Frame-Dependent Entanglement and Relativistic Quantum Information in Biquaternionic Form

Introduction

Entanglement is a property of a joint state, and a change of inertial frame is a change of the joint state. It is therefore not obvious that the entanglement of two particles should be the same for two observers in relative motion, and the relativistic quantum information literature has established that in general it is not: the reduced entropy of a particle's spin is frame-dependent when its momentum is uncertain, and the spin entanglement of a pair of wave packets changes under a boost. The mechanism is not a direct action of the boost on the spin. It is the momentum dependence of the Wigner rotation: at definite momentum a Lorentz transformation acts on the spin by a unitary element of the little group, and a momentum-independent unitary cannot change entanglement; when the momentum is a superposition, the Wigner rotation differs from branch to branch, the boost correlates spin with momentum, and tracing the momentum leaves a spin state whose entanglement has changed.

This article develops that mechanism inside the biquaternion framework. The joint state of two qubits is an element of the tensor-product algebra $\mathbb{B}\otimes\mathbb{B}\cong M_4(\mathbb{C})$; the reduced state of one qubit is a partial trace; the entanglement measures are the standard ones — the Schmidt coefficients, the concurrence, the entanglement entropy — computed with the algebra's trace pairing. The relativistic content is the action of the unit-norm group on the spin factor, which at sharp momentum is the tensor product of two unitary Wigner rotations and at uncertain momentum is a momentum-controlled rotation. The article derives both regimes, obtains the sharp-momentum invariance as a theorem, produces the frame dependence as an explicit computation on a state with a two-branch momentum distribution, and states what the framework adds to the standard account and what it does not.

The findings are stated in advance.

1. At sharp momenta the spin entanglement is Lorentz invariant. A boost acts on a definite-momentum product state by the little-group rotations $\tilde{W}_1\otimes\tilde{W}_2$, which are local unitaries; local unitaries preserve the Schmidt coefficients, the concurrence, and the entanglement entropy, and the reduced states are unchanged up to a rotation.
2. The frame dependence requires momentum uncertainty. A boost on a state whose momentum is a superposition acts by a branch-dependent rotation; the spin state after tracing the momentum is a mixture of rotated copies of the original, and neither the concurrence nor the reduced entropy is invariant.
3. The mechanism is spin–momentum entanglement generated by the boost. The boost is an entangling operation between the spin factor and the momentum factor whenever the Wigner rotation is momentum-dependent; the spin state loses purity exactly to the extent that this correlation is created, and the loss is the information-theoretic content of the boost.
4. The symmetric two-branch case is a dephasing channel. When the momentum branches are symmetric about the boost axis, the two Wigner rotations are rotations about a common axis by opposite angles, and their average is the dephasing of the spin about that axis: the Bloch component along the Wigner axis is preserved and the transverse components are damped by $\cos\alpha$, where $\alpha$ is the Wigner angle.
5. The $SL(2,\mathbb{C})$-invariant concurrence is a spinor concurrence. In the spinor description, $\tilde{\Lambda}\otimes\tilde{\Lambda}$ acts with unit determinant on each factor and leaves the symplectic concurrence $|\psi^{\mathsf T}(\epsilon\otimes\epsilon)\psi|$ exactly invariant; the frame dependence appears only after normalization, i.e. only for configurations with momentum spread, which is the algebraic statement that the effect is kinematic rather than dynamical.

The notation is that of the foundational articles: $\mathbb{B} = \mathbb{C}\otimes_\mathbb{R}\mathbb{H}$, quaternion units $e_0 = 1, e_1, e_2, e_3$ with $e_k^2 = -e_0$, central scalar $i$, trace $\mathrm{Tr}(\tilde{Q}) = 2\,\mathrm{Sc}(\tilde{Q})$, Hermitian sector $\mathbb{M}_+$, anti-Hermitian sector $\mathbb{M}_-$, and the matrix model $\Phi$. The two-qubit space is $\mathbb{B}\otimes\mathbb{B}\cong M_4(\mathbb{C})$ with $\mathrm{Tr}(x\otimes y) = \mathrm{Tr}_\mathbb{B}(x)\mathrm{Tr}_\mathbb{B}(y)$. The Wigner rotation of the relativistic-qubit companion is $\tilde{W}(\tilde{\Lambda},\tilde{U}) = \tilde{\Lambda}_{\tilde{U}'}\tilde{\Lambda}\tilde{\Lambda}_{\tilde{U}}^{-1}\in SU(2)$, acting on a spin state by conjugation.

