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An in-depth analysis of the energy eigenfunctions and eigenvalues of a finite quantum well. It covers the solution of the schrödinger equation for different potentials, the classification of states as bound or unbound, and the behavior of the wave function in different regions. The document also discusses the normalization of the wave function and the determinant condition for a solution.
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Solve the energy eigenvalue equation for different potentials
and for examples where there are many solutions with different
energies.
V 0
x > a
0 x < a
!
"
$
-a a → x
↑
Energy
V 0
0
E
E
Region 1 Region 2 Region 3
" (^) ( x ) =
1
x < # a
2
3
x > a
unbound states
(continuum states)
bound states
" 1
( x ) =^ Ce
1
x
C ' e
ik 1
x
Focus first on the case E < V 0
("bound states")
In regions 1 & 3, k is imaginary
! 3
( x ) =^ D^ ' e
" ik 3
x
De
ik 3
x
Imaginary k means
exponential growth or
exponential decay!
(classically forbidden region)
!
" 2
( x ) =^ Ae
2
x
Be
ik 2
x
Real k means oscillatory
behavior
(classically allowed region)
In region 2, k is real
It would not be physically reasonable to
allow an infinite probability of finding a
particle in a classically forbidden region.
-a a → x
↑
Energy
0
V 0
E
(^1 2 )
2
( x ) =^ Ae
ik 2
x
2
x
3
( x ) =^ D ' e
1
x
ik 1
x
1
( x ) =^ Ce
1
x
ik 1
x
" (^) ( x ) =
1
x < # a
2
3
x > a
ϕ is continuous everywhere
ϕ has a continuous derivative
ϕ goes to zero at ±∞
ϕ is normalized
0 0
k 1
= i
2 m
!
2
V 0
(!^ E )
k 2
2 mE
2
4 equations, 5 unknowns ( A , B , C , D, E ). ( E is buried in k 1
and k 2
)
Normalization gives fifth condition.
e
" ik 2
a
e
ik 2
a
" e
ik 1
a
e
ik 2
a
e
" ik 2
a
0 e
ik 1
a
ik 2
e
" ik 2
a
" ik 2
e
ik 2
a
" ik 1
e
ik 1
a
ik 2
e
ik 2
a
" ik 2
e
" ik 2
a
0 ik 1
e
ik 1
a
This set of equations has a solution when the determinant of the 4x
matrix is zero. Tedious! See Liboff for details. When the
determinant condition is set up, we get a condition on E! This
condition can be satisfied in 2 sets of ways. One set has A = B
(even solutions) and the other set has A = - B (odd solutions).
2 mEa
2
!
2
Here is one condition for the determinant to be zero
(Eqn 5.83 in McIntyre):
tan
2 mEa
2
!
2
=
2 m V 0
2
!
2
2 mEa
2
!
2
one value of E
3 values of E
-0.
0
1
-0.1 -0.05 0 0.05 0.
-0.
0
1
-0.1 -0.05 0 0.05 0.
-0.
0
1
-0.1 -0.05 0 0.05 0.
tan
2 mE 1
a
2
!
2
=
2 m V 0
! E 1
2
!
2
2 mE 1
a
2
!
2
tan
2 mE 3
a
2
!
2
=
2 m V 0
! E 3
2
!
2
2 mE 3
a
2
!
2
! cot
2 mE 2
a
2
!
2
=
2 m V 0
! E 2
2
!
2
2 mE 2
a
2
!
2
-0.
0
1
-0.1 -0.05 0 0.05 0.
-0.
0
1
-0.1 -0.05 0 0.05 0.
-0.
0
1
-0.1 -0.05 0 0.05 0.
This set corresponds to the green
curves on the previous graphs - the
value of V 0
that yields 3 solutions (
even and 1 odd).
Note the size of the decay length for
the state corresponding to each
energy. Wave function "leaks" into
forbidden region. We call this an
evanescent wave.
E n
=
n
2
!
2
!
2
2 m (^) ( 2 a )
2
-0.
0
1
-0.1 -0.05 0 0.05 0.
! n
( x ) =
2
2 a
cos
n " x
2 a
-0.
0
1
-0.1 -0.05 0 0.05 0.
! n
( x ) =
2
2 a
sin
n " x
2 a
n =1,3,5 (symmetric or even solutions)
n =2,4,6 (antisymmetric or odd solutions)
n
( x ) = (^0) n =1,2,3,4,5… for x<-a and x>a
x=-a x=a
x=-a x=a
E n
=
n
2
!
2
!
2
2 m (^) ( 2 a )
2
-0.
0
1
-0.1 -0.05 0 0.05 0.
-0.
0
1
-0.1 -0.05 0 0.05 0.
! n
( x ) =
2
2 a
sin
n " x
2 a
n =1,2,3,4,5… and 0 < x < 2a
(neither symmetric nor
antisymmetric solutions - about
x=0)
n
( x ) = (^0) n =1,2,3,4,5… for x < 0 and x > 2a
x=0 x=2a
x=0 x=2a