In the text "Introduction to Quantum Mechanics" by Griffiths the following is stated: The magnetic dipole moment $\vec{\mu}$ is proportional to its spin angular momentum $\vec{S}$: $$\vec{\mu} = \gamma \vec{S};$$ where the energy associated with the torque of a magnetic dipole in a uniform magnetic field $\vec{B}$ is $$H = – \vec{\mu} \cdot \vec{B}$$ so the Hamiltonian of a spinning charged particle at rest in a magnetic field $\vec{B}$ is $$H = -\gamma \vec{B} \cdot \vec{S}$$ Larmor precession: Imagine a particle of spin $\frac{1}{2}$ at rest in a uniform magnetic field, which points in the z-direction $$\vec{B} = B_0 \hat{k}.$$ The hamiltonian in matrix form is $$\hat{H} = -\gamma B_0 \hat{S_z} = -\frac{\gamma B_0 \hbar}{2} \begin{bmatrix}
1 & 0 \\
0 & -1 \\
\end{bmatrix} $$
The eigenstates of $\hat{H}$ are the same as those of $\hat{S_z}$:
$$\chi_+~~~~\text{with energy}~~~E_+ = -\frac{(\gamma B_0 \hbar)}{2}$$ or
$$\chi_-~~~~\text{with energy}~~~E_- = +\frac{(\gamma B_0 \hbar)}{2}$$
It is then stated "Evidently the energy is lowest when the dipole moment is parallel to the field". I assume they mean the energy is lowest when the particle is in the state $\chi_{+}$, but how does the particle being in this state correspond to the dipole moment being parallel to the magnetic field?
Thanks.
Best Answer
$\chi_+$ is the $z$ spin up state, because it is the eigenvector of $\sigma_z$ with eigenvalue $1$. Look at the definition of the magnetic field: $\boldsymbol{B} = B_0 \hat{k}$. We have defined the magnetic field pointing in the positive $z$ direction, that is, the same direction as the $\chi_+$ spin.
Therefore the $\chi_+$ state corresponds to a dipole moment with spin parallel to the magnetic field.