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?



$\chi_+$ 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.

  • $\begingroup$ Okay thanks but what exactly does it mean for a state to be in z spin up state $\chi_+ = \left( \begin{array}{c} 1\\ 0\\ \end{array} \right)$ where it has eigenvalue $\frac{\hbar}{2}$? Is the following the correct interpretation: Does it simply mean that the spin around the z axis can take only two values $\frac{\hbar}{2}$ and $-\frac{\hbar}{2}$ which corresponds to counter clockwise spin around the $\hat{z}$ axis (spin up) so that the vector points up the z axis and respectively counter clockwise spin around the $- \hat{z}$ axis (spin down) so the vector points down the z axis? $\endgroup$ – Alex Nov 22 '16 at 18:16
  • $\begingroup$ Basically, particles have spin angular momentum, which can be measured along a certain axis, for example with a Stern–Gerlach apparatus. When you measure the spin along the $z$ axis of a particle in the state $\chi_+$, for example by letting it pass through a Stern–Gerlach apparatus oriented along the $z$ axis, then you will measure $\hbar/2$. That's the meaning of the $\chi_+$ state. $\endgroup$ – jc315 Nov 22 '16 at 18:42
  • $\begingroup$ Also, I would be cautious of seeking a microscopic explanation of particle spin by imagining the particle actually spinning around a certain axis clockwise or counterclockwise. In nonrelativistic quantum mechanics I think it's best to think of spin as an entirely internal degree of freedom of a point particle, which is fundamentally different from orbital angular momentum associated with the physical rotation of an extended body. $\endgroup$ – jc315 Nov 22 '16 at 18:47
  • $\begingroup$ Okay thanks for you responses. In your answer to my original question you state "We have defined the magnetic field pointing in the positive $z$ direction, that is, the same direction as the $\chi_{+}$ spin." So you are interpreting the quantum spin (corresponding to the positive eigenvalue) as a vector up the z axis as well, this is looking at spin as rotation around an axis, so you are also looking at it in a classical sense based on your answer? $\endgroup$ – Alex Nov 22 '16 at 19:02
  • $\begingroup$ No problem. Even though there is no actual rotation about an axis, we can still associate spin with a direction. You can think of this as just an identifier of the corresponding spin operator and eigenvalue. So, for example, we associate $\chi_+$ with the $+z$ direction because it is an eigenvector of $S_z$ with eigenvalue $\hbar/2$. And we would associate to the $-x$ axis the eigenvector of $S_x$ with eigenvalue $-\hbar/2$. This is just a convenient identification and doesn't imply any actual classical rotation! But yes you are right, the terminology is inspired by the classical case. $\endgroup$ – jc315 Nov 22 '16 at 19:12

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