Hamiltonian of a particle in a magnetic field

[Kashmir]

I've just started Quantum mechanics
by McIntyre and have understood the following about operators which the author wrote till chapter 2:

1. Each observable has an operator
2. Operators act on kets to produce another kets.
3. Only eigenvalues of an operator are possible values of a measurement.

Now the author in Chapter 3 introduces the Hamiltonian operator $H$ as
"The eigenvalues of the Hamiltonian are the allowed energies of the quantum system, and the eigenstates
of $H$ are the energy eigenstates of the system"
I understood this.

Then the author discusses about a Spin 1/2 particle in a constant magnetic field along $z$ direction.

"The Hamiltonian operator represents the total energy of the system... So to begin, we consider the potential energy of a single magnetic dipole (e.g., in a silver atom) in a uniform magnetic field as the sole term in the Hamiltonian. Recalling that the magnetic dipole is given by
$\mu=g \frac{q}{2 m_{e}} \mathbf{S}$
the Hamiltonian is
$H =- \mathbf{ \mu }\cdot \mathbf{B}$"
I understand that if a particle having a magnetic moment $\mu$ is in a magnetic field $B$ then it has energy $E$ (a scalar) given as $E=-{\mu}. B$

Now in QM we have an operator relationship between the Hamiltonian (operator) and magnetic moment (operator) exactly in the same form as
$H=-{\mu}.B$
Why is that so?
Based on what the author has written so far as I've mentioned in starting of this post I cannot understand this correspondence.

Please help me.

 

[PeroK]

Now in QM we have an operator relationship between the Hamiltonian (operator) and magnetic moment (operator) exactly in the same form as
$H=-{\mu}.B$
Why is that so?
Based on what the author has written so far as I've mentioned in starting of this post I cannot understand this correspondence.

Perhaps it's down to good luck! It could have been something more complicated, but luckily the QM relationship was guessable from what we knew about classical EM.

It's not always the case. QCD and the quark model introduce concepts that are not derivable from classical physics: e.g. isospin and colour charge.

But, for an electron in a magnetic field we only had to guess that the law took the same form as the classical law. As Feynman said in his Cornell lectures: the way to discover a new law of physics is to guess!

 

[Kashmir]

Perhaps it's down to good luck! It could have been something more complicated, but luckily the QM relationship was guessable from what we knew about classical EM.

It's not always the case. QCD and the quark model introduce concepts that have are not derivable from classical physics: e.g. isospin and colour charge.

But, for an electron in a magnetic field we only had to guess that the law took the same form as the classical law. As Feynman said in his Cornell lectures: the way to discover a new law of physics is to guess!

Thank you once again :)
So basically I can't ask for a derivation?

 

[PeroK]

Thank you once again :)
So basically I can't ask for a derivation?

You'll notice that the electron dipole moment has a different gyromagnetic ratio from the classical case. That is derivable from QFT. To derive the law itself, you would need at least a QM model of the EM field. I.e. a quantised EM field.

 

[Kashmir]

You'll notice that the electron dipole moment has a different gyromagnetic ratio from the classical case. That is derivable from QFT. To derive the law itself, you would need at least a QM model of the EM field. I.e. a quantised EM field.

By law you mean the QM operator relation, here $\hat{H}=-\hat{\mu} \cdot B$

 

[PeroK]

By law you mean the QM operator relation, here $\hat{H}=-\hat{\mu} \cdot B$

Yes.

 

[vanhees71]

As this question occurs more often these days, here are my 2cts.

Of course you cannot derive fundamental physical laws somehow mathematically since they are based on observations. Nevertheless there are intuitive theoretical arguments based on fundamental principles that have been discovered by an interplay between observations/experiments and model/theory development.

For the here posed questions there are two empirical building blocks and the very fundamental concept of symmetry considerations.

