Vector addition in spin orbit coupling

[kelly0303]

Hello! I am reading about spin-orbit coupling in Griffiths book, and at a point he shows an image (section 6.4.1) of the vectors L and S coupled together to give J (figure 6.10) and he says that L and S precess rapidly around J. I am not totally sure I understand this. I know that in the presence of spin-orbit coupling $L^2$ and $S^2$ are conserved, so the magnitude of L and S must be constant but $L_z$ and $S_z$ are not good quantum numbers, so they must change. I assume that this precession around J is the only configuration consistent with these requirements. However, I am not sure I understand why does the precession need to be fast. $L_z$ and $S_z$ would be conserved only if you have no precession at all, so you would have consistent results no matter how fast they precess. Where is this "speed" of precession coming from? Thank you!

 

[MathematicalPhysicist]

Aren't $L_z$ and $S_z$ come in multiples of $\hbar$, thus they are also quantum numbers.

BTW, what is the rigorous definition of a quantum number?
https://en.wikipedia.org/wiki/Quantum_number

 

[vanhees71]

I think the Wikipedia explains it quite well. Quantum numbers are eigenvalues of conserved quantities, i.e., the operator representing such a quantity commutes with the Hamiltonian and thus can be diagonalized simultaneously with the Hamiltonian. If you are lucky, and your system has enough symmetries you can find a complete compatible set of conserved quantities. Then you have a complete orthonormal basis characterized by the energy eigenvalue and a set of "quantum numbers".

In the most simple case of the hydrogen atom the spin-orbit coupling occurs when you take into account the interaction of the electron with the magnetic field due to the proton's motion in the (momentaneous) restframe of the electron. Famously one has to take into account also the Thomas precession, which is a relativistic effect important also in the non-relativistic limit, correcting the naive spin-orbit coupling term by a factor 1/2 (solving the enigma of the anomalous Zeeman effect making it consistent with the gyro-factor 2 of the spin-magnetic eigenmoment of the electron on the one hand and the measured value of the Zeeman-doublet splitting of atomic spectral lines). So finally you get a term
$$\hat{H}_{\text{LS}} = \frac{\mu_{\text{B}}}{\hbar c^2 m e} \frac{1}{r} \partial_r U(r) \vec{L} \cdot \vec{S}.$$
Now the Hamiltonian does not commute anymore with $L_z$ and $S_z$ but only with the total angular-momentum component $J_z=L_z+S_z$, i.e., now "good quantum numbers" (which means they are eigenvalues of conserved quantities) are $\vec{L}^2$, $\vec{S}^2$, $\vec{J}^2$, and $J_z$, and of course the energy eigenvalue $E$ itself but no longer $L_z$ and $S_z$, because these operators cannot anymore diagonalized simultaneously with the Hamiltonian.

Note that for a complete treatment of the fine structure with leading-order perturbation theory you have to take into account also the leading-order corrections of the kinetic-energy piece of the Hamiltonian due to relativity as well as the "Darwin term". The correction is of order $\alpha^2$ relative to the "unperturbed" energy eigenvalues of the H atom, which are themselves of order $\alpha^2$.

 

[MathematicalPhysicist]

@vanhees71 for your last note, doesn't it use Relativistic QM in those calculations?
Which you once IIRC called it inconsistent theory.

 

[kelly0303]

I think the Wikipedia explains it quite well. Quantum numbers are eigenvalues of conserved quantities, i.e., the operator representing such a quantity commutes with the Hamiltonian and thus can be diagonalized simultaneously with the Hamiltonian. If you are lucky, and your system has enough symmetries you can find a complete compatible set of conserved quantities. Then you have a complete orthonormal basis characterized by the energy eigenvalue and a set of "quantum numbers".

In the most simple case of the hydrogen atom the spin-orbit coupling occurs when you take into account the interaction of the electron with the magnetic field due to the proton's motion in the (momentaneous) restframe of the electron. Famously one has to take into account also the Thomas precession, which is a relativistic effect important also in the non-relativistic limit, correcting the naive spin-orbit coupling term by a factor 1/2 (solving the enigma of the anomalous Zeeman effect making it consistent with the gyro-factor 2 of the spin-magnetic eigenmoment of the electron on the one hand and the measured value of the Zeeman-doublet splitting of atomic spectral lines). So finally you get a term
$$\hat{H}_{\text{LS}} = \frac{\mu_{\text{B}}}{\hbar c^2 m e} \frac{1}{r} \partial_r U(r) \vec{L} \cdot \vec{S}.$$
Now the Hamiltonian does not commute anymore with $L_z$ and $S_z$ but only with the total angular-momentum component $J_z=L_z+S_z$, i.e., now "good quantum numbers" (which means they are eigenvalues of conserved quantities) are $\vec{L}^2$, $\vec{S}^2$, $\vec{J}^2$, and $J_z$, and of course the energy eigenvalue $E$ itself but no longer $L_z$ and $S_z$, because these operators cannot anymore diagonalized simultaneously with the Hamiltonian.

Note that for a complete treatment of the fine structure with leading-order perturbation theory you have to take into account also the leading-order corrections of the kinetic-energy piece of the Hamiltonian due to relativity as well as the "Darwin term". The correction is of order $\alpha^2$ relative to the "unperturbed" energy eigenvalues of the H atom, which are themselves of order $\alpha^2$.

Thank you for this! I am pretty sure I understand why they are not conserved, what confuses me is the vectorial representation, in particular the fact that they move FAST around the the J vector. Why is it fast?

 

[vanhees71]

@vanhees71 for your last note, doesn't it use Relativistic QM in those calculations?
Which you once IIRC called it inconsistent theory.

It's only inconsistent if you try to do it in the 1st-quantization formalism and try to interpret a classical Dirac field as a "wave function" as the Schrödinger or Pauli field in non-relativistic QM. If you use it within QFT taking into account relativistic effects as small perturbations of the non-relativistic approximation all is fine (until the $Z$ of your atom gets too large, when relativistic effects are no small corrections anymore, but that's another story).

 

[vanhees71]

Thank you for this! I am pretty sure I understand why they are not conserved, what confuses me is the vectorial representation, in particular the fact that they move FAST around the the J vector. Why is it fast?

I'm not so sure I understand, what's meant by "fast" here. The pre-quantum ideas using this vector model should not be taken too seriously anymore anyway. Is this coming from Griffiths's book (if so, where is it stated and in which edition?). I think Griffiths's quantum-mechanics book is sometimes a bit confusing. I'd rather recommend Sakurai as an introductory grad-level text.

 

[kelly0303]

I'm not so sure I understand, what's meant by "fast" here. The pre-quantum ideas using this vector model should not be taken too seriously anymore anyway. Is this coming from Griffiths's book (if so, where is it stated and in which edition?). I think Griffiths's quantum-mechanics book is sometimes a bit confusing. I'd rather recommend Sakurai as an introductory grad-level text.

This is from the first edition of Griffiths, section 6.4.1 Weak-Field Zeeman Effect