How do you derive (9.38) of Vol. III of the Feynman lectures?

Question

The Feynman lectures volume 3 chapter 9 shows equation $(9.38)$, which is the equation I don't know how to derive, as follows,

$$i\hbar\frac{dC_{II}}{dt}=(E_0-A)C_{II}+\mu\mathcal{E}C_I.\tag{9.38}$$

To derive this he says to 'add the two equations, divide by the square root of 2' and then use, $(9.13)$, which is,

$C_I=\frac{1}{\sqrt2}[C_1-C_2] \ \ \ \ \ and \ \ \ \ \ C_{II}=\frac{1}{\sqrt2}[C_1+C_2]$

When he says to 'add the two equations' I don't know if he is referring to $C_1$ and $C_2$ or equation $(9.36)$, which is,

$i\hbar\frac{dC_{1}}{dt}=(E_0+\mu\mathcal{E})C_{1}-AC_2 \ \ \ \ \ and \ \ \ \ \ i\hbar\frac{dC_{2}}{dt}=-AC_1+(E_0-\mu\mathcal{E})C_{2}$

I am not exactly sure what should be used for $C_1$ and $C_2$ if these are the 'two equation' he is referring to. I tried these values I think should be used for the approximate energy to get,

$C_1=a_1e^{-(i/\hbar)(E_0+A+\frac{\mu^2\mathcal{E}^2}{2A})t} \ \ \ \ \ and \ \ \ \ \ C_2=a_2e^{-(i/\hbar)(E_0-A-\frac{\mu^2\mathcal{E}^2}{2A})t}$

However, I am still not able to derive $(9.38)$. Does anyone know how to do this?

Answer 1

By adding the two equations $(9.36)$ and dividing by the square root of two you get, \begin{align} \frac{1}{\sqrt2}[i\hbar\frac{dC_1}{dt}+i\hbar\frac{dC_2}{dt}] & = i\hbar\frac{d}{dt}\frac{1}{\sqrt2}[C_1+C_2] = i\hbar\frac{dC_{II}}{dt} \\ & =\frac{1}{\sqrt2}[(E_0+\mu\mathcal{E}-A)C_1+(E_0-\mu\mathcal{E}-A)C_2] \\ & =\frac{1}{\sqrt2}[E_0C_1+\mu\mathcal{E}C_1-AC_1+E_0C_2-\mu\mathcal{E}C_2-AC_2] \\ & =\frac{1}{\sqrt2}[E_0(C_1+C_2)+\mu\mathcal{E}(C_1-C_2)-A(C_1+C_2) \\ & =(E_0-A)C_{II}+\mu\mathcal{E}C_I \end{align}