Similitudes Between Biquaternion Rotors and Hamiltonian Flow

Introduction

A biquaternion rotor is a biquaternion $R$ of unit norm, $N(R) = R\bar{R} = e_0$, which acts on the algebra by the rotor conjugation

$$ \tilde{Q} \;\longmapsto\; R\,\tilde{Q}\,R^{*} , $$

where $R^{*} = \bar{R}^{\,*}$ is the Hermitian conjugate (so $R \tilde{Q} R^{*}$ is the map the brief writes as $R\,x\,\tilde{R}$). A Hamiltonian flow is the one-parameter family of canonical transformations generated by a function $H$ on a phase space, written in the compact form

$$ x' = \{x, H\} . $$

The two constructions look like the same idea twice. In each case a one-parameter group of transformations is produced from a single generating object, and in each case the generator is a derivation of an algebra: the inner derivation $\mathrm{ad}_G = [G,\cdot]$ for the rotor, the Poisson derivation $\{\cdot, H\}$ for the flow. The exponential map turns the generator into the transformation, and the composition of transformations into the addition of generators.

This article is an explicit study of that resemblance, of the place where it is exact, and of the places where it fails. The word similitude is the brief: what is claimed here is a structural resemblance, not an identity, and every parallel is stated together with the point at which it breaks. The essential difference, stated up front, is a difference of kind: a rotor is a group action on the algebra, while a Hamiltonian flow is the flow of a vector field on a phase space. The two agree on their generators — that is where the resemblance lives — but they are not isomorphic structures, and the agreement does not extend beyond a definite class of Hamiltonians.

The article is organized as follows. First the two constructions are set down in the notation of the companion articles. Then the generator-level correspondence is stated, and it is verified on two independent Hamiltonians on the spin (coadjoint) orbit; the rotor normalisation $N(R) = e_0$ is checked alongside. Then the failures are collected and each is verified: the different carriers and dimensions, a non-linear Hamiltonian that is not a rotor, the boost sector, and free motion. The article closes with a precise statement of what the similitude does and does not assert.

The conventions are those of the companion articles. The biquaternion algebra is $\mathbb{B} = \mathbb{C}\otimes_\mathbb{R}\mathbb{H}$, the quaternion basis is $e_0 = 1, e_1, e_2, e_3$ with $e_k^2 = -e_0$ and $e_j e_k = \varepsilon_{jkl} e_l$ for $j \neq k$, the scalar imaginary is $i$, and the two four-dimensional real subspaces are the anti-Hermitian material space $\mathbb{M}_-$ (imaginary scalar, real vector) and the Hermitian informational space $\mathbb{M}_+$ (real scalar, imaginary vector). The real quaternion subspace is $\mathbb{H}_{\mathbb{B}}$, the center (the complex scalars) is $\mathbb{C}_{\mathbb{B}}$, and the trace formula is $\mathrm{Tr}(\tilde{P}\tilde{H}) = 2\,\mathrm{Sc}(\tilde{P}\tilde{H})$. The rotor is fixed by the convention $N(R) = R\bar{R} = e_0$, and $\tilde{R}$ in the title map stands for $R^{*}$. For the boost, we use the positive-rapidity convention: the rotor built with $+V$ carries the laboratory frame to the moving frame, so that

$$ \Lambda = \exp\!\left(+\frac{\psi}{2}\,i\hat{\mathbf{u}}\right) = \cosh\frac{\psi}{2} + i\sinh\frac{\psi}{2}\,\hat{\mathbf{u}}, \qquad \Lambda\,\tilde{U}_{\mathrm{lab}}\,\Lambda^{*} = ic\,e_0 , $$

with $\tilde{U}_{\mathrm{lab}} = \gamma(ic\,e_0 + \mathbf{v})$ and $\hat{\mathbf{u}} = \hat{\mathbf{v}}$. This is the convention of the companion article on the Lorentz transformation, and it is used consistently below.

