Why does Hamiltonian follow the property $H^*_{ij} = H_{ji} $? I was reading Feynman's Lectures III's Hamiltonian Matrix. There I found this property of Hamiltonian Matrix:

The Hamiltonian has one property that can be deduced right away, namely, that
  $$H^*_{ij}  = H_{ji} $$This follows from the condition that the total probability that the system is in some state does not change. If you start with a particle—an object or the world—then you’ve still got it as time goes on. The total probability of finding it somewhere is $$\sum_i |C_i(t)|^2$$which must not vary with time. If this is to be true for any starting condition $ϕ$, then Eq. (8.40) must also be true.

I haven't understood Feynman's reasoning for proving the law; how "the total probability that the system is in some state does not change" assure the validity of the property? Can anyone explain me Feynman's reasoning?
 A: Okay, so, let's get this thing going right.
The property you're talking about is called by different names that you can google: e.g. that the Hamiltonian matrix must be self-adjoint (where the adjoint is the conjugate transpose) or Hermitian. 
The easiest property to prove about Hermitian matrices is that they have real eigenvalues, for their defining characteristic $\langle H \psi| \psi\rangle = \langle \psi| H \psi\rangle$ combines with the property $\langle a | b\rangle = \left(\langle b | a\rangle\right)^*$ to say that a certain number (an "expectation value") is equal to its own conjugate, which is only true for real numbers.
How does this connect to probability conservation? The total probability is$$P = \sum_i C_i^* ~ C_i$$and the Schrodinger equation 8.39 that he presents is:$$i\hbar\,\frac{\rm dC_i}{\rm dt}=\sum_jH_{ij}(t)C_j(t).$$The conjugate expression must therefore be:
$$-i\hbar\,\frac{\rm dC_i^*}{\rm dt}=\sum_jH_{ij}^*(t)C_j^*(t).$$ Now let us just take a derivative of $P$ with the product rule, as:$$\begin{align}i\hbar \frac{dP}{dt} &= i\hbar  \sum_i \left({\rm dC_i^*\over \rm dt} ~ C_i + C_i^* ~ {\rm dC_i\over \rm dt}\right)\\
&= \sum_{ij}\left(C_i^* ~H_{ij} ~C_j - H_{ij}^* ~C_j^*~ C_i\right)\end{align}$$We now perform a dirty trick: we split this sum into its two terms and we relabel each part of the sum differently. In the first term we take $i \mapsto m$ and $j \mapsto n$. But in the second term we take $i \mapsto n$ and $j \mapsto m$. We then combine both terms together to find:$$i\hbar \frac{\rm dP}{\rm dt} = \sum_{mn}\left(C_m^* ~H_{mn} ~C_n - H_{nm}^* ~C_m^*~ C_n\right),$$and, combining like terms:$$i\hbar \frac{\rm dP}{\rm dt} = \sum_{mn}C^*_m~C_n  \left(H_{mn} - H_{nm}^*\right).$$The implication is that if we want $\rm dP/\rm dt = 0$ for all $C_n$ (total probability does not escape the system) then we must have that the parenthesized matrix is the 0-matrix. 
Why? It's actually not too complicated to see. If we steal the $i$ from the left to define $$D_{mn} = -i \left(H_{mn} - H_{nm}^*\right),$$then we know for sure that $D_{mn}$ is Hermitian by construction. (Just take its conjugate transpose, I dare you.) Because it's Hermitian, it is never defective and we can diagonalize it with real eigenvalues down the diagonal. Now either one of these eigenvalues $\lambda$ is nonzero, in which case we can use the corresponding eigenvector as $C_n$ above and we get $\hbar \frac{\rm dP}{\rm dt} = \lambda \sum_m C_m^* C_m = \lambda P$, violating probability conservation, or else all of the eigenvalues are 0 and this expression is the 0-matrix. But the 0 matrix has all-zero components in all bases, not just its eigenbasis, so therefore $H_{mn} = H^*_{nm}$ in general.
A: If that property doesn't hold then $H$ is not Hermitian.  This means its eigenvalues could be complex and the time evolution operator won't be unitary:
$U(t)U(t)^\dagger=e^\frac{iHt-itH^\dagger }{\hbar} \neq1$
If it's not unitary, then it will change the length of your state vectors which you're evolving.  Since the length of these($\sum_i |C_i(t)|^2$) vectors are the sum of the probabilities, your total probability won't be conserved.
A: What he actually is referring to is the mathematical fact that in order for the standard norm (length) of $n$-tuple $[C_j]_{j=1,2,...n}$ to be preserved during evolution determined by the equation
$$
\frac{\rm dC_i}{\rm dt} = -\frac{i}{\hbar} \sum_j H_{ij} C_j
$$
where $H_{ij}$ is time-independent complex matrix, this matrix needs to be hermitian, i.e. obey the condition for all pairs $i,j$,
$$
H_{ij}^* = H_{ji}.
$$
This can be proven starting with the condition that norm squared is constant in time:
$$
\frac{\rm d}{\rm dt}\bigg(\sum_i C_i^*C_i\bigg)=0
$$
and using the above differential equation.
Feynman just rephrases this and uses term "total probability" instead of "norm squared", because it is common to interpret  $|C_i|^2$ as probability that $i^{\textrm{th}}$ eigenvalue of $H$ will be measured.
