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Reference to Problem 2, Chapter 2 in Modern Quantum Mechanics by JJ Sakurai,

Consider the following Hamiltonian of a two state system $$ H=H_{11}|1\rangle\langle1|+H_{22}|2\rangle\langle2|+H_{12}|1\rangle\langle2| \, , $$ or, written as a matrix as $$ H =\begin{pmatrix} H_{11} & H_{12} \\ 0 & H_{22} \\ \end{pmatrix} \, . $$ This Hamiltonian is not Hermitian and thus the time-evolution operator is not unitary. Hence the probability conservation is violated.

This is clear.

It is stated that, physically, the system can go from state 2 to state 1 but not from state 1 to state 2.

How do we reach the last statement?

What do we extract physically from the individual terms of a Hermitian operator with terms \begin{align} H_{11}&=\langle1|H|1\rangle \\ H_{12}&=\langle1|H|2\rangle \\ H_{22}&=\langle2|H|2\rangle \\ H_{21}&=\langle2|H|1\rangle \, ? \end{align}

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  • $\begingroup$ Hello. Please tell me if this post helps: physics.stackexchange.com/q/209350 Thank you. $\endgroup$ – Constantine Black Feb 24 '16 at 9:08
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    $\begingroup$ Just exponentiate them to get the components of the time evolution operator - then it should be clear why that "physical" statement is made. $\endgroup$ – ACuriousMind Feb 24 '16 at 12:09
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It works just like in classical mechanics: the Hamiltonian generates infinitesimal time translations. Take the Schrodinger equation, $$i \frac{d}{dt} | \psi \rangle = H |\psi \rangle$$ and expand it for small times. Then $$|\psi(t)\rangle \approx (1 - iHt) |\psi(0) \rangle.$$ That is, $H|\psi \rangle$ tells you what $|\psi \rangle$ will instantaneously evolve into. If, $\langle 2 | H | 1 \rangle = 0$, there is no evolution into state 2 from state 1.

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You can solve the problem directly. Assuming a Schrodinger-like equation with a very simple "Hamiltonian", $$i \frac{d}{dt} \left[ \begin{array}{} \psi_1(t) \\ \psi_2(t) \end{array} \right] = \left[ \begin{array}{} 1 & 1 \\ 0 & 1 \end{array} \right] \left[ \begin{array}{} \psi_1(t) \\ \psi_2(t) \end{array} \right], $$ it is straight-forward to show that the solution to this equation is $$ |\psi(t) \rangle \to e^{-it}\left[ \begin{array}{c} \psi_1(0) -it \psi_2(0) \\ \psi_2(0) \end{array} \right]. $$ Thus, if the system starts in the state $$ |\psi(0) \rangle \to \left[ \begin{array}{c} 1 \\ 0 \end{array} \right], $$ this means that $\psi_2(0) = 0$, and it is clear that the system stays in state 1.

This can easily be generalized to the general Hamiltonian in the OP, but this suffices to get the point across.

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