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There is no way you can derive it from non-relativistic mechanics from $\mathbf p,\mathbf r$ only. The spin is an intrinsic property of the electron and you have to postulate it. Here I propose three ways to convince yourself:

One option is to take it as it is. You can convince yourself in a heuristic manner that if you expand $(\mathbf p +e \mathbf A)^2$ with the gauge $\mathbf A = \mathbf B \times \mathbf r /2$ you get various terms, one of them is of the form:

$$H_L=\frac{\mu_{\rm B}}{\hbar}\mathbf B \cdot \mathbf L$$

you can say that the spin $\mathbf S$ is an angular momentum too so to obtain the true Hamiltonian, you have to do the following replacement:

$$H_L\to H_{L-S}=H_L+H_I=\frac{\mu_{\rm B}}{\hbar}\mathbf B \cdot (\mathbf L+ g \mathbf S)$$ where $g$ is some constant (the gyromagnetic factor).

Another possibility is to give up, go to relativistic Dirac equation and work your way to a non-relativistic equation, you will find that the true Hamiltonian (in the non-relativistic limit) is

$$H = \frac12 \frac{[2\mathbf S \cdot (\mathbf p + e\mathbf A)]^2}{2m\hbar}$$ if you expand that you get $H_I$ with $g=2$.

But there is a third possibility. If you have ever heuristically derived Dirac equation you'll know that it is derived by linearizing the relativistic dispersion relation $E^{\rm rel}= \sqrt{p^2c^2+m^2c^4}$ so that you only have linear derivatives ($p$ and not powers of it, $p^2,p^4,\cdots$). The third possibility consist on linearizing not the relativistic $E^{\rm rel}$ but linearizing your usual non-relativistic $E=p^2/2m$ instead. This is called Lévy-Leblond equation and leads to Pauli equation (Schrödinger+$H_I)$ when you try to solve it. Here is a link to some page that discusses it http://quantumchymist.blogspot.com/2014/04/is-spin-relativistic-effect-l-and-first.html but a formal derivation can be found in Greiner's Quantum Mechanics. Interestingly, it also provides $g=2$.

Which is the best? To me all are equally ad hoc and heuristic, the derivation should start from the most general Lagrangian and not a non-relativistic one. However, when one does not know the most general Lagrangian it is good to have different approaches to be convinced of the final result.

There is no way you can derive the spin interaction term $H_I$ from non-relativistic mechanics and using $\mathbf p,\mathbf r$ only. The spin is an intrinsic property of the electron and you have to postulate it. Here I propose three ways to convince yourself:

1. One option is to take it as it is, an interaction with an angular momentum. You can convince yourself that if you expand $(\mathbf p +e \mathbf A)^2$ with the gauge $\mathbf A = \mathbf B \times \mathbf r /2$ you get various terms, one of them is of the form:$$H_L=\frac{\mu_{\rm B}}{\hbar}\mathbf B \cdot \mathbf L\;.$$ As the spin $\mathbf S$ is also an angular momentum operator, to obtain the true Hamiltonian, you have to do the following replacement: $$H_L\to H_{L-S}=H_L+H_I=\frac{\mu_{\rm B}}{\hbar}\mathbf B \cdot (\mathbf L+ g \mathbf S)\;,$$ where $g$ is some constant (the gyromagnetic factor).

2. Another possibility is to give up the nonrelativistic origin. Take relativistic Dirac equation and work your way to a non-relativistic equation, you will find that the true Hamiltonian (in the non-relativistic limit) is $$H = \frac12 \frac{[2\mathbf S \cdot (\mathbf p + e\mathbf A)]^2}{2m\hbar}\,$$ if you expand that you get $H_I$ with $g=2$.

3. There is still a third possibility. If you have ever heuristically derived Dirac equation you'll know that it involves the linearization of the relativistic dispersion relation $E^{\rm rel}= \sqrt{p^2c^2+m^2c^4}$ imposing that you obtain an equation that only has linear derivatives ($p$ and not powers of it, $p^2,p^4,\cdots$). The alternative derivation of $H_I$ consists on linearizing, not the relativistic $E^{\rm rel}$, but your usual non-relativistic kinetic energy $E=p^2/2m$ instead. The resulting formula is called the Lévy-Leblond equation and naturally leads to Pauli equation (Schrödinger+$H_I)$ when you try to solve it. A step by step derivation can be found in Greiner's Quantum Mechanics. Interestingly, it also provides $g=2$.

Which is the best? In my opinion the derivation should start from the most general Lagrangian and not a non-relativistic one. However, when one does not know the most general Lagrangian it is good to have different approaches to be convinced of the final result.

