12_variation_perturbation_matrixhq

444 e 0790 1234 1777 find eigenvalues e sort

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: H E c2 eigenvec H E v Eigenvectors give the coefficients v 1 1.000 0.000 0.000 c0 0.000 0.000 0.000 v 2 0.000 1.000 0.000 c1 0.000 0.000 0.000 coefficients and basis functions give the wavefunctions 0( x) c0i 2 a 1( x) ( i 1) x a a 2 0( x) dx 1.000 0 c1i 2 a ( i 1) x a 0 a each eigenfunction has only one coefficient, i.e. that basis function is the eigenfunction a 2 1( x) dx 1.000 sin i 2( x) check normalization sin i 0.000 0.000 1.000 c2 0.000 0.000 0.000 0 c2i 2 a ( i 1) x a a 2 2( x) dx 1.000 sin i 0 0.6 x 0( x) .2 E v 1( x) .2 E v 10.4 Looks just like our analytical solution 2( x) .2 E v 2 Ev E v 1 0.2 E v 2 0 0 2 4 6 8 10 x 13 10/7/2013 Ramp Potential nmax 6 pick the max number of basis functions i 0 nmax 1 set up indices over the basis functions j 0 nmax 1 m 1 a 10 set constants for problem, atomic units hbar 1 set ramp to nonzero value Hamiltonian Matrix Elements - Note that the eigenvector and eigenfunction routines need the origin of matrices to be zero. Our first wavefunction has...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online