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I am reading Introduction to Quantum Mechanics 1st edition by David J. Griffiths and I have a couple questions about this section on page 160.

A spinning charged particle constitutes a magnetic dipole. Its magnetic dipole moment $\mu$ is proportional to its spin angular momentum S: $$ \mathbf{\mu} = \gamma\mathbf{S}$$ the proportionality constant $\gamma$ is called the gyromagnetic ratio.

Taking the magnetic dipole moment to be a vector in $\mathbb{R}^3$, what is S is referring to? I have not yet seen any vector in $\mathbb{R}^3$ defined as the spin angular momentum in the text, only spinors that give the general state of, for example, a spin-1/2 particle as $$\chi = \begin{pmatrix}a\\b\end{pmatrix} = a\chi_+ + b\chi_-$$ using the spin up and spin down eigenstates as basis vectors.

The section continues:

When a magnetic dipole is placed in a magnetic field $\mathbf{B}$, it experiences a torque, $\mathbf{\mu \times B}$, which tends to line it up parallel to the field (just like a compass needle). The energy associated with the torque is $$H = -\mu\cdot\mathbf{B}$$ so the Hamiltonian of a spinning charged particle, at rest in a magnetic field $\mathbf{B}$, becomes $$H = -\gamma\mathbf{B\cdot S}$$ where $\mathbf{S}$ is the appropriate spin matrix.

What is the mathematical meaning of this dot product $\mathbf{B\cdot S}$ of a vector in $\mathbb{R}^3$ with a 2x2 matrix (in the case of spin 1/2)?

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$\mathbf{S}$ is the spin operator. It is a vector operator that acts on spinors. It will have three components $(S_x, S_y, S_z)$ and for example if you take the $z$ axis as your spin measurement axis, you define spin up and down as the two eigenstates of $S_z$.

It can be shown that in matrix form $S_i$ is proportional to the Pauli matrix $\sigma_i$.

Finally, $\mathbf{S}\cdot\mathbf{B} = S_xB_x + S_yB_y + S_zB_z$. Note that in matrix form each component of $\mathbf{S}$ is a $2\times2$ matrix, so $\mathbf{S}\cdot\mathbf{B}$ is a $2\times2$ matrix too.

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  • $\begingroup$ I was just completing a basically identical answer. $\endgroup$ – Luboš Motl Jun 15 '16 at 16:43

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