V(x)=0 for L>x>0V(x)=∞ for x≥L, x≤0

Particle in a 1-Dimensional Box

ExVdx

d

m)(

2 2

22

Classical Physics: The particle can exist anywhere in the box and follow a path in accordance to Newton’s Laws.

Quantum Physics: The particle is expressed by a wave function and there are certain areas more likely to contain the particle within the box.

ExV

dx

xd

m

)(

)(

2 2

22

KE PE TE

Time Dependent Schrödinger Equation

)()(),( xtftx

Wave function is dependent on time and position function:1

Time Independent Schrödinger Equation

Applying boundary conditions:

E

dx

xd

m

*

)(

2 2

22Region I and III:

E

dx

xd

m

2

22 )(

2

Region II:

02

V(x)=0 V(x)=∞V(x)=∞

0 L x

Region I Region II Region III

Finding the Wave Function

E

dx

xd

m

2

22 )(

2

22

2 )(k

dx

xd

E

m

dx

xd22

2 2)(

This is similar to the general differential equation:

22 2

mE

k m

kE

2

22

kxBkxA cossin

So we can start applying boundary conditions:x=0 ψ=0

kBkA 0cos0sin0 01*00 BB

x=L ψ=00AkLAsin0 nkL where n= *

L

xnAII

sin

2

22

42 m

hkE

2

h

2

2

2

22

42

m

h

L

nE

2

22

8mL

hnE

Our new wave function:

But what is ‘A’?

Calculating Energy Levels:

Normalizing wave function:

1)sin(0

2 L

dxkxA

14

2sin

20

2

L

k

kxxA

14

2sin

2

2

Ln

LLn

LA

Since n= *

12

2

L

AL

A2

Our normalized wave function is:

L

xn

LII

sin2

E

x/L x/L

E

Particle in a 1-Dimensional Box

n=1

n=2

n=3

n=4

n=1

n=2

n=3

n=4

L

xn

LII

sin2

2

2sin

2

L

xn

LII

Applying the Born Interpretation

Particle in a 2-Dimensional BoxA similar argument can be made:

E

yxm

2

2

2

22

2

)()(),( yYxXyx

Lots of Boring Math

xEx

X

m

2

22

2

yEy

Y

m

2

22

2

Our Wave Equations:

Doing the same thing do these differential equations that we did in one dimension we get:

x

x

x L

xn

LX

sin

2

y

y

y L

xn

LY

sin

2

In one dimension we needed only one ‘n’But in two dimensions we need an ‘n’ for the x and y component.

Since )()( yYxX

y

y

x

x

yxnn L

xn

L

xn

LLyx

sinsin4

222

8 y

y

x

xnn L

n

L

n

m

hE

yx2

22

8mL

hnE

xn

L

xn

Lxn

sin2

For energy levels:

Particle in a 2-Dimensional Equilateral TriangleTypes of Symmetry:

C3

v

u

w

v

w u

σ1

C23

v u

w

u

v w

σ2

w

u v

σ2

,0),( yx

Let’s apply some Boundary Conditions:xy 3

a0)(3 xay

0y

Defining some more variables:

yAu )/2( )2/32/)(/2( xyAv

2)2/32/)(/2( xyAw2 wvu

So our new coordinate system:

,0),( yx wvu 2,0uwv 2,0vuw 2,0

Our 2-Dimensional Schrödinger Equation: E

yxm

2

2

2

22

2

)( 21 ycxcf

Substituting in our definitions of x and y in terms of u and v gives:

)( qvpufE

Where p and q are our nx and ny variables from the 2-D box!

Solution:

A

v

u w

E

3212331)( CCEAP

3212332 )( CCEAP

)( qvpufE

)(3 qwpvfC )(2

3 qwpvfC

)(1 qupvf )(2 qvpwf )(3 qwpuf

But what plugs into these?

So what is the wave equation?

It can be generated from a super position of all of the symmetry operations!

Finding the Wave Function

If you recall:

C3

v

u

w

C3

v

u

w

v

u w

v

u wu w

C23

v u

w

v

w u

σ1

v

w uw u

σ1u

v w

σ2

u

v w

σ2

w

u v

σ2

w

u v

σ2

So for example in C3 u v’s spot and v w’s spot

)( qvpufE Continuing with the others:

Substituting gives:

)()()(

)()()()( 1

qwpufqvpwfqupvf

qwpvfqwpvfqvpufAP

)()()(

)()()()( 2

qwpufqvpwfqupvf

qwpvfqwpvfqvpufAP

And we recall our original definitions:yAu )/2(

)2/32/)(/2( xyAv 2)2/32/)(/2( xyAw

Substituting and simplifying gives:

A

yqp

A

xqp

A

ypq

A

xp

A

yqp

A

xqAqp

)(sin

3)(cos

)2(sin

3cos

)2(sin

3cos)( 1,

A

yqp

A

xqp

A

ypq

A

xp

A

yqp

A

xqAqp

)(sin

3)(sin

)2(sin

3sin

)2(sin

3sin)( 2,

...3,2,1,0q ...2,1 qqp

...3,2,1q ...2,1 qqp

A1

A2

sinfSo if:

022

, )( EqpqpE qp Energy Levels:

p 4;

q 2;

A32

;

Plot3Dz ,x, 0, 1,y, 0, A, PlotPoints 100, Mesh False, BoxRatios 1, A, .4, ViewPoint0.000, 1.500, 3.384,ColorFunction Hue

Plotting in Mathematica

p 5;

q 0;

A32

;

ContourPlotz2,x, 0, 1,y, 0, A, ContourLines False, PlotPoints 100, Contours 20, ColorFunction Hue

0 0.25 0.5 0.75 1

0

0.2

0.4

0.6

0.8

-2

0

2

0 0.25 0.5 0.75 1

0

0.2

0.4

0.6

0.8

2 p=1 q=0

0 0.25 0.5 0.75 1

0

0.2

0.4

0.6

0.8

0

2

4

6

0 0.25 0.5 0.75 1

0

0.2

0.4

0.6

0.8

0 0.2 0.4 0.6 0.8 10

0.2

0.4

0.6

0.8

0 0.2 0.4 0.6 0.8 10

0.2

0.4

0.6

0.8

A1 2

A2 2