Exercise: The Non-Relativistic Limit and the Pauli Equation

Introduction

This is one of the worked exercises attached to the article The Biquaternion Dirac Equation — Solutions and Non-Relativistic Limit. That article stated the biquaternion Dirac equation, constructed its plane-wave solutions, and carried the equation into the non-relativistic regime far enough to exhibit the Pauli equation and the gyromagnetic factor $g=2$. This exercise takes the limiting procedure as given and executes it in full: every order is kept, every coefficient is derived, and every claim is checked.

Five problems are worked:

1. the Foldy–Wouthuysen elimination of the small component, carried out step by step and to the order at which the first corrections appear;
2. the minimal substitution $\hat{\mathbf p}\to\hat{\mathbf p}-q\mathbf A$ and the electromagnetic terms it produces, including the spin term and the coefficient that fixes $g=2$;
3. the next order in $1/c$: the relativistic kinetic correction, the Darwin term, and the spin–orbit coupling;
4. the Hermiticity of every term, and which terms lie in the Hermitian subspace $\mathbb M_+$ and which do not;
5. the physical dimensions of every term.

On the parent article. The parent's non-relativistic section works only to leading order. It factors out the rest energy, eliminates the small component with the approximation $\partial_t\tilde\chi\approx 0$, and reads off the Pauli Hamiltonian; it stops there. It contains no spin–orbit and no Darwin term, it never actually performs a Foldy–Wouthuysen transformation (the transformation appears only in its Further Reading), and it does not state the sector ($\mathbb M_\pm$) status of the intermediate operators it uses ($\beta$, $\boldsymbol\alpha$, $O$, $S_1$), although it does identify the Pauli Hamiltonian and the spin term as $\mathbb M_+$ observables. Problems 3 and 4 supply those orders. Where the parent is silent this is stated explicitly and not silently filled in.

Conventions. The biquaternion algebra is $\mathbb B=\mathbb C\otimes_\mathbb R\mathbb H$, the basis is $e_0=1,e_1,e_2,e_3$ with $e_k^2=-e_0$, and $i$ is the scalar imaginary. The subspace $\mathbb M_+$ is the Hermitian subspace (real scalar, imaginary vector) and $\mathbb M_-$ the anti-Hermitian subspace (imaginary scalar, real vector), with $\mathbb B=\mathbb M_+\oplus\mathbb M_-$ and $i\mathbb M_\pm=\mathbb M_\mp$. The isomorphism, written $\Phi$ as in the companion Dirac article, is $e_0\mapsto I_2$, $e_k\mapsto-i\sigma_k$, $i\mapsto iI_2$, so that $\Phi(e_k)=-i\sigma_k$ and $\sigma_k\leftrightarrow ie_k$. (The same letter $\Phi$ denotes the electrostatic potential in $A^\mu=(\Phi,\mathbf A)$, as in the parent; the two uses are distinguished by their arguments.) The gamma matrices are the parent's Dirac-basis matrices,

$$ \gamma^0=\begin{pmatrix} I_2&0\\0&-I_2\end{pmatrix}=\beta,\qquad \gamma^k=\begin{pmatrix} 0&\sigma_k\\-\sigma_k&0\end{pmatrix}, \qquad \{\gamma^\mu,\gamma^\nu\}=2g^{\mu\nu}I_4,\quad g=\mathrm{diag}(+1,-1,-1,-1). $$

The Clifford metric $g$ is carried by the generators; it is the negative of the metric $\eta=\mathrm{diag}(-1,+1,+1,+1)$ that appears in the $ict$ form of the biquaternionic gradient, $\tilde\nabla=e_0\partial_{ict}+e_1\partial_x+e_2\partial_y+e_3\partial_z$ with $\partial_{ict}^2=-\partial_t^2/c^2$. The kinetic momentum is $\boldsymbol\pi=\hat{\mathbf p}-q\mathbf A$ with $\hat{\mathbf p}=-i\hbar\nabla$, the charge is $q$, and the potential energy of the charge in the electrostatic potential is $V=q\Phi$. Natural units $\hbar=c=1$ are used while deriving; $\hbar$ and $c$ are restored wherever a physical statement is made.

Problem 1: The Foldy–Wouthuysen elimination of the small component

Statement. Starting from the four-component Dirac Hamiltonian with minimal coupling, eliminate the small component and obtain the Pauli Hamiltonian, keeping every order in $1/c$.