The 1st building block is Maxwell's electrodynamics, which has been found from considering all known experimental facts on electromagnetism and some "mechanical ideas" which however are quite obsolete today, as also Maxwell himself has started to figure out somewhat later. As it turned out Maxwell's electrodynamics is in fact the paradigmatic example of a relativistic classical field theory, which is however not so important in the here discussed context, and we can stay with the Newtonian description of matter for the purpose here but we need the 2nd building block, which is the discovery of modern quantum theory.

As you correctly say, a particle is described by a Hilbert space and self-adjoint operators, describing the observables. Among these operators is the Hamilton operator, which indeed (a) represents the total energy of a system and also (b) describes the time evolution of the system. This is a bit in analogy to classical mechanics in its formulation with the Hamilton canonical formalism.

Now the description of the dynamics of the system in terms of this Hamiltonian, is consistent with the spacetime model (here Galilei Newton spacetime) you can rely on its symmetry, described by the Galilei group.

One symmetry is the homogeneity of time, leading via Noether's theorem to energy conservation and that the Hamilton operator is the "generator" of time translations. This is the deeper reason, why the Hamilton operator has the properties you textbook describes: it's on one hand representing the total energy of the system and on the other provides the time evolution of the system.

Now you also need other operators for other observables you can build the Hamiltonian from, and the most important properties of these operators are the commutation relations among them, and again here the spacetime symmetries come to your help to define the operators representing the observables a particle can have. From classical mechanics again Noether's theorem tells you, what these observables are: from the symmetry of space under translations it follows that the total momentum of a system is conserved and that the corresponding operators thus must be the infinitesimal generators of translations. An analysis of this idea already provides the "Heisenberg algebra" of position and momentum,
$$[\hat{x}_j,\hat{p}_k]=\mathrm{i} \hbar \delta_{jk}, \quad [\hat{p}_j,\hat{p}_k]=0.$$
Then you have isotropy around any point of space. The corresponding conserved quantity is total angular momentum, and the properties of the rotation goup lead to the commutation relations between the angular momentum components as well as with position and momentum operators. Then you also have symmetry under changes from one inertial frame to another moving with constant velocity against the former, and this symmetry also helps to fix how the Hamiltonian of a free particle should look like, i.e.,
$$\hat{H}_0=\frac{1}{2m} \hat{\vec{p}}^2,$$
where $m$ is a parameter that specifies the socalled "irreducible representation of the Galilei group", and it's of course the mass of the particle.

This is of course a very condensed not very accurate summary of the idea. The details are pretty tricky but of amazing beauty. If you want to learn more about this approach to "derive" the algebra among the operators of non-relativistic QT, have a look at Ballentine, Quantum Mechanics.

This more detailed look also reveals that there is a different kind of angular momentum than you know from classical Newtonian physics, where all angular momenta are "orbital angular momenta", i.e., of the type $\vec{L}=\vec{x} \times \vec{p}$. In quantum mechanics there is an additional kind of intrinsic angular momentum called spin. Roughly speaking it's providing more intrinsic degrees of freedom to an elementary "point particle" than you have in classical physics. It's describing how the states describing a particle at rest ("zero-momentum eigenstates") transform under rotations. For particles without spin (socalled scalar particles) of course a rotation doesn't do anything to a zero-momentum state, as in classical mechanics. For particles with spin however, there are the additional angular-momentum like spin degrees of freedom. An elementary particle thus has a spin quantum number $s \in \{0,1/2,1,\ldots \}$.

$s=0$ are the scalar particles, described with a wave function $\psi(\vec{x})$ in the Hilbert space $\mathrm{L}^2(\mathbb{R}^3)$ of square-integrable functions.

To make the above point with the magnetic moment clear, we need particles with non-zero spin, and the most simple case is $s=1/2$, and the electron is such a particle with spin 1/2. As you can derive from the angular-momentum algebra, which the spin-degrees of freedom obey,
$$[\hat{s}_j,\hat{s}_k]=\mathrm{i} \hbar \epsilon_{jkl} \hat{s}_l,$$
where Einstein's summation convention is used (i.e., you sum over $l$ from 1 to 3, and $\epsilon_{jkl}$ is the Levi-Civita symbol, i.e., $\epsilon_{123}=1$ and totally antisymmetric under interchanges of indices).

