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.

• If diagonal entries exist, $|1 \rangle$ and $| 2 \rangle$ are not eigenstates so we have to worry about one transitioning to the other. Have you seen the fact that $\exp(-i \hat{H} t / \hbar)$ solves the Schroedinger equation? Dec 1 '21 at 18:04
• @ConnorBehan Yes! You are right! But I have another trouble: I do know this: $E_{21}=\langle2|exp(−i\hat Ht/\hbar)|1\rangle$ represent the transition transition amplitude from $\left| 1 \right\rangle$ to $\left| 2 \right\rangle$ , But I still have no idea about what's the relation between $\langle2|exp(−i\hat Ht/\hbar)|1\rangle$ and $\langle2|\hat H|1\rangle$, may I have more tips from you please? Dec 1 '21 at 18:56
• @ConnorBehan I tried use Taylor expansion to expand the evolution operator, but it looks can't help since $\left| 1 \right\rangle$ isn't the eigenstate of $\hat H$ as you said, so there is no such $\langle2|(\hat H)^n|1\rangle$=$\langle2|\hat H|1\rangle^n$, I think I still can't say something about it :( Dec 1 '21 at 19:01
• Dec 1 '21 at 20:19

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, $$1 = |1\rangle \langle 1| + |2\rangle \langle 2 |$$: $$|\psi(t) \rangle = |1\rangle -\frac{it}{\hbar} \left( \langle 1 | H | 1\rangle |1\rangle + \langle 2 | H | 1 \rangle |2 \rangle \right) + \ldots$$

• Thank you so much! I still got one last thing to confirm: since $\hat H$ isn't a diagonal matrix (which means $\left| 1 \right\rangle$ and $\left| 2 \right\rangle$ isn't the eigenstates of $\hat H$ ), this lead to: there is no such $\langle2|(\hat H)^n|1\rangle$=$[\langle2|\hat H|1\rangle]^n$ in Taylor expansion, so we do use some kind of approximation in order to better understanding, then we finally get this: $$\langle2|\hat H|1\rangle\propto \langle2|exp(-i\hat Ht/\hbar)|1\rangle$$ Am I right? Dec 1 '21 at 19:28
• The approximation of stopping at $n = 1$ yes. $\left < 2 | H | 1 \right >$ isn't literally the transition amplitude... just its first non-trivial term. Dec 1 '21 at 19:37
• @Connor Behan Great! Thanks a lot! Dec 2 '21 at 0:30

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.

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 $$|2\rangle$$.