Chapter 5 One Dimensional Potentials
Eigenvalue Equation in 1D
" h2 d 2
%
$ ! 2m dx 2 +V(x)' ( n (x) = E ( n (x)
#
&
!###"###$
time independent eigenvalue equation
H
( (x,t) = ) ( n (x)e
!i
En
%
t
general solution
n
5-2 Infinite Well
V(x) =
!
x"a
0 0<x<a
|ψ3>, 9Ε1
# 2m
&
! (x)''+ % 2 (E " V )( ! (x) = 0
$ !
'
# 2m &
k2 = % 2 E(
$ !
'
! (x) = A sin(kx) + B cos(kx)
|ψ2>, 4Ε1
! (0) = A ) 0 + B )1 = 0 * B = 0
! (a) = Asin(ka) = 0 * ka = n+ n = 1,2,3,...
! n (x) = Asin(kn x)
En =
2
! 2
k
2m
E=
! # n+ &
2m %$ a ('
2
|ψ1> , Ε1
2
0
a
Normalizing
a
A2
,
sin2 (kn x)dx = 1
0
1
A2
2
a
, (sin (k x) + cos(k x)) dx = A
2
n
n
0
2
sin(kn x)
a
! n (x) =
! # n+ &
E=
2m %$ a ('
2
2
1
2
a
,
dx = A2
0
a
=1
2
| φ3>, 9/4Ε1
2
|φ2>, Ε1
Assume well instaneoulsy expands a-> 2a
New eigenstates and energies
!n (x) =
2
n"
sin(
x)
2a
2a
|φ1> , 14 Ε1
0
2
!
" 2 # n" &
En =
2m %$ 2a ('
What is the probability that ) 1 * !1 ?
| ) 1 > = + an | !n >
n
a1n = < ) 1 | !n > =
P1*n = a1n
.
2
exp ansion of ground state ) 1 in new states in terms new states !n
2 2
a 2a
2a
,
0
"
n"
sin( x) sin(
x) dx
a
2a
2a
5-3 Simple Harmonic Oscillator potential V(x) =
1
k x2
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
1 d 2V
(x ! a)1 +
(x ! a)2 + .....
dx a
2 dx 2 a
!
=0
V(x)
Zero point energy
a
! ''+
2m
1
(E " m# 2 x 2 )! = 0
2
2
!
$=
m# d
! dy
x
k
m
#=
1) Switch to unitless var iables
d
=
dx
b
2E
!#
y=
and
m#
x
!
d2
m# d 2
=
! dy 2
dx 2
m# d 2
2m !#
1
! 2
+ 2 (
$ " m# 2
y )! = 0
2
! dy
2
2
m#
!
d2
! (y ) + ($ " y 2 )! (y ) = 0
dy 2
unitless equation
2) Assymptotic solution y % ±&
! ''" y 2! = 0
% ! (y ) = A e
"
1 2
y
2
+
+
1 2
y
2
B e
"#
$ $
%
diverges at &
g=0
Let ! (y ) = h(y ) e
"
1 2
y
2
h ''" 2yh '+ ($ " 1)h = 0
! '(y ) = h ' e
"
1 2
y
2
" yh e
"
1 2
y
2
Re sulting equation for h(y )
! ''(y ) = h '' e
"
1 2
y
2
" yhe
"
1 2
y
2
" y 'h e
"
1 2
y
2
+ y 2he
"
1 2
y
2
h(y ) = ! a n y n+"
3) Power series solution
n=0
h(y ) = ! an y n+"
n=0
h '(y ) = ! (n + " )an y n+" #1
n=1
h ''(y ) = ! (n + " # 1)(n + " )an y n+" #2
con tan t " to be det ermined
n=2
! (n + " # 1)(n + " )a y
n+" #2
n
n=0
! {(n + " # 1)(n + " )a y
(" # 1)(" )a0 y " #2 = 0
n=0
For " = 0
n=0
n=1
(
n+" #2
{(n # 1)(n)a y
n=0
}
)
# 2(n + " ) # ($ # 1) an y n+" = 0
n
n=0
!
