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## Main Question or Discussion Point

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 [itex] \mu B[/itex] parts not diagonal? If the Hamiltonian is [itex]\vec{\mu} \cdot \vec{B}[/itex] , why aren't all the components just diagonal? How is this matrix built systematically? Can someone please explain?

The following part is from the book:

We now consider the effect of a uniform magnetic field [itex]\mathbf{B} = B\hat{z}[/itex] on the hyperfine levels of the [itex] {}^2 S_{1/2}[/itex] 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 [itex] \{F^2, F_z, I^2, S^2\}[/itex] . Therefor if we write out a matrix for the hyperfine Hamiltonian [itex] H_\text{hf}[/itex] in the coupled basis [itex] \lvert Fm_F\rangle[/itex] , it is diagonal. However, the Hamiltonian [itex] H_B[/itex] 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 [itex] \lvert(SI)m_S, m_I\rangle[/itex] , made up of eigenstates of the operators [itex] \{I^2, I_z, S^2, S_z\}[/itex] . 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 [itex] \lvert Fm_F\rangle[/itex] and uncoupled [itex] \lvert(SI)m_Sm_I\rangle[/itex] 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 [itex] H_\text{hf} + H_B[/itex] 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 [itex] (\lvert 1,1\rangle, \lvert 1,-1\rangle, \lvert 1,0\rangle, \lvert 0,0\rangle)[/itex] .

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

The following part is from the book:

We now consider the effect of a uniform magnetic field [itex]\mathbf{B} = B\hat{z}[/itex] on the hyperfine levels of the [itex] {}^2 S_{1/2}[/itex] 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 [itex] \{F^2, F_z, I^2, S^2\}[/itex] . Therefor if we write out a matrix for the hyperfine Hamiltonian [itex] H_\text{hf}[/itex] in the coupled basis [itex] \lvert Fm_F\rangle[/itex] , it is diagonal. However, the Hamiltonian [itex] H_B[/itex] 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 [itex] \lvert(SI)m_S, m_I\rangle[/itex] , made up of eigenstates of the operators [itex] \{I^2, I_z, S^2, S_z\}[/itex] . 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 [itex] \lvert Fm_F\rangle[/itex] and uncoupled [itex] \lvert(SI)m_Sm_I\rangle[/itex] 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 [itex] H_\text{hf} + H_B[/itex] 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 [itex] (\lvert 1,1\rangle, \lvert 1,-1\rangle, \lvert 1,0\rangle, \lvert 0,0\rangle)[/itex] .

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