# Deuteron Ground State / Perturbation Theory

1. Sep 25, 2012

### EEWannabe

1. The problem statement, all variables and given/known data
The deuteron ground state is made up of l = 0 and l = 2 states;
a)Show this mixture cannot be an eigenstate of a central potential Hamiltonian
b)Using first-order time independent perturbation theory, argue the potential must contain a term proportional to some combination of Y2m for this mixing to be possible.
c) However Y2m are not scalars, they're components of rank 2 tensors. The Hamiltonian must be a scalar so how can one construct a perturbation that is a scalar yet contain this rank 2 tensor?

2. Relevant equations

$H = (\frac{p^{2}}{2m} + V(r))$
$\psi_{total} = c\psi_{l=0} + d\psi_{l=2}$
$\psi_{l=0} = \frac{u_{0}(r)}{r} Y_{0m}$
$\psi_{l=2} = \frac{u_{2}(r)}{r} Y_{2m}$

3. The attempt at a solution

a) Since r is the same for both of them the fact that $H \psi = E \psi$, this implies that

$\frac{p^{2}}{2m}c \psi_{l=0} + \frac{p^{2}}{2m}c \psi_{l=2} = D (c \psi_{l=0} + d \psi_{l=2}) = (E - V(r))(c \psi_{l=0} + d \psi_{l=2})$

where D is a constant, however this is not true as as the two psi functions are not of the same form, hence they are not an eigenstate if the potential is constant.

b) I'm really stuck on this bit, i'm not really very familiar with time independant pertubation theory but i've been trying to teach myself, so I let the hamiltonian equal sometime of the form;

$H_{new} = H_{0} + V$

I've tried playing around with the algebra trying to figure out why V(r) would be proportional to Y2m but i'm really stuck, and I can't envisage why it'd be proportional to Y2m but not Y0m anyway.

I'm sorry but I really don't know where to go, if anyone could point me in the right direction that'd be great thanks.