# This step in Griffiths' Introduction to Quantum Mechanics book

I am working my way through time-dependent perturbation theory at the moment. There is a derivation of the formula for determining the time-dependent coefficients, $$c_a(t)$$, $$c_b(t)$$, which I am stuck on:

Assuming that there are two orthonormal states, $$\psi_a$$ and $$\psi_b$$ (that are eigenfunctions of the Hamiltonian), we solve for $$c_a(t)$$ and $$c_b(t)$$ by demanding that $$\Psi(t)$$ satisfy the time-dependent Schrodinger equation, $$H\Psi=i\hbar\frac{\partial\Psi}{\partial t}, H=H^0+H'(t).$$We find that $$c_a[H^0\psi_a]e^{-E_at/\hbar}+c_b[H^0\psi_b]e^{-E_bt/\hbar}+c_a[H'\psi_a]e^{-E_at/\hbar}+c_b[H'\psi_b]e^{-E_bt/\hbar}$$ $$=i\hbar\biggl[\dot c_a\psi_ae^{-iE_at\hbar}+\dot c_b\psi_be^{-iE_bt/\hbar}+c_a\psi_a\biggl(-\frac{iE_a}{\hbar}\biggr)e^{-iE_at/\hbar}+c_b\psi_b\biggl(-\frac{iE_b}{\hbar}\biggr)e^{-iE_bt\hbar}\biggr].$$

The first two terms on the left cancel out the last two terms on the right, and hence..."

I am not sure how we are doing this step above. It looks like the product rule since the coefficients and the exponential are both time dependent, but there are only four terms. Does it have something to do with the coefficients of the non-time dependent Hamiltonian not being functions of time?

• Yes, the eigenfunctions $\psi_a$ and $\psi_b$ are time-independent. This comes from how we solve the Schroedinger equation---by assuming a separability of the space and time variables, so that the solution looks like $\Psi(x,t)=\psi(x)\phi(t)$. – hiccups Mar 15 at 14:38
• I see that $\psi_a$ and $\psi_b$ are time-independent, does that mean the first two terms in the LHS vanish because we're taking the time derivative of $\psi_a$ and $\psi_b$? If that is the case why do the third and fourth terms not vanish too? – Charlie Mar 15 at 14:53
• The $\psi$ are eigenfunctions of $H_0$, so the first two terms on the LHS (that are in square brackets) become $E_a\psi_a$ and $E_b\psi_b$. These are equal to the final two terms on the RHS, so they cancel, leaving the final two terms on LHS = first two terms on RHS. – hiccups Mar 15 at 15:02
• Substitute $H^0\psi_a$ with $E_a\psi_a$ on the left since they’re eigenfunctions. (Similarly for $b$) – Antonino Travia Mar 16 at 12:58
• I still don't have this clear sorry. Why, in the first term on the RHS does it appear that the time derivative of the coefficient $c_a$ has been taken but not the time derivative of the exponential? However in the third and fourth terms on the RHS the coefficient has not been derived but the exponential has. I wondered if this was the product rule being used somewhere but the number of terms doesn't seem to add up if this is the case. – Charlie Mar 16 at 14:15