GREPhysics.NET: GR8677 #96

Well let's TeX out the derivation in case somebody needs it. We are supposing that our Hamiltonian is a sum of an unperturbed hamiltonian which is $H_0$ and a small pertubation $V$ . Then our stationary Schrodinger equation becomes $H\psi=H_0\psi + V\psi=E\psi$ . As the solutions of the unperturbed system make a complete set of functions our function of the perturbed system must be a superposition of the functions $\psi^{(0)}_n$ , i.e. solutions of the equation $H_0\psi=E\psi$ . In other words $\psi=\sum_n a_n\psi^{(0)}_n$ . Replacing this into our initial Schrodinger equation we get $H_0\sum_n a_n\psi^{(0)}_n+ V\sum_n a_n\psi^{(0)}_n=E\sum_n a_n\psi^{(0)}_n$ or $\sum_n a_nE_n\psi^{(0)}_n+ \sum_n a_nV\psi^{(0)}_n=E\sum_n a_n\psi^{(0)}_n$ , where $E_n$ are energy levels of unperturbed system. We now multiply both sides by $\psi^{(0)}_m^*$ and integrate to get $a_mE_m+ \sum_n a_n\int\psi_m^{(0)*}V\psi^{(0)}_ndV=Ea_m$ or $a_m(E_m+V_{mm}-E)+ \sum_{n\ne m} a_nV_{mn}=0$ . where $V_{mn}=\int\psi_m^{(0)*}V\psi_n^{(0)}dV$ It would be beneficial to memorize this result for the future as from it is easier to proceed with almost any problem conserning perturbation. Note that had the problem been the one with degenerate eigenstates the result would be somewhat different, as our development would have to include eigenfunctions with the same eigenvalue of energy. Now we expand $a_m$ and $E$ into series, i.e. $a_m=a_m^{(0)}+a_m^{(1)}+...$ and $E=E^{(0)}+E^{(1)}+...$ . Note that our perturbation can be expanded into series where the zeroth term would be zero, as the zeroth term corresponds to no perturbation. For the first order perturbation we have: $a_m^{(0)}(E_m-E^{(0)}+a_m^{(1)}(E_m-E^{(0)})+a_mV_{mm})+\sum_{n\ne m +\sum_{n\ne m}a_n^{(0)} V_{mn}} =0$ . The level of interest is the ground level state. For a ground state we then have that the zeroth order coefficients are zero for every $m$ except $m=0$ , i.e. that our unperturbed ground state wavefunction has only one term in the expansion, namely itself. Now it follows that for $m\ne s$ we have $a_m^{(1)}(E_m-E^{(0)})+ V_{m0}}=0$ or that $a_m^{(1)}=\frac{V_{m0}}{(E_m-E^{(0)})}=0$ . As the unperturbed ground state is an even function and the perturbation is an even function we have that $V_{m0}$ are zero for $m$ such that $\psi_{m}^{(0)}$ is odd. This is true for $m=1, 3,5...$ , therefore the coefficients vanish for odd $m$ .

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