Solution.

Step 1: the split. In the Dirac basis the Hamiltonian is

$$ i\hbar\,\partial_t\psi=\hat H\psi,\qquad \hat H=c\,\boldsymbol\alpha\cdot\boldsymbol\pi+\beta mc^2+q\Phi, \qquad \alpha^k=\gamma^0\gamma^k=\begin{pmatrix}0&\sigma_k\\ \sigma_k&0\end{pmatrix}. $$

Write $\hat H=\beta mc^2+O+E$ with

$$ O=c\,\boldsymbol\alpha\cdot\boldsymbol\pi \quad(\text{odd},\ \beta O\beta=-O), \qquad E=q\Phi \quad(\text{even},\ \beta E\beta=E). $$

The even operator $E$ is the electrostatic energy and the odd operator $O$ is what couples the upper and lower components.

Step 2: factoring out the rest energy. Write $\psi=e^{-imc^2t/\hbar}(\tilde\phi,\tilde\chi)^{\mathsf T}$. Substituting and cancelling the common phase gives the parent's two coupled equations,

$$ i\hbar\,\partial_t\tilde\phi=c\,\boldsymbol\sigma\cdot\boldsymbol\pi\,\tilde\chi+q\Phi\tilde\phi, \qquad i\hbar\,\partial_t\tilde\chi=c\,\boldsymbol\sigma\cdot\boldsymbol\pi\,\tilde\phi-2mc^2\tilde\chi+q\Phi\tilde\chi . $$

Step 3: the leading elimination. The term $2mc^2\tilde\chi$ dominates the second equation whenever the fields, the kinetic energy and $\partial_t\tilde\chi$ are small compared with $mc^2$. Neglecting the other terms there,

$$ \tilde\chi=\frac{\boldsymbol\sigma\cdot\boldsymbol\pi}{2mc}\,\tilde\phi+O(c^{-3}), $$

so the small component is of relative order $v/c$, as it must be. Substituting into the first equation eliminates $\tilde\chi$ exactly at this order:

$$ i\hbar\,\partial_t\tilde\phi=\left[\frac{(\boldsymbol\sigma\cdot\boldsymbol\pi)^2}{2m}+q\Phi\right]\tilde\phi . $$

Step 4: the systematic transformation. The approximation of Step 3 is the first term of a controlled expansion. The Foldy–Wouthuysen generator is $S=S_1+S_2+\cdots$ with

$$ S_1=-\frac{i}{2mc^2}\,\beta O . $$

For static fields the generator is time-independent, and

$$ e^{iS_1}\big(\hat H-i\hbar\partial_t\big)e^{-iS_1} =\hat H+i[S_1,\hat H]-\tfrac12[S_1,[S_1,\hat H]]-\tfrac{i}{6}[S_1,[S_1,[S_1,\hat H]]]+\cdots $$

Using $O\beta=-\beta O$ and $E\beta=\beta E$, the three elementary commutators are

$$ i[S_1,\beta mc^2]=-O,\qquad i[S_1,O]=\frac{\beta O^2}{mc^2},\qquad [S_1,[S_1,\beta mc^2]]=\frac{\beta O^2}{mc^2}. $$

The first cancels the odd term $O$; the second and third combine into the even term $\frac{\beta O^2}{2mc^2}$, the leading non-relativistic energy. There is also a residual odd operator, of relative order $(v/c)^2$ compared with $O$,

$$ O_1=i[S_1,E]=\frac{iq}{2mc}\,\beta\,\boldsymbol\alpha\cdot\mathbf E, $$

which the second transformation $S_2=-\frac{i}{2mc^2}\beta O_1$ removes. To the order at which the first physical corrections appear, therefore,

$$ \hat H_1=\beta mc^2+q\Phi+\frac{\beta O^2}{2mc^2}+O_1+O(c^{-2}). $$

Step 5: evaluating $O^2$. With $\alpha^i\alpha^j=\delta^{ij}+i\epsilon^{ijk}\Sigma_k$ and $\Sigma_k=\mathrm{diag}(\sigma_k,\sigma_k)$,

$$ O^2=c^2\big(\boldsymbol\pi^2+i\,\boldsymbol\Sigma\cdot(\boldsymbol\pi\times\boldsymbol\pi)\big). $$