Spin 1/2 is now a realization of this angular-momentum algebra in a two-dimensional Hilbert space, the spinor space. It's given by the set of Pauli matrices via $\hat{s}_j=\hbar/2 \hat{\sigma}_j$. Thus you have
$$\hat{\vec{s}}^2=\hbar^2 s(s+1) \hat{1} \quad \text{with} \quad s=1/2$$
and $\hat{s}_3$ is diagonal, i.e., our chosen standard basis of the spinor states are the eigenstates of $\hat{s}_3$ with eigenvalues $\pm \hbar/2$.

Further it turns out that in non-relativistic physics the spin commutes with both $\hat{\vec{x}}$ and $\hat{\vec{p}}$. So you can take as basis set the common eigenvectors of position and $\hat{s}_3$. The states are then wave functions with values in spinor space,
$$\psi_{\sigma}(\vec{x})=\langle \vec{x},\sigma|\psi \rangle.$$
Here $\sigma \in \{1,-1\}$ denoting the eigenvalues of $\hat{s}_3$ as $\hbar \sigma/2$. The scalar product is
$$\langle \psi_1|\psi_2 \rangle=\int_{\mathbb{R}^3} \mathrm{d}^3 x \sum_{\sigma} \psi_{1 \sigma}^*(\vec{x}) \psi_{2 \sigma}(\vec{x}).$$
The Hamiltonian for the free particle still is
$$\hat{H}_0 = \frac{1}{2m} \hat{\vec{p}}^2=-\frac{\hbar^2}{2m} \Delta.$$
We note that we can write this alternatatively in the a bit strange form
$$\hat{H}_0=-\frac{\hbar^2}{2m} (\hat{\vec{\sigma}} \cdot \vec{\nabla})^2, \qquad (*)$$
and this will become important in a moment.

Now we need the other fundamental property of electrodynamics, being a gauge theory. In classical electrodynamics that comes only into the game when you introduce the electromagnetic potentials to express the electromagnetic field via (working with Heaviside-Lorentz units, which are more convenient than SI units here)
$$\vec{E}=-\frac{1}{c} \partial_t \vec{A} - \vec{\nabla} \Phi, \quad \vec{B}=\vec{\nabla} \times \vec{A},$$
where for given $\vec{E}$ and $\vec{B}$ the potentials are only defined up to a "gauge transformation", i.e., if $\Phi$ and $\vec{A}$ describe the fields $\vec{E}$ and $\vec{B}$ via the formulas above, then so does
$$\vec{A}'=\vec{A}-\vec{\nabla} \chi, \quad \Phi'=\Phi+frac{1}{c} \partial_t \chi.$$

In quantum mechanics there is a beautiful idea, how to establish gauge invariance. The idea is that the whole formalism of QT is invariant under multiplications of the vectors (or wave functions when using the position representation/aka the formulation of QM as "wave mechanics") with constant phase factors, i.e., if $\psi_{\sigma}(t,\vec{x})$ describes the state of the electron, then $\psi_{\sigma}'(t,\vec{x})=\exp(\mathrm{i} \alpha) \psi_{\sigma}(\vec{x})$ describes the same state, and the entire formalism, including Schrödingers time-evolution equation (which just says that the Hamilton operator is the generator of time translation) is invariant under this transformation.

The idea is now to make this symmetry "local", i.e., you look for a modified Schrödinger equation that is invariant under multiplication of the wave function with an arbitrary space-time dependent parameter $\alpha(t,\vec{x})$, i.e.,
$$\psi_{\sigma}'(t,\vec{x}) = \exp[\mathrm{i} \alpha(t,\vec{x})] \psi_{\sigma}(t,\vec{x}).$$
The original Schrödinger equation with the free-particle Hamiltonian obviously does not fulfill this, because you get additional terms when taking time derivatives and use the $\vec{\nabla}$-operator.