# 2! (n + " )an y n+" +($ # 1)! an y n+" = 0
% " = 1 ,0
(
lowest power
}
)
# 2(n) # ($ # 1) an y n = 0
n#2
n
y #2 n = 0
(#1)(0)a0 y #2 = 0
let a0 = cons tan t TBD
y #1 n = 1
(0)(1)a1y #1 = 0
let a1 = cons tan t TBD
(
)
# ( 2(1) # ($ # 1)) a y
(1)(2)a2 y # 2(0) # ($ # 1) a0 y 0 = 0
n=2
y
1
n=3
(2)(3)a3 y
nth
(n + 1)(n + 2)an+2 # 2n # ($ # 1) an = 0
0
1
1
!
(
1
=0
φ6
φ5
φ4
2a2 = ($ # 1)a0
y
0
yn
V(x)
φ3
φ2
φ1
6a3 = ($ # 3)a1
)
an+2 =
4) Recurrsion relationship
( 2n ! (" ! 1)) a
(n + 2)(n + 1) n
All coefficients can be det er min ed in terms of a0 and a1
h+ (y ) = a0 (1 + a2 y 2 + a4 y 4 +.......)
even series
h! (y ) = a1 (y + a3 y + a5 y ....... )
odd series
1
3
5
5) Check for convergence ! d ' Alembert 's RatioTest
lim
an+2 y n+2
n#$
an y
Trouble!
R2 =
n
( 2n ! (" ! 1))
= lim
(n + 2)(n + 1)
n#$
y2 = 0
ratio of con sec utive terms test
For n >> " ratio of con sec utive terms R2
( 2n ! (" ! 1)) y
2
(n + 2)(n + 1)
$
$
k
y
y
~%
n=0 n!
k =0 (k / 2)!
same as e y = %
2
2n
2 2
y
n
#
even
y k +2
(k / 2)!
2
(k + 2 / 2)!
R2 =
=
y2 = y2
k
k
y
( k + 2 / 2)!
(k / 2)!
(
)
2
Series diverges as e y and not a permissible wave function!
But the series truncates to a polynomial when 2n ! (" ! 1) = 0 " = 2n + 1
or
E = (n + 1/ 2)!#
Hermite Polynomials
H0 (y) = 1
H1(y) = 2y
H2 (y) = 4y 2 ! 2
H3 (y) = 8y 3 ! 12y .........
Full Solution
$ n (y) =
1
Hn (y) e
2 n! %
n
!
1 2
y
2
y=
m#
x
!
< $ n | $ m > = & nm
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
The golden property
(
)
| n ± 1>
|n >
Let !"H,A± #$ % n = ±& A±% n
HA±% n = A± H% n + !"H,A± #$ % n
HA±% n = A± ' n% n ± & A±% n
a±
HA±% n = ' n ± & A±% n
(
)
H A±% n = ' n ± & A±% n )
!"
# #
$
!"
# #
$
(% n±1
A±% n also an eigenfunction of H with * = En ± '
(% n±1
% 1 ( A+ % 0 , % 2 ( A+ % 1 , ..........
% n ( A+ % n+1
|ψ5>
|ψ4>
|ψ3>
|ψ2>
|ψ0>
Harmonic Oscillator – Algebraic Method
Consider the reduced hamiltonian equation for the harmonic oscillator
+
/
- 1 " !d
%
"
%
1 +d
1 -*
,
$ dy + y '
$ dy + y ' + 2 0 ( = )(
#
&
#
&
- !2##
"##$ !2##"##$
-.
-1
+
!