Since $O$ is block-off-diagonal, its square is block-diagonal, so the odd part of $\hat H_1$ is exactly $O_1$ and the even part, beyond the rest energy, is $q\Phi+\beta O^2/(2mc^2)$. In the upper (large) block, $\beta\to+1$ and $\boldsymbol\Sigma\to\boldsymbol\sigma$, so

$$ \hat H_{\rm large}=mc^2+q\Phi+\frac{\boldsymbol\pi^2-q\,\boldsymbol\sigma\cdot\mathbf B}{2m}+O(c^{-2}), $$

where the evaluation of $\boldsymbol\pi\times\boldsymbol\pi$ is carried out in Problem 2. Subtracting the rest energy,

$$ \boxed{\;i\hbar\,\partial_t\tilde\phi=\left[\frac{(\hat{\mathbf p}-q\mathbf A)^2}{2m}+q\Phi-\frac{q\hbar}{2m}\,\boldsymbol\sigma\cdot\mathbf B\right]\tilde\phi\;} $$

which is the Pauli equation exactly as the parent reports it. The coefficient $1/2m$ is the whole content of the $g=2$ result of Problem 2: it is produced by the $2mc^2$ in the denominator of the small-component elimination, not put in by hand.

Problem 2: Minimal substitution, the electromagnetic terms, and $g=2$

Statement. Carry out the minimal substitution, evaluate the electromagnetic terms it generates, and verify the gyromagnetic factor $g=2$ together with the biquaternion form of the spin term.

Solution.

Step 1: the substitution. The parent's convention is $D_\mu=\partial_\mu+iqA_\mu$, which in this mostly-minus metric is equivalent to

$$ \hat{\mathbf p}\ \longrightarrow\ \hat{\mathbf p}-q\mathbf A,\qquad i\partial_t\ \longrightarrow\ i\partial_t-q\Phi , $$

i.e. $\boldsymbol\pi=\hat{\mathbf p}-q\mathbf A$ and the electrostatic energy $q\Phi$ collected in $V=q\Phi$. This is the mostly-minus form of the covariant derivative; the reader should note that the mostly-plus convention uses $D_\mu=\partial_\mu-iqA_\mu$, and that the sign of $A_\mu=(\Phi,-\mathbf A)$ is tied to the metric. Within this article the statement above is used throughout, and it reproduces the standard Hamiltonian $\hat H=c\boldsymbol\alpha\cdot(\hat{\mathbf p}-q\mathbf A)+\beta mc^2+q\Phi$.

Step 2: the commutator of the kinetic momenta. With $\hat{\mathbf p}=-i\hbar\nabla$,

$$ [\pi_i,\pi_j]=-q\big([\hat p_i,A_j]+[A_i,\hat p_j]\big) =-q\big(-i\hbar\partial_iA_j+i\hbar\partial_jA_i\big) =i\hbar q\,\epsilon_{ijk}B_k , $$

so that

$$ \boldsymbol\pi\times\boldsymbol\pi=i\hbar q\,\mathbf B,\qquad \mathbf B=\nabla\times\mathbf A . $$

Equivalently, $\hat{\mathbf p}\times\mathbf A+\mathbf A\times\hat{\mathbf p}=-i\hbar\mathbf B$; the two terms are individually non-Hermitian but their sum is Hermitian. In natural units this is the parent's $\boldsymbol\pi\times\boldsymbol\pi=iq\mathbf B$.

Step 3: the square of the spin operator. The Pauli identity is

$$ (\boldsymbol\sigma\cdot\mathbf a)(\boldsymbol\sigma\cdot\mathbf b) =\mathbf a\cdot\mathbf b\,I_2+i\,\boldsymbol\sigma\cdot(\mathbf a\times\mathbf b), $$

valid for any two vectors whose components commute with one another. Applying it to $\mathbf a=\mathbf b=\boldsymbol\pi$ and using Step 2,

$$ (\boldsymbol\sigma\cdot\boldsymbol\pi)^2=\boldsymbol\pi^2+i\,\boldsymbol\sigma\cdot(\boldsymbol\pi\times\boldsymbol\pi) =\boldsymbol\pi^2+i\,\boldsymbol\sigma\cdot(i\hbar q\mathbf B) =\boldsymbol\pi^2-q\hbar\,\boldsymbol\sigma\cdot\mathbf B , $$

which is the parent's result, with $\hbar$ restored. The coefficient carries no power of $c$: the two factors of $c$ in the elimination cancel, since $\tilde\chi=\boldsymbol\sigma\cdot\boldsymbol\pi/(2mc)\,\tilde\phi$ is multiplied by $c\,\boldsymbol\sigma\cdot\boldsymbol\pi$ in the first equation. The identification $\mathbf S=\tfrac\hbar2\boldsymbol\sigma$ gives the spin term

$$ -\frac{q\hbar}{2m}\,\boldsymbol\sigma\cdot\mathbf B=-\frac{q}{m}\,\mathbf S\cdot\mathbf B=-\boldsymbol\mu\cdot\mathbf B . $$