The idea to "cure" this is that one introduces "covariant derivatives" via
$$\frac{1}{c} \mathrm{D}_t \psi_{\sigma}=\left(\frac{1}{c} \hbar \partial_t +\mathrm{i} q/c \Phi \right), \quad \mathrm{D}_j = (\hbar \partial_j-\mathrm{i} q/c A_j).$$
Then one substitutes $\hbar \partial_t \Rightarrow \mathrm{D}_t$ and $\hbar \partial_j \rightarrow \mathrm{D}_j$.

Using this in (*) leads after some algebra to the correct Pauli equation, including the gyrofactor of 2. For this it was important to write $\hat{H}_0$ in the somewhat tricky form (*). Defining the spinor field $\Psi \in\mathbb{C}^2$ as
$$\Psi(t,\vec{x})=\begin{pmatrix} \psi_1(t,\vec{x}) \\ \psi_{-1}(t,\vec{x}) \end{pmatrix}$$
it reads
$$\mathrm{i} \hbar \partial_t \Psi = -\frac{\hbar^2}{2m} (\vec{\nabla}-\mathrm{i} q \vec{A}/c)^2 \Psi+ q \Phi \Psi -\frac{q}{2mc} g_s (\hat{\vec{s}} \cdot \vec{B}) \Psi \quad \text{with} \quad g_s=2.$$

 

[Kashmir]

As this question occurs more often these days, here are my 2cts.

Of course you cannot derive fundamental physical laws somehow mathematically since they are based on observations. Nevertheless there are intuitive theoretical arguments based on fundamental principles that have been discovered by an interplay between observations/experiments and model/theory development.

For the here posed questions there are two empirical building blocks and the very fundamental concept of symmetry considerations.

The 1st building block is Maxwell's electrodynamics, which has been found from considering all known experimental facts on electromagnetism and some "mechanical ideas" which however are quite obsolete today, as also Maxwell himself has started to figure out somewhat later. As it turned out Maxwell's electrodynamics is in fact the paradigmatic example of a relativistic classical field theory, which is however not so important in the here discussed context, and we can stay with the Newtonian description of matter for the purpose here but we need the 2nd building block, which is the discovery of modern quantum theory.

As you correctly say, a particle is described by a Hilbert space and self-adjoint operators, describing the observables. Among these operators is the Hamilton operator, which indeed (a) represents the total energy of a system and also (b) describes the time evolution of the system. This is a bit in analogy to classical mechanics in its formulation with the Hamilton canonical formalism.

Now the description of the dynamics of the system in terms of this Hamiltonian, is consistent with the spacetime model (here Galilei Newton spacetime) you can rely on its symmetry, described by the Galilei group.

One symmetry is the homogeneity of time, leading via Noether's theorem to energy conservation and that the Hamilton operator is the "generator" of time translations. This is the deeper reason, why the Hamilton operator has the properties you textbook describes: it's on one hand representing the total energy of the system and on the other provides the time evolution of the system.

Now you also need other operators for other observables you can build the Hamiltonian from, and the most important properties of these operators are the commutation relations among them, and again here the spacetime symmetries come to your help to define the operators representing the observables a particle can have. From classical mechanics again Noether's theorem tells you, what these observables are: from the symmetry of space under translations it follows that the total momentum of a system is conserved and that the corresponding operators thus must be the infinitesimal generators of translations. An analysis of this idea already provides the "Heisenberg algebra" of position and momentum,
$$[\hat{x}_j,\hat{p}_k]=\mathrm{i} \hbar \delta_{jk}, \quad [\hat{p}_j,\hat{p}_k]=0.$$
Then you have isotropy around any point of space. The corresponding conserved quantity is total angular momentum, and the properties of the rotation goup lead to the commutation relations between the angular momentum components as well as with position and momentum operators. Then you also have symmetry under changes from one inertial frame to another moving with constant velocity against the former, and this symmetry also helps to fix how the Hamiltonian of a free particle should look like, i.e.,
$$\hat{H}_0=\frac{1}{2m} \hat{\vec{p}}^2,$$
where $m$ is a parameter that specifies the socalled "irreducible representation of the Galilei group", and it's of course the mass of the particle.

