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)?


1 Answer 1


$\mathbf{S}$ 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.

  • $\begingroup$ I was just completing a basically identical answer. $\endgroup$ Jun 15, 2016 at 16:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.