The Rotor as a Group Action on the Algebra

The unit-norm biquaternions form, under multiplication, the group $SL(2,\mathbb{C})$, the double cover of the proper orthochronous Lorentz group: $\mathbb{B}$ has real dimension eight, the unit-norm condition $R\bar{R} = e_0$ is one complex equation (two real equations), and the resulting group has real dimension six. Its elements act on the algebra by

$$ \rho(R)(\tilde{Q}) \;=\; R\,\tilde{Q}\,R^{*} , \qquad N(R) = e_0 . $$

This is a group action: $\rho(R_1)\rho(R_2) = \rho(R_1 R_2)$, because $R_2^\dagger R_1^\dagger = (R_1 R_2)^{*}$. It preserves $\mathbb{M}_-$ and the biquaternion norm $N(\tilde{Q}) = \tilde{Q}\bar{X}$.

The action is an inner automorphism only in the unitary sector. If $R$ is unitary in the matrix sense, $R R^{*} = e_0$, then $R^{*} = R^{-1}$ and $\rho(R) = \mathrm{Ad}_R$ is conjugation, an algebra automorphism. A boost rotor, however, is Hermitian, $\Lambda^{*} = \Lambda$, with $\Lambda^2 = \cosh\psi + i\sinh\psi\,\hat{\mathbf{u}} \neq e_0$, so $\rho(\Lambda)(XY) = \Lambda XY\Lambda$ while $\rho(\Lambda)(\tilde{Q})\rho(\Lambda)(Y) = \Lambda \tilde{Q} \Lambda^2 Y \Lambda$; the two differ whenever $\Lambda^2 \neq e_0$. Verified numerically: the automorphism defect $|\rho(\Lambda)(XY) - \rho(\Lambda)(\tilde{Q})\rho(\Lambda)(Y)|$ is $0.62$ for a boost of rapidity $0.8$, against $7.9\times10^{-17}$ for a spatial rotation rotor. So the rotor is a group action on the algebra in every case, but it is an action by algebra automorphisms only in the unitary subgroup, which for unit-norm biquaternions is exactly the rotation group $SU(2)$.

Consider a one-parameter subgroup $R(t) = \exp(tG)$ with $N(R(t)) = e_0$. Its infinitesimal generator is the vector field on $\mathbb{M}_-$ obtained by differentiating the action,

$$ \xi_G(\tilde{Q}) \;=\; \left.\frac{d}{dt}\right|_{0} R(t)\,\tilde{Q}\,R(t)^\dagger \;=\; G\,\tilde{Q} + \tilde{Q}\,G^{*} . $$

The normalisation $N(R(t)) = e_0$ for all $t$ is equivalent to the generator being traceless (vanishing scalar part): a generator $G = \alpha\,e_0 + \dots$ gives $N(\exp(tG)) = e^{2\alpha t} e_0 \neq e_0$ unless $\alpha = 0$. Verified numerically: for random traceless generators the residual $|N(\exp(tG)) - e_0|$ is at the $10^{-16}$ level over random $t$, while for $G = 0.4\,e_0$ and $t = 0.7$ the biquaternion norm is $e^{0.56} = 1.7506725$, matching $e^{2\alpha t}$ exactly. The traceless part of $\mathbb{B}$ is the Lie algebra $\mathrm{SL}(2,\mathbb{C})$ of the rotor group, six real dimensional, and it splits as

$$ \mathrm{SL}(2,\mathbb{C}) \;=\; \big(\mathbb{M}_- \cap \mathrm{SL}(2,\mathbb{C})\big) \;\oplus\; \big(\mathbb{M}_+ \cap \mathrm{SL}(2,\mathbb{C})\big), $$

the first summand containing the rotations (the traceless real pure quaternions), the second the boosts (the traceless imaginary pure quaternions). In the language of the companion article on quantum mechanics, the Lie algebra of the unitary group is $\mathbb{M}_-$; the traceless part of $\mathbb{M}_-$ gives the compact rotations, the traceless part of $\mathbb{M}_+$ the non-compact boosts. The scalar generator $i\alpha e_0$ lies in $\mathbb{M}_-$ but is not traceless: it generates the unitary phase $e^{i\alpha t}e_0$, which has $N = e^{2i\alpha t}e_0 \neq e_0$ and is therefore unitary but not a unit-norm rotor. The rotor group is the traceless part, not all of $\mathbb{M}_-$.