This is of course a very condensed not very accurate summary of the idea. The details are pretty tricky but of amazing beauty. If you want to learn more about this approach to "derive" the algebra among the operators of non-relativistic QT, have a look at Ballentine, Quantum Mechanics.

This more detailed look also reveals that there is a different kind of angular momentum than you know from classical Newtonian physics, where all angular momenta are "orbital angular momenta", i.e., of the type $\vec{L}=\vec{x} \times \vec{p}$. In quantum mechanics there is an additional kind of intrinsic angular momentum called spin. Roughly speaking it's providing more intrinsic degrees of freedom to an elementary "point particle" than you have in classical physics. It's describing how the states describing a particle at rest ("zero-momentum eigenstates") transform under rotations. For particles without spin (socalled scalar particles) of course a rotation doesn't do anything to a zero-momentum state, as in classical mechanics. For particles with spin however, there are the additional angular-momentum like spin degrees of freedom. An elementary particle thus has a spin quantum number $s \in \{0,1/2,1,\ldots \}$.

$s=0$ are the scalar particles, described with a wave function $\psi(\vec{x})$ in the Hilbert space $\mathrm{L}^2(\mathbb{R}^3)$ of square-integrable functions.

To make the above point with the magnetic moment clear, we need particles with non-zero spin, and the most simple case is $s=1/2$, and the electron is such a particle with spin 1/2. As you can derive from the angular-momentum algebra, which the spin-degrees of freedom obey,
$$[\hat{s}_j,\hat{s}_k]=\mathrm{i} \hbar \epsilon_{jkl} \hat{s}_l,$$
where Einstein's summation convention is used (i.e., you sum over $l$ from 1 to 3, and $\epsilon_{jkl} is the Levi-Civita symbol, i.e.,$\epsilon_{123}=1$and totally antisymmetric under interchanges of indices). Spin 1/2 is now a realization of this angular-momentum algebra in a two-dimensional Hilbert space, the spinor space. It's given by the set of Pauli matrices via$\hat{s}_j=\hbar/2 \hat{\sigma}_j$. Thus you have \hat{\vec{s}}^2=\hbar^2 s(s+1) \hat{1} \quad \text{with} \quad s=1/2 and$\hat{s}_3$is diagonal, i.e., our chosen standard basis of the spinor states are the eigenstates of$\hat{s}_3$with eigenvalues$\pm \hbar/2$. Further it turns out that in non-relativistic physics the spin commutes with both$\hat{\vec{x}}$and$\hat{\vec{p}}$. So you can take as basis set the common eigenvectors of position and$\hat{s}_3$. The states are then wave functions with values in spinor space, \psi_{\sigma}(\vec{x})=\langle \vec{x},\sigma|\psi \rangle. Here$\sigma \in \{1,-1}$denoting the eigenvalues of$\hat{s}_3$as$\hbar \sigma/2$. The scalar product is \langle \psi_1|\psi_2 \rangle=\int_{\mathbb{R}^3} \mathrm{d}^3 x \sum_{\sigma} \psi_{1 \sigma}^*(\vec{x}) \psi_{2 \sigma}(\vec{x}). The Hamiltonian for the free particle still is \hat{H}_0 = \frac{1}{2m} \hat{\vec{p}}^2=-\frac{\hbar^2}{2m} \Delta. We note that we can write this alternatatively in the a bit strange form \hat{H}_0=-\frac{\hbar^2}{2m} (\hat{\vec{sigma}} \cdot \vec{\nabla})^2, and this will become important in a moment. Now we need the other fundamental property of electrodynamics, being a gauge theory. In classical electrodynamics that comes only into the game when you introduce the electromagnetic potentials to express the electromagnetic field via (working with Heaviside-Lorentz units, which are more convenient than SI units here) \vec{E}=-\frac{1}{c} \partial_t \vec{A} - \vec{\nabla} \Phi, \quad \vec{B}=\vec{\nabla} \times \vec{A}, where for given$\vec{E}$and$\vec{B}$the potentials are only defined up to a "gauge transformation", i.e., if$\Phi$and$\vec{A}$describe the fields$\vec{E}$and$\vec{B}$via the formulas above, then so does \vec{A}'=\vec{A}-\vec{\nabla} \chi, \quad \Phi'=\Phi+frac{1}{c} \partial_t \chi. In quantum mechanics there is a beautiful idea, how to establish gauge invariance. The idea is that the whole formalism of QT is invariant under multiplications of the vectors (or wave functions when using the position representation/aka the formulation of QM as "wave mechanics") with constant phase factors, i.e., if$\psi_{\sigma}(t,\vec{x})$describes the state of the electron, then$\psi_{\sigma}'(t,\vec{x})=\exp(\mathrm{i} \alpha) \psi_{\sigma}(\vec{x})$describes the same state, and the entire formalism, including Schrödingers time-evolution equation (which just says that the Hamilton operator is the generator of time translation) is invariant under this transformation. The idea is now to make this symmetry "local", i.e., you look for a modified Schrödinger equation that is invariant under multiplication of the wave function with an arbitrary space-time dependent parameter$\alpha(t,\vec{x})$, i.e., \psi_{\sigma}'(t,\vec{x}) = \exp[\mathrm{i} \alpha(t,\vec{x})] \psi_{\sigma}(t,\vec{x}). The original Schrödinger equation with the free-particle Hamiltonian obviously does not fulfill this, because you get additional terms when taking time derivatives and use the$\vec{\nabla}## \operator.

