How do you derive (9.10) from the Feynman lectures Vol. III?

The Feynman lectures volume 3 chapter 9 analyzes the ammonia molecule. It assumes there are only two base states. The amplitude, $C_1$, to be in the first state, $lvert1rangle$, in which the nitrogen atom is above the plane of the three hydrogen atoms is given by,

$$C_1=frac{a}{2}e^{-(i/hbar)(E_0-A)t}+frac{b}{2}e^{-(i/hbar)(E_0+A)t}.tag{9.2}$$

and the amplitude, $C_2$, for the nitrogen atom to be in the second base state, $lvert2rangle$, where the nitrogen atom is below this plane is,

$$C_2=frac{a}{2}e^{-(i/hbar)(E_0-A)t}-frac{b}{2}e^{-(i/hbar)(E_0+A)t}.tag{9.3}$$

There are two stationary states. The first one is $lvertpsi_Irangle$ and corresponds to when $a=0$ in $(9.2)$ and $(9.3)$. The other one is $lvertpsi_{II}rangle$ and corresponds to when $b=0$ in $(9.2)$ and $(9.3)$. The amplitude to be in the new base states $C_I$ and $C_{II}$ are given as,

$$C_I=frac{1}{sqrt2}[C_1-C_2], C_{II}=frac{1}{sqrt2}[C_1+C_2].tag{9.13}$$

Equation $(9.10)$ for the state $lvertpsi_Irangle$, also says these amplitudes should be,

$$C_I=e^{-(i/hbar)(E_0+A)t}, C_{II}=0.tag{9.10}$$

When I try to calculate $C_{II}$ for state $lvertpsi_Irangle$ for when $a=0$, I get,

$C_{II}=frac{1}{sqrt2}[C_1+C_2]=frac{1}{sqrt2}[frac{b}{2}e^{-(i/hbar)(E_0-A)t}-frac{b}{2}e^{-(i/hbar)(E_0-A)t}]=0$

which is the same $C_{II}$ as given in $(9.10)$, but when I try to calculate $C_{I}$ I get,

$C_I=frac{1}{sqrt2}[C_1-C_2]=frac{1}{sqrt2}[frac{b}{2}e^{-(i/hbar)(E_0-A)t}+frac{b}{2}e^{-(i/hbar)(E_0-A)t}]=frac{b}{sqrt2}e^{-(i/hbar)(E_0+A)t}$

This is exactly the same as $C_{I}$ in $(9.10)$ except for the factor $frac{b}{sqrt2}$. Are you able to just assume that $b=sqrt2$ to eliminate this factor and make $C_I$ equal to what is shown in $(9.10)$, or is this incorrect?