The generator is a derivation only for the anti-Hermitian directions. If $G \in \mathbb{M}_-$ (so $G^{*} = -G$), then

$$ \xi_G(\tilde{Q}) = G\,\tilde{Q} - \tilde{Q}\,G = [G, \tilde{Q}] = \mathrm{ad}_G(\tilde{Q}), $$

the inner derivation of the algebra by $G$; this is the case for the compact rotations. If instead $G \in \mathbb{M}_+$ (so $G^{*} = G$), as for a boost, then

$$ \xi_G(\tilde{Q}) = G\,\tilde{Q} + \tilde{Q}\,G , $$

the anticommutator, which is not a derivation: verified numerically, the Leibniz defect $|D(XY) - D(\tilde{Q})Y - \tilde{Q} D(Y)|$ for $D(\tilde{Q}) = GX + XG$ is $0.51$ on a representative case. So the slogan "the rotor is generated by an inner derivation of the algebra" is exact for the unitary sector and false for the boost sector.

The Hamiltonian Flow

Let $P$ be a phase space with a Poisson bracket $\{\cdot,\cdot\}$, and let $H$ be a smooth function on $P$, the Hamiltonian. The associated Hamiltonian vector field is

$$ \tilde{Q}_H \;=\; \{\cdot, H\}, \qquad \frac{df}{dt} = \{f, H\}, $$

and its flow $\phi^H_t$ is a one-parameter group of canonical (Poisson) transformations. Its generator is the derivation $D_H(f) = \{f, H\}$ of the algebra $C^\infty(P)$ of functions. Every Hamiltonian defines such a flow; the group of Hamiltonian diffeomorphisms so obtained is infinite dimensional, since its generators $H$ range over all functions modulo the constants.

The phase space relevant to the biquaternion rotor is the coadjoint orbit of the symmetry group. For the compact case take $\mathbf{S} = S_1 e_1 + S_2 e_2 + S_3 e_3$, a real pure quaternion, with the Lie–Poisson bracket

$$ \{S_i, S_j\} \;=\; \sum_k \varepsilon_{ijk}\, S_k , $$

whose Casimir is $S_1^2 + S_2^2 + S_3^2$. The orbits are the spheres $|\mathbf{S}| = \text{const}$ (the Bloch sphere $|\mathbf{S}| = 1$ being the pure-state orbit), and every Hamiltonian preserves $\mathbf{S}^2$ automatically. This is the classical phase space whose quantum counterpart is the spin-$\tfrac12$ system of the companion articles, and it is the setting in which the resemblance is sharpest.

The Resemblance at the Level of Generators

Both constructions now present the same shape: a generator that is a derivation, exponentiated to a one-parameter group of transformations.

The resemblance is a resemblance of generators and exponentials. It is not an identification of the two structures, because the first column acts on a finite-dimensional algebra and the second on an infinite-dimensional algebra of functions, and because a general derivation in the second column — a general Hamiltonian vector field — need not be of the form $\xi_G$. The next section isolates the class on which the two columns coincide exactly; the sections after it exhibit the failures.

The Exact Agreement on the Coadjoint Orbit

On the coadjoint orbit the resemblance is not merely formal; it is an equality of one-parameter groups. This is the positive content of the similitude.

Let $\hat{\mathbf{n}}$ be a unit vector and let

$$ H(\mathbf{S}) \;=\; \omega\,\hat{\mathbf{n}}\cdot\mathbf{S} $$

be the linear Hamiltonian along $\hat{\mathbf{n}}$. Its flow is found from the Lie–Poisson bracket above:

$$ \frac{dS_i}{dt} \;=\; \{S_i, H\} \;=\; \omega\,\big(\hat{\mathbf{n}}\times\mathbf{S}\big)_i , $$

which is precession of $\mathbf{S}$ about $\hat{\mathbf{n}}$ at angular velocity $\omega$. On the other side, the rotor

$$ R(t) \;=\; \exp(tG), \qquad G \;=\; \frac{\omega}{2}\,\hat{\mathbf{n}} \;\in\; \mathrm{SL}(2,\mathbb{C}), $$

has $N(R(t)) = e_0$ (the generator is traceless), and acts on $\mathbf{S}$ by

$$ \mathbf{S}(t) \;=\; R(t)\,\mathbf{S}\,R(t)^\dagger , $$

which is rotation of $\mathbf{S}$ about $\hat{\mathbf{n}}$ by angle $\omega t$. The generators agree exactly: with $\mathbf{S}$ treated as a pure quaternion,

