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.
Quantum Mechanics – What Do the Off-Diagonal Elements of the Hamiltonian Matrix Represent?
hamiltonianhilbert-spaceperturbation-theoryquantum mechanicsquantum-states
Related Solutions
Let me give it a shot:
If I interpret this correctly, $\mathbf{F}$ will be the operator for the full spin of the coupled system, $\mathbf{S}$ will be the operator of the electron spin (usually, one would consider $\mathbf{J}$, the spin containing also spin-orbit coupling, but we are on the S-shell, hence no angular momentum) and $\mathbf{I}$ will be the nuclear spin. Then it should hold that $\mathbf{F}=\mathbf{S}+\mathbf{I}$, right?
First, let's have a look at the hyperfine structure Hamiltonian $\mathbf{H}_{hf}$. By construction of $\mathbf{F}$, the eigenstates of $\mathbf{H}_{hf}$ will be eigenstates of $F^2$ and $F_z$. This is just the same as for angular momentum and electron spin (and we construct $\mathbf{F}$ to have this property - this lets us label the eigenstates by the quantum number corresponding to $\mathbf{F}$). Hence the Hamiltonian must be diagonal in the $|F^2,m_F\rangle$-basis. One can also see that $F^2$ commutes with $I^2$ and $S^2$ (and so does $F_z$), since $\mathbf{F}=\mathbf{I}+\mathbf{S}$.
Now we have a look at $\mathbf{H}_B$, the interaction Hamiltonian with a constant magnetic field. We can see that (up to some prefactor) $\mathbf{H}_B=S_z$. Hence the eigenstates of $\mathbf{H}_B$ must be eigenstates of $S_z$ and thus also of $S^2$ and, since the two operators are independent (they relate to two different types of spins, hence the operators should better commute) also to $I^2$ and $I_z$, if you want.
The crucial problem is that $S_z$ and $F^2$ do not commute. Why? Well: $\mathbf{F}=\mathbf{I}+\mathbf{S}$ hence $F^2=S^2+I^2+2\mathbf{S}\cdot \mathbf{I}$. Now $S_z$ and $\mathbf{S}$ do not commute, because $S_z$ does not commute with e.g. $S_x$, which is part of $\mathbf{S}$. Since $F^2$ commutes with $\mathbf{H}_{hf}$ and $S_z$ commutes with $\mathbf{H}_B$, but not with $F^2$, we have that $\mathbf{H}_{hf}$ does not commute with $\mathbf{H}_B$. This means that $\mathbf{H}_B$ and $\mathbf{H}_{hf}$ cannot be diagonal in the same basis, hence you need to have off-diagonal elements.
In order to see how the matrix representing $\mathbf{H}_B$ looks like in the $|F^2,m_F\rangle$-basis, you can express the $|m_I,m_S\rangle$-basis (in which $\mathbf{H}_B$ is diagonal) in terms of the other basis. This is exactly what equations (4.21) do. These are obtained by ordinary addition of angular momenta. From there, you can construct the unitary transforming the basis $|m_I,m_S\rangle$ into $|F^2,m_F\rangle$ and $\mathbf{H}_B$ will be the diagonal matrix in the basis $|m_I,m_S\rangle$ conjugated with this unitary.
EDIT: I'm not quite sure whether I understand correctly what your problem is, but let me elaborate: We want to find the Hamiltonian $\mathbf{H}_B$ in the $|m_Im_S\rangle$ basis. In this basis, it is diagonal, because $\mathbf{H}_B$ is essentially $S_z$ (hence commutes with $S_z$) and it must also commute with $I_z$ since $S_z$ and $I_z$ are independent.
If we order the basis according to $|\frac{1}{2},\frac{1}{2}\rangle,|-\frac{1}{2},-\frac{1}{2}\rangle,|\frac{1}{2},-\frac{1}{2}\rangle,|-\frac{1}{2},\frac{1}{2}\rangle$, then, we can just read off the Hamiltonian: The first and fourth vector are eigenvectors to eigenvalue $\mu B$, the others of $-\mu B$ (by definition of $S_z$, since the second component in $|m_Im_S\rangle=|(SI)I_z,S_z\rangle$ tells us the eigenvalue of $S_z$ that the basis vector corresponds to), i.e. $$ \mathbf{H}_B=\begin{pmatrix}{} \mu B & 0 & 0 & 0 \\ 0 & -\mu B & 0 & 0 \\ 0 & 0 & -\mu B & 0 \\ 0 & 0 & 0 & \mu B\end{pmatrix}$$ Now, as I said, you just have to change the basis. The matrix transforming the above basis into the new basis is given by eqn. (4.21a-d): $$U:=\begin{pmatrix}{} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & \frac{1}{2} & \frac{1}{2} \\ 0 & 0 & -\frac{1}{2} & \frac{1}{2} \end{pmatrix}$$ where the ordering of the $|Fm_F\rangle$-basis is as for $\mathbf{H}$ in your text.
Now calculate $U\mathbf{H}_B U^{\dagger}$ and that should give you the part of $\mathbf{H}$ coming from $\mathbf{H}_B$ in the $|F,m_F\rangle$-basis and this will be exactly what is written in your book.
EDIT 2: I sort of suspected this, so here is some more linear algebra for the problem. I'll use Dirac notion since I suspect you are more familiar with this:
Now suppose you have given two bases $|e_i\rangle$ and $|f_i\rangle$ and suppose they are orthonormal bases. What we want is a matrix $U$ that transforms one basis into the other (I'll call it $U$, since it'll be a unitary - if the bases are not orthonormal, it'll only be an invertible matrix). So we want a matrix such that $$ |f_i\rangle:=U|e_i\rangle \qquad \forall i$$ How to construct this matrix? Well, given an equation for $|f_i\rangle$ in terms of the $|e_i\rangle$ will give you the i-th row of the matrix. You can also see the matrix elements in Dirac notation: $$ \langle e_j|U|e_i\rangle=\langle e_j|f_i\rangle $$
In your case, $|e_i\rangle=|m_Im_S\rangle$ and $|f_i\rangle=|F^2,m_F\rangle$. Hence equation (4.21a) will give you the first row of the matrix (the ordering of the basis vectors $|m_Im_S\rangle$ as I proposed above), (4.21c) the second (notice the basis ordering in the matrix $\mathbf{H}$!) (4.21b) the second and (4.21d) the last row of the matrix. Using the equation for the matrix elements above, you should be able to check that with not too much trouble. You can also easily check that $U$ is indeed a unitary (i.e. $UU^{\dagger}=U^{\dagger}U=\mathbb{1}$.
Then we can calculate the matrix elements: $$ \langle e_i |\mathbf{H}|e_j\rangle=\langle e_i|U^{\dagger}U\mathbf{H}U^{\dagger}U|e_j\rangle=\langle f_i| U\mathbf{H}U^{\dagger}|f_j\rangle $$, which tells you how the matrix looks like in the other basis.
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 $A$.
Best Answer
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, $1 = |1\rangle \langle 1| + |2\rangle \langle 2 |$: $$ |\psi(t) \rangle = |1\rangle -\frac{it}{\hbar} \left( \langle 1 | H | 1\rangle |1\rangle + \langle 2 | H | 1 \rangle |2 \rangle \right) + \ldots $$