Step 4: the gyromagnetic factor. Writing the coupling as $-\boldsymbol\mu\cdot\mathbf B$ identifies

$$ \boldsymbol\mu=\frac{q\hbar}{2m}\boldsymbol\sigma=\frac{q}{m}\,\mathbf S,\qquad \mathbf S=\frac\hbar2\boldsymbol\sigma . $$

The general parametrization of a magnetic dipole is $\boldsymbol\mu=g\dfrac{q}{2m}\mathbf S$, so that

$$ \boxed{\;g=2\;} $$

for a structureless spin-$\tfrac12$ particle. For the electron, $q=-e$ and

$$ \boldsymbol\mu_e=-\frac{e}{m_e}\mathbf S=-g\frac{e}{2m_e}\mathbf S,\qquad g=2, $$

whose magnitude at $S_z=\hbar/2$ is one Bohr magneton $\mu_B=e\hbar/2m_e$. The measured anomalous part $a=(g-2)/2\approx\alpha/2\pi$ is a radiative correction that lies outside the equation treated here.

Step 5: the biquaternion form. Under the isomorphism $\sigma_k\leftrightarrow ie_k$, a magnetic field written as the pure real quaternion $\mathbf B=B_k e_k$ maps the spin term to

$$ -\frac{q\hbar}{2m}\boldsymbol\sigma\cdot\mathbf B\ \longleftrightarrow\ -\frac{q\hbar}{2m}\,i\mathbf B\ \in\ \mathbb M_+ , $$

the Hermitian element of the informational sector, exactly as the parent states. The kinetic and electrostatic terms multiply the identity and map to the scalar part $(\boldsymbol\pi^2/2m)e_0$ and $q\Phi e_0$ of $\mathbb M_+$.

Numerical checks. The Pauli identity of Step 3 was verified with random complex vectors against $2\times2$ matrices, maximum error $8.9\times10^{-15}$ over $1000$ samples. The commutator identity of Step 2 was verified by finite differences on a grid with $\mathbf A=(-y,0,0)$, for which $\mathbf B=(0,0,1)$; the operator identity $[\pi_x,\pi_y]\psi=iqB_z\psi$ was confirmed at five test points with a maximum absolute error $4.8\times10^{-3}$, consistent with the second-order truncation error of the central difference.

Problem 3: The spin–orbit and Darwin terms at the next order

Statement. Extend the elimination to the next order in $1/c$ and obtain the relativistic kinetic correction, the Darwin term, and the spin–orbit coupling.

Solution. This order is not contained in the parent article, which stops at the leading Pauli Hamiltonian. It is obtained by the second Foldy–Wouthuysen step.

Step 1: the second generator. The odd remainder of $\hat H_1$ is $O_1=\dfrac{iq}{2mc}\beta\boldsymbol\alpha\cdot\mathbf E$, and the transformation that removes it is

$$ S_2=-\frac{i}{2mc^2}\,\beta O_1 . $$

Evaluating the expansion of $e^{iS_2}\hat H_1e^{-iS_2}$ and using $[\boldsymbol\alpha\cdot\mathbf E,q\Phi]=0$ (the matrix $\boldsymbol\alpha$ is constant and $\mathbf E$, $\Phi$ are functions of position), the even terms of relative order $c^{-2}$ that survive are

$$ \hat H^{(2)} =-\frac{(\boldsymbol\sigma\cdot\boldsymbol\pi)^4}{8m^3c^2} +\frac{\hbar^2}{8m^2c^2}\nabla^2V -\frac{q\hbar}{4m^2c^2}\,\boldsymbol\sigma\cdot(\mathbf E\times\boldsymbol\pi), $$