Hello Vanhees. I think the Latex code at the end of your post is somewhere misplaced.

 

[Kashmir]

As this question occurs more often these days, here are my 2cts.

Of course you cannot derive fundamental physical laws somehow mathematically since they are based on observations. Nevertheless there are intuitive theoretical arguments based on fundamental principles that have been discovered by an interplay between observations/experiments and model/theory development.

For the here posed questions there are two empirical building blocks and the very fundamental concept of symmetry considerations.

The 1st building block is Maxwell's electrodynamics, which has been found from considering all known experimental facts on electromagnetism and some "mechanical ideas" which however are quite obsolete today, as also Maxwell himself has started to figure out somewhat later. As it turned out Maxwell's electrodynamics is in fact the paradigmatic example of a relativistic classical field theory, which is however not so important in the here discussed context, and we can stay with the Newtonian description of matter for the purpose here but we need the 2nd building block, which is the discovery of modern quantum theory.

As you correctly say, a particle is described by a Hilbert space and self-adjoint operators, describing the observables. Among these operators is the Hamilton operator, which indeed (a) represents the total energy of a system and also (b) describes the time evolution of the system. This is a bit in analogy to classical mechanics in its formulation with the Hamilton canonical formalism.

Now the description of the dynamics of the system in terms of this Hamiltonian, is consistent with the spacetime model (here Galilei Newton spacetime) you can rely on its symmetry, described by the Galilei group.

One symmetry is the homogeneity of time, leading via Noether's theorem to energy conservation and that the Hamilton operator is the "generator" of time translations. This is the deeper reason, why the Hamilton operator has the properties you textbook describes: it's on one hand representing the total energy of the system and on the other provides the time evolution of the system.

Now you also need other operators for other observables you can build the Hamiltonian from, and the most important properties of these operators are the commutation relations among them, and again here the spacetime symmetries come to your help to define the operators representing the observables a particle can have. From classical mechanics again Noether's theorem tells you, what these observables are: from the symmetry of space under translations it follows that the total momentum of a system is conserved and that the corresponding operators thus must be the infinitesimal generators of translations. An analysis of this idea already provides the "Heisenberg algebra" of position and momentum,
$$[\hat{x}_j,\hat{p}_k]=\mathrm{i} \hbar \delta_{jk}, \quad [\hat{p}_j,\hat{p}_k]=0.$$
Then you have isotropy around any point of space. The corresponding conserved quantity is total angular momentum, and the properties of the rotation goup lead to the commutation relations between the angular momentum components as well as with position and momentum operators. Then you also have symmetry under changes from one inertial frame to another moving with constant velocity against the former, and this symmetry also helps to fix how the Hamiltonian of a free particle should look like, i.e.,
$$\hat{H}_0=\frac{1}{2m} \hat{\vec{p}}^2,$$
where $m$ is a parameter that specifies the socalled "irreducible representation of the Galilei group", and it's of course the mass of the particle.

