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v m n,( )0
L
xV x( ) m x,( ) n x,( )
d:=
V x( )1
2k x xe( )
2:=
Here, we define the potential energy function for the harmonic
oscillator and then supply the formula for
evaluating matrix elements of the potential.
Ebox n( )hbar
2n
2 2
2 L2:= n x,( )
2
Lsin
n xL
:=
x.exe 2.5:=hbar 1:=L 5:=k 0.2:= 1500:=
This MATHCAD worksheet uses "particle-in-a-box" wave functions
as a basis set to solve the harmonic
oscillator problem.
= reduced mass L = width of the box k = force constant xe =
equilibrium bond length
The values selected are appropriate for HCl. All quantities are
expressed in "atomic units".
herm n x,( ) 1 n 0=if
2 x( ) n 1=if
2 x herm n 1 x,( ) 2 n 1( ) herm n 2 x,( )[ ] otherwis
:=
This defines the Hermite polynomial, Hn(x).
eigensort S,( ) M augment ST,( )M csort M 1,( )
M1
S submatrix M 1, rows M( ), 2, cols M( ),( )T
S( )
:=
This defines the function "eigensort" which you will use to sort
the eigenvectors.
The assignment template will define this for you. Need we say,
"Ne touche pas"?!
Chemistry 450
Mathcad Demonstration 1
ORIGIN 1:=
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E n( )
0.00577
0.01732
0.02887
0.04041
0.05196
0.06351
0.07506
0.08660
0.09815
0.10970
=
E v( ) hbar v1
2+
:=
Evn 1+0.00727
0.02302
0.05334
0.07821
0.14000
0.16879
0.26983
0.29558
0.44299
0.45867
=n0
1
2
3
4
5
6
7
8
9
=
k
:=
n 0 9..:=|
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Ev are the approximate energies for the harmonic oscillator and
E(n) are the exact energies. How well
do they compare?
Ev C( ) eigensort Ev C,( ):=
C eigenvecs H( ):=Ev eigenvals H( ):=
.... and now we find the eigenvalues and eigenvectors of the
hamiltonian matrix. A quick "call" to
the "eigensort" function re-arranges the eigenvalues and
eigenvectors so that they are energy-ordered.
H1 1, 0.082=m,n : h(m,n)HHm n, h m n,( ):=
n 1 nmax..:=m 1 nmax..:=
nmax 10:=
Here, we construct the hamiltonian matrix.
h 1 1,( ) 0.082=h m n,( ) Ebox n( ) m n,( ) v m n,( )+:=
m n,( ) 1 m n=if
0 otherwise
:=
Here we define the formula for evaluating matrix elements of the
harmonic oscillator hamiltonian.
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(v,x) are the approximate wave functions. ho(v,x) are the exact
wave functions.
v x,( )
1
nmax
m
Cm v 1+, m x,( )=
:=
_____________________________________________________________________________________
z x( ) hbar
1
2
x xe( ):=
ho v x,( )1
2v
v!
1
2 hbar
1
4
herm v z x( ),( ) e
z x( )2
2
:=
Here, we plot the approximate wave function and the exact wave
function on the same set of axes.
How do they compare?
x 0 0.01, L..:=
0 1 2 3 4 51
0
1
2
0 x,( )
ho 0 x,( )
x
_____________________________________________________________________________________
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