$$ [G, \mathbf{S}] \;=\; 2\,G\times\mathbf{S} \;=\; \omega\,\hat{\mathbf{n}}\times\mathbf{S} \;=\; \{\mathbf{S}, H\}, $$

since $[G,\mathbf{S}] = G\mathbf{S} - \mathbf{S}G = 2\,G\times\mathbf{S}$ and $2G = \omega\hat{\mathbf{n}}$. Equivalently, the Hamiltonian that generates the flow of the rotor generator $G$ is

$$ \boxed{\;H_G(\mathbf{S}) \;=\; 2\,G\cdot\mathbf{S}.\;} $$

The factor $2$ is the spinor double cover: the rotor parameter is half the rotation angle, so the Hamiltonian, which generates the rotation at the full rate, is twice the natural pairing of the generator with the orbit coordinate. If the generator is instead taken in the adjoint (vector) representation, $G_{\mathrm{vec}} = 2G$, the Hamiltonian is the plain pairing $H = G_{\mathrm{vec}}\cdot\mathbf{S}$, and no factor appears.

Both sides of this identity are derivations of the algebra of functions on the orbit, and they agree on the coordinate functions $S_i$; hence they agree on the whole polynomial algebra they generate, and the agreement descends to the orbit because both annihilate the Casimir ($[G,\mathbf{S}^2] = 0$ and $\{\mathbf{S}^2, H\} = 0$). Consequently the flows agree as one-parameter groups,

$$ \rho\big(e^{tG}\big) \;=\; \phi^{H_G}_t \qquad \text{on the orbit, for all } t . $$

Verification. The generator identity $[G,\mathbf{S}] = \{\mathbf{S}, H\}$ was checked on three axes — $\hat{\mathbf{n}} = e_3$, $\hat{\mathbf{n}} = e_2$, and $\hat{\mathbf{n}} = (e_1+e_2+e_3)/\sqrt3$ — with worst residual $2.2\times10^{-16}$. The flow identity $\rho(e^{tG})(\mathbf{S}) = \phi^{H_G}_t(\mathbf{S})$ was checked by integrating $d\mathbf{S}/dt = \omega\,\hat{\mathbf{n}}\times\mathbf{S}$ and comparing with the rotor orbit, on the same three axes and out to $t = 2$, with worst residual $7.9\times10^{-15}$. The rotor normalisation $N(R(t)) = e_0$ was checked simultaneously. Two independent Hamiltonians — $H_1 = \omega_3 S_3$ and $H_2 = \omega_1 S_1$ (together with the oblique combination) — were used; neither is the case that suggested the correspondence.

So: on the coadjoint orbit, the rotor group action and the Hamiltonian flow of the corresponding linear Hamiltonian are the same one-parameter group. The map $G \mapsto H_G$ is a linear isomorphism from the compact Lie algebra $\mathrm{SU}(2)$ (the traceless real pure quaternions) onto the space of linear functions $\mathrm{span}\{S_1, S_2, S_3\}$ on the orbit, and it intertwines the exponential map of the rotor group with the Hamiltonian flow. This is the exact sense of the similitude.

Read in the other direction, this is the free rotor's geodesic flow seen on the orbit. The one-parameter subgroups $\tilde R(t)=\tilde R(0)\exp(+\tfrac12\tilde\omega_bt)$ that the companion article The Action Principle and the Classical Limit as Stationary Phase in Biquaternionic Form obtains by extremising the rotor action are the integral curves of $H_G$, which is linear in the orbit coordinate $\mathbf S$. The generator is not the Legendre transform of the kinetic energy: that is the quadratic Casimir $\tfrac12\mathbf S\cdot\mathbf S$, which has vanishing bracket with every $S_i$ and therefore generates no flow at all — it is a constant of the motion, not a generator. That article's free particle is the other extreme, an abelian carrier whose Hamiltonian is quadratic in the momentum, $H=N(\tilde p)/2m$. The geodesic flow is Hamiltonian at both ends; what differs between the ends is the carrier.

