# Individual terms in a Hamiltonian matrix

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}

• Hello. Please tell me if this post helps: physics.stackexchange.com/q/209350 Thank you. – Constantine Black Feb 24 '16 at 9:08
• Just exponentiate them to get the components of the time evolution operator - then it should be clear why that "physical" statement is made. – ACuriousMind Feb 24 '16 at 12:09

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.
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.