Particle in a 1-Dimensional Box
Time Dependent Schrödinger Equation
  2 d 2
 V ( x)  E
2m dx 2
Region I
Region II
Region III
KE
PE
TE
Wave function is dependent on time and position function:
V(x)=∞
V(x)=0
V(x)=∞
1
 ( x, t )  f (t ) ( x)
Time Independent Schrödinger Equation
L
0
V(x)=0 for L>x>0
V(x)=∞ for x≥L, x≤0
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.
x
  2 d 2 ( x)
 V ( x)  E
2
2m dx
Region I and III:
Applying boundary conditions:
  2 d 2 ( x)
  *  E
2
2m dx
Region II:
  2 d 2 ( x)
 E
2m dx 2
 0
2
Finding the Wave Function
d 2 ( x) 2m

 2 E
dx 2

  2 d 2 ( x)
 E
2m dx 2
This is similar to the general differential equation:
d 2 ( x)

 k 2
2
dx
  A sin kx  B cos kx
So we can start applying boundary conditions:
x=0 ψ=0
0  Asin 0k  B cos 0k
x=L ψ=0
0  Asin kL A  0
Calculating Energy Levels:
2mE
k  2

2
k 2 2
E
2m
h

2
n 2 2 h 2
E 2
L 2m4 2
k 2h2
E
2m4 2
n2h2
E
8mL2
 II  A sin
nx
L
But what is ‘A’?
Normalizing wave function:
L
2
(
A
sin
kx
)
dx  1

0
L
 x sin 2kx 
A  
1
4k  0
2
2
0  0  B *1 B  0
kL  n where n=
Our new wave function:
*
n 

sin
2
L
2L
L
A  
1
n 
2


4
L


Since n=
2 L 
A   1
2
*
A
2
L
Our normalized wave function is:
 II 
2
nx
sin
L
L
Particle in a 1-Dimensional Box
2
nx
 II 
sin
L
L
Applying the
Born Interpretation
 II
2
2
nx 
  sin

L
L 
n=4
n=3
E
x/L
2
n=4
E
n=3
n=2
n=2
n=1
n=1
x/L
Particle in a 2-Dimensional Box
Doing the same thing do these differential
equations that we did in one dimension we get:
A similar argument can be made:
  2   2  2
 2  2
2m  x
y

  E

X
 ( x, y )  X ( x)Y ( y )
Y
In one dimension we needed only one ‘n’
But in two dimensions we need an ‘n’ for the x and y component.
Lots of Boring Math
Our Wave Equations:
 2
2m
 2
2m
2 X
 E x
2
x
 2Y
 E y
2
y
n x
2
sin x
Lx
Lx
n yx
2
sin
Ly
Ly
  X ( x)Y ( y )
Since
n n 
x y
n yx
nxx
4
sin
sin
Lx Ly
Lx
Ly
n 
x
2
nx
sin
L
L
For energy levels:
En x 
2
2
nh
8mL2
2
2



h  nx   n y  
  

8m  Lx   Ly  


2
En x n y
Particle in a 2-Dimensional Equilateral Triangle
Let’s apply some Boundary Conditions:
Types of Symmetry:
y  3x
v
E
 ( x, y )  0,
u
C3
0
w
u
w
w
v
C23
v
v
u
y  3 ( a  x)
u  (2 / A) y
v  (2 / A)(  y / 2  3 x / 2)
w  (2 / A)(  y / 2  3 x / 2)  2
u  v  w  2
 ( x, y )  0,
u
w
a
So our new coordinate system:
σ2
σ1
y0
Defining some more variables:
u
v
A
w
u  0, v  2  w
v  0, w  2  u
w  0, u  2  v
Our 2-Dimensional Schrödinger Equation:
  2   2  2


2m  x 2 y 2

  E

Solution:   f (c1 x  c2 y)
σ2
w
Substituting in our definitions of x and y in terms of u and v gives:
E  f ( pu  qv)
u
v
Where p and q are our nx and ny variables from the 2-D box!
Finding the Wave Function
Substituting gives:
So what is the wave equation?
P( A1 )  f ( pu  qv)  f ( pv  qw)  f ( pv  qw)
It can be generated from a super position
of all of the symmetry operations!
 f ( pv  qu )  f ( pw  qv)  f ( pu  qw)
P( A2 )  f ( pu  qv)  f ( pv  qw)  f ( pv  qw)
P( A1 )  E  C3  C32   1   2   3
 f ( pv  qu )  f ( pw  qv)  f ( pu  qw)
f  sin
So if:
And we recall our original definitions:
P( A2 )  E  C3  C32   1   2   3
E  f ( pu  qv)
But what plugs into these?
u  (2 / A) y
v  (2 / A)(  y / 2  3 x / 2)
w  (2 / A)(  y / 2  3 x / 2)  2
If you recall:
w
w
w
C2
u
w
v
u
v
u
C3
σ2
σ1
v
v
v
u
Substituting and simplifying gives:
 q 3x   (2 p  q)y 
 p 3x   (2q  p)y 
p ,q ( A1 )  cos 
 cos 
 sin 
 sin 


A
A


 A  
 A  
w
σ2
3
w
u
u
So for example in C3 u v’s spot and v w’s spot
Continuing with the others:
E  f ( pu  qv)
 1  f ( pv  qu)
C3  f ( pv  qw)
 2  f ( pw  qv)
C32  f ( pv  qw)
 3  f ( pu  qw)
 ( p  q) 3x   ( p  q)y 
 cos 
 sin 

A
A


 
v
A1
q  0,1,2,3...
p  q  1, q  2...
 q 3x   (2 p  q)y 
 p 3x   (2q  p)y 
p ,q ( A2 )  sin 
 sin 
 sin 
 sin 


A
A


 A  
 A  
 ( p  q) 3x   ( p  q)y 
 sin 
 sin 

A
A


 
A
q  1,2,3...
2
Energy Levels:
p  q  1, q  2...
E p,q  ( p 2  pq  q 2 ) E0
Plotting in Mathematica
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0.2
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p
4;
q
2;
0
2
0.25
0.5
,
x, 0, 1 ,
ColorFunction
p
5;
q
0;
0.75

;
Plot3D z
A
2
-2
3
A
4
3
2
0
1
0
0.25
0.5

p=1 q=0
y, 0, A , PlotPoints
100, Mesh
False, BoxRatios
0.75
1
2
1, A, .4 , ViewPoint
0.000, 1.500, 3.384 ,
Hue
0.8
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;
ContourPlot z
0
0
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,
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x, 0, 1 ,
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1
y, 0, A , ContourLines
0
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False, PlotPoints
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100, Contours
1
20, ColorFunction
Hue

A1

2

A2

2