Transcript Particle in a Box - Tufts University

Particle in a 1-Dimensional Box
Time Dependent Schrödinger Equation
  2 d 2
 V ( x)  E
2m dx2
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 dx2
 0
2
Finding the Wave Function
d 2 ( x) 2m

 2 E
dx2

  2 d 2 ( x)
 E
2m dx2
This is similar to the general differential equation:
d 2 ( x)

 k 2
2
dx
  A sin kx  B coskx
So we can start applying boundary conditions:
x=0 ψ=0
0  A sin 0k  B cos 0k
x=L ψ=0
0  Asin kL
A0
where n=
Calculating Energy Levels:
2mE
k  2

2
k 2 2
E
2m
k 2h2
E
2m4 2
h

2
n 2 2 h 2
E 2
L 2m4 2
n2h2
E
8m L2
 II  A sin
nx
L
But what is ‘A’?
Normalizing wave function:
L
 ( A sin kx )
2
dx  1
0
L
 x sin 2kx 
A  
1
4k  0
2
2
0  0  B *1 B  0
kL  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
 II 
2
nx
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
 Ex
x 2
 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 L y
Lx
Ly
n 
x
2
nx
sin
L
L
For energy levels:
2
2
nh
En x 
8m L2
2
h  nx   n y
  

8m  Lx   L y

2
En x n y




2




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  3x / 2)
w  (2 / A)( y / 2  3x / 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  3x / 2)
w  (2 / A)( y / 2  3x / 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  
 ( p  q) 3x   ( p  q)y 
 cos
 sin 

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)
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...
Ep,q  ( p2  pq  q2 )E0
Plotting in Mathematica
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x, 0, 1 ,
ColorFunction
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Plot3D z
A
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3
A
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3
2
0
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
p=1 q=0
y, 0, A , PlotPoints
100, Mesh
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0.000, 1.500, 3.384 ,
Hue
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ContourPlot z
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20, ColorFunction
Hue
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
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
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