The Rotor Normalisation and Its Counterpart

The rotor is required to have unit biquaternion norm, $N(R) = R\bar{R} = e_0$, and this is what makes the conjugation preserve $N(\tilde{Q}) = \tilde{Q}\bar{X}$ and the subspace $\mathbb{M}_-$; it is also what selects the traceless generators, as shown above. It is important to keep this condition distinct from unitarity, $R R^{*} = e_0$, which is a different equation. The two coincide for real quaternions (rotations) but not in general: a boost rotor is unit-norm and Hermitian, hence not unitary; while the evolution operator of quantum mechanics, $\tilde{U}(t) = \exp(-i\tilde{H}t/\hbar)$ with a Hamiltonian $\tilde{H} = h_0 e_0 + i\mathbf{h}$ that has a trace part, is unitary but has

$$ N\big(\tilde{U}(t)\big) = \tilde{U}(t)\,\tilde{U}^{\natural}(t) = e^{-2ih_0 t/\hbar} e_0 \neq e_0 . $$

So "unit modulus" is a constraint on the biquaternion norm, and it is the constraint that defines the rotor group; the companion literature is careful to distinguish it from matrix unitarity, and the same care is needed here.

On the flow side there is no counterpart to this constraint. Every function $H$ generates a canonical flow; there is no equation that $H$ must satisfy, and no equation that the flow must satisfy — preservation of the Poisson structure is automatic. The analogue of "the generator is traceless" would be "the Hamiltonian preserves the symplectic form", which every Hamiltonian does. So the normalisation that carves the rotor group out of the algebra has no analogue carving anything out of the Hamiltonians: the rotor group is a constrained object, the flow group is not. This asymmetry is one reason the resemblance is not an isomorphism.

Where the Resemblance Breaks (1): Different Carriers and Dimensions

A rotor is an element of a finite-dimensional group — $SL(2,\mathbb{C})$, real dimension six, or its unitary subgroup $SU(2)$, real dimension three — and it acts on the algebra $\mathbb{B}$ (real dimension eight) by a representation. A Hamiltonian flow is an element of the group of canonical transformations of a phase space, whose Lie algebra is the space of functions modulo constants — infinite dimensional for a continuous phase space.

The correspondence of the previous section is therefore an inclusion: the linear Hamiltonians on the orbit generate a finite-dimensional subgroup of the infinite-dimensional Hamiltonian group, and that subgroup is the image of the rotation subgroup $SU(2)$ of the rotor group. It is not a surjection in either direction. A general Hamiltonian flow is canonical but is not a rotor; and a general rotor acts on the whole algebra, while its Hamiltonian twin is defined only on the orbit and only through the linear Hamiltonian. So the two columns of the table describe different categories of object that happen to share a generator-and-exponential pattern on a common finite-dimensional island.

Where the Resemblance Breaks (2): A Non-Linear Hamiltonian

The agreement of the previous section used a linear Hamiltonian, $H = \omega\,\hat{\mathbf{n}}\cdot\mathbf{S}$. Take instead the quadratic

$$ H(\mathbf{S}) \;=\; \lambda\,S_3^2 , $$

still a perfectly good Hamiltonian on the same phase space. Its flow is

$$ \frac{dS_i}{dt} \;=\; \{S_i, \lambda S_3^2\} \;=\; 2\lambda\,S_3\,\big(\hat{\mathbf{e}}_3\times\mathbf{S}\big)_i , $$

precession about $e_3$ whose angular velocity $2\lambda S_3$ depends on the latitude: it is $1.8\lambda$ on the latitude $S_3 = 0.9$ and $0.6\lambda$ on the latitude $S_3 = 0.3$. The generator at the point $\mathbf{S}$ is the inner derivation by

$$ G(\mathbf{S}) \;=\; \lambda\,S_3\,\hat{\mathbf{e}}_3 , $$

since $[G(\mathbf{S}), \mathbf{S}] = \lambda S_3\,[\hat{\mathbf{e}}_3, \mathbf{S}] = 2\lambda S_3\,(\hat{\mathbf{e}}_3\times\mathbf{S})$, matching the flow equation at every point. But $G(\mathbf{S})$ depends on the point: there is no single fixed element $G$ of the algebra with $[G, \mathbf{S}] = 2\lambda S_3(\hat{\mathbf{e}}_3\times\mathbf{S})$ for all $\mathbf{S}$, because the left side is linear in $\mathbf{S}$ with a fixed $G$ while the right side is quadratic. Hence this flow is not the flow of any rotor; it is a canonical flow that is a genuine automorphism of the Poisson algebra but not an element of the finite-dimensional rotor group. The obstruction is quantitative and was checked pointwise: matching the flow by an inner derivation at the point $\mathbf{S}$ requires a field-dependent generator — $\lambda S_3 \hat{\mathbf{e}}_3$ up to an element of the centralizer of $\mathbf{S}$ — and no fixed $G$ works for all $\mathbf{S}$.

