summer12 lectures lect11

7/30/2019 .. Summer12 Lectures Lect11

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Particles in 3D Potentials

and the Hydrogen Atom

P(r)

0 4a00

1

r

r = a0z

x

L

LL

2222

2

8zyxnnn nnn

mL

hE

zyx

)()()(),,( zyxzyx oa/r

3o

ea

1)r(

2

613

n

eV.En


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Overview

3-Dimensional Potential Well Product Wavefunctions

Concept of degeneracy

Early (Incorrect!) Models of the Hydrogen Atom Planetary Model

Schrdingers Equation for the Hydrogen Atom Semi-quantitative picture from uncertainty principle

Ground state solution* Spherically-symmetric excited states (s-states)*

*contain details beyond what we expect you to learn here


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Quantum Particles in 3D Potentials

One consequence of confining a quantum particle in two orthree dimensions is degeneracy -- the occurrence ofseveral quantum states at the same energy level.

So far, we have considered quantum particlesbound in one-dimensional potentials. Thissituation can be applicable to certain physicalsystems but it lacks some of the features ofmany real 3D quantum systems, such asatoms and artificial quantum structures:

(www.kfa-juelich.de/isi/)

A real (3D)quantum dot

To illustrate this important point in a simple system, weextend our favorite potential -- the infinite square well --

to three dimensions.


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Particle in a 3D Box (1)

outside box, x or y or z < 0

outside box, x or y or z > L

0 inside boxU(x,y,z) =

Lets solve this SEQ for the particle in a 3D box:

y

The extension of the Schrdinger Equation (SEQ) to 3Dis straightforward in cartesian (x,y,z) coordinates:

E)z,y,x(Udz

d

dy

d

dx

dm

2

2

2

2

2

22

2

),,( zyx where

This simple U(x,y,z) can be separatedA special feature

U(x,y,z) = U(x) + U(y) + U(z)

z

x

L

LL

Kinetic energy term in theSchrdinger Equation

2222

1zyx ppp

mlike

Easy to see in this case since U = 0 or

Very useful in general as illustratedin the solution on the following slide


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Particle in a 3D Box (3) So, finally, the eigenstates and associated

energies for a particle in a 3D box are: z

x

y L

L

L

zL

ny

L

nx

L

nN z

yx sinsinsin

where nx,ny, and nzcan each have values 1,2,3,.

This problem illustrates 2 important new points.

(1) Three quantum numbers (nx,n

y,n

z) are needed to completely

identify the state of this three-dimensional system.

(2) More than one state can have the same energy:Degeneracy.

Degeneracy reflects an underlying symmetry in U(x,y,z)

3 equivalent directions

2222

2

8zyxnnn nnn

mL

hE

zyx


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Consider a particle in a two-dimensional (infinite) well, with Lx = Ly.

1. Compare the energies of the (2,2), (1,3), and (3,1) states?

a. E(2,2) > E(1,3) = E(3,1)b. E(2,2) = E(1,3) = E(3,1)

c. E(1,3) = E(3,1) > E(2,2)

2. If we squeeze the box in the x-direction (i.e., Lx < Ly) compareE(1,3) with E(3,1):

a. E(1,3) < E(3,1)

b. E(1,3) = E(3,1)c. E(1,3) > E(3,1)

Lecture 11, Act 1


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Consider a particle in a two-dimensional (infinite) well, with Lx = Ly.

1. Compare the energies of the (2,2), (1,3), and (3,1) states?

a. E(2,2) > E(1,3) = E(3,1)b. E(2,2) = E(1,3) = E(3,1)

c. E(1,3) = E(3,1) > E(2,2)

2. If we squeeze the box in the x-direction (i.e., Lx < Ly) compareE(1,3) with E(3,1):

a. E(1,3) < E(3,1)

b. E(1,3) = E(3,1)c. E(1,3) > E(3,1)

Lecture 11, Act 1 - Solution

E(1,3) = E(1,3) = E0 (12 + 32) = 10 E0

The tighter confinement along x will increasethe contribution to E. The effect will be

greatest on states with greatest nx:

E(2,2) = E0 (22 + 22) = 8 E0

2

2 2

( ) (1,3) (3,1)2 4 4 9; 36 18x y

n n x y

y

hE n n E E

mL

Example: Lx = Ly/2

222

2 28x y

yxn n

x y

nnhE

m L L


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z

x

yL

L

L

D=1

6Eo (2,1,1) (1,2,1) (1,1,2) D=3

(1,1,1)3Eo

(nx,nynz) 2z2y2x2

2

nnn nnnmL8

hE

zyx

nx,ny,nz = 1,2,3,...

Energy levels (1) Now back to a 3D cubic box:

Show energies and label (nx,ny,nz) for the first

11 states of the particle in the 3D box, andwrite the degeneracy D for each allowed energy.

Use Eo= h2/8mL2.

E


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z

x

yL

L

L

D=1

6Eo (2,1,1) (1,2,1) (1,1,2) D=3

(1,1,1)3Eo

(nx,nynz) 2z2y2x2

2

nnn nnnmL8

hE

zyx

nx,ny,nz = 1,2,3,...

Energy levels (1) Now back to a 3D cubic box:

Show energies and label (nx,ny,nz) for the first

11 states of the particle in the 3D box, andwrite the degeneracy D for each allowed energy.

Use Eo= h2/8mL2.

E

9Eo

11Eo

12Eo


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For a symmetric cube infinite box, we just saw that the 5th energystate has an energy of 12 E0 and a degeneracy of 1, with quantumnumbers (2,2,2).

1. What is the energy of the next energy level?

a. 13E0

b. 14E0

c. 15E0

2. What is the degeneracy of this energy level?

a. 2

b. 4

c. 6

Lecture 11, Act 2

L 11 2 l


12/33

For a symmetric cube infinite box, we just saw that the 5th energystate has an energy of 12 E0 and a degeneracy of 1, with quantumnumbers (2,2,2).

1. What is the energy of the next energy level?

a. 13E0

b. 14E0

c. 15E0

2. What is the degeneracy of this energy level?

a. 2

b. 4

c. 6


E(1,2,3) = E0 (12 + 22 + 32) = 14 E0

Any ordering of the quantum numbers1, 2, and 3 will give the same total

energy. There are 6 possible ways toorder them: (1,2,3), (1,3,2), (2,1,3),(3,1,2), (2,3,1), (3,2,1).

Note: For this system, 6 is the

maximum possible degeneracy.

E l l (2)


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E

3Eo

6Eo

9Eo

11Eo

(nx,ny,nz)

z

x

y

L1

L2 > L1

L1

Energy levels (2)

Now consider a non-cubic box:

Assume that the box is stretched only along

the y-direction. What do you think willhappen to the cubes energy levels below?

2y22

22

z

2

x21

2

nnn nmL8

hnn

mL8

hE

zyx

E l l (2)


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(1) The symmetry of U isbroken for y, so the three-fold degeneracy is loweredatwo-fold degeneracy remains

due to 2 remaining equivalentdirections, x and z.

(1,1,1)D=1

(1,2,1) D=1D=2(2,1,1) (1,1,2)

(2) There is an overall loweringof energies due to decreased

confinement along y.

E

3Eo

6Eo

9Eo

11Eo

(nx,ny,nz)

Energy levels (2) Now consider a non-cubic box:

Assume that the box is stretched only along

the y-direction. What do you think willhappen to the cubes energy levels below?

z

x

y

L1

L2 > L1

L1

2y22

22

z

2

x21

2

nnn nmL8

hnn

mL8

hE

zyx

Figure outnext level.


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Another 3D System: The Atom-electrons confined in Coulomb field of a nucleus

Geiger-Marsden (Rutherford) Experiment (1911):

Measured angular dependence of particles(He ions) scattered from gold foil.

Mostly scattering at small angles supported the plum pudding model. But

Occasional scatterings at large angles Something massive in there!

Au

v

a nucleus!