The companion articles supply the pieces: - Companion article The Relativistic Qubit in Biquaternionic Form, for the little group, the Wigner rotation, and the distinction between the spinor action and the spin-state rotation. - Companion article Entangled Subsystems in the Biquaternion Framework, for the singlet idempotent, the partial trace, the correlation function, and no-signaling. - Companion article Exercise: The Reduced State of an Entangled Subsystem, for the partial trace and the reduced state. - Companion article Exercise: Entanglement Entropy and the Partial Trace, for the Schmidt decomposition and the entropy of a reduced state. - Companion article The Bell Basis as the Idempotent Basis of $\mathbb{B}\otimes\mathbb{B}$, for the tensor-product basis and the Bell idempotents. - Companion article The Lorentz Group in Biquaternionic Form — Structure and Representations, for the little-group rotations and the composition of boosts.

Entangled States in the Tensor Product

The joint state and the reduced state

The state space of two qubits is the tensor product $\mathbb{B}\otimes\mathbb{B}$, isomorphic to $M_4(\mathbb{C})$ as a complex algebra, with the tensor trace $\mathrm{Tr}(x\otimes y) = \mathrm{Tr}_\mathbb{B}(x)\mathrm{Tr}_\mathbb{B}(y)$. A pure two-qubit state is a rank-one idempotent, $$ \tilde\Pi = |\Psi\rangle\langle\Psi| \quad\text{with}\quad |\Psi\rangle\in S\otimes S, $$ and the reduced state of the first qubit is the partial trace $$ \mathrm{Tr}_2:\ \mathbb{B}\otimes\mathbb{B}\to\mathbb{B}, \qquad \mathrm{Tr}_2(\tilde{A}\otimes\tilde{B}) = \tilde{A}\,\mathrm{Tr}_\mathbb{B}(\tilde{B}), $$ extended linearly. On the singlet, $$ \tilde\Pi_{\mathrm{singlet}} = \tfrac14\bigl(e_0\otimes e_0 + e_1\otimes e_1 + e_2\otimes e_2 + e_3\otimes e_3\bigr), $$ the partial trace kills the traceless tensor terms and returns the maximally mixed state, $$ \mathrm{Tr}_2\bigl(\tilde\Pi_{\mathrm{singlet}}\bigr) = \tfrac12 e_0 , $$ whose von Neumann entropy is $\log 2$: the mixedness of the reduced state is the algebraic residue of the entanglement of the joint state.

Entanglement measures

Three measures are used below, all standard.

Schmidt coefficients. A pure two-qubit state admits the decomposition $|\Psi\rangle = \sum_{i=1}^2 s_i\,|a_i\rangle|b_i\rangle$ with $s_i\geq0$ and $s_1^2+s_2^2=1$; the $s_i^2$ are the eigenvalues of either reduced state, and the state is a product if and only if one coefficient vanishes.

Concurrence. With the two-qubit spin-flip $\tilde{\Psi} = (i\sigma_2)\otimes(i\sigma_2)$ in the matrix model, the concurrence of a pure state is $$ C(|\Psi\rangle) = \bigl|\langle\Psi|\,\tilde{\Psi}\,|\Psi^{*}\rangle\bigr| = 2\,s_1s_2 , $$ vanishing on product states and equal to $1$ on the Bell states; for a mixed state $\tilde{\rho}$ it is $C(\tilde{\rho}) = \max(0,\lambda_1-\lambda_2-\lambda_3-\lambda_4)$, where $\lambda_1\geq\lambda_2\geq\lambda_3\geq\lambda_4$ are the square roots of the eigenvalues of the Hermitian positive-semidefinite matrix $\tilde{\rho}^{1/2}\,\tilde{\Psi}\tilde{\rho}^{*}\tilde{\Psi}\,\tilde{\rho}^{1/2}$.

Entanglement entropy. For a pure joint state it is the von Neumann entropy of either reduced state, $E = S(\mathrm{Tr}_2\tilde\Pi) = -s_1^2\log s_1^2 - s_2^2\log s_2^2$, and it is a function of the single invariant $s_1^2$. For mixed joint states the partial-trace entropy is a marginal entropy and not an entanglement measure, as the companion exercise records; the concurrence is used for mixed states below.

The framework inherits all three unchanged: they are computed from the trace and the tensor product, and the algebra's trace pairing is the Born pairing. The relativistic question is what the unit-norm group does to them.