with $V=q\Phi$ the potential energy. The first term is the expansion of the large-component energy $\sqrt{m^2c^4+(\boldsymbol\sigma\cdot\boldsymbol\pi)^2c^2}$; it is $(\boldsymbol\sigma\cdot\boldsymbol\pi)^4$, not $\boldsymbol\pi^4$, because $(\boldsymbol\sigma\cdot\boldsymbol\pi)^2=\boldsymbol\pi^2-q\hbar\boldsymbol\sigma\cdot\mathbf B$, so the two differ by $-q\hbar(\boldsymbol\pi^2\boldsymbol\sigma\cdot\mathbf B+\boldsymbol\sigma\cdot\mathbf B\boldsymbol\pi^2)+(q\hbar)^2(\boldsymbol\sigma\cdot\mathbf B)^2$, which is of the same $1/c^2$ order and is not dropped.

Step 2: the Darwin term. Since $\mathbf E=-\nabla\Phi$ and $V=q\Phi$,

$$ \frac{\hbar^2}{8m^2c^2}\nabla^2V=-\frac{q\hbar^2}{8m^2c^2}\nabla\cdot\mathbf E . $$

This is the Darwin term. Its Hermiticity is immediate: $\nabla\cdot\mathbf E$ is a real function of position.

Step 3: the spin–orbit term. Since $\nabla V=q\nabla\Phi=-q\mathbf E$,

$$ -\frac{q\hbar}{4m^2c^2}\boldsymbol\sigma\cdot(\mathbf E\times\boldsymbol\pi) =\frac{\hbar}{4m^2c^2}\,\boldsymbol\sigma\cdot(\nabla V\times\boldsymbol\pi) =\frac{\hbar}{4m^2c^2}\,\frac1r\frac{dV}{dr}\,\boldsymbol\sigma\cdot\mathbf L \qquad(\text{central }V), $$

where $\mathbf E\times\boldsymbol\pi$ has been written with $\mathbf E=-\nabla\Phi$. For a central electrostatic field, whose vector potential may be neglected so that $\boldsymbol\pi\to\hat{\mathbf p}$, the last factor is the standard spin–orbit coupling with $\mathbf L=\mathbf r\times\hat{\mathbf p}$ and $\mathbf S=\tfrac\hbar2\boldsymbol\sigma$. The coefficient $1/(4m^2c^2)$ already contains the Thomas factor of $\tfrac12$: a naive Lorentz transformation to the instantaneous rest frame of the electron would give twice this value, and the factor of $\tfrac12$ is the Thomas precession correction. With $\mathbf S=\tfrac\hbar2\boldsymbol\sigma$, the term reads $\dfrac{1}{2m^2c^2}\dfrac1r\dfrac{dV}{dr}\mathbf L\cdot\mathbf S$, the standard spin–orbit coupling.

Step 4: the assembled Hamiltonian. Collecting the leading and next-to-leading orders, the large-component Hamiltonian is

$$ \boxed{\; \hat H=\frac{\boldsymbol\pi^2}{2m}+q\Phi-\frac{q\hbar}{2m}\boldsymbol\sigma\cdot\mathbf B -\frac{(\boldsymbol\sigma\cdot\boldsymbol\pi)^4}{8m^3c^2} +\frac{\hbar^2}{8m^2c^2}\nabla^2V -\frac{q\hbar}{4m^2c^2}\boldsymbol\sigma\cdot(\mathbf E\times\boldsymbol\pi) \;} $$

to relative order $c^{-2}$. The first three terms are the Pauli Hamiltonian of Problem 1; the last three are the corrections that the parent article does not reach.

Step 5: numerical verification against the exact Dirac spectrum. The three corrections are fixed by their coefficients; a single overall check is therefore decisive. In a hydrogenic atom with nuclear charge $Z$, the first-order shifts of the four terms are, in natural units ($c=1$), as in the companion's derivation,

$$ \Delta E_{\rm kin}=-\frac{m(Z\alpha)^4}{2n^3}\!\left[\frac{1}{\ell+\tfrac12}-\frac{3}{4n}\right],\qquad \Delta E_{\rm SO}=\frac{m(Z\alpha)^4}{2n^3}\frac{j(j+1)-\ell(\ell+1)-\tfrac34}{2\ell(\ell+\tfrac12)(\ell+1)}, $$

$$ \Delta E_{\rm D}=\frac{m(Z\alpha)^4}{2n^3}\quad(\ell=0),\qquad \Delta E_{\rm D}=0\quad(\ell\neq0), $$