Discrete Emission and Absorption spectra

When excited in an electrical discharge, atomsemitted radiation only at discrete wavelengths

Different emission spectra for different atoms

Early hints of the quantum nature of atoms:

(nm)

Atomic hydrogen

Conclusion: Most of atomic mass isconcentrated in a small region of the atom

R th f d E i t


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Rutherford Experiment

At Cl i l Pl t M d l


17/33

Atoms: Classical Planetary Model

Classical picture: negatively charged objects(electrons) orbit positively charged nucleusdue to Coulomb force.

There is a BIG PROBLEM with this:

As the electron moves in its circular orbit,it is ACCELERATING.

As you learned in Physics 212, acceleratingcharges radiate electromagnetic energy.

Consequently, an electron wouldcontinuously lose energy and spiral into thenucleus in about 10-9 sec.

+Ze

-eF

(An early model of the atom)

The planetary model doesnt lead to stable atoms.


18/33


19/33


20/33


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P t ti l E f th H d At


22/33

How do we describe the hydrogenatom quantum mechanically?

We need to specify U, the potentialenergy of the electron:

We assume that the Coulomb forcebetween the electron and the nucleusis the force responsible for binding

the electron in the atom

This spherically symmetric problem can be solved in

spherical coordinates.

Potential Energy for the Hydrogen Atom

We cannot separate this potential energy in xyz coordinatesas we did for the 3D box,U(x,y,z) = U(x) + U(y) + U(z)

,which led to the product wavefunctions: x,y,z x y z

r

e)r(U

2

r

U(r)

0

229

0

C/Nm1094

1

B t fi t: H d t lit ti l


23/33

But first: Hydrogen atom qualitatively

Why doesnt the quantum electron collapse into thenucleus, where its potential energy is lowest?

The more confined it gets, the bigger p spread it has, fromHeisenberg Uncertainty. More p2/2m means more KE. So theres a tradeoff between lowering PE and raising KE.

About what spread a0 minimizes KE+PE? Roughly,

Take derivative,find a0 tominimize E:

2 2 2

202 2 o

p e

E KE PE PEm ma a

2 2 4

2 2

2 4

2

so 2

so 2

o

m ea PE KE

m e

m e

E

virial theoremExact, generally

true for atoms

Doing the same thing carefully ends upwith the same result!

So the general energy and size scale

of an atom are fixed bythe Heisenberg Uncertainty relation.

Lecture 11 Act 3


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Consider an electron around a nucleus that has two protons(and two neutrons, like an ionized Helium atom).

1. Compare the effective Bohr radius a0,He with the usual Bohrradius for hydrogen, a0:

a. a0,He > a0

b. a0,He = a0

c. a0,He < a0

2. What is the ratio of ground state energies E0,He/E0,H?

a. E0,He/E0,H = 1

b. E0,He/E0,H = 2c. E0,He/E0,H = 4

Lecture 11, Act 3

Lecture 11 Act 3 Solution


25/33

Consider an electron around a nucleus that has two protons(and two neutrons, like an ionized Helium atom).

1. Compare the effective Bohr radius a0,He

with the usual Bohr radiufor hydrogen, a0:

a. a0,He > a0

b. a0,He = a0

c. a0,He < a0

2. What is the ratio of ground state energies E0,He/E0,H?

a. E0,He/E0,H = 1

b. E0,He/E0,H = 2c. E0,He/E0,H = 4


Look at how a0 depends on the charge:

Clearly the electron will be more tightly bound, so|E0,He| > |E0,H| . How muchmore tightly?

