V(x)=0for L>x>0 V(x)=∞for x≥L, x≤0
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Transcript of V(x)=0for L>x>0 V(x)=∞for x≥L, x≤0
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
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