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.

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    $\begingroup$ 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? $\endgroup$ Dec 1 '21 at 18:04
  • $\begingroup$ @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? $\endgroup$ Dec 1 '21 at 18:56
  • $\begingroup$ @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 :( $\endgroup$ Dec 1 '21 at 19:01

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

  • $\begingroup$ 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? $\endgroup$ Dec 1 '21 at 19:28
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    $\begingroup$ 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. $\endgroup$ Dec 1 '21 at 19:37
  • $\begingroup$ @Connor Behan Great! Thanks a lot! $\endgroup$ 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 $|2\rangle$.


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