This is of course a very condensed not very accurate summary of the idea. The details are pretty tricky but of amazing beauty. If you want to learn more about this approach to "derive" the algebra among the operators of non-relativistic QT, have a look at Ballentine, Quantum Mechanics.

This more detailed look also reveals that there is a different kind of angular momentum than you know from classical Newtonian physics, where all angular momenta are "orbital angular momenta", i.e., of the type $\vec{L}=\vec{x} \times \vec{p}$. In quantum mechanics there is an additional kind of intrinsic angular momentum called spin. Roughly speaking it's providing more intrinsic degrees of freedom to an elementary "point particle" than you have in classical physics. It's describing how the states describing a particle at rest ("zero-momentum eigenstates") transform under rotations. For particles without spin (socalled scalar particles) of course a rotation doesn't do anything to a zero-momentum state, as in classical mechanics. For particles with spin however, there are the additional angular-momentum like spin degrees of freedom. An elementary particle thus has a spin quantum number $s \in \{0,1/2,1,\ldots \}$.

$s=0$ are the scalar particles, described with a wave function $\psi(\vec{x})$ in the Hilbert space $\mathrm{L}^2(\mathbb{R}^3)$ of square-integrable functions.

To make the above point with the magnetic moment clear, we need particles with non-zero spin, and the most simple case is $s=1/2$, and the electron is such a particle with spin 1/2. As you can derive from the angular-momentum algebra, which the spin-degrees of freedom obey,
$$[\hat{s}_j,\hat{s}_k]=\mathrm{i} \hbar \epsilon_{jkl} \hat{s}_l,$$
where Einstein's summation convention is used (i.e., you sum over $l$ from 1 to 3, and $\epsilon_{jkl}$ is the Levi-Civita symbol, i.e., $\epsilon_{123}=1$ and totally antisymmetric under interchanges of indices).

Spin 1/2 is now a realization of this angular-momentum algebra in a two-dimensional Hilbert space, the spinor space. It's given by the set of Pauli matrices via $\hat{s}_j=\hbar/2 \hat{\sigma}_j$. Thus you have
$$\hat{\vec{s}}^2=\hbar^2 s(s+1) \hat{1} \quad \text{with} \quad s=1/2$$
and $\hat{s}_3$ is diagonal, i.e., our chosen standard basis of the spinor states are the eigenstates of $\hat{s}_3$ with eigenvalues $\pm \hbar/2$.

Further it turns out that in non-relativistic physics the spin commutes with both $\hat{\vec{x}}$ and $\hat{\vec{p}}$. So you can take as basis set the common eigenvectors of position and $\hat{s}_3$. The states are then wave functions with values in spinor space,
$$\psi_{\sigma}(\vec{x})=\langle \vec{x},\sigma|\psi \rangle.$$
Here $\sigma \in \{1,-1\}$ denoting the eigenvalues of $\hat{s}_3$ as $\hbar \sigma/2$. The scalar product is
$$\langle \psi_1|\psi_2 \rangle=\int_{\mathbb{R}^3} \mathrm{d}^3 x \sum_{\sigma} \psi_{1 \sigma}^*(\vec{x}) \psi_{2 \sigma}(\vec{x}).$$
The Hamiltonian for the free particle still is
$$\hat{H}_0 = \frac{1}{2m} \hat{\vec{p}}^2=-\frac{\hbar^2}{2m} \Delta.$$
We note that we can write this alternatatively in the a bit strange form
$$\hat{H}_0=-\frac{\hbar^2}{2m} (\hat{\vec{\sigma}} \cdot \vec{\nabla})^2, \qquad (*)$$
and this will become important in a moment.

