Off-diagonal elements of Hamiltonian matrix $H_{12}$ & $H_{21}$: energy of transition from $|1\rangle$ to $|2\rangle$ or amplitude of transition?

Question

$ \newcommand{\k}[1]{\left| #1 \right\rangle} \newcommand{\dd}[1]{\frac{d #1}{dt}} $In a Hamiltonian Matrix like this:

$$H = \begin{pmatrix} E_{11} & E_{12} \\ E_{21} & E_{22} \end{pmatrix}$$, $E_{11}\;\&\;E_{22}$ represents the energy of states $\k 1$ & $\k 2$ which are the base states of a system. But what about $E_{12} \;\&\; E_{21}$? They are the off-diagonal elements; are they energy for transition $\k1 \to \k2$ & $\k2\to\k1$ respectively?

Feynman, in his lectures always called these elements the amplitude to move from $\k1$ to $\k2$ & vice-versa.

$\bullet$ 1st reference:

The ammonia molecule has one nitrogen atom and three hydrogen atoms located in a plane below the nitrogen so that the molecule has the form of a pyramid[...]We will say that the molecule is in the state $|1⟩$ when the nitrogen is “up,” and is in the state $|2⟩$ when the nitrogen is “down,”[...]$$iℏ\dd{C_1}=H_{11}C_1\;\;,\;\;iℏ\dd{C_2}=H_{22}C_2.$$ We can easily solve these two equations; we get $$C_1=(\text{const})e^{−(i/ℏ)H_{11}t}\;\;,\;\;C_2=(\text{const})e^{−(i/ℏ)H_{22}t}.$$ These are just the amplitudes for stationary states with the energies $E_1=H_{11}$ and $E_2=H_{22}.$ It turns out that it is possible for the nitrogen to push its way through the three hydrogens and flip to the other side.There is, therefore, some small amplitude that a molecule which starts in $|1⟩$ will get to the state $|2⟩.$ The coefficients $H_{12}$ and $H_{21}$ are not really zero. Again, by symmetry, they should both be the same—at least in magnitude. In fact, we already know that, in general, $H_{ij}$ must be equal to the complex conjugate of $H_{ji}$, so they can differ only by a phase. It turns out, as you will see, that there is no loss of generality if we take them equal to each other. For later convenience we set them equal to a negative number; we take $H_{12}=H_{21}=−A.$

$\bullet$ 2nd reference:

[...]If we hadn’t allowed for the possibility of the nitrogen flipping back and forth, we would have taken $A$ equal to zero and the two energy levels would be on top of each other at energy $E_0.$

$\bullet$ 3rd reference:

So long as the two protons of the hydrogen molecular ion are far apart, it still requires about this much energy—which is for our present considerations a great deal of energy—to get the electron somewhere near the midpoint between the protons. So it is impossible, classically, for the electron to jump from one proton to the other. However, in quantum mechanics it is possible—though not very likely. There is some small amplitude for the electron to move from one proton to the other. As a first approximation, then, each of our base states $|1⟩$ and $|2⟩$ will have the energy $E_0$, which is just the energy of one hydrogen atom plus one proton. We can take that the Hamiltonian matrix elements $H_{11}$ and $H_{22}$ are both approximately equal to $E_0.$ The other matrix elements $H_{12}$ and $H_{21}$, which are the amplitudes for the electron to go back and forth, we will again write as $−A.$

$\bullet$ 4th reference:

Now the amplitude $A$ for an electron which is near one proton to get to the other one depends on the separation between the protons. The closer the protons are together, the larger the amplitude.

So, Feynman always considers $H_{12}$ and $H_{21}$ as the amplitude for transition. Though I wouldn't call $H$ here an operator but rather generator of translation of time, but it is not the energy of transition.

But why are they not energy of transition? After all they are the elements of Hamiltonian Matrix, isn't it?

I asked the same in my previous query & got this:

Is $A$ the energy to go from $|1⟩$ to $|2⟩$? Is it the amplitude to go from $|1⟩$ to $|2⟩$? I'm not understanding this as Feynman is referring it as the amplitude but since it is an element of the Hamiltonian matrix, it should be the energy to go from $|1⟩$ to $|2⟩$ like $H_{11}$ being the energy of $|1⟩.$ So, is $H_{ij}$ the energy or the amplitude to go from $|1⟩$ or $|2⟩$? – user36790

@user36790 It has units of energy, but it's an off-diagonal term in the Hamiltonian, so it doesn't represent the energy of a state. I would call it an amplitude or a coupling. – zeldredge

From the reply, I could know that off-diagonal elements are not energy of transition.

