Physics 2D Lecture SlidesDec 1

Vivek SharmaUCSD Physics

QM in 3 Dimensions • Learn to extend S. Eq and its

solutions from “toy” examples in 1-Dimension (x) → three orthogonal dimensions (r ≡x,y,z)

• Then transform the systems – Particle in 1D rigid box 3D

rigid box– 1D Harmonic Oscillator 3D

Harmonic Oscillator • Keep an eye on the number

of different integers needed to specify system 1 3 (corresponding to 3 available degrees of freedom x,y,z)

y

z

x

ˆˆ ˆr ix jy kz= + +

Quantum Mechanics In 3D: Particle in 3D BoxExtension of a Particle In a Box with rigid walls

1D → 3D⇒ Box with Rigid Walls (U=∞) in

X,Y,Z dimensions

yy=0

y=L

z=L

z

Ask same questions:• Location of particle in 3d Box• Momentum • Kinetic Energy, Total Energy• Expectation values in 3D

To find the Wavefunction and various expectation values, we must first set up the appropriate TDSE & TISE

U(r)=0 for (0<x,y,z,<L)

x

The Schrodinger Equation in 3 Dimensions: Cartesian Coordinates

2 2 22

2 2

2 2 2 2 2

22

2

22

2

2 2

Time Dependent Schrodinger Eqn:( , , , )( , , , ) ( , , ) ( , ) .....In 3D

2

2 22

2

x y z tx y

x y

z t U x y z x t im t

m x m y m

z

mSo

∂ ∂ ∂∇

∂Ψ− ∇ Ψ + Ψ =

∂

− ∇ = +⎛ ⎞ ⎛ ⎞∂ ∂ ∂− − + −⎜ ⎟ ⎜ ⎟∂ ∂ ∂⎝ ⎠ ⎝

+∂

⎠

= +∂ ∂

x

2

x x [K ] + [K ] + [K ] [ ] ( , ) [ ] ( , ) is still the Energy Conservation Eq

Stationary states are those for which all proba

[ ]

=

bilities

so H x t E

K

x t

z⎛ ⎞

=

Ψ

⎟⎠

=

⎜⎝

Ψ

-i t

are and are given by the solution of the TDSE in seperable form: = (r)eThis statement is simply an ext

constant in time

(ension of what we

, derive

, , ) ( , )d in case of

x y z t r t ωψΨ = Ψ1D

time-independent potential

y

z

x

Particle in 3D Rigid Box : Separation of Orthogonal Spatial (x,y,z) Variables

1 2 3

1

2

2 3

2

in 3D:

x,y,z independent of each ( , , ) ( ) ( ) ( )and substitute in the master TISE, after dividing thruout by = ( ) ( ) (

- ( , , ) ( ,

other , wr

, ) ( , ,

)and

) ( ,

ite

, )

n

2mx y z

TISE x y z U x y z x y z E x y

x y zx y

z

zψ ψ ψ ψ

ψ ψ ψ

ψ

ψ

ψ ψ∇

=

+ =

221

21

22222

232

3

21

2

2

( )12 ( )

This can only be true if each term is c

oting that U(r)=0 fo

onstant for all x,y,z

( )12 ( )

(2

r (0<x,y,z,<L)

( )12 (

)

z E Constm z z

ym

xm x x

xm

y yψ

ψψ

ψ

ψ

ψψ⎛ ⎞∂

− +⎜ ⎟⎛ ⎞∂

+ − =

⇒

⎛ ⎞∂−⎜ ⎟∂⎝ ⎠∂⎝ ⎠

=

⇒

−

⎜ ⎟⎝ ⎠

∂

∂

223

3 32

222

2 21 12 2

1 2 3

) ( ) ;

(Total Energy of 3D system)

Each term looks like

( ) ( ) ;2

With E

particle in

E E E=Constan

1D box (just a different dimension)

