Kydt

2

22

d y K y ydt M

2 /K M

/K M

siny t

S

Mass on a spring

From Newton’s second law

i.e.,

where we definewe have oscillatory solutions of angular frequencye.g.,

2

2

d yF Ma M Kydt

2

22

d y K y ydt M

2 /K M

/K M

siny t

S

Mass on a spring

From Newton’s second law

i.e.,

where we definewe have oscillatory solutions of angular frequencye.g.,

2

2

d yF Ma M Kydt

2

22

d y K y ydt M

2 /K M

/K M

siny t

S

Mass on a spring

From Newton’s second law

i.e.,

where we definewe have oscillatory solutions of angular frequencye.g.,

2

2

d yF Ma M Kydt

2

22

d y K y ydt M

2 /K M

/K M

siny t

Mass on a spring

From Newton’s second law

i.e.,

where we definewe have oscillatory solutions of angular frequencye.g., S

2

2

d yF Ma M Kydt

2

22

d y K y ydt M

2 /K M

/K M

siny t

Potential energy

The potential from the restoring force F is S

z

Mass mForce F

0

z

oV z F dz

212

V z Kz

z

0

z

o oKz dz 212

Kz 2 212

m z

Harmonic oscillator Schrödinger equation

With this potential energythe Schrödinger equation is

For convenience, we define a dimensionless distance unit

so the Schrödinger equation becomes

2 212

V z m z

2 22 2

2

12 2

d m z Em dz

m z

2 2

2

12 2

d Ed

Harmonic oscillator Schrödinger equation

One specific solution to this equation

is

with a corresponding energy This suggests we look for solutions of the form

where is some set of functions still to be determined

2exp( / 2)

/ 2E

2exp / 2n n nA H

nH

2 2

2

12 2

d Ed

Harmonic oscillator Schrödinger equation

Substitutinginto the Schrödinger equation

gives

This is the defining differential equation for the Hermite polynomials

2exp / 2n n nA H

2

2

22 1 0n nn

d H dH E Hd d

2 2

2

12 2

d Ed

Harmonic oscillator Schrödinger equation

Solutions to

exist provided

that is,

2

2

22 1 0n nn

d H dH E Hd d

2 1 2E n

0, 1, 2,n

12nE n

0, 1, 2,n

Harmonic oscillator Schrödinger equation

The allowed energy levels are equally spaced separated by an amount

where is the classical oscillation frequency Like the potential well

there is a “zero point energy” here

/ 2

12nE n

0, 1, 2,n

Hermite polynomials

The first Hermite polynomials areNote they are either

odd or eveni.e., they have a definite parity

They satisfy a “recurrence relation”

successive Hermite polynomialscan be calculated from the previous two

0 1H

1 2H

22 4 2H

33 8 12H

4 24 16 48 12H 1 22 2 1n n nH H n H

12nE n

0, 1, 2,n

m z

Harmonic oscillator solutions

Normalizing

gives

2exp / 2n n nA H

12 !n n

An

Harmonic oscillator solutions

Normalizing

gives

or

2exp / 2n n nA H

12 !n n

An

21 exp2 ! 2n nn

m m mz z H zn

12nE n

0, 1, 2,n

m z

Harmonic oscillator eigensolutions

4 2 0 2 4

2

4

20 exp / 2A

/ 2

21 exp / 2 2A

3 / 2

2 22 exp / 2 4 2A

5 / 2

2 33 exp / 2 8 12A

7 / 2

2 4 24 exp / 2 16 48 12A

9 / 2Energy 2 / 2

Classical turning points

The intersections of the parabola

and the dashed lines

give the “classical turning points”

where a classical mass of that energy turns round and goes back downhill

4 2 0 2 4

2

4

/ 2

3 / 2

5 / 2

7 / 2

9 / 2Energy 2 / 2