Physics 2D Lecture SlidesMar 10

Vivek SharmaUCSD Physics

• 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)

QM in 3 Dimensions

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

x

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)

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

y z [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 2 22 2 2 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

( ) kZeU rr

=

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

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