Hydrogen Atom and QM Hydrogen Atom and QM in 3-Din 3-D

1.1. HW 8, problem 6.32 and A HW 8, problem 6.32 and A review of the hydrogen atomreview of the hydrogen atom

2.2. Quiz 10.30Quiz 10.30

3.3. Topics in this chapter:Topics in this chapter: The hydrogen atom The hydrogen atom The Schrödinger equation in 3-D The Schrödinger equation in 3-D The Schrödinger equation for The Schrödinger equation for

central forcecentral force

Today

HW 8, problem 6.32HW 8, problem 6.32Following discussion in the textbook, page 196 – 199, to the left of the step ( x < 0 ):

Apply smoothness condition:

k k 'B

k k '

So:

To the left of the step ( x < 0 ):

1 1 ikx ikxinc refl e Be

1 B C

Solve for B and C:

with

342 and 2k mE k' m E E

2ik' x

trans Ce

1 20 0x x

1 2

0 0x x

d d

dx dx

1k B k' C

1 3 and 4 3B C

1

3ikx

refl e 4

3ik' x

trans e

2 2

2 2

1 3 1

1 9refl

inc

R

HW 9, problem 6.32HW 9, problem 6.32Conditions given:

Because you will have to “tunnel” to Jupiter.

02 2

0 0

16 1L m E UE E

T eU U

116 10 mL Rectangular barrier :

80 4 10 J/kgU m

The particle (you):

65 kgm 4 m/sv

21520 J

2E mv 8 10

0 4 10 J/kg 2 6 10 JU m .

0E U

The probability that you end up there is:

Which is:522 10 0T e

The hydrogen atomThe hydrogen atomThe electrical potential:

To solve the Schrödinger equation in a 3-D polar system is trivial. Let’s start from one of its solutions, the energy level:

4

2 220

2

1, 1 2 3

2 4

13 6 eV

n

meE n , , ,...

n

.

n

2 2

1 1 13 6 eV

photon atom,i atom, f

f i

E E E

.n n

2 2

7 12 2

1

13 6 eV 1 1

1 1 1 097 10 m

photon

f i

f i

E

hc

.

hc n n

.n n

2

0

1

4

eU r

r

Example

7.2

The Schrödinger equation, from 1-The Schrödinger equation, from 1-D to 3-DD to 3-D

1-D:

22

22

d xU x x E x

m dx

3-D:

2 2 2 2

2 2 22x, y,z U x, y,z x, y,z E x, y,z

m x y z

If choose Cartesian coordinates:

2 2 22

2 2 2x y z

Or from:

2

2

2x, y,z U x, y,z x, y,z E x, y,z

m

In a more general case, the coordinate is represented by a vector

r

The Schrödinger equation in 3-D:

2

2 r r r r2

U Em

Now the normalization condition is

2

all space

r 1dV

For bound states, the standing wave is a 3-D standing wave, with energy quantized by 3 quantum numbers, each for one dimension.

The 3-D infinite well, just as an The 3-D infinite well, just as an exampleexample

2 2 2 2

2 2 22x, y,z U x, y,z x, y,z E x, y,z

m x y z

and because

2 2 22

2 2 2x y z

Inside the box, if we express the wave function like this:

This leads to three solutions:

The Schrödinger equation becomes:

And the energy quantization:

This is only possible if

0 0 0 0r

otherwisex y zx L , x L , x L

U

Here Cartesian coordinates are natural choice, so:

x, y,z F x G y H z

2 2 2

2 2 2 2

1 1 1 2F x F y H z mE

F x x G y y H z z

0x, y,z

r 0U

2

2

1x

d F xC

F x dx

2

2

1y

d F yC

G y dy

2

2

1z

d H zC

H z dzan

dand

sin xx

x

nF x A x

L

sin y

yy

nG y A y

L

sin z

zz

nH z A z

L

2 22 2 2 2

2 2 2 2

2yx z

x y z

nn n mE

L L L

22 2 2 2

2 2 2 2x y z

yx zn ,n ,n

x y z

nn nE

L L L m

Discussion about the energy levels and their Discussion about the energy levels and their wave functionswave functions

Energy levels:

There are three quantum number that define an energy state and its wave function.

sin sin sinx y z

yx zn ,n ,n

x y z

nn nx, y,z A x y z

L L L

The ground state:

22 2 2 2

2 2 2 2x y z

yx zn ,n ,n

x y z

nn nE

L L L m

Eave function:

x y zn ,n ,n

11 1x y zn ,n ,n , , Can anyone be 0?2 2

111 2 2 2

1 1 1

2, ,x y z

EL L L m

111 sin sin sin, ,x y z

x, y,z A x y zL L L

Now a special symmetric case, a cube:

x y zL L L L

2 2

2 2 222x y zn ,n ,n x y zE n n n

mL

2 2

111 23

2, ,EmL

and

Now take

2 2

2 2 222x y zn ,n ,n x y zE n n n

mL

for 2 11x y zn ,n ,n , , 1 2 1x y zn ,n ,n , ,

and

11 2x y zn ,n ,n , ,

The energy levels are the same: 2 2

2 11 1 2 1 11 2 26

2, , , , , ,E E EmL

While the wave functions are not the same:

2 11

2sin sin sin, , A x y z

L L L

1 2 1

2sin sin sin, , A x y z

L L L

11 2

2sin sin sin, , A x y z

L L L

This is called degeneracy. The energy levels are called degenerated.

