Download Quantum Physics 2005 Notes-5 Solving the Time Independent Schrodinger Equation

Quantum Physics 2005
Notes-5
Solving the
Time Independent Schrodinger Equation
(Chapters 8-9)
Notes 5
Quantum Physics F2005
1
The Schrodinger Equation
Hˆ ! = Eˆ!
2
2
h "
"!
#
! + V ( x, t )! = ih
2
"t
2m "x
Notes 5
Quantum Physics F2005
2
TISE reminder
h 2 " 2! ( x)
#
+ V ( x)! ( x) = E! ( x)
2
2m "x
h 2 " 2! ( x)
+ ( E # V )! ( x) = 0
2
2m "x
Notes 5
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3
Piecewise potentials 1
Solution to the TISE in a region where V is a constant
and E>V:
h 2 " 2! ( x)
+ ( E # V )! ( x) = 0
2
2m "x
Guess: ! = eikx
h2k 2
#
= ( E #V )
2m
! = Aeikx + Be# ikx
Notes 5
⇒ k=±
2m ( E # V )
Quantum Physics F2005
h2
4
Piecewise potentials 2
Solution to the TISE in a region where V is a constant
and E<V:
h 2 " 2! ( x)
# E # V ! ( x) = 0
2
2m "x
Guess: ! = eikx
h2k 2
= # E #V
2m
Let K = ik = ±
⇒ k = ±i
2m E # V
! = Ae Kx + Be # Kx
Notes 5
2
2m E # V
h2
( $ K is a real number)
h
(exponential growth or decay)
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Piecewise potentials 3
• Many quantum physics problems can be
solved by using the solutions for constant
potential in each region and then matching up
the solutions and derivatives at the
boundaries.
– Wavefunctions must always be matched.
– Derivatives must be matched if the are no infinite
potential steps in the problem.
Notes 5
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The “Local” wavelength
• The basic solution in a constant potential
always results in a constant k, related to the
energies E and V. It is useful to think of the
wavefunction as having a wavelength in each
region,
2m( E # V )
! =e ; k =
h2
ikx
! =e
# Kx
Notes 5
; K=
2m ( E # V )
h
2
in a classically allowed region
; in a classically forbidden region
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The Step Potential
U
E
0
x
x=0
• State general solutions in each region.
• Carefully eliminate non-physical possibilities.
• Match wavefunctions the boundary.
– Two possibilities E>V and E<V.
Notes 5
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Solutions in two regions
! 1 ( x, t ) = ( Aeik x + Be#ik x )
1
1
! 2 ( x, t ) = Ceik x
2
1)
A + B = C (continuity)
2) ik1 A # ik1 B = ik2C
( "smoothity" )
B = C # A ⇒ i 2k1 A = i ( k2 + k1 ) C ⇒ C =
B=
2k1
A
k2 + k1
k1 # k2
A
k2 + k1
C
2 2mE / h 2
=
=
In terms of energy:
2
2
A
2mE / h + 2m( E # V ) / h
2 E
E + (E #V )
For E>>V, C/A ⇒ 1. For E=V, C/A ⇒ 2/ V
Notes 5
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Probability current in two regions
#i h  * "
" *
%
%
#
%
% 

2m 
"x
"x

which for a pure momentum state is:
j ( x, t ) =
j = k %*%
hk1 2
A
m
hk
jreflected = # 1 B 2
m
hk
jtransmitted = + 2 C 2
m
jincident = +
2
2k1k2
jtransmitted k2C 2 k2  2k1 
T&
=
=
=


jincident
k1 A2 k1  k2 + k1  ( k2 + k1 ) 2
Notes 5
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10
and for E<V
1)
A + B = C (continuity)
2) ik1 A # ik1 B = # KC
( "smoothity" )
B = C # A ⇒ i 2k1 A = ( # K + ik1 ) C ⇒ C =
B=
ik1 + K
A
ik1 # K
2k1
A
# K + ik1
hk1 2
A
m
hk
hk
jreflected = # 1 B 2 = # 1 A2
m
m
#i h  * "!
"! * 
!
jtransmitted =
#!
=0


2m 
"x
"x 
jincident = +
2
jtransmitted k2 C 2 k2  2k1 
2k1 k2
T&
=
=
=


2
jincident
k1 A2 k1  k2 + k1 
( k2 + k1 )
R =1
Notes 5
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Notes 5
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12
R = 1 for E/Vo <1
For E/Vo > 1
R=1–T