H
a
a
% "
1
1"
1%
a + a !( = (!2 + y)(+2 + y)( = $ !2 2( +y2( ! 2y
( + y 2( ' = $ H ! ' (
%
'
2
2 $#
2
#
&
&
! y2( !(
" !d 2
%
2
$ dy 2 + y ' ( = 2)(
#
&
!#
#"##
$
a ! a +( =
(a a
+
!
% "
1
1"
1%
(+2 + y)(!2 + y)( = $ !2 2( !y2( + 2y
( + y 2( ' = $ H + ' (
%
'& #
2
2 $#
2&
+ y2( +(
)
! a ! a + ( = ( * [a + ,a ! ] = !1
Some Commutator Algebra
[A,A] = 0
[A,B + C] = [A,B] + [A,C]
[AB,C] = [A,C]B + A[B,C]
[A,BC] = [A,B]C + B[A,C]
[A,B n ] = nAB n!1[A,B]
H = a ± a & ± 1/ 2
H = 1/ 2(a + a ! + a ! a + )
H = a + a ! + 1/ 2
Golden Property of Hamiltonian Operator
#
[a + a ! , a + ] " n = % [a + ,a + ] a ! ! a + [a ! ,a +
# #
$
%$ !"
=0
A± = a ±
and
&
]( " n + = !a + [a + ,a ! ] " n = a + " n
!"
# #
$
('
!1
[a ! a + , a + ] " n = [a ! ,a + ]a + ! a ! [a + ,a + ] " n + = [a ! ,a + ] a + " n = a + " n
!"
# #
$
+1
[H,a ± ] "
= ±a ± " n
the golden property
) H a ±" n = (* n ± 1) a ±" n
|!0 >
# "d
&
a |! 0 > = %
+ y ( |! 0 > = 0
$ dy
'
"
|! 0 > = )
* 0 = < ! 0 | a + a " + 1/ 2 | ! 0 > = < ! 0 | a + a " | ! 0 > = +1/ 2
"1/4
e
"
y2
2
E0 = 1/ 2 !+
| !1 >
# "d
&
a |! 0 > = %
+ y ( | ! 0 > = C 2y | ! 0 > = C | ! 1 >
$ dy
'
| ! 1 > = C)
*1 = < ! 1 | a + a " + 1/ 2 | ! 1 > = 1+ 1/ 2
E1 = 3 / 2 !+
+
a+ | ! n > =
a" | ! n >=
n + 1 | ! n+1 > ,
n | ! n"1 >
,
"1/4
(2y)e
"
y2
2
# d
&
%$ " dy + y (' | ! n >
n +1
n +1 2
&
1
1 1 # d
| ! n"1 > =
a" | ! n > =
%$ + dy + y (' | ! n >
n
n 2
| ! n+1 > =
1
a+ | ! n > =
1
1
5-3 Potential Barrier Problem (E<Vo)
1
V=Vo
R
T
E<Vo
I
III
II
x=0
V=0
x=a
General Solutions of the 1D Schrodinger Equation
!! 2
" ''+V(x)" = E"
2m
2m
(E !V(x))" = 0
let k =
!2
" ''+ k 2" = 0
Harmonic equation
2m
(E !V(x)) the particle wave number
!2
" ''+
E < V0
E >V = 0
1) Re gion I and III
" (x) = A e +$ x + B e !$ x
2) Re gion II E < V
! III (x) = T e +ikx
q = 2m E "V0 / !
2mE / !
k=
Apply Boundary Conditions
1) ! I (0) = ! II (0)
1+ R = A + B
2) ! 'I (0) = ! 'II (0)
Ae
4) ! 'II (a) = ! 'III (a)
q(A e
+B e
+qa
"qa
"B e
(
4k q + k + q
2
2
2
2
!k
(1! | R |2 )
m
)
2
T =
(
4k 2q 2 + k 2 ! q 2
)
2
sinh (qa)
j x>a =
sin2 (qa)
) = ikT e +ika
R =
2
4k 2q 2 sinh(qa)
+ika
(k
2
!k
(|T |2 )
m
E > V0 Let q = iq !
