An Electron Trapped in A Potential Well Probability densities for an infinite well Solve...

An Electron Trapped in A Potential Well

Probability densities for an infinite well

2/2/,0)( LxLxU

2/,2/,)( LxLxxU

)()()(d

)(d

2 2

22

xExxUx

x

m

Solve Schrödinger equation

2/ ,2/)()(d

)(d

2 2

22

LxLxxExx

x

m

outside the well

2/ ,2/0)( LxLxx

)(xU

x2/Lo2/L

2 2

2

d ( )( ) / 2 / 2

2 d

xE x L x L

m x

inside the well

22

2

d ( )( ) / 2 / 2

d

xk x L x L

x

2

2

mE

k

An Electron Trapped in A Potential Well

kxBkxAx cossin)( The general solution is

)(xU

x2/Lo2/L

By the boundary conditions 0)2/cos()2/sin( kLBkLA 0)2/cos()2/sin( kLBkLA

0)2/cos()2/sin( kLBkLA

0)2/cos(2 kLB

0)2/sin(2 kLA nkL 0 ,6,4,2 Bn x)

L

nπ(Asin

x)L

nπ(Bcos 0 ,5,3,1 An

0)2/cos(or

0)2/sin(

kL

kL

An Electron Trapped in A Potential Well)(xU

x2/Lo2/L

0 ,6,4,2 Bn x)L

nπ(Asin

x)L

nπ(Bcos 0 ,5,3,1 An

Normalization

2/

2/1d)(*)(

L

Lxxx

For odd wave function )cos()( xL

nBxodd

/2 2 2

/2cos ( ) 1

L

L

nB x dx

L

L

B2

For even wave function

/2 2 2

/2sin ( ) 1

L

L

nA x dx

L

2A

L

2sin( )

nx

L L

2cos( )

nx

L L

1

2( ) cos

xx

L L

3

2 3( ) cos

xx

L L

2

2 2( ) sin

xx

L L

probability

2 2

2

d ( )( ) / 2 / 2

2 d

xE x L x L

m x

inside the well

22

2

d ( )( ) / 2 / 2

d

xk x L x L

x

2

2

mE

k

An Electron Trapped in A Potential Well)(xU

x2/Lo2/L

0)2/cos(2 kLB

0)2/sin(2 kLA nkL

L

nk

2

22222

22 mL

n

m

kE

1

2( ) cos

xx

L L

3

2 3( ) cos

xx

L L

2

2 2( ) sin

xx

L L

2

22

1 2mLE

2

22

2

2

mLE

2

22

3 2

9

mLE

energy distribution )(xU

x2/Lo2/L

0U

Barrier TunnelingBarrier Tunneling

Dividing the space into three parts: I) to the left of the barrier II) within the barrier; and III) to the right of the barrier

)0( )( 21 xeex ikxikxI

)( )( 5 Lxex ikxIII

22

02

d ( )( ) ( )

2 dII

II II

xU x E x

m x

2 0

2 2

2 ( )d ( )( ) 0

dII

II

m U Exx

x

20 /)(2 EUmII

0)(d

)(d 22

2

xkx

xII

II

xixiII

IIII eex 43)(IIi

xxII eex 43)(

The conditions for wave functions at the boundary are continuity.

Lx

III

Lx

II

IIIII

x

II

x

I

III

x

x

x

x

LL

x

x

x

x

d

)(d

d

)(d

)()(

d

)(d

d

)(d

)0()0(

00

LeU

E

U

ET 2

00

)1(16

Barrier TunnelingBarrier Tunneling

LeU

E

U

ET 2

00

)1(16

Tunneling current

Si(111) Surface

mm 11

Tetracene/Ag(110)

[001]

0.00

0.05

0.10

Hig

ht(

nm

)

0.04 nm

Wave (matter wave)

Uncertainty Principle

Wave Function

SchrSchrödinger’södinger’s EquationEquation

probability

Free ElectronsHydrogen atomHydrogen atom

p

h

Ev

h de Broglie relation,

de Broglie wavelength.

/ 2xx p / 2E t

)(0),( tkxieψtxΨ

),(*),(),(2

txψtxψtxΨ

Potential Well

Particle

The Nature of The Nature of MatterMatter

Energy quantization Probability

Barrier TunnelingBarrier Tunneling STM

Chapter 48

Atomic Structure

r

SchrSchrödinger’södinger’s EquationEquation

)()()()())(1

(2 2

22

2

2

rErrUrr

l

rr

rrm

r

zeU

2

04

1

2

0

1( ) ( )

4

zer r

r

0l

0/

30

1)( are

ar

2

20

0 me

ha

=0.529ÅBohr radius

0/230

2 1)( are

ar

drre

adVr ar 2/2

30

2 41

)( 0

drrp )(

0)41

( 2/230

0 readr

d ar

0ar

ground state (n=1)n=2

1l

4

2 2 20

1 1, 2, 3,

8n

meE n

h n

11, 13.6eVn E

12n

EE

n

ground state

Energy quantization

Principle quantum numberr

zeU

2

04

1

r

SchrSchrödinger’södinger’s EquationEquation

)()()()())(1

(2 2

22

2

2

rErrUrr

l

rr

rrm

2

0

1( ) ( )

4

zer r

r

The Uncertainty Principle

2xp x

2L

rpL

rΦx

The angular momentum-angle Uncertainty Relationship

L p r

r

SchrSchrödinger’södinger’s EquationEquation

Angular Momentum of Electrons in Atoms

)1( llL

)1( ,3 ,2 ,1 ,0 nlAngular momentum quantum number

)()()()())(1

(2 2

22

2

2

rErrUrr

l

rr

rrm

Angular momentum quantization

l 0 1 2 3 4 5

code s p d f g h

Using this labeling, we can express 1s for ground state (n=1, l=0). The first excited state has two designations: 2s (n=2, l=0) and 2p (n=2, l=1).