The Lorentz Transformation of a Two-Particle State

Sharp momenta: local unitary rotations

Consider two massive particles with sharp four-velocities $\tilde{U}_1,\tilde{U}_2$. Their joint state is a product of a momentum part and a spin part, $$ |\Psi\rangle = |\tilde{U}_1\rangle\otimes|\tilde{U}_2\rangle\otimes|\chi\rangle , \qquad |\chi\rangle\in S\otimes S , $$ and a Lorentz transformation $\tilde{\Lambda}$ maps the momentum labels to $\tilde{U}_j' = \tilde{\Lambda}\tilde{U}_j\tilde{\Lambda}^{*}$ and applies to the spin the Wigner rotations of the two particles, $$ |\Psi\rangle \ \longmapsto\ |\tilde{U}_1'\rangle\otimes|\tilde{U}_2'\rangle\otimes\bigl(\tilde{W}_1\otimes\tilde{W}_2\bigr)|\chi\rangle , \qquad \tilde{W}_j = \tilde{W}(\tilde{\Lambda},\tilde{U}_j)\in SU(2). $$ This is the standard action of the Lorentz group on definite-momentum one-particle states, and it is the content of the Wigner construction: the momentum labels move, the spin is rotated by an element of the little group, and the transformation is unitary on the full Hilbert space.

The spin state is therefore transformed by a local unitary $\tilde{W}_1\otimes\tilde{W}_2$. Local unitaries preserve every entanglement measure: $$ C\bigl((\tilde{W}_1\otimes\tilde{W}_2)|\chi\rangle\bigr) = C(|\chi\rangle), \qquad S\bigl(\mathrm{Tr}_2\,\tilde\Pi_{(\tilde{W}_1\otimes\tilde{W}_2)\chi}\bigr) = S\bigl(\mathrm{Tr}_2\,\tilde\Pi_{\chi}\bigr), $$ because $\tilde{W}_1\otimes\tilde{W}_2$ preserves the tensor product structure and the trace pairing, and because the Schmidt coefficients are the singular values of the coefficient matrix, which a pair of unitaries leaves unchanged. Hence:

Theorem (sharp-momentum invariance). For two particles of definite four-velocity, and for any Lorentz transformation, the spin entanglement of the pair — its Schmidt coefficients, its concurrence, and its entanglement entropy — is invariant.

The theorem is worth stating because it is the exact sense in which the framework reproduces the standard result, and because it isolates the only remaining source of frame dependence. A sharp momentum cannot produce a frame-dependent spin entanglement, because a momentum-independent unitary on the spin factor cannot create or destroy entanglement. If the framework is to exhibit frame dependence, it must come from a state that is not a momentum eigenstate.

A worked sharp-momentum instance

The theorem is easily checked in an instance that also supplies the Wigner rotor used below. Let particle $A$ have speed $0.5c$ and momentum making angle $\theta = 0.8$ with the boost axis $\hat{\mathbf{z}}$, and let particle $B$ have momentum along $\hat{\mathbf{z}}$, so that $\tilde{W}_B = e_0$. For a boost of rapidity $\psi = 1$ the Wigner rotor of $A$ is $$ \tilde{W}_A = \cos\frac{\alpha}{2}\,e_0 + \sin\frac{\alpha}{2}\,\hat{\mathbf{y}} = 0.995308\,e_0 + 0.096756\,\hat{\mathbf{y}}, \qquad \alpha = 0.193815\ \text{rad} = 11.1048^\circ , $$ a rotation about $\hat{\mathbf{y}}\propto\hat{\mathbf{z}}\times\hat{\mathbf{u}}$, as the geometry requires. Starting from $|\chi\rangle = (|\!\uparrow\uparrow\rangle+|\!\downarrow\downarrow\rangle)/\sqrt2$, the spin state is rotated by $\tilde{W}_A\otimes e_0$ and its concurrence is $1.00000000$; the reduced state of $A$ is the maximally mixed state carried to a rotated maximally mixed state, hence unchanged, and the reduced entropy remains $\log 2$. This is the theorem in one instance: at sharp momentum the boost is a local unitary and every entanglement measure is invariant.