( ) ( )2

So wavefunctions

t

z E zm z

yy

E x Ex

ym

ψ ψ ψ ψ ψ∂− =

∂−

∂

=

==∂

+ +

∂

3 31 2 21must be like ,( ) sin x ,( ) s ) s nin ( iyy kx k z k zψψ ψ∝ ∝∝

Particle in 3D Rigid Box : Separation of Orthogonal Variables

1 1 2 2 3 3

i

Wavefunctions are like , ( ) sin

Continuity Conditions for and its fi

( ) sin y

Leads to usual Quantization of Linear Momentum p= k .....in 3D

rst spatial derivative

( )

s

sin x ,

x

i i

z k z

n k

x

L

yk

p

k ψ

ψ π

ψ

π

ψ∝ ∝

⇒

=

∝

=

1 2 3

2

2

1 3 1

2 2 22

2

2 3 ; ;

Note: by usual Uncertainty Principle argumen

(n ,n ,n 1,2,3,.. )

t neither of n , n ,n 0! ( ?)

1Particle Energy E = K+U = K +0 = )2

(m 2

(

zy

x y z

n

why

p nL

nmL

p nL L

p p p

π

π

π⎛ ⎞ ⎛ ⎞= = ∞⎜ ⎟ ⎜⎛ ⎞= ⎜ ⎟⎝ ⎠

⎟⎝ ⎠ ⎝

=

⎠

+ + = 2 2 21 2 3

2

1 2 3

2

1

Ei

3

-

3

1

)

Energy is again quantized and brought to you by integers (independent)and (r)=A sin (A = Overall Normalization Cosin y

(r)

nstant)

(r,t)= e [ si

n ,n , nsin x

sin x ysn in ]t

k

n n

k

A k k

k

k

z

z

ψ

ψ

+ +

=ΨE-i

et

Particle in 3D Box :Wave function Normalization Condition

3

*

1

1

21

x,y,

E E-i -i

2

E Ei i*2

2 22

2

3

*3

z

2

(r) e [ sin y e

(r) e [ s

(r,t)= sin ]

(r,t)= sin ]

(r,t)

sin x

sin x

sin x

in y e

[ si

Normalization Co

(r,t)= sin ]

ndition : 1 = P(r)dx

n y

dyd

1

z

t t

t t

k z

k

k

k

A k

A k

A k zk k

A

z

ψ

ψ

Ψ

Ψ

Ψ

⇒

Ψ

=

=

=

∫∫∫L L L

2

3 3E

2 2 21 2 3

x=0 y=

2 2 -

1

z

3

i

0

2

=0

sin x dx

s

sin y dy sin z dz =

(

2 2 2

2 2 an r,t)=d [ s sinii ex yn ] nt

L

k

L LA

A kL

k k k

k zL

⎛ ⎞⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠⎝ ⎠

⎡ ⎤ ⎡ ⎤⇒ = ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦Ψ

∫ ∫ ∫

Particle in 3D Box : Energy Spectrum & Degeneracy

1 2 3

2 22 2 2

n ,n ,n 1 2 3 i

2 2

111 2

2 2

211 121 112 2

2

3Ground State Energy E2

6Next level 3 Ex

E ( ); n 1, 2, 3... , 02

s

cited states E = E E2

configurations of (r)= (x,y,z) have Different ame energy d

i

mL

mL

n n n nmL

π

π

ψ ψ

π= + + = ∞ ≠

=

⇒ = =

⇒ egeneracy

yy=L

z=Lz

xx=L

2 2

211 121 112 2

Degenerate States6E

= E E2mLπ

= =

x

y

z

E211 E121 E112ψ

E111

x

y

z

ψ

Ground State

Probability Density Functions for Particle in 3D Box

Same Energy Degenerate StatesCant tell by measuring energy if particle is in

211, 121, 112 quantum State

Source of Degeneracy: How to “Lift” Degeneracy • Degeneracy came from the

threefold symmetry of a CUBICAL Box (Lx= Ly= Lz=L)

• To Lift (remove) degeneracy change each dimension such that CUBICAL box Rectangular Box

• (Lx≠ Ly ≠ Lz)• Then

2 22 2 2 231 2

2 2 22 2 2x y z

nn nEmL mL mL

ππ π⎛ ⎞⎛ ⎞ ⎛ ⎞= + +⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠ ⎝ ⎠

Ener

gy

The Hydrogen Atom In Its Full Quantum Mechanical Glory

2 2 2

By example of particle in 3D box, need to use seperation of variables(x,y,z) to derive 3 in

1 1

This approach willdependent d

( ) M

iffer

ore compli

ential. eq

cated form of U than bo

get

x

ns.

