I'm trying to understand how Hamiltonian matrices are built for optical applications. In the excerpts below, from the book "Optically polarized atoms: understanding light-atom interaction", what I don't understand is: Why are the $\mu B$ parts not diagonal? If the Hamiltonian is $\vec{\mu} \cdot \vec{B}$, why aren't all the components just diagonal? How is this matrix built systematically? Can someone please explain?

We now consider the effect of a uniform magnetic field $\mathbf{B} = B\hat{z}$ on the hyperfine levels of the ${}^2 S_{1/2}$ ground state of hydrogen. Initially, we will neglect the effect of the nuclear (proton) magnetic moment. The energy eigenstates for the Hamiltonian describing the hyperfine interaction are also eigenstates of the operators $\{F^2, F_z, I^2, S^2\}$. Therefor if we write out a matrix for the hyperfine Hamiltonian $H_\text{hf}$ in the coupled basis $\lvert Fm_F\rangle$, it is diagonal. However, the Hamiltonian $H_B$ for the interaction of the magnetic moment of the electron with the external magnetic field,

$$H_B = -\mathbf{\mu}_e\cdot\mathbf{B} = 2\mu_B B S_z/\hbar,\tag{4.20}$$

is diagonal in the uncoupled basis $\lvert(SI)m_S, m_I\rangle$, made up of eigenstates of the operators $\{I^2, I_z, S^2, S_z\}$. We can write the matrix elements of the Hamiltonian in the coupled basis by relating the uncoupled to the coupled basis. (We could also carry out the analysis in the uncoupled basis, if we so chose.)

The relationship between the coupled $\lvert Fm_F\rangle$ and uncoupled $\lvert(SI)m_Sm_I\rangle$ bases (see the discussion of the Clebsch-Gordan expansions in Chapter 3) is

$$\begin{align} \lvert 1,1\rangle &= \lvert\bigl(\tfrac{1}{2}\tfrac{1}{2}\bigr)\tfrac{1}{2}\tfrac{1}{2}\rangle,\tag{4.21a} \\ \lvert 1,0\rangle &= \frac{1}{\sqrt{2}}\biggl(\lvert\bigl(\tfrac{1}{2}\tfrac{1}{2}\bigr)\tfrac{1}{2},-\tfrac{1}{2}\rangle + \lvert\bigl(\tfrac{1}{2}\tfrac{1}{2}\bigr),-\tfrac{1}{2}\tfrac{1}{2}\rangle\biggr),\tag{4.21b} \\ \lvert 1,-1\rangle &= \lvert\bigl(\tfrac{1}{2}\tfrac{1}{2}\bigr),-\tfrac{1}{2},-\tfrac{1}{2}\rangle,\tag{4.21c} \\ \lvert 0,0\rangle &= \frac{1}{\sqrt{2}}\biggl(\lvert\bigl(\tfrac{1}{2}\tfrac{1}{2}\bigr)\tfrac{1}{2},-\tfrac{1}{2}\rangle - \lvert\bigl(\tfrac{1}{2}\tfrac{1}{2}\bigr),-\tfrac{1}{2}\tfrac{1}{2}\rangle\biggr),\tag{4.21d} \end{align}$$

Employing the hyperfine energy shift formula (2.28) and Eq. (4.20), one finds for the matrix of the overall Hamiltonian $H_\text{hf} + H_B$ in the coupled basis

$$H = \begin{pmatrix} \frac{A}{4} + \mu_B B & 0 & 0 & 0 \\ 0 & \frac{A}{4} - \mu_B B & 0 & 0 \\ 0 & 0 & \frac{A}{4} & \mu_B B \\ 0 & 0 & \mu_B B & -\frac{3A}{4} \end{pmatrix},\tag{4.22}$$

where we order the states $(\lvert 1,1\rangle, \lvert 1,-1\rangle, \lvert 1,0\rangle, \lvert 0,0\rangle)$.

And for Eq. (2.28) the other part is

$$\Delta E_F = \frac{1}{2}AK + B\frac{\frac{3}{2}K(K + 1) - 2I(I + 1)J(J + 1)}{2I(2I - 1)2J(2J - 1)},\tag{2.28}$$

where $K = F(F + 1) - I(I + 1) - J(J + 1)$. Here the constants $A$ and $B$ characterize the strengths of the magnetic-dipole and the electric-quadrupole interaction, respectively. $B$ is zero unless $I$ and $J$ are both greater than $1/2$.

  • $\begingroup$ Could you please edit your question to quote the relevant passages from the book, rather than just including images of text? That's a general rule for whenever you're including text from another source: retype it, don't use screenshots. $\endgroup$
    – David Z
    Commented Oct 19, 2013 at 22:35
  • $\begingroup$ @DavidZ I'm sorry, but the whole paragraph is relevant. I can't remove any part, because otherwise the problem in the book won't be clear. If I'm mistaken, please clarify what should be cut. $\endgroup$ Commented Oct 19, 2013 at 22:36
  • $\begingroup$ I didn't say you have to remove anything. I just said you should type it out, rather than including an image. $\endgroup$
    – David Z
    Commented Oct 19, 2013 at 22:43
  • 1
    $\begingroup$ @DavidZ Ah, come on! That's gonna take me half an hour to to style it with latex and all those equations! I'm including it this time as a screen shot because it's gonna be clearer to the reader and simpler for me to post the question... there's no redundancy in this at all! $\endgroup$ Commented Oct 19, 2013 at 22:51
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    $\begingroup$ It's not searchable or selectable as an image. This time I'll do it for you (and I bet it won't take half an hour), but a good rule of thumb to keep in mind for the future is that you shouldn't include more text in the question than you're willing to type out. $\endgroup$
    – David Z
    Commented Oct 19, 2013 at 23:26

1 Answer 1


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.

  • $\begingroup$ Thank you for your reply. The problem is exactly the way you understood it. Actually my problem is how to create the Hamiltonian matrix in the $ \left| {{I^2}{I_z}{S^2}{S_z}} \right\rangle $ basis. Can you please explain how to create that hamiltonian? Please add it to your answer, if you could. Thanks again. $\endgroup$ Commented Oct 20, 2013 at 17:47
  • $\begingroup$ I extended my answer - not sure though whether this is, what you are looking for. $\endgroup$
    – Martin
    Commented Oct 20, 2013 at 19:26
  • $\begingroup$ Thank you so much for your reply. Sorry for the dull questions, but this really helps. I had calculated the CG coeffs before posting the question, but I didn't know how to proceed and calculate the matrix you called $U$. So my CG coeffs are like the book found out exactly, which I constructed from $ \left\langle {{F^2}{F_z}} \right|\left| {{I^2}{I_z}{S^2}{S_z}} \right\rangle $, but how do you make that a matrix? Yes it sounds trivial, but please explain that part in more details. $\endgroup$ Commented Oct 20, 2013 at 20:57
  • $\begingroup$ I edited some more linear algebra at the end (general and special for the problem). In principle, it is only changing bases from linear algebra. It's okay though - I remember well when I learned that stuff a couple of years ago, it took me some time to transfer the necessary linear algebra from the maths to the physics courses, too. $\endgroup$
    – Martin
    Commented Oct 20, 2013 at 22:53

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