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?

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Question 1) I am not sure to understand it, but the mathematical explanation for the construction of the solutions is quite simple. The system of equations becomes diagonal in the eigen basis spanned by the eigen vectors of H. You can indeed write your system in a matrix notation:

dC/dt= H*C ,

If H is diagonalizable then: H=P*D*P^(-1) dC/dt=P*D*P^(-1)*C

P^(-1)*dC/dt=D*P^(-1)*C d(P^(-1)*C)/dt=D*P^(-1)*C

And setting: Y=P^(-1)*C you get : dY/dt=D*Y whose solution is: Y=Y(0)*exp(D)*t Returning to the C variables: C=P*Y gives: C(t)=C(0)*exp(D)*t

Question 2: The system is either in [1> or [2> state. And the sum of the corresponding probabilities must be 1 otherwise it would mean some part of the system is missing.

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