Look at E0:

In general, for a hydrogenic

atom with Z protons:

2 2

0

0 0 ,2(2 ) 2

He

aa a

m e m e e

This makes sense more charge

stronger attraction electron sitscloser to the nucleus

2

0, 0 , 0 ,

(2 )4

2 2( / 2)H He H

o o

e e eE E E

a a

20,2

, 2 213.6H

n Z

E Z

E Z eVn n

Depends

on

whichstate

Product Wavefunction in Spherical Coordinates


26/33

Product Wavefunction in Spherical Coordinates The potential is spherically symmetric, i.e., U(r )

depends only on the radius. Therefore theproblem is separable and the solution to the SEQin spherical coordinates is a product wave

function of the form:

x

yzr

)r(RE)r(Rr

er

rr

1

m20nn0n

2

2

22

KE term PE term

)r(R),,r( 0nf

s-stateWavefunction:

),()(),,( ff lmn ln lm YrRr with quantum numbers: n l andm

principal orbital magnetic(angular momentum)

Here Rnlis the radial part, andYlmare spherical harmonics. TheSchrdinger equation in spherical coordinates is complicated. Anappreciation of the problem can be gained by considering only sphericallsymmetric states (wavefunctions with no angular dependence). For thes

s-states, l = 0 and m = 0, and the radial SEQ takes the form:

Radial Eigenstates of H:


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Radial Eigenstates of H:

s-state wavefunctionsThe zeros in the subscripts below are a reminder that these are

states with zero angular momentum.

The radial eigenstates, Rno(r), for the electron in theCoulomb potential of the proton are plotted below:

03/

2

00

0,33

223)(ar

ear

arrR

02/

00,2

21)(

are

a

rrR

0/

0,1 )(ar

erR

r

R30

r 15a000

r

R20

10a000

R10

00

4a0

22o

2

n

n

eV6.13

n

1

a2

eE

22

0em

a

= Bohr radius = 0.053 nm

Plug these into radial SEQ (Appendix)

Probability Density of Electrons


28/33

Probability Density of Electronsrobability density = Probability per unit volume = 2 Rn02 for s-state

The density of dots plotted below is proportional to Rn02.

r

R10

0

0

4a0 r

R20

10a0

0

0

1s state 2s state

R di l P b bilit D iti f t t


29/33

Radial Probability Densities for s-states Summary of wave functions and radial probability

densities for some s-states:

r

R20

0 10a0

0

0 10a00 r

P20

P10

0 4a00 r

r0 4a0

0

R10

radial wave functions radial probability densities, P(r)

R30

r0 15a00

0 20a0r

P30

These P(r) include both |2(r)| and a factor of r2

because bigger r shells have more volume.

Optical Transitions in H Example


30/33

Optical Transitions in H -- Example

An electron, initially excited to the n = 3energy level of the hydrogen atom, falls

to the n = 2 level, emitting a photon inthe process. What are the energy andwavelength of the photon emitted?

-15

-10

-5

0

0 5 10 15 20

r/a0

E (eV)U(r)

-15

0

0 20

E2

E1

E3

Answer:

(nm)

Atomic hydrogen

Discharge tubes

Optical Transitions in H Example


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-15

-10

-5

0

0 5 10 15 20

r/a0

E (eV)U(r)

-15

0

0 20

E2

E1

E3

2

13.6 eVn

En

2 2

1 113.6 eV

i fn n

i f

En n

Therefore:

3 2

1 1

13.6 eV 1.9eV9 4ph otonE E

1240eV nm656nm

1.9eVphoton

hc

E

(nm)

Atomic hydrogen

Optical Transitions in H --Example

An electron, initially excited to the n = 3energy level of the hydrogen atom, falls

to the n = 2 level, emitting a photon inthe process. What are the energy andwavelength of the photon emitted?

You will experimentally measure several transitions in Lab.


32/33

The Radial Equation

222

2 2

2 2 20 0

110

22 1, , , 1,2,3,

2

n

n n

l ldR adr b R

dr rrdrr

mEZ Za b E n

a ma n

22

2

00

sin sin 1 sin 0

, ,

0,1, 2, , 1 , 1, , 1,

0 0; Constant

lm

Y Yl l Y

Y Y

l n m l l l l

l m Y

The Angular Equation


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3/ 2

0

00 0

12 1

1

, , ,

2 1,

!

nlm nl lm

Zr

nal

nl nl nl n

n ln l xl x

nl n l

r R Y

Z ZR r C r e L r C

na nan

dL x x e x e

dx

The Wave Function

summer12 lectures lect11

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