leads to energy levels degeneration. For example, for:

Energy levels degenerated and splitting Energy levels degenerated and splitting (symmetry broken)(symmetry broken)

The symmetric case:x y zL L L L

2 11x y zn ,n ,n , , 1 2 1x y zn ,n ,n , , 11 2x y zn ,n ,n , ,

2 2

2 11 1 2 1 11 2 26

2, , , , , ,E E EmL

2 11

2sin sin sin, , A x y z

L L L

1 2 1

2sin sin sin, , A x y z

L L L

11 2

2sin sin sin, , A x y z

L L L

The wave functions are different, but symmetric.

x y y z

E

11 1, ,

2 11, , 1 2 1, , 11 2, ,

Plot the energy levels

When the symmetry is broken: for example: energy level splits

0 9x y zL L L, L . L

Energy levels degenerated and splitting Energy levels degenerated and splitting (symmetry broken)(symmetry broken)

E

11 1, ,

2 11, , 1 2 1, , 11 2, ,

splitting

0 9x y zL L L, L . L x y zL L L L Symmetric: Symmetry broken:

2 2

111 23

2, ,EmL

2 2

2 11 1 2 1 11 2 26

2, , , , , ,E E EmL

2 2 2 2

111 2 2 2

12 3

0 9 2 2, ,E. mL mL

2 2 2 2

2 11 1 2 1 2 2 2

15 6

0 9 2 2, , , ,E E. mL mL

2 2 2 2 2

11 2 2 2 2

22 6

0 9 2 2, ,E. mL mL

11 2 2 11 1 2 1, , , , , ,E E E

11 1, ,

2 11, , 1 2 1, ,

11 2, ,

degenerated

Reveals more details

The Schrödinger equation for The Schrödinger equation for central forcecentral force

U U rCentral force:For example:

2

0

1

4

eU r

r

the force:2

20

1F r

4

e ˆr

Polar coordinates are a natural choice:

2

sin cos

sin sin

cos

sin

x r

y r

z r

dV r drd d

22 2 2

2 2

1csc sin cscr

r r r

2

2

2r, , U r r, , E r, ,

m

with:

Solve Schrödinger equation for Solve Schrödinger equation for central forcecentral force

Separate variables:

2

22

2R R R

d d mrr r E U r r C r

dr dr

2

2

2r, , U r r, , E r, ,

m

Becomes three equations:

R Θ Φr, , r

radial equation:

polar equation:

2Θsin sin sin Θ Θ

ddC D

d d

azimuthal equation:

2

2

ΦΦ

dD

d

Both and are

constants.C D

The solution to the azimuthal equation is (the simplest): Φ 0 1 2 3lim

le m , , , ,...

The z component of the particle’s angular moment is quantized:

zL

z lL m

L

0 1 2 3lm , , , ,... is called the magnetic quantum number.

2lm D

Where is called

the orbital quantum number, and

The angular momentum and its The angular momentum and its quantum numbersquantum numbers

Leads to the quantize the magnitude of the particle’s momentum:

The solution to the polar equation: 2 2Θsin sin sin Θ Θl

ddC m

d d

1L l l

Because

0 1 2 3l , , , ,...

zL L we have 0 1 2 3lm , , , ,..., l

l

L

lm

zL

0 1 2

20

0

0

0 1,

0,

60 1 2, , 0 2, ,

Example 7.3, 7.4 on black board.

1C l l

The shape of an atom of central forceThe shape of an atom of central forceThe angular probability density for a central force:

The radial equation of a central The radial equation of a central forceforce

2

22

2R R R

d d mrr r E U r r C r

dr dr

radial equation:

This leads to the solution for energy levels, and the principal quantum number. Degenerate (only depends on n, not l and ml )

2

0

1

4

eU r

r

Here one needs to know the potential explicitly. Assume hydrogen atom:

4

2 220

2

1, 1 2 3

2 4

13 6 eV

n

meE n , , ,...

n

.

n

The relationship of the three quantum numbers (magnetic, orbital and principal):

0 1 2 3lm , , , ,..., l

0 1 2 3 1l , , , ,...,n

1 2 3n , , ,...

rearrange: 22

22 2

1R R R R

2 2

l ld dr r r U r r E r

mr dr dr mr

KEradial KErotationPE

The electron “cloud” in the The electron “cloud” in the hydrogen atomhydrogen atom

0l 1l

1l

2l

Electron probability density.

Surfaces of constant probability density.

Radial probability: the “size” of Radial probability: the “size” of the atomthe atom

2 2P r r R r

Radial probability:

The Bohr radius: 2

00 2

40 0529 nma .

me

Example 7.6

Review questionsReview questions

What are the steps in working out the What are the steps in working out the Schrödinger equation for hydrogen Schrödinger equation for hydrogen atom? atom?

How do you connect the quantum How do you connect the quantum numbers introduced in the solutions with numbers introduced in the solutions with those learned from a chemistry class?those learned from a chemistry class?

Preview for the next class Preview for the next class (10/28)(10/28)

Text to be read:Text to be read: 8.1, 8.2 and 8.38.1, 8.2 and 8.3

Questions:Questions: What had the Stern-Gerlach experiment been What had the Stern-Gerlach experiment been

designed to prove? What it actually proved? designed to prove? What it actually proved?

Homework 11, due by 11/6Homework 11, due by 11/6

Problems 7.37, 7.38 and 7.45 on page Problems 7.37, 7.38 and 7.45 on page 281.281.