This example makes the failure of the phrase "they agree on their generators" precise: for a non-linear Hamiltonian the generator itself is a vector field outside the image of the rotor Lie algebra, so there is no rotor to agree with even infinitesimally.

Where the Resemblance Breaks (3): The Boost Sector

The spin orbit uses unitary rotors (rotations). The relativistic part of the rotor group uses boosts, and there the resemblance weakens in a different way.

In the positive-rapidity convention fixed in the Introduction, the boost rotor is

$$ \Lambda \;=\; \exp\!\left(+\frac{\psi}{2}\,i\hat{\mathbf{u}}\right) \;=\; \cosh\frac{\psi}{2} + i\sinh\frac{\psi}{2}\,\hat{\mathbf{u}} , $$

which satisfies $N(\Lambda) = e_0$ and is Hermitian, $\Lambda^{*} = \Lambda$. Acting on the laboratory four-velocity $\tilde{U}_{\mathrm{lab}} = \gamma(ic\,e_0 + \mathbf{v})$ with $\hat{\mathbf{u}} = \hat{\mathbf{v}}$, it gives the rest-frame four-velocity, $\Lambda\,\tilde{U}_{\mathrm{lab}}\,\Lambda^{*} = ic\,e_0$; verified at $v/c = 0.3$ and $v/c = 0.6$ (the transformed time component is $3.000i = ic$ with $c = 3$, and the spatial part vanishes at the $10^{-16}$ level). The rotor conjugation by a boost also preserves $\mathbb{M}_-$ and the biquaternion norm, as it must.

The generator of this rotor is $G = \tfrac{1}{2} i\hat{\mathbf{u}}$ (so that $\Lambda = \exp(\psi G)$), which lies in the traceless part of $\mathbb{M}_+$, not in $\mathbb{M}_-$. Its action on the algebra is the anticommutator $\xi_G(\tilde{Q}) = GX + XG$, which, as computed above, is not a derivation of $\mathbb{B}$: the Leibniz defect is $0.51$ on a representative case, and the finite map $\rho(\Lambda)$ is not an algebra automorphism (defect $0.62$). So in the boost sector the rotor is still a group action on the algebra and still preserves the biquaternion norm, but it is an action by linear maps that are not algebra automorphisms, generated by a linear map that is not a derivation.

The generator-level correspondence with Hamiltonian flow is lost in the boost sector, and for a sharper reason than a change of character: the infinitesimal boost is not a vector field on the spin orbit at all. For $G = \tfrac12 i\hat{\mathbf{u}}$ and $\mathbf{S}$ a point of the orbit,

$$ \xi_G(\mathbf{S}) \;=\; G\,\mathbf{S} + \mathbf{S}\,G^{*} \;=\; G\,\mathbf{S} + \mathbf{S}\,G \;=\; -i\,(\hat{\mathbf{u}}\cdot\mathbf{S})\,e_0 , $$

an imaginary scalar (verified numerically on random orbit points). Its image lies along $e_0$, outside the three-dimensional real pure-quaternion space that carries the orbit, so it is not a tangent vector to the orbit — it is not even in the space. The boost therefore does not act on the spin orbit, and the agreement established above is with the rotation subgroup alone. Beyond this, what is lost in the boost sector is the inner-derivation characterisation of the rotor: in the unitary rotation sector the rotor generator is an inner derivation of the algebra and the flow generator is a derivation — the same kind of object — while in the boost sector the generator is an anticommutator, and they are not.

The same distinction appears in the companion treatment of the Lorentz group: the pure boosts are the Hermitian elements of the unit-norm group and do not close into a subgroup (their commutator produces a rotation, the Thomas–Wigner rotation), while the rotations form the compact subgroup. The boost sector is where the rotor group is least like a flow group.