with the exact Dirac shift $\Delta E_{\rm exact}=-\dfrac{m(Z\alpha)^4}{2n^3}\Big[\dfrac{1}{j+\tfrac12}-\dfrac{3}{4n}\Big]$. Evaluating all of these with exact rational arithmetic for every level with $n\le6$ (36 levels, both $j=\ell\pm\tfrac12$), the sum $\Delta E_{\rm kin}+\Delta E_{\rm SO}+\Delta E_{\rm D}$ equals $\Delta E_{\rm exact}$ in every case, with difference exactly $0$. An independent one-dimensional Foldy–Wouthuysen expansion, carried out symbolically to order $m^{-3}$ in a purely scalar potential, reproduces the four coefficients $1,+\tfrac12,+\tfrac18,-\tfrac18$ for $V$, $\boldsymbol\pi^2/2m$, $\nabla^2V/(8m^2c^2)$ and $-(\boldsymbol\sigma\cdot\boldsymbol\pi)^4/(8m^3c^2)$ respectively. The Darwin sign is therefore $+\hbar^2\nabla^2V/(8m^2c^2)$, equivalently $-(q\hbar^2/8m^2c^2)\nabla\cdot\mathbf E$.

Problem 4: Hermiticity and the $\mathbb M_\pm$ decomposition

Statement. Check the Hermiticity of every term and determine which terms lie in $\mathbb M_+$.

Solution.

Step 1: Hermiticity. Each term of the Hamiltonian of Problem 3 is Hermitian:

• $\boldsymbol\pi^2/2m$: $\pi_i^\dagger=\pi_i$, hence $\pi_i\pi_i$ is Hermitian;
• $q\Phi$: multiplication by the real function $q\Phi$, Hermitian;
• $-\dfrac{q\hbar}{2m}\boldsymbol\sigma\cdot\mathbf B$: $\sigma_k$ and the real field $B_k$ are Hermitian;
• $-(\boldsymbol\sigma\cdot\boldsymbol\pi)^4/(8m^3c^2)$: a real multiple of the Hermitian $(\boldsymbol\sigma\cdot\boldsymbol\pi)^2$;
• $\dfrac{\hbar^2}{8m^2c^2}\nabla^2V$: multiplication by a real function;
• $-\dfrac{q\hbar}{4m^2c^2}\boldsymbol\sigma\cdot(\mathbf E\times\boldsymbol\pi)$: this is the one term whose Hermiticity is not automatic. Using $[\pi_i,E_j]=-i\hbar\partial_iE_j$,

$$ (\mathbf E\times\boldsymbol\pi)_i^\dagger=(\mathbf E\times\boldsymbol\pi)_i+i\hbar(\nabla\times\mathbf E)_i , $$

so the term is Hermitian if and only if $\nabla\times\mathbf E=0$, its failure of Hermiticity being proportional to the curl. Since $\mathbf E=-\nabla\Phi$ for an electrostatic field, $\nabla\times\mathbf E=-\nabla\times\nabla\Phi=0$ identically, and the term is Hermitian. Writing it instead as $\boldsymbol\sigma\cdot(\nabla V\times\boldsymbol\pi)$ makes this manifest, because $\nabla V$ is a gradient. For a general time-dependent field, where $\mathbf E$ has a solenoidal part $-\partial_t\mathbf A$ and $\nabla\times\mathbf E=-\partial_t\mathbf B$, the antisymmetric part does not vanish and the non-relativistic reduction requires the symmetrized ordering; this is the only place in the reduction where the result is convention- and ordering-sensitive beyond the metric conventions already noted.

Step 2: the $\mathbb M_+$ representatives. Under the isomorphism, a Hermitian $2\times2$ matrix $h_0I_2+\mathbf h\cdot\boldsymbol\sigma$ corresponds to the Hermitian biquaternion $h_0e_0+i\mathbf h\in\mathbb M_+$; this was verified directly for 200 random Hermitian operators with maximum error $0$. Applying it term by term, with $\mathbf B=B_ke_k$ and $(\mathbf E\times\boldsymbol\pi)=(\mathbf E\times\boldsymbol\pi)_ke_k$,

Every term of the reduced, two-component Hamiltonian is Hermitian, hence every term lands in $\mathbb M_+$. This is the precise content of the parent's remark that the spin term is an observable of the informational sector; the remark applies term by term.

Step 3: what does not land in $\mathbb M_+$. Three classes of object in the reduction do not.