Vo 


1
1
#
#
=
E 


Vo 
1+ 1# 
E 

Notes 5
Quantum Physics F2005
2
13
The Step Barrier
(Much of this section is from Prof. Wang’s notes (2004))
Notes 5
Quantum Physics F2005
14
Barrier Potential
0, x < 0
V ( x) = V0, 0 < x < a
0, x > a







Notes 5
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15
V(x)
Case 1 E < V0
Vo
E
0
! ( x) =
{
Ae ik 1 x + Be # ik 1
Ce # k 2 x + De k 2 x
Aˆ e ik 1 x + Bˆ e # ik 1 x
a
x
x<0
0<x<a
x>a
(5-73)
k = 2mE
h
where 1
2m(V # E )
0
k =
2
h
Notes 5
Quantum Physics F2005
16
use
Bˆ = 0
for
x>a
! (o) continuous  at x = 0

! '(o)
"

! (a) continuous  at x = a

"
! '(a)

Express B, C , D, Aˆ in terms of A.
Notes 5
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17
x=0
! (x ) :
! (0) :
! ' ( x) :
! ' (0) :
= Ce # k 2 x + De k 2 x
Ae ik1 x + Be # ik1 x
A
=C
B
+
ik1 Ae ik1 x # ik1 Be # ik1 x
! (a) :
! ' ( x) :
! ' (a ) :
Notes 5
Note
(
# k 2 Ce # k 2 x # De k 2 x
(
Bˆ = 0
for
ik1 x
ˆ
= Ae
= Aˆ e ik1a
Ce # k 2 x + De k 2 x
Ce # k 2 a + De k 2 a
)
# k 2 Ce # k 2 a # De k 2 a
)
(1)
= #k2Ce# k2 x +k 2Dek2 x
= # k 2 (C # D )
ik1 ( A # B )
x=a
! (x ) :
+ D
= ik1 Aˆ e ik1 x
ik1a
ˆ
= ik1 Ae
Quantum Physics F2005
(2)
x>a
(3)
(4)
18
2
2
A
1  k2 k1 
= 1 +  +  sinh k2 a
4  k1 k2 
Aˆ
2
2
B
1  k2 k1 
=  +  sinh k2 a
4  k1 k2 
Aˆ
Transmission coefficient T
2
ˆA
=
T=
A
Notes 5
1



k
k
1
 2
1+  + 1 sinh 2 k a
2
4  k k 
2
 1
Quantum Physics F2005
1
=
1+
sinh 2 k 2 a
E 
E 
4
1#

V0  V0 
19
Reflection coefficient R
2 ˆ2
B
B A
R=
=
A
Aˆ A
2
2
2


k
k
ˆ
1 2
A
2
1

sinh k2a '
=
+
4  k1 k2 
A
Recall
Note:
Notes 5
2
ˆA
T=
=
A
1



k
k
1
 2
1+  + 1 sinh 2 k a
2
4  k k 
2
 1
T + R =1
Quantum Physics F2005
20
Approximate transmission coefficient when
1
a〉〉
k2
E 
E  #2k2a
T = 16
1#
e
V0  V0 
1 u #u
sinh u = e # e
2
(
)
and
u = k2a
This form is frequently used as an approximation in tunnelling
calculations.
Examples: STM, Quantum wells and barriers in
semiconductors, nuclear decay
Notes 5
Quantum Physics F2005
21
Case 2 E > V0
E
V(x)
Vo
0
x
a
 Aeik1x + Be#ik1x
x<0


! ( x) =  ik x
#ik3 x
3
0 < x < a note k3
+ De
Ce
 Ae
#ik1x
ˆ
ˆ
1 x + Be
ik
x >a

Notes 5

2mE
 k =
1
h
where 
2m( E # V )

0
k
=
 3
h
Quantum Physics F2005
22
(
Follow the method used in case 1 E < V0
)
M
T=
1
sin 2 k3a
1+

