Q:

Base states with hamiltonian matrix

It is better for you to have studied "Feynman lectures on Physics Vol.3", because I cannot distinguish whether the words or expressions are what Feynman uses only or not and in order to summarize my questions here, I have to just quote the contents of the book.
However, one thing I notice is that "base state" that Feynman explains seems to be "basic orthonomal state vector"...
With a pair of hamiltonian matrix equation
$i\hbar \frac{d{C}_{1}}{dt} =  {H}_{11}{C}_{1} + {H}_{12}{C}_{2}$
$i\hbar \frac{d{C}_{2}}{dt} =  {H}_{21}{C}_{1} + {H}_{22}{C}_{2}$
where ${C}_{x} = |\psi\rangle$ , $\psi =$ arbitrary state, the book set the states 1 and 2 as "base states". There are only two base states for some particle. Base states have a condition - $\langle i|j\rangle = {\delta}_{ij}$.
I think the "kronecker delta" means that once the particle is in the state of j, we will not be able to find the state i, so if we suppose all the components of hamiltonian are constant, we can say ${H}_{12}$ and ${H}_{21}$ should be zero.   ..............(1)
However the book says that states 1 and 2 are base states and ${H}_{12}$ and ${H}_{21}$ can be nonzero at the same time (if you have the book, refer equ. (9.2) and (9.3) and page 9-3.). There can be probability to transform from state 1 to state 2 and vice versa.....
Then, the relationship that I think like (1) between the "Kronecker delta" and the components of hamiltonian is not correct at all?? Or, is my thought that ${H}_{12}$ and ${H}_{21}$ should be zero true?

A:

The equation you wrote down is 
$$i\hbar\ \dot C(t)=HC(t),$$
where $C(t)$ is a vector with two components $H$ is a constant hermitean $2\times2$ matrix. 
You're right, you can diagonalize it and find associated orthogonal base states $e_1^{\text{diag}_H},e_2^{\text{diag}_H}$. However, say you've chosen the diagonalization, there is still a whole bunch of other unrelated base states $e_1^\text{others},e_2^\text{others}$, which are also orthogonal: 
If $$e_i^*e_j=\delta_{ij},$$ then $$(Ue_i)^*(Ue_j)=\delta_{ij}$$ holds too, where $U\in SU(2)$. These $Ue_i$ are also a valid base and as $SU(2)$ is a three parameter Lie group, there are a bunch of orthogonal vector pairs. So $H$ can be diagonal, while your vectors looks wild and conversely, if you choose the representation in which your vectors are $\{1,0\},\{0,1\}$, then $H$ might not be diagonal there (because it was diagonal w.r.t. another base).

"There can be probability to transform from state 1 to state 2 and vice versa....."
Notice that if you have a finite dimensional matrix with constant coefficients, then you can actually integrate/exponentiate the Schrödinger equation. That's a one-liner in Mathematica actually, they have "MatrixExp". If $H$ is diagonal, then the time evolution will also be a diagonal matrix.
http://www.wolframalpha.com/input/?i=MatrixExp[{{0%2C-1}%2C{1%2C0}}*t]
http://www.wolframalpha.com/input/?i=MatrixExp[{{x%2Cy}%2C{0%2Cz}}*t]
http://www.wolframalpha.com/input/?i=MatrixExp[i*{{H11%2CH12}%2C{H21%2CH22}}*t]%2F%2FSimplify
So to sum it up (I hope I didn't confuse you): If you're actually in an $H$-Eigenstate, then you can diagonalize the matrix and you will never switch from $e_1^{\text{diag}_H}$ to $e_2^{\text{diag}_H}$. If you're in just any state called $e_i$, then it doesn't help that there is a state called $e_2$, which is orthogonal to it.
As an example, consider the time evolution $$U(t)=\{\{\cos(t),-\sin(t)\},\{\sin(t),\cos(t)\}\},$$ which is $$R\in SO(2)\in SU(2)$$. You might happen to be in the state $\{1,0\}$ and so want to choose the basis $\{1,0\},\{0,1\}$ (so your state fullfills the orthogonality relation), but the Hamiltonian associate with the above time evolution is $$H=U'(0)=\{\{0,-1\},\{1,0\}\}.$$ (it's not an accident that this is a Pauli matrix, as these are the generators of $SU(2)$).

