0
$\begingroup$

In time-dependent pertubation theory we can denote the Schrödinger equation by a set of two equations $$\dot{c_a} = -\frac{i}{\hbar}\Big[c_aH'_{aa}+c_bH'_{ab}e^{-i(E_b-E_a)t/\hbar}\Big] \\ \dot{c_b} =-\frac{i}{\hbar}\Big[c_aH'_{bb}+c_bH'_{ba}e^{-i(E_b-E_a)t/\hbar}\Big] $$ with matrix elements $H'_{ij} = \langle \psi_i|H'|\psi_j\rangle $ and a time-dependent pertubation $H'(t)$ for a two-level system where we denote our wave function by $$\Psi(t) = c_a(t)\psi_ae^{-iE_a/\hbar}+c_b(t)\psi_be^{iE_b/\hbar} $$as a linear combination of two orthonormal eigenstates $\psi_a,\psi_b$ of our unperturbed Hamiltonian $H^0$.

Why do we (does Griffiths) assume that the diagnoal matrix elements of $H'$ typically vanish?

$\endgroup$
1
$\begingroup$

Intuitively, $\langle i| H^{\prime} | j \rangle$ gives the interaction between states $i$ and $j$. For most physical systems, adding a perturbed potential, is same as introducing this interaction between different energy eigenstates. In other words, you can also say that the diagonal element of full Hamiltonian come only from the original Hamiltonian.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.