E  E
4
# 1
V0  V0 
Ramsauer effect:
Note if k a =( , 2( , 3( ...,
3
then
T =1 and R = 0.
Notes 5
Quantum Physics F2005
23
This happens when the length of a of the barrier region
is equal to an integral or half integral number of
wavelength )3 = 2( / k3 in that region.
This is a result of destructive interference between
reflections at x = 0 and x = a.
Notes 5
Quantum Physics F2005
24
from Eisberg and Resnick
Notes 5
Quantum Physics F2005
25
The Finite Square Well
Notes 5
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26
V ( x) =
! (x) =
With
Notes 5
{
{
0
#
V0
a
a
< x<
2
2
x >
a
2
A sin kx + B cos kx
Kx
Ce + De
# Kx
Fe Kx + Ge # Kx
2mE
k=
h
And
Quantum Physics F2005
a
a
# <x<
2
2
a
x<#
2
a
x>
2
2m(V0 # E )
K=
h
27
• It is frequently best to start such a problem by
sketching the form of the solutions.
• We can do this by analogy with the infinite box,
which we have previously solved.
• The solutions are sin(nk0x) and cos (nk0x), which
come to zero at the boundaries.
• Our guessed solution should look like this, but be
relaxed at the boundaries to allow tunnelling into
the barriers.
• Solutions will be even (cos) or odd (sin). We will
want to use this symmetry to do less work to find
the solutions.
Notes 5
Quantum Physics F2005
28
The finite values of !(x) for all x requires
D = 0 and F = 0
! ( x) =
Notes 5
{
A sin k 1 x + B cos k 1 x
Ce k 2 x
Ge
#k2x
a
a
< x<
2
2
a
x< #
2
#
x>
Quantum Physics F2005
a
2
29
We can use symmetry to reduce the number of parameters
For the even case:
 Ae Kx
x < -L / 2

! n ( x) =  B cos kx
 Ae # Kx
x > L/2

and for odd:
 Ae Kx
x < -L / 2

! n ( x) =  B sin kx
 # Ae # Kx
x > L/2

Notes 5
Quantum Physics F2005
30
Symmetry, continuity and smoothity
Then, for the odd parity solutions
Notes 5
Quantum Physics F2005
31
From these equations alone, we have found the allowed energies:
K
 kL 
= tan   for odd n (even parity)
k
 2 
K
 kL 
= # cot   for even n (odd parity)
k
 2 
The even parity state will be the lowest energy state.
These equations cannot be solved in closed form, but
we can find the numerical solutions using "solve" in Maple
or by plotting each side as a function of k.
On the next slide we'll switch to terms of energy
because everyone does.
Notes 5
Quantum Physics F2005
32
In terms of the energies:
for even parity
 L 2m(V # En )   V

tan 
 =  # 1
2
h
 2
  En 
#1/ 2
#1/ 2
 L 2m(V # En ) 
V

for odd parity cot 
# 1
 = #
2
h
 2

 En 
In the text he sets a potential strength parameter:
mL2
* =
V and uses
2
2h
2
mL
+ = * 2 # 2 En as the variable.
2h
Notes 5
Quantum Physics F2005
33
Notes 5
Quantum Physics F2005
34
For even parity
The eigenfunctions are
! (x) =
Notes 5
a
k

a  2 2  k2 x
 B cos k1 e  e
2


B cos k1 x
a
k

a  2 2  #k2 x
 B cos k1 e  e
2


Quantum Physics F2005
a
x<#
2
a
a
# <x<
2
2
a
x>
2
35
Notes 5
Quantum Physics F2005
36
For odd parity solutions
The eigenfunctions are
! (x) =
Notes 5
a
k

a  2 2  k2 x
 A sin k1 e  e
2


A sin k1 x
a
k

a  2 2  #k2 x
 A sin k1 e  e
2


Quantum Physics F2005
a
x<#
2
a
a
# <x<
2
2
a
x>
2
37
• Summary of eigenfunctions and eigenvalues
Notes 5
Quantum Physics F2005
38
The Harmonic Oscillator
Notes 5
Quantum Physics F2005
39
5-5 The Harmonic Oscillator
This applies to a diatomic
molecule and any system
with a stable equilibrium
configuration.
Notes 5
Quantum Physics F2005
40
•
Step 1
1 2
V (x) = kx
2
A particle experiences
a force
dV
F =#
= # kx
dx
At classical turning point x
2E
x = ±A = ±
Notes 5
k
Quantum Physics F2005
41
• Step 2
Write
h2 d 2
#
! (x ) + V (x )! ( x ) = E! (x )
2
2m dx
! ( x)
as
!
2
d
2m  k 2