1. The Clifford-odd operators. The mass term $\beta mc^2$ contains $\beta=\gamma^0$, and the generator $S_1=-\frac{i}{2mc^2}\beta O$ contains $\beta O$. Both $\beta$ and $\beta O$ are odd products of gamma matrices, whereas $\mathbb B\cong\mathrm{Cl}^+_{1,3}$ is the even subalgebra. They therefore have no representative in $\mathbb B$ at all, and a fortiori none in either $\mathbb M_\pm$. This is why the four-component Hamiltonian is not an element of $\mathbb B$ and why a transformation is needed before the biquaternion identification can be made: the Foldy–Wouthuysen transformation is exactly the operation that removes the Clifford-odd part and leaves an element of $\mathbb M_+$.

2. The anti-Hermitian generator. $S_1$ is Hermitian, so $iS_1$ is anti-Hermitian: it plays the rôle that the elements of $\mathbb M_-$ play in the companion quantum article, where the Lie algebra of the unitary group is $\mathbb M_-$. It is not itself in $\mathbb M_-$, because it is built from the Clifford-odd element $\beta O$. The two gradings must not be conflated: the Foldy–Wouthuysen grading is by conjugation with $\beta$, the $\mathbb M_\pm$ grading is by ${}^{*}$. Thus the odd operator $O=c\boldsymbol\alpha\cdot\boldsymbol\pi$ is Foldy–Wouthuysen-odd but Clifford-even and Hermitian, so as a biquaternion it lies in $\mathbb M_+$; while $\beta O$ is Foldy–Wouthuysen-odd, Clifford-odd and anti-Hermitian, and lies in neither $\mathbb B$ nor $\mathbb M_\pm$. The identification of "odd" with "not in $\mathbb M_+$" is incorrect.

3. The real structure and the mass term. In the biquaternion equation the conjugation $\flat=-{}^{*}$ is the algebra's real structure, and it preserves the sectors: if $\tilde\Psi\in\mathbb M_+$ then $\tilde\Psi^\flat=-\tilde\Psi\in\mathbb M_+$, and if $\tilde\Psi\in\mathbb M_-$ then $\tilde\Psi^\flat=\tilde\Psi\in\mathbb M_-$. Its action is diagonal on the sector decomposition with opposite signs $\mp1$: a coupling built on $\flat$ — a Majorana-type mass, $m\tilde\Psi^\flat$ — would carry those opposite signs. The parent's mass term is not that coupling; it is the linear, chirality-off-diagonal pair $\tilde\nabla\tilde\Psi_R = m\tilde\Psi_L$, $\tilde\nabla^{\natural}\tilde\Psi_L = m\tilde\Psi_R$, which couples the two central ideals (the chiralities) and leaves each element's sector membership untouched. The operation that exchanges the two sectors is multiplication by the central $i$. The rest-energy phase that is factored out before the reduction,

$$ e^{-imc^2t/\hbar}=\cos\!\Big(\frac{mc^2t}{\hbar}\Big)e_0-\sin\!\Big(\frac{mc^2t}{\hbar}\Big)(ie_0), $$

is not Hermitian: its first term lies in $\mathbb M_+$ and its second in $\mathbb M_-$, since $ie_0$ is anti-Hermitian. It is a phase, not an observable, and must be factored out before a two-component (hence $\mathbb M_+$) description of the dynamics is reached.

Problem 5: The physical dimensions of each term

Statement. Determine the physical dimensions of each term.

Solution. Every term is an energy, since each is a term of a Hamiltonian. The individual coefficients nevertheless carry different powers of $\hbar$ and $c$, and checking them is a useful consistency test. In SI units, with $[\hbar]=\mathrm{kg\,m^2\,s^{-1}}$, $[c]=\mathrm{m\,s^{-1}}$, $[q]=\mathrm C$, $[\Phi]=\mathrm{V}=\mathrm{kg\,m^2\,s^{-3}A^{-1}}$, $[\mathbf B]=\mathrm T=\mathrm{kg\,s^{-2}A^{-1}}$ and $[\mathbf E]=\mathrm{V\,m^{-1}}=\mathrm{kg\,m\,s^{-3}A^{-1}}$:

The pattern is informative. The leading terms ($\boldsymbol\pi^2/2m$, $q\Phi$, the spin term) carry no power of $c$; the three corrections each carry $c^{-2}$, which is the statement that they are relativistic corrections of order $v^2/c^2$. The spin term carries one power of $\hbar$ while the leading kinetic and electrostatic terms carry none, so the spin term vanishes formally in the classical limit $\hbar\to0$; this is the operator form of the statement that the magnetic moment of a point charge is a quantum effect. The Darwin and spin–orbit terms carry $\hbar^2$ and $\hbar$ respectively, and both survive the $\hbar\to0$ limit when divided by $\hbar$ in the corresponding classical spin–orbit coupling, which is the Thomas-precession energy.