U rr x y z

∝ = ⇒+ +

2 22

2

To simplify the situation, use appropriate variablesIndependent Cartesian (x,y,z) Inde. Spherical Polar (r, ,

very ugly since we have a "conjoined triplet"

Instead of writing Laplacian

)

x y

θ φ

∂ ∂∇ = +

∂ ∂

→

2

2

2 2

2

2 2 2

2

2

2

22

2 2 2

22

2

, write

1sin

TISE for (x,y,z)= (r, , ) become

1r

1 (r, , ) (r, , )r

s

1 2m+

1= s

(E-U(r))

insi

si

n

1 sins

(r, , )n

in

r

r

rr

z

r

r

rr r

r θθ

θ θ θ

θθ θ

φ

ψ ψ

θψ θ φ ψ θ φ

ψ θ φ

θ φ

θ φ

∂ ∂⎛ ⎞⎜ ⎟∂ ∂⎝ ⎠

∂ ∂⎛

∂ ∂⎛

⎞⎜ ⎟∂ ∂

∂+

∂

∂∂

∂∂

⎞∇ + +⎜ ⎟∂ ∂⎝ ⎠

∂ ∂⎛ ⎞+ +⎜ ∂ ⎠⎝⎟∂ ⎝⎠ !!!! fun!!!

(r, , ) =0 ψ θ φ

r

2

( ) kZeU rr

=

Spherical Polar Coordinate System

2

( sin )Vol

( )( ) = r si

ume Element dV

ndV r d rd dr

drd dθ φ θ

θ θ φ

=

Don

’t P

anic

: Its

sim

pler

than

you

thin

k ! 2

2

2 2 22

2

2 2

2 2

2

1 2m+ (E-U (r))sin

T ry to free up las

1 (r, , ) =0 r

all except

T his requires m ulti

t term fro

plying thruout by sin

si

1 sinsi

si si

m

n

n

n

rr

r r

r

r

rr r

ψ ψ ψ ψ θ φ

φ

θ

ψ

θθ θ θ

θθ

θ

θ

φ∂ ∂⎛ ⎞

⎜ ⎟∂ ∂⎝ ⎠

⇒

∂ ∂⎛ ⎞⎜ ⎟∂ ∂⎝ ⎠

∂ ∂⎛ ⎞+ +⎜ ⎟∂ ∂⎝ ⎠

∂+

∂

∂∂

2

2

2 2 2

2

2m ke+ (E+ )r

(r, , )=R (r). ( ). ( ) P lug it in to the T ISE above & divide thruout by (r, , )=R (r). ( ).

sin =0

For Seperation of V ariables, W rite

( , , )r

N ote tha

(

t :

)

n r

r

φψ θ φ θ φ

ψ ψ θ ψ

ψ θ φ

φ

θ

θ

θ

φ

θ∂⎛ ⎞ +⎜ ⎟∂⎝ ⎠

∂

∂∂

Θ

Ψ

Θ ΦΦ

∂

2 2 22 2

2 22

( ). ( )

( , , ) ( ) ( )

( , , ) ( ) ( )

s

R (r) r

( ) w hen substitu ted in T I

in sin =0

R earrange by ta

sin

king the

sin

SE

( )

1 2m ke+ (E+

)r

r R r

r R r

R rrR r r

θ φ

θ φ φθθ φ θ

θ

θ

θφ

φ

θ θφ

θθ θ

θ

= Θ Φ

∂Ψ= Φ

∂∂Ψ

= Θ∂

∂ ∂⎛ ∂ ∂Θ⎛ ⎞+ +⎜ ⎟Θ⎞

⎜ ∂ ∂⎝⎟∂ ∂

∂∂

∂Θ⇒

∂∂Φ

∂

∂∂⎝ Φ⎠⎠

Φ

2 2

2

2 2 2

22

2

2m ke 1+ (E+ )r

LH S is fn . of r, & R H S is fn of only , for equality to be true for all r, ,

LH S= constant = R H

term on R H S

sin s

S =

sin sin

m

in =-

l

R rrR r r φ

θ φ θ φ

θ θ

φ

θθ θ

θ∂ ∂⎛ ⎞⎜ ⎟

∂ ΦΦ ∂

⇒

∂ ∂Θ⎛ ⎞+ ⎜ ⎟Θ ∂ ∂⎝⎝ ⎠ ⎠∂ ∂

2 2 22 2

2

2

2

2

sin sin =m

Divide Thruout by sin and arrange all terms with r aw

Now go break up LHS to seperate the terms...r .. 2m keLHS: + (E+ )