! = 2  x # E !
2
dx
h 2

d2
mk  2 2 E 
! = 2 x #
!
2
dx
k 
h 
2
 mk 2 2 E
h d
! = 
x #
2
h
mk dx
 h
Notes 5
Quantum Physics F2005
m
!
k 
42
Change variable
{
Let
Then
mk 2
, =
x
h
2E m 2E
=
)=
h k h- o
2
d2
2
!
=
,
#) !
2
d,
(
)
Assume ground state only
! (, ) = e
Notes 5
#
,2
2
d2
2
!
=
,
# 1!
Provided
2
d,
Quantum Physics F2005
(
)
)0 = 1
43
• Step 3
Continuity of !(,) at x = ± A
Recall
Classically
,max
Notes 5
2
mk 2
, =
x
h
2
1 2
E = kA
2
, max
2
mk 2
=
A
h
2E
A =
k
2
mk 2E 2E m 2E
=
=
=
=)
h k
h k h-0
Recall
k
m
Physics F2005
-o Quantum
=
44
Therefore
2 E0
= )0
h-0
Ground state
h-0
1
E0 =
)0 = h-0
2
2
See Fig. 5-9
Notes 5
Quantum Physics F2005
45
Notes 5
Quantum Physics F2005
46
• Step 4
! 0 ( x) = Ae
#
,2
2
See Fig. 5-9
• Step 5
Notes 5
%0 ( x, t ) = Ae
Quantum Physics F2005
mk 2
#
x
2h
e
iE0t
#
h
47
• Step 6
.
∫
#.
.
∫
#.
%0 ( x, t ) dx = 1
mk 2
x
#
2
h
Ae
2
dx = 2 A
2
.
∫
0
e
mk 2
x
#
h
dx
From integration table
.
∫
e
#a 2 x2
#.
.
∫
0
e
dx =
#a 2 x2
(
.
Change lower limit
a
dx =
e
#.
.
dx = 2 ∫ e
#a 2 x2
0
(
2a
1
4
2 2

(
h 
2
2
 & 1
2A
= A 
1
 (mk )4 
 mk 
⇒

2
Notes 5 
Quantum Physics F2005
h 


(
∫
#a 2 x2
 mk 
A= 2 2 
( h 
1
8
48
dx
Substitute the normalized A in to the wavefunction
Then the ground state eigenfunction is
1
8
 mk 
%0 ( x, t ) =  2 2  e
( h 
Notes 5
Quantum Physics F2005
mk 2
x
#
2h
e
iE0t
#
h
49
Figure 5-9
Energy levels and eigenfunctions for the first four stationary states of
the harmonic oscillator.
Shaded areas represent
the penetration of the
wave function into regions
where classical motion
is forbidden.
Eigenvalue:
Notes 5
1
h-o

En =
)n = h-o  n + 
2
2

Quantum Physics F2005
50
Quantum:
Probability densities for bound states
* ground-state n = 0
Notes 5
Quantum Physics F2005
51
N = 10
Notes 5
Quantum Physics F2005
52
Classical probability:
n/.
Notes 5
Quantum Physics F2005
53
Eigenfunction other than the ground state
! (, ) = H (, )e
Where
d
! = 000
d,
2
d
! = 000
2
d,
}
We got
Notes
Try
to5 solve H?
#
,2
2
H (, ) is Hermite polynomial
2
substitute into
d
2
! = , #)!
2
d,
(
)
2
d
d
H + () # 1)H = 0
H # 2,
2
d,
d,
Quantum Physics F2005
54
Since V(x) is a symmetric function of x, one expects both even and odd
functions in H(,).
n
k
a
,
∑ k
H n (, ) =
even k =0
Even n
n
k
a
,
∑ k
Odd n
odd k =1
Notes 5
Quantum Physics F2005
55
Take the even-order polynomial as an example,
2
4
H n = a0 + a2, + a4, + 0 0 0 + an,
Where
n
an 1 0
d
Hn = 000
d,
d2
Hn = 000
2
d,
}
into
2
d
d
H + () # 1)H = 0
H # 2,
2
d,
d,Quantum Physics F2005
Notes 5
(1)
56
Sort out the coefficients of each power of ,.
For , n the coefficient is
# 2nan + () #1)an = 0
() # 2n # 1)an = 0
⇒ ) = 2n + 1
Notes 5
Quantum Physics F2005
(2)
57
Recursion relation
ak + 2
2(k # n)
ak
=
(k + 1)(k + 2)
Combine equations (1) & (2) which gives the Hermite
differential equation.
d2
d
H n # 2,
H n + 2nH n = 0
2
d,
d,
First few Hermite Polynomials
H 0 (, ) = 1
H1 (, ) = 2,
H 2(, ) = 4, 2 # 2
H 3(, ) = 8, 3 # 12,
Notes 5
0
Quantum Physics F2005
58
The 3D Box – Quantum well
Notes 5
Quantum Physics F2005
59
5-11 The 3-D Box
a a


x in # , 
0
 2 2

b b


V ( x, y , z ) = 
y in # , 
 2 2


c c

. otherwise x in # 2 , 2 



The 3D Box problem is a straightforward extension of the
1D infinite well or 1D box problem where