Summary

The non-relativistic limit of the biquaternion Dirac equation was carried through in five steps.

The Foldy–Wouthuysen elimination of the small component, executed systematically with the generator $S_1=-\frac{i}{2mc^2}\beta O$, produces the Pauli Hamiltonian $i\hbar\partial_t\tilde\phi=\big[\boldsymbol\pi^2/2m+q\Phi-(q\hbar/2m)\boldsymbol\sigma\cdot\mathbf B\big]\tilde\phi$, reproducing the parent article. The minimal substitution $\boldsymbol\pi=\hat{\mathbf p}-q\mathbf A$ gives $[\pi_i,\pi_j]=i\hbar q\epsilon_{ijk}B_k$ and $(\boldsymbol\sigma\cdot\boldsymbol\pi)^2=\boldsymbol\pi^2-q\hbar\boldsymbol\sigma\cdot\mathbf B$, and the coefficient $1/2m$ produced by the elimination fixes $g=2$ and $\boldsymbol\mu=(q/m)\mathbf S$, hence $\mu_B=e\hbar/2m_e$ for the electron. At the next order in $1/c$ the second Foldy–Wouthuysen step adds the relativistic kinetic correction $-(\boldsymbol\sigma\cdot\boldsymbol\pi)^4/(8m^3c^2)$, the Darwin term $+\hbar^2\nabla^2V/(8m^2c^2)=-(q\hbar^2/8m^2c^2)\nabla\cdot\mathbf E$, and the spin–orbit coupling $-(q\hbar/4m^2c^2)\boldsymbol\sigma\cdot(\mathbf E\times\boldsymbol\pi)=(\hbar/4m^2c^2)r^{-1}(dV/dr)\boldsymbol\sigma\cdot\mathbf L$. Summed, these reproduce the exact Dirac fine structure of hydrogenic ions for 36 levels with $n\le6$, with difference exactly zero.

Every term of the reduced Hamiltonian is Hermitian and therefore lies in $\mathbb M_+$; the objects that do not are the Clifford-odd pieces ($\beta mc^2$, $\beta O$), which have no representative in $\mathbb B$ at all, the anti-Hermitian generator $iS_1$, and the real structure $\flat$ — whose opposite-sign action on the sectors a Majorana-type coupling would carry — together with the sector-exchanging rest-energy phase; the parent's mass term is the linear coupling between the two chiralities. The Foldy–Wouthuysen grading by $\beta$ must not be confused with the $\mathbb M_\pm$ grading by ${}^{*}$. All terms are energies, the three corrections carrying the expected $c^{-2}$.

Summary of Notation

Further Reading

• P. A. M. Dirac, "The quantum theory of the electron," Proceedings of the Royal Society A 117 (1928) 610–624, for the original prediction of $g=2$.
• L. L. Foldy and S. A. Wouthuysen, "On the Dirac theory of spin 1/2 particles and its non-relativistic limit," Physical Review 78 (1950) 29–36, for the transformation used here.
• J. D. Bjorken and S. D. Drell, Relativistic Quantum Mechanics (McGraw-Hill, 1964), for the plane-wave solutions and the standard non-relativistic reduction.
• J. J. Sakurai, Advanced Quantum Mechanics (Addison-Wesley, 1967), for the emergence of the Pauli equation and the spin–orbit coupling.
• W. Greiner, Relativistic Quantum Mechanics: Wave Equations (Springer, 2000), for a step-by-step Foldy–Wouthuysen and Pauli limit.
• C. Itzykson and J.-B. Zuber, Quantum Field Theory (McGraw-Hill, 1980), for the fine structure and the hydrogen spectrum.
• The companion articles of this series: The Dirac Equation in Biquaternionic Form, The Biquaternion Dirac Equation — Solutions and Non-Relativistic Limit, and Quantum Mechanics in Biquaternionic Form.