But what are the energies of stationary states? They are: $E_0-A$ & $E_0 +A;$ if $-A$ is not the energy of transition, how could it contribute to the stationary state's energy: $E_0 \pm A$? I'm not understanding this. $-A$ is the generator of time-evolution for the transition from $\k 1$ to $\k 2$, then how could it be taken as part of the energies in the stationary states? Also, I would want to know why off-diagonal elements aren't representing energy. Can anyone please explain?

Answer 1

I know this is an old question, but I'm not sure all parts of the OP's question have been completely addressed.

The off-diagonal elements of the Hamiltonian, as @Discovery pointed out, are indicative of the coupling between the two energy states. If they are zero, there is zero probability of a transition between $\mid 1 \rangle$ and $\mid 2 \rangle$. However, since here they are nonzero, there is a nonzero probability of transition.

I believe the reason Feynman refers to these as transition amplitudes is that they dictate how the amplitude changes in time. I've normally heard of this being referred to as the transition frequency.

To illustrate this, note that the probability amplitude of $\mid 1 \rangle \rightarrow \mid 2 \rangle$ after time $t$ is

$$\langle 2 \mid U^{\dagger} U \mid 1 \rangle = \langle 2 \mid e^{iHt/\hbar}e^{-iHt/\hbar} \mid 1 \rangle$$ which can be expressed as

$$\langle 2 \mid e^{iHt/\hbar} \mid 1 \rangle \langle 1 \mid e^{-iHt/\hbar} \mid 1 \rangle ~+ ~\langle 2 \mid e^{iHt/\hbar} \mid 2 \rangle \langle 2 \mid e^{-iHt/\hbar} \mid 1 \rangle = $$

$$e^{iE_{12}t/\hbar} e^{-iE_{11}t/\hbar} + e^{iE_{22}t/\hbar} e^{-iE_{12}t/\hbar} = $$

$$e^{-i(E_{11}-A)t/\hbar} + e^{i(E_{22}-A)t/\hbar}$$

So you can see that the transition amplitude oscillates with frequency proportional to an energy difference which depends on $A$.

Answer 2

$ \newcommand{\k}[1]{\left| #1 \right\rangle} $Let me consider the problem here briefly. Set a basis of two-dimensional Hilbert space consisting of two vectors,

$|1\rangle = \begin{pmatrix} 1 \\ 0 \end{pmatrix},\,|2\rangle=\begin{pmatrix} 0 \\ 1 \end{pmatrix}$

assuming that they are eigenvectors of given Hamiltonian at first. From the general postulate of quantum mechanics, any other states under same Hamiltonian can be described by

$|\psi\rangle=C_1|1\rangle +C_2|2\rangle= C_1\begin{pmatrix} 1 \\ 0 \end{pmatrix}+C_2\begin{pmatrix} 0 \\ 1 \end{pmatrix}$.

Then we can see that

$H|\psi\rangle=C_1\begin{pmatrix} E_{11} \\ E_{21} \end{pmatrix}+C_2\begin{pmatrix} E_{12} \\ E_{22} \end{pmatrix}$ and

$$\langle\psi|H|\psi\rangle=|C_1|^2E_{11}+C_1^*C_2E_{12}+C_1C_2^*E_{21}+|C_2|^2E_{22}\tag{1}$$

However, when $E_{12}$ and $E_{21}$ are not zero, the two basis vectors are not eigenvectors since $H|1,2\rangle \not=C|1,2\rangle$. So we should treat them as a basis merely and eigenvectors of Hamiltonian have the form of $|\psi\rangle$. In the true eigenvector-basis, Hamiltonian matrix can be described by block-diagonal form.

Anyway assuming that $E_{11}=E_{22},\,E_{12}=E_{21}$, diagonalized Hamiltonian matrix is

$H = \begin{pmatrix} E_{11}-E_{12} & 0 \\ 0 & E_{11}+E_{12} \end{pmatrix}$

and corresponding eigenvectors are

$$\k I=\frac{1}{\sqrt2}[-\k 1+\k 2],\,\\ \k {II}= \frac{1}{\sqrt2}[|1\rangle+|2\rangle]. \tag{2}$$

From equation $(1)$, we can see that any arbitrary state $|\psi\rangle$ have energy $E_{11}+(\text{deviation})$ and from equation $(2)$, true eigenvectors are mixed states of discrete up and down states.

We cannot describe states in eq. $(2)$ neither up, down or middle-state. States in eq. $(2)$ "deviates" from up and down states equally and so the deviation energy in eq. $(1)$ is $+E_{12}$ or $-E_{12}$ as seen in diagonalized Hamiltonian. They are not energy of discrete transition but the degree of deviation or mixing of states.