There is no way you can derive it from non-relativistic mechanics from $\mathbf p,\mathbf r$ only. The spin is an intrinsic property of the electron and you have to postulate it. Here I propose three ways to convince yourself:

One option is to take it as it is. You can convince yourself in a heuristic manner that if you expand $(\mathbf p +e \mathbf A)^2$ you get various terms, one of them is of the form:

$$H_L=\frac{\mu_{\rm B}}{\hbar}\mathbf B \cdot \mathbf L$$

you can say that the spin $\mathbf S$ is an angular momentum so to obtain the true Hamiltonian, you have to do the following replacement:

$$H_L\to H_{L-S}=H_L+H_I=\frac{\mu_{\rm B}}{\hbar}\mathbf B \cdot (\mathbf L+ g \mathbf S)$$ where $g$ is some constant (the gyromagnetic factor).

Another possibility is to give up, go to relativistic Dirac equation and work your way to a non-relativistic equation, you will find that the true Hamiltonian (in the non-relativistic limit) is

$$H = \frac12 \frac{[2\mathbf S \cdot (\mathbf p + e\mathbf A)]^2}{2m\hbar}$$ if you expand that you get $H_I$ with $g=2$.

But there is a third possibility. If you have ever heuristically derived Dirac equation you'll know that it is derived by linearizing the relativistic dispersion relation $E^{\rm rel}= \sqrt{p^2c^2+m^2c^4}$ so that you only have linear derivatives ($p$ and not powers of it, $p^2,p^4,\cdots$). The third possibility consist on linearizing not the relativistic $E^{\rm rel}$ but linearizing your usual non-relativistic $E=p^2/2m$ instead. This is called Lévy-Leblond equation and leads to Pauli equation (Schrödinger+$H_I)$ when you try to solve it. Here is a link to some page that discusses it http://quantumchymist.blogspot.com/2014/04/is-spin-relativistic-effect-l-and-first.html but a formal derivation can be found in Greiner's Quantum Mechanics. Interestingly, it also provides $g=2$.

Which is the best? To me all are equally ad hoc and heuristic, the derivation should start from the most general Lagrangian and not a non-relativistic one. However, when one does not know the most general Lagrangian it is good to have different approaches to be convinced of the final result.

There is no way you can derive it from non-relativistic mechanics from $\mathbf p,\mathbf r$ only. The spin is an intrinsic property of the electron and you have to postulate it. Here I propose three ways to convince yourself:

One option is to take it as it is. You can convince yourself in a heuristic manner that if you expand $(\mathbf p +e \mathbf A)^2$ with the gauge $\mathbf A = \mathbf B \times \mathbf r /2$ you get various terms, one of them is of the form:

$$H_L=\frac{\mu_{\rm B}}{\hbar}\mathbf B \cdot \mathbf L$$

you can say that the spin $\mathbf S$ is an angular momentum too so to obtain the true Hamiltonian, you have to do the following replacement:

$$H_L\to H_{L-S}=H_L+H_I=\frac{\mu_{\rm B}}{\hbar}\mathbf B \cdot (\mathbf L+ g \mathbf S)$$ where $g$ is some constant (the gyromagnetic factor).

Another possibility is to give up, go to relativistic Dirac equation and work your way to a non-relativistic equation, you will find that the true Hamiltonian (in the non-relativistic limit) is

$$H = \frac12 \frac{[2\mathbf S \cdot (\mathbf p + e\mathbf A)]^2}{2m\hbar}$$ if you expand that you get $H_I$ with $g=2$.

But there is a third possibility. If you have ever heuristically derived Dirac equation you'll know that it is derived by linearizing the relativistic dispersion relation $E^{\rm rel}= \sqrt{p^2c^2+m^2c^4}$ so that you only have linear derivatives ($p$ and not powers of it, $p^2,p^4,\cdots$). The third possibility consist on linearizing not the relativistic $E^{\rm rel}$ but linearizing your usual non-relativistic $E=p^2/2m$ instead. This is called Lévy-Leblond equation and leads to Pauli equation (Schrödinger+$H_I)$ when you try to solve it. Here is a link to some page that discusses it http://quantumchymist.blogspot.com/2014/04/is-spin-relativistic-effect-l-and-first.html but a formal derivation can be found in Greiner's Quantum Mechanics. Interestingly, it also provides $g=2$.

Which is the best? To me all are equally ad hoc and heuristic, the derivation should start from the most general Lagrangian and not a non-relativistic one. However, when one does not know the most general Lagrangian it is good to have different approaches to be convinced of the final result.

