approximation methods

How does the variational method help us find a solution

It restricts us to a subspace of the hilbert space, when we minimise the energy it is alwys >= Eg. By adding more basis functions we expand the subspace and the minima of our energy functional decreases or stays the same

How can symmetry be used to simplify linear variational problems

If the basis is non-orthogonal, it still holds that basis functions having different eigenvalues under the symmetry-operator R are orthogonal and that the matrix elements of the Hamiltonian between them vanishes. This means that the Hamiltonian matrix and the overlap matrix have the same block-diagonal structure. Therefore, the respective generalised eigenvalues for the blocks can be dealt with independently of each other and we achieve the same speed-up.

What characteristics should the trial wave function exhibit

1. It should be normalizable, so that total probability is 1

2. It should fit the potential, for instance it should obey the same symmetries as the potential, have the maxima of phi at enrgy minima, go to zero at boundaries

3. Ground state has no nodes→ should not cross y=0

what happens when a potential encounters a sharp change in potential

part of it gets reflected back

what is the relation between the differential crossection and the amplitud of the scattered wave

dsigma/domega = |f|²

For scattering if q=(k-k’) and k and k’ are of equal magnitud, what is q²

q² = (k-k’)*(k-k’) = k² + k’² -2kk’cos(a) = k²(1-cos(a)) q points in the z direction.

what is domega

sin(theta) dtheta dfi

when doing partial wave analysis in what case is it valid

The potential has spherical symmetry. As a result, in the sum given by Eq. (7),

only the m = 0 terms are kept at each value l. The scattered states l, m̸ = 0 are

discarded.

This can be understood as a consequence of angular momentum conservation. The

incident plane wave indeed carries no orbital momentum, so its m = 0. Because the

potential is spherically symmetric, m is conserved upon scattering. The scattering

states then need to have m = 0

When working in the low energy limit for scattering and the energy is given by h²k²/2m what can be done to simplify integration

Low energy limit means k is small and therefore q, this means we can taylor expand