3
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

$\chi_+$ 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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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