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An atom has an electromagnetic moment, $\mu = -g\mu_B S$ where S is the electronic spin operator ($S=S_x,S_y.S_z$) and $S_i$ are the Pauli matrices, given below. The atom has a spin $\frac{1}{2}$ nuclear magnetic moment and the Hamiltonian of the system is

\begin{gather*} H = -\mu .B + \frac{1}{2}A_0S_z \end{gather*}

The first term is the Zeeman term, the second is the Fermi contact term and $A_0$ is a real number. Obtain the Hamiltonian in matrix form for a magnetic field, $B=B_x,B_y,B_z$. Show that when the atom is placed in a magnetic field of strength B, aligned with the z axis, transitions between the ground and excited states of the atom occur at energies:

\begin{gather*} E= g\mu_B B + \frac{1}{2}A_0 \end{gather*}

The Pauli Matrices are:

\begin{gather*} S_x = \frac{1}{2} \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} , \ S_y = \frac{1}{2} \begin{bmatrix} 0 & -i \\ i & 0 \end{bmatrix} , \ S_z = \frac{1}{2} \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} \end{gather*}

Where do I even start for a solution to this problem I am unclear as to how to formulate the B matrix. If I can get that hopefully the second part will become apparent to prove

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  • $\begingroup$ S is a vector where each component of the vector is a matrix. B is just a normal vector of numbers. You know how to take the scalar product of two vectors S.B; just multiply components and add up. But here each multiplication is a matrix times a number which just another matrix. $\endgroup$ – Bruce Greetham Nov 11 '18 at 20:22
  • $\begingroup$ So is it just HB = S.B? $\endgroup$ – Oisin Brannock Nov 11 '18 at 20:25
  • $\begingroup$ No, you are just asked to work out H. But you are told $H = \mu . B + ... $ and you are told $\mu$ is proportional to S. So to work out H you need to first work out S.B as a single matrix. $\endgroup$ – Bruce Greetham Nov 11 '18 at 20:29
  • $\begingroup$ I'm still confused though as I have no indication as to what value I should assign B in the product $\endgroup$ – Oisin Brannock Nov 11 '18 at 20:33
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    $\begingroup$ Yes exactly right. $\endgroup$ – Bruce Greetham Nov 11 '18 at 21:11
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The Hamiltonain is calculated as \begin{align} H =& \, g \mu_B \, \left(B_x S_x + B_y S_y + B_z S_z\right) \, + \, \frac{1}{2}A_0 S_z = \\ =& \, \frac{g \mu_B}{2} \, \left(B_x \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} + B_y \begin{bmatrix} 0 & -i \\ i & 0 \end{bmatrix} + B_z \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}\right) \, + \, \frac{1}{4}A_0 \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} \\ =& \, \frac{g \mu_B}{2} \, \begin{bmatrix} B_z & B_x-iB_y \\ B_x+iB_y & -B_z \end{bmatrix} \, + \, \frac{1}{4} \begin{bmatrix} A_0 & 0 \\ 0 & -A_0 \end{bmatrix} \\ \end{align} In the case of a constant magnetic field aligned with the $z-$axis, $B_x = B_y=0$ and $B_z = B$. Then $$H = \, \frac{g \mu_B}{2} \, \begin{bmatrix} B_z & 0 \\ 0 & -B_z \end{bmatrix} \, + \, \frac{1}{4} \begin{bmatrix} A_0 & 0 \\ 0 & -A_0 \end{bmatrix} = \frac{1}{2}\begin{bmatrix} g\mu_B\, B_z+\frac{1}{2}A_0 & 0 \\ 0 & - \, g\mu_B\,B_z-\frac{1}{2}A_0 \end{bmatrix} $$ By solving the linear eigenvalue equations $$H \, | \psi \rangle = \lambda\, | \psi \rangle $$ you would get the basis energy states (the eignevectors $| \psi \rangle$) and their energy levels (the eigenvalyes $\lambda$). Since $H$ is a 2 by 2 matrix, so $$ | \psi \rangle = \begin{bmatrix}\psi_1 \\ \psi_2 \end{bmatrix}$$ the equation is $$\begin{bmatrix} \frac{1}{2} g\mu_B\, B_z+\frac{1}{4}A_0 & 0 \\ 0 & - \, \frac{1}{2} g\mu_B\,B_z-\frac{1}{4}A_0 \end{bmatrix} \, \begin{bmatrix}\psi_1 \\ \psi_2 \end{bmatrix} = \lambda \, \begin{bmatrix}\psi_1 \\ \psi_2 \end{bmatrix}$$ so it is easy to see that the egienvectros are $$\begin{bmatrix} 1 \\ 0 \end{bmatrix} \text { and } \begin{bmatrix} 0 \\ 1\end{bmatrix}$$ with energy levels $$ \frac{1}{2} g\mu_B\, B_z+\frac{1}{4}A_0 \,\, \text { and }\,\, - \frac{1}{2} g\mu_B\, B_z-\frac{1}{4}A_0$$ respectively. There are only two eigenstates and the transition from on to the other happens when the energy is equal to the difference of the energy levels, i.e. $$\left(\frac{1}{2} g\mu_B\, B_z+\frac{1}{4}A_0 \right) - \left( - \frac{1}{2} g\mu_B\, B_z-\frac{1}{4}A_0\right) = g\mu_B\, B_z+\frac{1}{2}A_0$$

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