Momentum Superposition and the Frame Dependence

The boost as a momentum-controlled rotation

Let the momentum of each particle be a superposition. Write the total state as $$ |\Psi\rangle = \sum_{i}\sqrt{q_i}\,|\tilde{U}_{1,i}\rangle\otimes|\tilde{U}_{2,i}\rangle\otimes|\chi_i\rangle , \qquad \sum_i q_i = 1 , $$ where the spin state may itself be correlated with the momentum branch. The Lorentz transformation acts branch by branch, $$ |\Psi\rangle \ \longmapsto\ \sum_i\sqrt{q_i}\,|\tilde{U}_{1,i}'\rangle\otimes|\tilde{U}_{2,i}'\rangle\otimes \bigl(\tilde{W}_{1,i}\otimes\tilde{W}_{2,i}\bigr)|\chi_i\rangle , \qquad \tilde{W}_{j,i} = \tilde{W}(\tilde{\Lambda},\tilde{U}_{j,i}), $$ so the boost is a controlled rotation: it rotates the spin by an amount that depends on the momentum. When the states $|\tilde{U}_{j,i}\rangle$ are orthogonal, tracing the momentum leaves the spin state $$ \tilde{\rho}_{\mathrm{spin}} \ \longmapsto\ \sum_i q_i\,\bigl(\tilde{W}_{1,i}\otimes\tilde{W}_{2,i}\bigr)\,\tilde{\rho}_{\mathrm{spin},i}\,\bigl(\tilde{W}_{1,i}\otimes\tilde{W}_{2,i}\bigr)^\dagger , $$ a mixed-unitary channel on the spin factor whose Kraus operators are the branch Wigner rotations. Two features are immediate and are the whole of the mechanism.

First, the channel is unitary — hence entanglement-preserving — if and only if the Wigner rotation is the same on every branch, which is the sharp-momentum case of the theorem. Second, when the rotations differ across branches the channel is a non-trivial average of local unitaries; it can and generally does change both the reduced state and the entanglement, and it is unital (it preserves the maximally mixed state) but not unitary.

The symmetric two-branch case: a dephasing channel

The simplest instance is sharp enough to be worked in closed form. Let particle $A$ have a momentum in a symmetric superposition of two directions making angles $\pm\theta$ with the boost axis in a common plane, and let particle $B$ have a sharp momentum parallel to the boost axis, so that $\tilde{W}_B = e_0$ and the boost acts on $A$ alone. By the symmetry, the two branch rotations are $$ \tilde{W}_{A,\pm} = \cos\frac{\alpha}{2}\,e_0 \pm \sin\frac{\alpha}{2}\,\hat{\mathbf{n}}, \qquad \alpha = \alpha(\psi,\theta), $$ rotations about a common axis $\hat{\mathbf{n}}$ by opposite angles, and the channel is $$ \tilde{\rho}_A \ \longmapsto\ \tfrac12\bigl(\tilde{W}_{A,+}\tilde{\rho}_A\tilde{W}_{A,+}^{*} + \tilde{W}_{A,-}\tilde{\rho}_A\tilde{W}_{A,-}^{*}\bigr). $$ Writing the Bloch vector of $\tilde{\rho}_A$ as $\mathbf{r} = r_\parallel\hat{\mathbf{n}} + \mathbf{r}_\perp$, the two rotated copies differ only in the sign of the rotation and their average is $$ r_\parallel \ \longmapsto\ r_\parallel, \qquad \mathbf{r}_\perp \ \longmapsto\ \cos\alpha\ \mathbf{r}_\perp , $$ so the channel is the dephasing about the Wigner axis: the Bloch component along $\hat{\mathbf{n}}$ survives and the transverse components are damped by $\cos\alpha$. The Bloch radius of a pure initial state becomes $r' = \sqrt{\cos^2\alpha\,\sin^2\Theta + \cos^2\Theta}$, where $\Theta$ is the angle between $\mathbf{r}$ and $\hat{\mathbf{n}}$; for $\mathbf{r}\perp\hat{\mathbf{n}}$ it is exactly $\cos\alpha$, and for $\mathbf{r}\parallel\hat{\mathbf{n}}$ it is $1$ and the state is unchanged. The von Neumann entropy of the dephased single-qubit state is the binary entropy of the reduced radius, $$ S(\tilde{\rho}_A') = h\!\left(\frac{1+r'}{2}\right), \qquad h(p) = -p\log_2 p - (1-p)\log_2(1-p), $$ which for a purely transverse initial Bloch vector is $h((1+\cos\alpha)/2)$, and which is the entropy of the spin–momentum correlation the boost has created.