There is no way you can derive it from non-relativistic mechanics from $\mathbf p,\mathbf r$ only. The spin is an intrinsic property of the electron and you have to postulate it. Here I propose three ways to convince yourself:

One option is to take it as it is. You can convince yourself in a heuristic manner that if you expand $(\mathbf p +e \mathbf A)^2$ you get various terms, one of them is of the form:

$$H_L=\frac{\mu_{\rm B}}{\hbar}\mathbf B \cdot \mathbf L$$

you can say that the spin $\mathbf S$ is an angular momentum so to obtain the true Hamiltonian, you have to do the following replacement:

$$H_L\to H_{L-S}=H_L+H_I=\frac{\mu_{\rm B}}{\hbar}\mathbf B \cdot (\mathbf L+ g \mathbf S)$$ where $g$ is some constant (the gyromagnetic factor).

Another possibility is to give up, go to relativistic Dirac equation and work your way to a non-relativistic equation, you will find that the true Hamiltonian (in the non-relativistic limit) is

$$H = \frac12 \frac{[2\mathbf S \cdot (\mathbf p + e\mathbf A)]^2}{2m\hbar}$$ if you expand that you get $H_I$ with $g=2$.

But there is a third possibility. If you have ever heuristically derived Dirac equation you'll know that it is derived by linearizing the relativistic dispersion relation $E^{\rm rel}= \sqrt{p^2c^2+m^2c^4}$ so that you only have linear derivatives ($p$ and not powers of it, $p^2,p^4,\cdots$). The third possibility consist on linearizing not the relativistic $E^{\rm rel}$ but linearizing your usual non-relativistic $E=p^2/2m$ instead. This is called Lévy-Leblond equation and leads to Pauli equation (Schrödinger+$H_I)$ when you try to solve it. Here is a link to some page that discusses it http://quantumchymist.blogspot.com/2014/04/is-spin-relativistic-effect-l-and-first.html but a formal derivation can be found in Greiner of Quantum Mechanics. Interestingly, it also provides $g=2$.

Which is the best? To me all are equally ad hoc and heuristic, the derivation should start from the most general Lagrangian and not a non-relativistic one. However, when one does not know the most general Lagrangian it is good to have different approaches to be convinced of the final result.

There is no way you can derive it from non-relativistic mechanics from $\mathbf p,\mathbf r$ only. The spin is an intrinsic property of the electron and you have to postulate it. Here I propose three ways to convince yourself:

One option is to take it as it is. You can convince yourself in a heuristic manner that if you expand $(\mathbf p +e \mathbf A)^2$ you get various terms, one of them is of the form:

$$H_L=\frac{\mu_{\rm B}}{\hbar}\mathbf B \cdot \mathbf L$$

you can say that the spin $\mathbf S$ is an angular momentum so to obtain the true Hamiltonian, you have to do the following replacement:

$$H_L\to H_{L-S}=H_L+H_I=\frac{\mu_{\rm B}}{\hbar}\mathbf B \cdot (\mathbf L+ g \mathbf S)$$ where $g$ is some constant (the gyromagnetic factor).

Another possibility is to give up, go to relativistic Dirac equation and work your way to a non-relativistic equation, you will find that the true Hamiltonian (in the non-relativistic limit) is

$$H = \frac12 \frac{[2\mathbf S \cdot (\mathbf p + e\mathbf A)]^2}{2m\hbar}$$ if you expand that you get $H_I$ with $g=2$.

But there is a third possibility. If you have ever heuristically derived Dirac equation you'll know that it is derived by linearizing the relativistic dispersion relation $E^{\rm rel}= \sqrt{p^2c^2+m^2c^4}$ so that you only have linear derivatives ($p$ and not powers of it, $p^2,p^4,\cdots$). The third possibility consist on linearizing not the relativistic $E^{\rm rel}$ but linearizing your usual non-relativistic $E=p^2/2m$ instead. This is called Lévy-Leblond equation and leads to Pauli equation (Schrödinger+$H_I)$ when you try to solve it. Here is a link to some page that discusses it http://quantumchymist.blogspot.com/2014/04/is-spin-relativistic-effect-l-and-first.html but a formal derivation can be found in Greiner's Quantum Mechanics. Interestingly, it also provides $g=2$.

Which is the best? To me all are equally ad hoc and heuristic, the derivation should start from the most general Lagrangian and not a non-relativistic one. However, when one does not know the most general Lagrangian it is good to have different approaches to be convinced of the final result.