a

& sin si

y fromr

1

n lR r

r

rr

r

R r

R

θ θ

θ

θ θθ θ

θ

θ

∂ ∂Θ⎛ ⎞+ ⎜ ⎟Θ ∂ ∂⎝ ⎠∂ ∂⎛

∂

⎞

∂∂

⎜ ⎟∂ ∂⎝ ⎠⇒

2

2

2 2

2

m 1 sinsin sin

Same argument : LHS is fn of r, RHS is fn of , for them to be equal for a

LHS = const =

2m ke(E+ )=r

What do we have after shuffl

ll r,

= ( in1) g RHS

l

l l

R rr

θθ θ θ θ

θ θ

∂ ∂Θ⎛ ⎞− ⎜ ⎟⎛ ⎞ +

Θ ∂ ∂⎝ ⎠

⇒ +

⎜ ⎟∂⎝ ⎠

22

2

2

2

2 22

2 2 2

!

m1 sin ( 1) ( ) 0.....(2)sin sin

...............

d ..(1)

1 2m ke ( 1)(E+ )- ( ) 0....(

m 0.

3

.

)r

T

l

l

d R r l lr R rr dr

d d ld

d

r

ld

r

θ

φ

θθ θ θ θ

⎡ ⎤Θ⎛ ⎞ + + − Θ

⎡ ⎤∂ +⎛ ⎞ + =⎜ ⎟ ⎢ ⎥∂⎝ ⎠ ⎣ ⎦

=⎜ ⎟ ⎢ ⎥⎝ ⎠ ⎣ ⎦

Φ+ Φ =

hese 3 "simple" diff. eqn describe the physics of the Hydrogen atom.All we need to do now is guess the solutions of the diff. equationsEach of them, clearly, has a different functional form

Solutions of The S. Eq for Hydrogen Atom2

22

dThe Azimuthal Diff. Equation : m 0

Solution : ( ) = A e but need to check "Good Wavefunction Condition"Wave Function must be Single Valued for all ( )= ( 2 )

( ) = A e

l

l

l

im

im

dφ

φ

φ

φφ φ φ π

φ

Φ+ Φ =

Φ⇒ Φ Φ +

⇒ Φ ( 2 )

2

2

A e 0, 1, 2, 3....( )

The Polar Diff. Eq:

Solutions : go by the name of "Associated Legendre Functions"

1 msin

( 1) ( ) 0sin si

Quantum #

n

l

l

iml

d d

Magnetic

ld

m

ld

φ π

θ θθ θ θ θ

+

⎡ ⎤Θ⎛ ⎞ + + − Θ =⎜ ⎟ ⎢ ⎥⎝ ⎠

=

⎣ ⎦

⇒ = ± ± ±

only exist when the integers and are related as follows 0, 1, 2, 3....

: Orbital Q

;

ua

p

nt

osit

um N

ive numb

umber

er

1For 0, =0 ( ) = ; 2

For

l

l

l

l ml l

l m

l

m

θ

= ± ± ± ± =

= ⇒ Θ

2

1, =0, 1 Three Possibilities for the Orbital part of wavefunction

6 3[ 1, 0] ( ) = cos [ 1, 1] ( ) = sin 2 2

10[ 2, 0] ( ) = (3cos 1).... so on and so forth (see book) 4

l

l l

l

l m

l m l m

l m and

θ θ θ θ

θ θ

= ± ⇒

= = ⇒ Θ = = ± ⇒ Θ

= = ⇒ Θ −

Φ

Solutions of The S. Eq for Hydrogen Atom2 2

22 2 2

2

20

1 2m ke ( 1)The Radial Diff. Eqn: (E+ )- ( ) 0r

: Associated Laguerre Functions R(r), Solutions exist only if:

