# Stationary states with a pair of hamiltonian equations

I read some derivation related with probability amplitudes and hamiltonian matrix in some book, and have a few questions.

Here what the book says is.

We want the general solution of the pair of Hamiltonian equations,

$i\hbar \frac {d{C}_{1}}{dt } = {H}_{11}{C}_{1} + {H}_{12}{C}_{2}$

$i\hbar \frac {d{C}_{2}}{dt } = {H}_{21}{C}_{1} + {H}_{22}{C}_{2}$

where ${C}_{x} = <x|\psi> , \psi =$ arbitrary state.

Suppose that the coefficients (hamiltonians) are constant, and we can use the trial functions

${C}_{1} = {a}_{1}{e}^{-iwt} , {C}_{2} = {a}_{2}{e}^{-iwt} , w = E/\hbar$

Substituting ${C}_{1} , {C}_{2}$ into the equations, we get

$E{a}_{1} = {H}_{11}{a}_{1} + {H}_{12}{a}_{2} \ ,\ E{a}_{2} = {H}_{21}{a}_{1} + {H}_{22}{a}_{2}$

By using a condition that ${a}_{1}$ and ${a}_{2}$ cannot be zero, and setting ${H}_{11} = {H}_{22}={E}_{0}$ and ${H}_{12} = {H}_{21}=-A$ , we can conclude that $E$ should be ${E}_{0} + A = {E}_{Ⅰ}$ or ${E}_{0} - A = {E}_{Ⅱ}$. (and by linear algebra, we can know $\frac {{a}_{1}}{{a}_{2}} = \frac {{H}_{12}}{E-{H}_{11}}$)

For general case, the two solutions ${E}_{Ⅰ}$ and ${E}_{Ⅱ}$ refer to two stationary states,

$|{\psi}_{Ⅰ}> = |Ⅰ>{e}^{-(i/\hbar){E}_{Ⅰ}t}$ and $|{\psi}_{Ⅱ}> = |Ⅱ>{e}^{-(i/\hbar){E}_{Ⅱ}t}$

with

$|Ⅰ> = |1>{a}_{1}' + |2>{a}_{2}'$ and $|Ⅱ> = |1>{a}_{1}'' + |2>{a}_{2}''$*................(1)*

where $\frac {{a}_{1}'}{{a}_{2}'} = \frac {{H}_{12}}{{E}_{Ⅰ}-{H}_{11}}$ and $\frac {{a}_{1}''}{{a}_{2}''} = \frac {{H}_{12}}{{E}_{Ⅱ}-{H}_{11}}$.

If the system is known to be in one of the stationary states, the sum of the probabilities that it will be found in $|1>$ or $|2>$ must equal one - ${|{C}_{1}|}^{2} +{|{C}_{2}|}^{2} = 1$*.....................(2)*

Questions))

1. Why should we struct two stationary states by setting states $|Ⅰ>$and $|Ⅱ>$ in the way of (1) ? How can we express the states $|Ⅰ>$and $|Ⅱ>$ like (1)?

2. Why should the sum of probabilities in (2) equal one?