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Chapter 5 One Dimensional Potentials
We must solve the steady-stae eigenvalue equation in 1D.
h
2
2 m
d
2
dx
2
H
n
( x ) = E ( n
( x ) time independent eigenvalue equation
( ( x , t ) =
n
n
( x ) e
! i
E n
% t general solution
5 - 2 Infinite Well V ( x ) =
! x " a
0 0 < x < a
! ( x )''+
2 m
2
E
! ( x ) = 0
! ( x ) = A sin( kx ) + B cos( kx ) k
2
2 m
2
E
! (0) = A ( 0 + B ( 1 = 0 ) B = 0
! ( a ) = A sin( ka ) = 0 ) ka = n * n = 1 , 2 , 3 ,...
n
( x ) = A sin( k n
x ) E n
2
2 m
k
2 E =
2
2 m
n *
a
2
Normalizing
A
2
0
a
sin
2 ( k n
x ) dx = 1
A
2
0
a
sin
2 ( k n
x ) + cos( k n
( x )) dx^ =^ A
2
0
a
dx = A
2 a
n
( x ) =
a
sin( k n
x ) E =
2
2 m
n *
a
2
Assume well instaneoulsy expands a-> 2a
New eigenstates and energies! n
( x ) =
a
sin(
n "
2 a
x )
E
n
2
2 m
n "
2 a
2
1
n
a n
n
> exp ansion of ground state ) 0
in new states in terms new states! n
a 1 n
1
n
a
a 0
2 a
sin(
a
x ) sin(
n "
2 a
x ) dx
P
1 , n
= a 1 n
2
|φ1> , 14 Ε 1
|φ2>, Ε 1
| φ3>, 9/4Ε 1
|ψ1> , Ε 1
|ψ2>, 4 Ε 1
|ψ3>, 9 Ε 1
0 a
0 2a
5 - 3 Simple Harmonic Oscillator potential V ( x ) =
k x
2
If we expand any potential about its minima to second order we arrive at solving the
harmonic oscillator problem in those regions, thus the importance.
V ( x ) = V ( a ) +
dV
dx a
= 0
( x! a )
1
d
2 V
dx
2
a
( x! a )
2
2 m
2
( E "
m #
2 x
2 )! = 0 # =
k
m
1 ) Switch to unitless var iables $ =
2 E
and y =
m #
x
d
dx
m #
d
dy
d
2
dx
2
m #
d
2
dy
2
m #
d
2
dy
2
2 m
2
m #
m #
y
2 )! = 0
d
2
dy
2
! ( y ) + ( $ " y
2 )! ( y ) = 0 unitless equation
2 ) Assymptotic solution y % ±&
! ''" y
2 ! = 0 %! ( y ) = h ( y ) e
"
1
2
y
2
1
2
y
2
diverges at & g = 0
! ( y ) = h ( y ) e
"
1
2
y 2
! '( y ) = h ' e
"
1
2
y 2
" yh e
"
1
2
y 2
! ''( y ) = h '' e
"
1
2
y 2
" yhe
"
1
2
y 2
" y ' h e
"
1
2
y 2
2 he
"
1
2
y 2
h ''" 2 yh '+ ( $ " 1 ) h = 0 Re sulting equation for h ( y )
a
b
V(x)
x
Zero point energy
Series diverges as e
y
2
and not a permissible wave function!
But the series truncates to a polynomial when 2 n! ( "! 1 ) = 0 " = 2 n + 1 or E = ( n + 1 / 2 )!#
Hermite Polynomials
H
0
( y ) = 1 H 1
( y ) = 2 y H 2
( y ) = 4 y
2 ! 2 H 3
( y ) = 8 y
3 ! 12 x .........
Full Solution
n
( y ) =
n n! %
H
n
( y ) e
!
1
2
y 2
2 E
and y =
m #
x
Solution by Raising and lowering Operators
The harmonic oscillator Hamiltonian has a particularly symmetric form. It can be shown that so-called
raising and lowering operators to build the Hamiltonian. http://en.wikipedia.org/wiki/Ladder_operator
Let H , A
± ! "
n
= ± & A %
n
HA
± % n
= A
± H % n
+ H , A
± ! "
n
HA
± % n
= A
± E n
n
± & A
± % n
HA
± % n
= E
n ( ±^ &) A
± % n
H A
± % n
% n ± 1
= E
n ( ±^ &) A
± % n
% n ± 1
H = a
a ! ( +^1 /^2 ) !"
Let a ±
d
dy
and A
± = a ±
a "
d
2
dy
2
2
n
( y ) = (^) ( 2 n + (^1) )) n
( y ) where x = ay and a =
m "
| n ± 1 >
a
±
| n >
a
, a !
a
a !
n
> = n | & n
a !
a
n
> = n + 1 ( )
n
a
n
> = n | & n
a
n
> = n + 1 | & n + 1
n + 1
n + 1
a
n
n + 1
d
dy
n
a !
n
>= n | & n! 1
n! 1
n
a !
n
n
d
dy
n
**5 - 3 Potential Barrier Problem (E V " ( x ) = A e
ikx
+ B e
! ikx or " ( x ) = A sin( kx ) + B cos( kx ) k real
Damped solutions
If E < V " ( x ) = A e
+ B e
! # x or " ( x ) = A sinh( # x ) + B cosh( # x ) k = i # complex
I
( x ) = I e
+ R e
" ikx ! II
( x ) = A e
+ B e
" qx ! III
( x ) = T e
Boundary Conditions
I
II
(0) 1 + R = A + B k ( 1 + R ) = k ( A + B )
I
II
(0) k ( 1 " R ) = q ( A " B ) k ( 1 " R ) = q ( A " B )
II
( a ) =! III
( a ) A e
" qa = T e
II
( a ) =! ' III
( a ) q ( A e
" qa ) = ikT e
x=0 (^) x=a
V=Vo
V=
E
1 R
T
I
II
III
5 - 4 WKB Approximation - Barrier Transmission
In the barrier transmission problem we are asked what is the probability that a matter wave of energy
E =
p
2
2 m
will transdfer through a potiential barrier V(x). Classically the wave would be reflected.
But QM allows the wave to tunnel through the barrier.
Approximation of tunneling probability T
2
P
T
= T
2 ~ e
! 2
q ( x ) dx
=
h
p
<< barrier width
q ( x ) =
2 m
2
( V ( x )! E )
Cold Emission Alpha Decay
V ( x ) = W! e " x V
Coulomb
( x ) =
Z
1
Z
2
e
2
0
r
V(x)
V=
E
T = e
!
a
b
"
q ( x ) dx
Tunneling
wave
a b
! e " x
W
V = V
Nuclear
+ V
Coulomb
