What do the off-diagonal elements of Hamiltonian matrix physically represent? A briefly question: what's the "physical meaning" of the off-diagonal elements of Hamiltonian matrix? Such as an Hamiltonian Matraix looks like:
$$\hat H = \begin{pmatrix} E_{11} & E_{12} \\ E_{21} & E_{22} \end{pmatrix}$$
My teacher told me such a matrix element :
$$E_{21}=\langle2|\hat H|1\rangle$$
Corresponding to the transition amplitude from $\left| 1 \right\rangle $ to $ 
\left| 2 \right\rangle$. I thought about it for days, but I just can't figure it out.
 A: This is similar to Zack's answer, but on a more elementary level.
You need to begin with the time-dependent Schrödinger equation
$$i\hbar\frac{d}{dt}|\psi(t)\rangle=\hat{H}|\psi(t)\rangle$$
Using your given Hamiltonian matrix and writing the state
$|\psi(t)\rangle$ as a column vector this becomes
$$\begin{align}
i\hbar \dot{\psi}_1(t)=E_{11} \psi_1(t) + E_{12} \psi_2(t) \\
i\hbar \dot{\psi}_2(t)=E_{21} \psi_1(t) + E_{22} \psi_2(t)
\end{align}.$$
Now let's assume the system starts in state $|1\rangle$.
That means the starting condition is $|\psi(0)\rangle=|1\rangle$
or
$$\begin{align}
\psi_1(0) &= 1 \\
\psi_2(0) &= 0.
\end{align}$$
Then the solution for small $t$ is
$$\begin{align}
\psi_1(t) &= 1 &-i\frac{E_{11}t}{\hbar} &+ O(t^2) \\
\psi_2(t) &=   &-i\frac{E_{21}t}{\hbar} &+ O(t^2)
\end{align}$$
Here you see, it is the matrix element $E_{21}$ determining
how fast the $\psi_2$ component grows from zero to bigger values.
A: The off-diagonal elements represent the "coupling" between those basis states. I believe it is equal to the transition amplitude within the perturbative approximation.  To understand the off-diagonal elements, consider what would happen if they were zero.  Then the diagonal Hamiltonian matrix is already expressed in the eigenstates of the Hamiltonian.  $\hat{H} |1\rangle = E_{11} |1\rangle$ and $\hat{H}|2\rangle = E_{22} |2\rangle$.  This only occurs when $\langle 1|\hat{H}|2\rangle=0$.  If $\langle 1|\hat{H}|2\rangle\ne 0$, then the states $|1\rangle$ and $|2\rangle$ are coupled together by that $\hat{H}$, and the eigenstates of $\hat{H}$ will be some superposition of $|1\rangle$ and $|2\rangle$.
A: Remember, the meaning of the Hamiltonian in the first place is that it generates time translations via the Schrodinger equation:
$$
i \hbar \frac{\partial}{\partial t} |\psi(t) \rangle = \hat{H} | \psi(t) \rangle
$$
You can formally solve the Schrodinger equation of a time independent Hamiltonian as $| \psi(t) \rangle = e^{-i H t / \hbar} | \psi(0) \rangle$. To gain some intuition, expand the exponential in power series:
$$
|\psi(t) \rangle = | \psi(0) \rangle - \frac{i t}{\hbar} H | \psi(0) \rangle - \frac{t^2}{2\hbar^2} H^2 | \psi(0) \rangle + \ldots
$$
Now, imagine starting off your system in state $|1\rangle$. Then, according to the above equation, if $H$ has off-diagonal elements connecting the state $|1\rangle$ to the state $|2\rangle$, then the Schrodinger equation will generate some amplitude for the system at a later time to be in state $|2\rangle$. The rate at which the state transitions from $|1\rangle$ to $|2\rangle$ will be proportional to $\langle 2 | H | 1 \rangle$, at least to first order in $t$. You can see this by simply using a resolution of the identity, $1 = |1\rangle \langle 1| + |2\rangle \langle 2 |$:
$$
|\psi(t) \rangle = |1\rangle -\frac{it}{\hbar} \left( \langle 1 | H | 1\rangle |1\rangle + \langle 2 | H | 1 \rangle |2 \rangle \right) + \ldots
$$