1. E>0 or has negtive values given by ke 1E=-2a

d R r l lr R rr dr r r

Solutions

n

⎡ ⎤∂ +⎛ ⎞ + =⎜ ⎟ ⎢ ⎥∂⎝ ⎠ ⎦

⎝

⎣

⎛ 2

0 2

2. And when n = integer such that 0,1,2,3,4,, , ( 1)

n = principal Quantum # or the "big daddy" qunatuTo

; Bohr

Summa

m # : The hy

Rad

r drogeize n

ius

atom

a

l

mke

n= −

⎞ = =⎜ ⎟⎠

n = 1,2,3,4,5,....0,1,2,3,,4....( 1)

m 0, 1, 2, 3,.Quantum # appear only in Trappe

is brought to you

d systems

The Spati

by the letters

al Wave Function o

f the

Hydrogen Atom

..

l

l nl

∞= −

= ± ± ± ±

lm( , , ) ( ) . ( ) . ( ) Y (Spherical Harmonics)

l

l

mnl lm nl lr R r Rθ φ θ φΨ = Θ Φ =

Radial Wave Functions & Radial Prob Distributions

0

0

-r/a3/20

r-2

a3/2

00

23

23/20 00

R(r)= 2 e

a

1 r(2

n

1 0 0

2 0 0

3 0 0

- )e a2 2a

2 r(27 18 2 )a81 3a

l

rar e

a

l m

−

− +

n=1 K shelln=2 L Shell n=3 M shelln=4 N Shell……

l=0 s(harp) sub shelll=1 p(rincipal) sub shelll=2 d(iffuse) sub shelll=3 f(undamental) ssl=4 g sub shell……..

Symbolic Notation of Atomic States in Hydrogen

2 2

4 4

2

( 0) ( 1) ( 2) ( 3) ( 4

3 3 3

) .....

1

4

3

1s p

s l p l d l f l g l

s

l

sd

n

ps p

= = =→ = =

↓

5 5 5 5 4

5 4

5s p d f gd f

Note that: •n =1 non-degenerate system •n1>1 are all degenerate in l and ml.

All states have same energyBut different spatial configuration

2

20

ke 1E=-2a n

⎛ ⎞⎜ ⎟⎝ ⎠

Facts About Ground State of H Atom -r/a

3/20

-r/a100

0

2 1 1( ) e ; ( ) ; ( )a 2 2

1 ( , , ) e ......look at it caref

1. Spherically s

1, 0,

ymmetric no , dependence (structure)

2. Probab

0

ully

i

ln l r

ra

m R θ φπ

θ φπ

θ φ

⇒ = Θ = Φ =

Ψ =

⇒

= = =

22

100 30

Likelihood of finding the electron is same at all , and depends only on the radial seperation (r) between elect

1lity Per Unit Volume : ( ,

ron & the nucleus.

3 Energy

,

of Ground ta

)

S

rar e

aθ

πθ φ

φ−

Ψ =

2

0

kete =- 13.62a

Overall The Ground state wavefunction of the hydrogen atom is quite Not much chemistry or Biology could develop if there was only the ground state of the Hydrogen Atom!

We nee

boring

eV= −

d structure, we need variety, we need some curves!

Interpreting Orbital Quantum Number (l)

2

RADIAL ORBITAL

RADIAL ORBI

2 22

2 2 2

22

TAL2 2

K

1 2m ke ( 1)Radial part of S.Eq

K ;substitute this form for

n: (E+ )- ( ) 0r

For H Atom: E = E

K K

K + U =

1 2m ( 1)-2m

d dR r l lr R rr dr dr r

d dR l lrr dr dr r

ker

⎡ ⎤+⎛ ⎞ + =⎜ ⎟ ⎢ ⎥⎝ ⎠ ⎣ ⎦

+⎛ ⎞ +⎜

+

⎟

−

+⎝ ⎠

[ ]