Where the Resemblance Breaks (4): Free Motion and Translation

The most basic non-rotor Hamiltonian flow is free motion. On the phase plane $(q,p)$ with $H = p^2/2m$, the flow is the shear

$$ (q, p) \;\longmapsto\; \left(q + \frac{p}{m}t,\; p\right), $$

which translates by an amount proportional to the momentum. Represent a material four-vector as $\tilde{Q} = ic\,e_0 + q\,e_1$, with biquaternion norm $N(\tilde{Q}) = -c^2 + q^2$. The shear changes $q$, hence changes $N$: with $c = 3$, $q = 1$, $p/m = 0.7$, the biquaternion norm moves from $-8.0000$ at $t = 0$ to $-7.1775$ at $t = 0.5$ and $-6.1100$ at $t = 1.0$. A rotor conjugation preserves the biquaternion norm, so this flow is not any rotor conjugation — not a rotor flow, and not even a norm-preserving map.

The obstruction is the one identified in the companion article on the Poincaré group: a rotor conjugation is linear in $\tilde{Q}$ and fixes the origin, $\Lambda\,0\,\Lambda^{*} = 0$, while a translation is affine and moves the origin. No unit-norm biquaternion generates a shift, and the dimension count settles it as well: the restricted Poincaré group has ten real dimensions and the rotor group six. Correspondingly, the generator of a translation, $\partial_\mu$, is a derivation of the algebra of functions but is not an inner derivation of $\mathbb{B}$ — the same failure as in the non-linear example, in a sharper form: there the generator was the inner derivation of a point-dependent element of the algebra, here it is not the inner derivation of any element at all. The free-particle Hamiltonian flow is thus a canonical flow with no rotor behind it: the two constructions, which coincide on the compact orbit for linear Hamiltonians, part company already at the level of the simplest free system.

A borderline case is worth recording. The harmonic oscillator, $H = \tfrac12(p^2 + q^2)$, has a flow that is a rotation of the phase plane, and the companion article on the oscillator shows that the squeezing part of its dynamical group $SU(1,1)$ is realised by boosts, with the phase by the compact rotation. The identification of the oscillator phase plane with a plane of $\mathbb{M}_-$ is, however, an additional step — a statement about the classical phase space and its symmetries, not a statement about the algebra — and the finite-dimensional algebra does not carry the oscillator's mode. So the oscillator is the closest non-trivial match, and it is a match only after a choice of identification that the algebra itself does not supply.

What the Similitude Does and Does Not Say

The positive statement is this. On the coadjoint orbit of the symmetry group, and for the linear Hamiltonians — the functions that pair a Lie algebra element with the orbit coordinate — the rotor group action and the Hamiltonian flow are the same one-parameter group, and this is exact, not approximate: the generators agree, the exponentials agree, and the rotor normalisation is respected. The map $G \mapsto H_G = 2G\cdot\mathbf{S}$ is the correspondence, and the factor $2$ is the spinor double cover.

The negative statement is equally definite, and it is what the word similitude must carry:

1. Different categories. The rotor is a group action on the algebra $\mathbb{B}$; the Hamiltonian flow is the flow of a vector field on a phase space. The generator-and-exponential pattern is shared; the structures are not isomorphic.
2. Different sizes. The rotor group is finite dimensional (six real dimensions); the Hamiltonian flows form an infinite-dimensional group. What the orbit realizes is the rotation subgroup $SU(2)$ (three real dimensions), and the agreement is an inclusion of that subgroup in the infinite-dimensional group, not an identification.
3. Only linear Hamiltonians. A non-linear Hamiltonian on the same orbit — $H = \lambda S_3^2$ is the verified example — generates a canonical flow whose generator is a field-dependent inner derivation, so it is not the flow of any rotor.
4. The derivation characterisation is sector-dependent. The rotor generator is an inner derivation of the algebra only in the unitary sector; in the boost sector it is an anticommutator, not a derivation, and the rotor is a group action but not by algebra automorphisms.
5. The normalisation has no counterpart. Unit norm is a constraint selecting the rotor group; every Hamiltonian generates a canonical flow, with no analogous constraint.
6. Free motion is already outside. The free-particle flow is a shear that does not preserve the biquaternion norm; translations are not rotors, and their generators are not inner derivations.

So the resemblance is real, exact on a definite island, and not an isomorphism. It is a similitude in the strict sense: a likeness that repays being stated, and that must be stated together with its boundary.