# a , a 
0
x
in

 2 2 
V ( x) = 
. x > a

2
Notes 5
Quantum Physics F2005
60
Px
1- D
V (x)
Px / h"
i"x
E / ih
"
"t
2
Px
+ V ( x) = E
2m





Notes 5
2
h" %
i "x
+V ( x)% = ih " %
2m
"t





2 "2
2
i
"
h
h
%
#
% +V ( x)% =
"t
2m Quantum
"x2 Physics F2005
61
3 #D
Px, Py , Pz
V ( x, y , z )
Px / h " , Py / h " , Pz / h "
i "z
i "x
i "y
E / ih
2
2
Px + Py + Pz
2m
"
"t
2
+ V ( x, y , z ) = E
"
h 2  " 2
"2
" 2 
#
%+
%+
% + V ( x , y , z ) % = ih %
"t
2m  "x 2
"y 2
"z 2 
Notes 5
h2 2
"
)% =F2005
#
2 % + V (Quantum
x, y, zPhysics
ih %
2m
"t
62
1- D
• Volume element d3 = dx
• Probability of finding a particle in d3 at time t
2
.
P( x, t ) = ∫#. % ( x, t ) dx
• Normalization condition
2
.
∫#. % ( x, t ) dx = 1
• Stationary State wave function
# iEt
%( x,t) =! ( x)e h
&&dinger equation
• Time independen t Schro
2 2
# h " ! +V ( x)! = E!
2m "x2
• Eigenfunction
Notes 5
! (x )
Quantum
n Physics F2005
63
• Volume d3 = dx dy dz
3 -D
• Probabilit y of finding a particle in d3 at time t
2
P( x, y, z,t) = %( x, y, z,t) dx dy dz
• Normlization conditon
.
#.
.
#.
2
.
#.
∫ dx ∫ dy ∫ dz % ( x, y, z , t ) = 1
• Stationary state wave function
% ( x, y , z , t ) = ! ( x , y , z ) e
#
iEt
h
• Eigenfunction ! n( x)
&&dinger equation
• Time independen t Shcro
2
# h 22! ( x, y, z) +V ( x, y, z)! = E!
2m
• Eigenfunction ! n ( x), ! n ( y), ! n ( z)
Notes 5
1 Quantum2 Physics F20053
64
1D
! n ( x)
a
x4
2
! n ( x) =
2 cos  n( x
a sin  a
x>
a
2
! n ( x) = 0
Subsitute ! back into time independent Schr&o&dinger
eq. to obtain allowed energies for the particle
2
2 h2(F2005
Notes 5
Quantum
En = nPhysics
2ma2
65
3D
! n n n ( x, y, z) =! n ( x) ! n ( x) ! n ( z)
1 2 3
1
a
x4 ,
2
! n n n ( x, y, z) =
1 2 3
2
b
c
y 4 ,z 4
2
2
2  cos  n1( x
a sin  a
3
n(


2
cos
•   2 y
b sin  b
n(


2
• c  cos  3c z
 sin 
Notes 5
Quantum Physics F2005
66
x > a, y > b, z > c
2
2
2
! n n n ( x, y, z) = 0
1 2 3

2
2
2
n   n   n   h 2( 2
En n n = 1  +  2  +  3  
1 2 3 a   b   c   2m








Notes 5

Quantum Physics F2005
67
Degeneracy & symmetry
A degeneracy always reflects the existence of some
symmetry in a given problem.
E.g. 1 - 3D box with a = b 1 c

2
2
2
+
n1 n2 n3  h2( 2
+ 
En n n =
a2
c2  2m
1 2 3









Exchange n1 & n2, En1n2n3 stays the SAME.
But
! n n n &! n n n are DISTINCT.
1 2 3
because
!n n n
213
( x, y, z,t ) &
!n n n
( x, y, z,t )
1 2 3
213
Notes 5
Quantum Physics F2005
differ in their dependence
on x & y.
68
• Symmetry here is interchang e coordinate s x & y.
•Degeneracy here is two DISTINCT eigenfunctions
having the SAME energy value.
See Fig. 5-27
112
a = b 1 c box case #
#211,121
E.g. 2
Fig. 5-27
a=b=c
box case # 211,121,112
See Figure 5 - 27 a 1 b 1 c box
Notes 5
# 112
# 211
# 121Quantum Physics F2005
69
Notes 5
Quantum Physics F2005
70