Now we need the other fundamental property of electrodynamics, being a gauge theory. In classical electrodynamics that comes only into the game when you introduce the electromagnetic potentials to express the electromagnetic field via (working with Heaviside-Lorentz units, which are more convenient than SI units here)
$$\vec{E}=-\frac{1}{c} \partial_t \vec{A} - \vec{\nabla} \Phi, \quad \vec{B}=\vec{\nabla} \times \vec{A},$$
where for given $\vec{E}$ and $\vec{B}$ the potentials are only defined up to a "gauge transformation", i.e., if $\Phi$ and $\vec{A}$ describe the fields $\vec{E}$ and $\vec{B}$ via the formulas above, then so does
$$\vec{A}'=\vec{A}-\vec{\nabla} \chi, \quad \Phi'=\Phi+frac{1}{c} \partial_t \chi.$$

In quantum mechanics there is a beautiful idea, how to establish gauge invariance. The idea is that the whole formalism of QT is invariant under multiplications of the vectors (or wave functions when using the position representation/aka the formulation of QM as "wave mechanics") with constant phase factors, i.e., if $\psi_{\sigma}(t,\vec{x})$ describes the state of the electron, then $\psi_{\sigma}'(t,\vec{x})=\exp(\mathrm{i} \alpha) \psi_{\sigma}(\vec{x})$ describes the same state, and the entire formalism, including Schrödingers time-evolution equation (which just says that the Hamilton operator is the generator of time translation) is invariant under this transformation.

The idea is now to make this symmetry "local", i.e., you look for a modified Schrödinger equation that is invariant under multiplication of the wave function with an arbitrary space-time dependent parameter $\alpha(t,\vec{x})$, i.e.,
$$\psi_{\sigma}'(t,\vec{x}) = \exp[\mathrm{i} \alpha(t,\vec{x})] \psi_{\sigma}(t,\vec{x}).$$
The original Schrödinger equation with the free-particle Hamiltonian obviously does not fulfill this, because you get additional terms when taking time derivatives and use the $\vec{\nabla}$-operator.

The idea to "cure" this is that one introduces "covariant derivatives" via
$$\frac{1}{c} \mathrm{D}_t \psi_{\sigma}=\left(\frac{1}{c} \hbar \partial_t +\mathrm{i} q/c \Phi \right), \quad \mathrm{D}_j = (\hbar \partial_j-\mathrm{i} q/c A_j).$$
Then one substitutes $\hbar \partial_t \Rightarrow \mathrm{D}_t$ and $\hbar \partial_j \rightarrow \mathrm{D}_j$.

Using this in (*) leads after some algebra to the correct Pauli equation, including the gyrofactor of 2. For this it was important to write $\hat{H}_0$ in the somewhat tricky form (*). Defining the spinor field $\Psi \in\mathbb{C}^2$ as
$$\Psi(t,\vec{x})=\begin{pmatrix} \psi_1(t,\vec{x}) \\ \psi_{-1}(t,\vec{x}) \end{pmatrix}$$
it reads
$$\mathrm{i} \hbar \partial_t \Psi = -\frac{\hbar^2}{2m} (\vec{\nabla}-\mathrm{i} q \vec{A}/c)^2 \Psi+ q \Phi \Psi -\frac{q}{2mc} g_s (\hat{\vec{s}} \cdot \vec{B}) \Psi \quad \text{with} \quad g_s=2.$$

ThankYou, but currently it's out of my scope right now. I'll come back to it soon. :)

 

[HomogenousCow]

Well you could argue that to preserve spatial parity the Hamiltonian has to be some function of $S\cdot B$ since both $S$ and $B$ change sign under spatial reflection. Taking a linear approximation gives you your Hamiltonian.

 

[HomogenousCow]

Uhh correction to my earlier post, $S$ and $B$ don’t change sign under parity so if you want a Hamiltonian linear in $S$ you have to dot it with another pseudo-vector, in this case $B$.