Angular momentum quantum number

l=0, 1, 2, …, n-1. Ex. n=1, l=0 for s state

n=2, l=0, 1. for l=0, s state and l=1, p state

Space quantization

)1(coscos 11

ll

m

L

L lz

2)1( ,1 llLl

θml=-l, -(l-1), …-1, 0, 1, …,(l-1), l

magnetic quantum number

0)1( ,0 llLl

z

,3 ,2 ,1 1

8 2220

4

nnh

meEn

n Principle quantum number

)1( llL )1( ,3 ,2 ,1 ,0 nl

l Angular momentum quantum number

ml=-l, -(l-1), …-1, 0, 1, …,(l-1), l

ml magnetic quantum number

)1(coscos 11

ll

m

L

L lz

,....,,, NMLK

,....,,, fdps

(n=1, l=0)The Ground State

ground state (n=1)

ml=0, is spherically symmetric.

drrea

dVr ar 2/230

2 41

)( 0

0)41

( 2/230

0 readr

d ar

0ar

The 2s State (n=2, l=0)

ml=0, is spherically symmetric.

An Excited State Of The Hydrogen Atom

The 2p State (n=2, l=1)

ml=-1, 0, +1 is not spherically symmetric.

Electron Spin

Pauli pointed out the need for a 4th quantum number in 1924—Spin quantum number

2

1sand)1( ssS

)1( llL

l Angular momentum quantum number

Spin magnetic quantum number2

1sm

)1( ,3 ,2 ,1 ,0 nl

s Spin momentum quantum number

The States of Atomic Hydrogen

The assembly of all hydrogen-atom states with the same principal quantum number n are said to form a shell. The collection of all states with the same value of the orbital angular momentum quantum number l is called a subshell.

For a certain angular momentum l, there are 2l+1 states, and consider spin there are 2(2l+1) states.

ml=-l, -(l-1), …-1, 0, 1, …,(l-1), l

12 l

)1(,2 ,1 ,0 nl

2

1sm

)1(

0

)12(2n

ln lZ

2[1 3 5 (2 1)]n 22n

)0(1 ,1 lsnstates 2

)0(2 ,2 lsn

states 2 6

)0(3 ,3 lsn2 6

)2,1,0,1,2(),2(3 lmld

10

states

Example

)1 ,0,1(),1 (2 lmlp

)1,0,1(),1(3 lmlp

ml=-l, -(l-1), …-1, 0, 1, …,(l-1), l

12 l

states

2

1sm

)1(

0

)12(2n

ln lZ

2[1 3 5 (2 1)]n 22n

The X-Ray Spectrum of Atoms

The Characteristic X-ray Spectrum

Atomic Magnetism

How to study the angular momentum properties of the atom ?

2

2r

el

L is the orbital angular momentum vector of the electron.

The component of z directionzlz L

m

e

2

L

l

2

2mr

m

e L

m

e

2

)1( llL

Lz=mlħ, ml=-l, -(l-1), ...-1, 0, +1, ..., +l

Atomic Magnetism

Lz=mlħ, ml=-l, -(l-1), ...-1, 0, +1, ..., +l

m

ehmm

m

eL

m

ellzlz

422

It can be expressed by Bohr magnetron µB

J/T10274.94

24m

ehB

Bllz m

We can express the magnetic dipole moment in terms of the Bohr magnetron

The energy of electron

,3 ,2 ,1

1

8 2220

4

n

nh

meEn

i fn nE E hv

If we were to place an atom having a magnetic dipole moment in a magnetic field , which we assume is in the z direction, the energy associated with the interaction between the atom and the magnetic field is

l

B

l lz z l B zU B B m B

That is, atoms with different values of ml have different energies in the field, which provides a way to determine their orbital angular momentum.

Atomic Magnetism

dBdW l )(

00

sin)(

dBdBW Ll

cosBL BL

The Stern-Gerlach ExperimentBU

zz BU

z

B

z

UF z

z d

d

d

d

there is no semiclassical corresponding. The full quantum mechanics gives

sz s Bm 0002319305.2sg

Atomic Magnetism

spin angular momentum

Bllz m angular momentum

Sodium atom (Z=11)

2

1 lslj

jmmJ jj ,1 ,0 ,

The Zeeman effect

Bsssz mg

Inverted population

Lasers and Laser Light

Stimulated

emissionEmission

Lasers and Laser Light

Four properties

1) Laser light is highly monochromatic.

2) Laser light is highly coherent

3) Laser light is highly directional

4) Laser light can be sharply focused.

ExercisesP1099 15, 16, 17P1100 28, 30