Download Quantum Physics 2005 Notes-6 Solving the Time Independent Schrodinger Equation

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Transcript
Quantum Physics 2005
Notes-6
Solving the
Time Independent Schrodinger Equation
(A Numerical Approach)
Notes 6
Quantum Physics F2005
1
The Schrodinger Equation
Hˆ ! = Eˆ!
h2 " 2
"!
#
! + V ( x, t )! = ih
2
"t
2m "x
2
" ! ( x)
2m
= # 2 ( E # V ( x))! ( x)
2
"x
h
Notes 6
Quantum Physics F2005
2
A dimensionless form for calculations
#The coefficient
2m
38
has
a
value
of
~
2
$
10
for an electron
2
h
in MKS units .
#Big and small numbers are problems in repetitive
numerical calculations, so we will scale it out by redefining
making the original length and energy variables dimensionless.
#The choice of scale is arbitrary, so we can make it
to fit the problem.
Notes 6
Quantum Physics F2005
3
A specific example
#For a finite square well of width L, we expect the energies to be
h2$ 2
of order: E0 =
and useful lengths to be of order L.
2
2mL
#As a concrete example, let's take a finite square well:
L = 10 nm and m = 9 x10-31 kg
1.242
Thus, E0 =
~ 0.00375 eV
4.09 •100
We will also take V0 = 4 E0 to finish setting up the example.
Notes 6
Quantum Physics F2005
4
Letting z = x / L and scaling % =E / E0 and W = V / E0 we have:
" 2! ( z )
2 (V ( x ) # E )
2
=
(
$
)
!
(
)
=
(
$
)
z
(W # % )! .
2
"z
E0
We are now ready to solve this TISE numerically.
Convert the TISE to a finite difference equation.
a) z ⇒ z j = j &z (step size chosen for accuracy and time)
b) ! ( z ) ⇒ ! ( z j ) ' ! j
c) W ( z ) ⇒ W ( z j ) ' W j
!
'
j+1/2
"
j
! =
Notes 6
!
(
'
j +1
#! j )
&z
'
'
! j+1/2
#! j-1/2
&z
=
! j +1 # 2! j +! j #1
&z
Quantum Physics F2005
5
Making a difference equation
" 2! ( z )
2
Plugging the differences into
=
(
$
)
(W # % )! .
2
"z
We get:
! j+1 =  2 # &z 2 ($ ) 2 ( % # W j ) ! j #! j #1
If we know the values of ! j and ! j #1 near some point, we
can solve for ! j+1.
We can usually get ! j and ! j #1 from the continuity or symmetry
conditions at a point.
The only parameter with which to achieve agreement of
the wavefunction with expected behavior is % .
Notes 6
Quantum Physics F2005
6
The starting point
• We can choose the starting point based on
symmetry.
• For the finite well, even solutions have zero
slope at the center. (Let’s take !0= !1=1.)
• Expected asymptotic behavior: For z growing
to large values (outside the well) we expect
the wavefunction to go to zero.
Notes 6
Quantum Physics F2005
7
A spreadsheet example
Position !j
Wj
0
1
0
1
1
0
2
 2 # &z 2 ($ ) 2 ( % # W j ) ! j #! j #1


0
3
..
0
..
..
4
n
Vary E to get the right
behavior out here.
4
file: C:\Documents and Settings\Peter Persans\My
Documents\classes\quantumphysics\lectures\numerical_integration_TISE_even_square_well.xls
Notes 6
Quantum Physics F2005
8