2
k(1" R) = q(A " B)
=T e
"qa
4k 2q 2 sinh(qa)
2
j x<0 =
+qa
2mE / !
k(1+ R) = k(A + B)
k(1" R) = q(A " B)
3) ! II (a) = ! III (a)
T =
( k is real )
( k = i$ complex)
! II (x) = A e +qx + B e "qx
! I (x) = I e +ikx + R e "ikx
k=
" (x) = A e ikx + B e !ikx
#
2
R =
+ q2
(
)
2
4k q + k + q
2
2
2
2
sinh(qa)
2
)
2
! q2
(
)
2
sinh(qa)
4k 2q 2 + k 2 ! q 2
)
2
2
2
sinh (qa)
j x<0 = j x>a "
and
(k
2
sin2 (qa)
2
1= T + R
1= T + R
2
2
1= T + R
2
2
5-4 WKB Approximation
If the potential is slowly varying wrt incident wavelength λ=2π/k then we can make an approximate
solution to the Schrodinger equation called the WKB approximation.
2
k (x)
"$$
#$$
%
2m
SOLUTION
! ''+ 2 (E " V (x))! = 0!!!!!! #PROPOSED!!
######
$
!
i
! (x)' =!!ik(x)!e %
k(x)!dx
i
!!! WKB (x) =!!e %
k(x)!dx
!!!!!!!!!!!!
=!ik(x)!! (x)
! (x)'' =!ik(x)'!! (x) + ik(x)! (x)'!!!!!!$!!!!!!!!!! (x)''" !ik(x)'! (x)!+k 2 (x)! (x) = 0!
dk(x)
'x
!<<!!k 2 !!OR!!!! & 2 <<!
!!!!!Wavelength!small!wrt !size / strength!of ! potential
dx
'V
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Barrier Transmission Problem
!!
In the barrier transmission problem we are asked what is the probability that a matter wave of energy
E=
p2
2m
will transdfer through a potiential barrier V(x). Classically the wave would be reflected.
But QM allows the wave to tunnel through the barrier.
V(x)
b
!
! T =!e a
a
E
V=0
b
Tunneling
wave
k(x)!dx
2m
(V (x) " E)
!2
Tunneling!probality
k(x) =
b
PT = T = e
2
!
2
!
a
k(x)!dx
!
Cold Emission
V(x) = W ! e
Alpha Decay
" x
VCoulomb (x) =
V = VNuclear +VCoulomb
W
!e
"x
0
W/eε
In cold emission electrons emerge from a
metals surface when a large electrostatic
potential gradient V=-qEx is applied. These
electrons tunnel through the potential
barrier. W=work function.
Fowler-Nordheim Formula
$ 2 2m W /e"
2
T =!exp & !
#0
!
%
Z1 Z2 e 2
4#" 0 r
'
W ! e" x )
(
$ 4 2m1/2W 3/2 '
!!!!!!!=!exp & !
)
3!e"
%
(
In alpha decay alphas emerge from a nucleus
by tunneling through the the Coulomb barrier.
The tunneling rate depends on the apha energy
and nuclear radius.
$ 2 2 $ Z Z e2 mR ' 1/2 Ec / E 1 E '
2
T =!exp & ! 2 & 1 2
*1 y ! Ec dy)(
)
% ! % 4"# 0 (
$ 2 2 $ Z Z e2 mR ' 1/2 $ "
!!!!!!=!!exp & ! 2 & 1 2
) &
% ! % 4"# 0 ( % 2
''
Ec
! 2) )
E
((
2
dN = !N+ T dt !!!,!!!N(t) = N(0)e!+ T t !!!!
2
!!!!!!
- = + T .!decay!rate!!!!!!/ = 1 / - .!lifetime!
2
STM – Scanning Tunneling Microscope
-Vo
gnd