ORBITAL

RADIAL

OR

2

2

22

B

2

2

( ) 0

( 1)Examine the equation, if we set get a diff. eq. in r2m

1 2m ( ) 0 which

Further, we also kno

depends only on radi

K

K us

w

r of orb

K tha

i

t

t

R r

l l thenr

d dRr R rr dr dr

⎡ ⎤=⎢ ⎥

⎣ ⎦

+

⎛ ⎞ + =⎜ ⎟⎝ ⎠

=

22

ITAL ORBITorb

2

AL 2

2

ORBITAL 22

L= r p ; |L| =mv r

( 1

1 ; K2 2

Putting it all togather: K magnitude of) | | ( 1)2m

Since integer=0,1,2,3...(

A

n-1) angular mome

ng.2

nt

Mom

orbit

l l L l lr

l

Lmvmr

L

posi

mr

tive

× ⇒

+= = +

=

=

= ⇒

=

⇒

um| | ( 1)

| | ( 1) : QUANTIZATION OF Electron's Angular Momentum

L l l discrete values

L l l

= + =

= +

pr

L

Magnetic Quantum Number ml (Right Hand Rule)

QM: Can/Does L have a definite direction

Classically, direction & Magnitud

? Proof by NegatˆSuppose L was precisely known/defined (L || z)

e of L

S

always well defi

n

ed

:

n

io

L r p= ×

2

z

z z

Electron MUST be in x-y orbit plane

z = 0 ;

, in Hydrogen atom, L can not have precise measurable

ince

Uncertainty Principle & An

p p ; !!!

gular Momentum

value

: L

2

pz E

L r p

Som

φ

= × ⇒

⇒ ∆ ∆ ∆ ⇒ ∆ ∞

∆

∞

∆

=∼ ∼ ∼

∼

Z

Z

The Z compo

Arbitararily

nent of

L vector spins around Z axis (precesses).

|L | ; 1, 2, 3...

( 1)

It

picking Z axis as a referen

L

:| L |

ca

| | (always)

s

n

in

ce directi

c

o

:

e

n

l l

l

m

Note L

m l

m l l

<

= = ± ± ± ±

< +

Z never be that |L | ( 1) (breaks Uncertainty Principle)So you see, the dance has begun !

lm l l= = +

L=2, ml=0,±1, ± 2 : Pictorially

Sweeps Conical paths of different ϑ: Cos ϑ = LZ/L and average

<LX> = 0 <LY> = 0

What’s So “Magnetic” ?

Precessing electron Current in loop Magnetic Dipole moment µ

More in this in Tomorrow’s lecture when we look at Energy States

Radial Probability Densities l

2* 2 2

m

2

( , , ) ( ) . ( ) . ( )

( , , ) |

Y

Probability Density Function in 3D:

P(r, , ) = =| | Y |

: 3D Volume element dV= r .sin . . .

Prob. of finding parti

| | .

cle in a ti

n

l

l

l

nlnl lml

n

m

mll

Note d

r R r

r

r d

R

d

R

θ φ θ φ

θθ φ

φ

φ

θ θ

Ψ = Θ Φ =

ΨΨ Ψ =

l

l

2 2

22

m0

2

2

0

2 2

m

y volume dV isP.dV = | Y | .r .sin . . .The Radial part of Prob. distribution: P(r)dr

P(r)dr= | ( ) |

When

| ( ) |

( ) & ( ) are auto-normalized then

P(r)dr

| | .

| |

=

.

|

l

l

l

lm

l

m

m

n

nl

n

llR

R r d

R

dr

d

d

r

d

dπ π

θ

θ φ

φ φ

θ φ

θ

θ

Θ

Θ

Φ

Φ

∫ ∫

2 2

2 2nl

2 2

0

0

in other words

Normalization Condition: 1 = r |R | dr

Expectation Values

P(r)=r |

<f(

| |. . ;

r)>= f(r).P(r)dr

nll r r Rd

∞

∞

∫

∫

Ground State: Radial Probability Density

0

0

0

0

2 2

22

3

22

30

2

2

0

0

0

( ) | ( ) | .4

4( )

Probability of finding Electron for r>a

To solve, employ change of variable

2rDefine z= ; limits of integra

4

12

tiona

ra

r aa

r a

ra

r e dr

P r dr r r dr

P r dr r ea

change

P a

P z

ψ π

−∞

>

∞

>

−

=

⇒ =

⎡ ⎤⎢ ⎥

=

=

⎣ ⎦

∫

2 22

(such integrals called Error. Fn)

1 66.7% =- [ 2 2] | 5 0.6672

!!

z

z

e

z e

z

e

d

z −

−

∞+ + = = ⇒

∫