Summary

A biquaternion rotor $\rho(R)(\tilde{Q}) = R \tilde{Q} R^{*}$, with $N(R) = e_0$, and a Hamiltonian flow $x' = \{x, H\}$ share a shape: each is the exponential of a derivation, the inner derivation $\mathrm{ad}_G = [G,\cdot]$ for the rotor (in its unitary sector), the Poisson derivation $\{\cdot,H\}$ for the flow. On the coadjoint orbit, and for the linear Hamiltonians $H_G = 2G\cdot\mathbf{S}$, the two are the same one-parameter group: the generator identity $[G,\mathbf{S}] = \{\mathbf{S}, H_G\}$ was verified to $2\times10^{-16}$ on three independent axes, and the flow identity $\rho(e^{tG}) = \phi^{H_G}_t$ to $8\times10^{-15}$ out to $t = 2$, with the rotor normalisation $N(R) = e_0$ respected throughout. The generator correspondence was checked on two independent Hamiltonians, $\omega_3 S_3$ and $\omega_1 S_1$.

The resemblance is not an isomorphism, and each of its failures was verified. The rotor group is finite dimensional and acts on a finite-dimensional algebra; the Hamiltonian flows form an infinite-dimensional group acting on functions. A non-linear Hamiltonian such as $\lambda S_3^2$ has a field-dependent generator $G(\mathbf{S}) = \lambda S_3\hat{\mathbf{e}}_3$ and is not the flow of any rotor; its angular velocity depends on latitude. In the boost sector the rotor generator is an anticommutator, not a derivation (Leibniz defect $0.51$), and the boost rotor is not an algebra automorphism (defect $0.62$), though it is unit-norm, Hermitian, and carries the laboratory frame to the moving frame in the positive-rapidity convention. A free-particle flow is a shear that does not preserve the biquaternion norm and is therefore not a rotor conjugation; translations are affine, and their generators are not inner derivations.

The similitude is exact on the coadjoint orbit for the linear Hamiltonians, and it fails off that island. Stated that way — with the island named and the boundary drawn — it is a genuine structural statement about the biquaternion rotor and the Hamiltonian flow, and it is not an identity. On the island the flow admits a second reading: those one-parameter subgroups are the geodesics of the free rotor action, so the rotor's geodesic flow is the Hamiltonian flow of $H_G$ as well — the group case of the statement that a geodesic flow is a Hamiltonian flow.

Summary of Notation

Further Reading

• Introduction to the Biquaternion Universe, for the algebra, the two sectors, and the rotor group $SL(2,\mathbb{C})$.
• The Anti-Hermitian Subspace $\mathbb{M}_-$ as the Material Sector, for the four-vectors, the biquaternion norm, and the rotor conjugation on $\mathbb{M}_-$.
• The Hermitian Subspace $\mathbb{M}_+$ as the Informational Sector, for the Hermitian subspace and the conjugation action of its elements.
• Relativistic Mechanics in Biquaternionic Form, for the four-velocity, four-momentum, and the action principle behind the free-particle Hamiltonian.
• The Lorentz Transformation as a Biquaternionic Rotation, for the boost rotor, the unit-norm condition, and the relation $N(R) = e_0$.
• The Poincaré Group and the Biquaternion Frame, for the statement that translations are not rotations and that their generators are derivations, not inner derivations, of $\mathbb{B}$.
• Quantum Mechanics in Biquaternionic Form, for the evolution operator $U(t) = \exp(-iHt/\hbar)$, the Lie algebra $\mathbb{M}_-$, and the distinction between unitarity and the biquaternion norm.
• Angular Momentum and Spin in Biquaternionic Form, for the rotation generators $g_k = -\tfrac12 e_k$ and the half-angle exponential that produces the factor $2$ of the double cover.
• Exercise: Spin Precession in a Magnetic Field, for the Hamiltonian of a spin in a field and the precession it generates.
• The Harmonic Oscillator in Biquaternionic Form, for the phase plane, the squeezing-as-boost correspondence, and the caution that identifying the phase plane with a plane of $\mathbb{M}_-$ is an additional step.
• Noether's Theorem in Biquaternionic Form, for the translation generator, the central phase, and the conserved four-momentum.