Explicit frame dependence of the concurrence

The two-particle effect follows from the single-particle channel. Take the spin state $$ |\chi\rangle = \frac{1}{\sqrt2}\bigl(|\!\uparrow\uparrow\rangle + |\!\downarrow\downarrow\rangle\bigr) $$ to be a maximally entangled state of $A$ and $B$, let $A$ carry the symmetric two-branch momentum superposition with $\theta = 0.8$ and speed $0.5c$, and let $B$ carry a sharp momentum along the boost axis. Boosting with rapidity $\psi$ gives, for the spin state after tracing the momentum,

where $C$ is the concurrence and $S$ is the von Neumann entropy of the two-qubit spin state. The concurrence decreases monotonically with the boost rapidity and the entropy increases from zero, because the boost has turned the pure spin state into an equal mixture of two differently rotated copies. For this configuration the concurrence is exactly $\cos\alpha$, where $\alpha$ is the Wigner angle of a branch, so the entanglement loss is the same dephasing factor that the single-particle channel produced; the branch angles are $\alpha = 0.098585$, $0.193815$, $0.275265$, $0.382723$ at the four non-zero rapidities. At $\psi = 1$ the concurrence has fallen by about two per cent and at $\psi = 2.5$ by about seven per cent. The numbers are the exact output of the channel above, computed from the Wigner rotations of the biquaternion construction, and the effect vanishes at $\psi = 0$: the frame dependence is created by the boost and by the momentum spread together, and by neither alone.

If both particles carry momentum superpositions whose branch directions differ, the effect is larger, because then the local channel is non-trivial on both factors. For $A$ superposed in the boost–$x$ plane at $\theta_A = 0.8$ and $B$ superposed in the boost–$y$ plane at $\theta_B = 2.0$, also at speed $0.5c$, the same initial state gives $$ \psi = 1:\ C = 0.95881842,\quad S = 0.166051; \qquad \psi = 2:\ C = 0.88816953,\quad S = 0.372196 . $$ The comparison of the two tables isolates the origin of the effect: what matters is not the boost alone nor the momentum spread alone, but the mismatch between the Wigner rotations of the branches that are traced over.

The unital channel and the invisibility of the effect in the marginals

The channel of the previous subsection is unital: each branch rotates the identity into the identity, $$ \sum_i q_i\,\tilde{W}_i\,e_0\,\tilde{W}_i^{*} = \sum_i q_i\,e_0 = e_0 , $$ because every Wigner rotor is unitary. The maximally mixed spin state is therefore fixed, and the fully mixed state is a fixed point of every frame transformation. A consequence is worth recording: for the state $|\chi\rangle = (|\!\uparrow\uparrow\rangle+|\!\downarrow\downarrow\rangle)/\sqrt2$ the marginal of each particle is maximally mixed, and it remains maximally mixed under the boost, so the entire frame dependence of the concurrence is invisible in the single-particle marginals. The effect is a change in the correlations of the pair, not in the statistics of either particle alone. This is also why the boost cannot be used to signal: the marginal of the un-boosted particle is untouched, by the cyclicity of the partial trace, and the change in the joint state is only visible to an observer who has access to both particles and knows the momentum distribution.

The invariant spinor concurrence and where the invariance goes

The spinor description makes the invariant and the non-invariant parts of the concurrence visible. On the spinor module, a pure two-qubit state is a four-component object $\psi\in S\otimes S$, and the Lorentz group acts by $\tilde{\Lambda}\otimes\tilde{\Lambda}$ with $\det\tilde{\Lambda} = 1$. With the symplectic form $$ \epsilon = \begin{pmatrix}0 & 1\\ -1 & 0\end{pmatrix}, \qquad \tilde{\Lambda}^{\mathsf T}\epsilon\,\tilde{\Lambda} = \det(\tilde{\Lambda})\,\epsilon = \epsilon , $$ the symplectic concurrence $$ \mathcal{C}(\psi) = \bigl|\psi^{\mathsf T}(\epsilon\otimes\epsilon)\psi\bigr| $$ satisfies $$ \mathcal{C}\bigl((\tilde{\Lambda}\otimes\tilde{\Lambda})\psi\bigr) = \bigl|\det\tilde{\Lambda}\bigr|^2\,\mathcal{C}(\psi) = \mathcal{C}(\psi). $$ The unnormalized concurrence is therefore exactly $SL(2,\mathbb{C})\otimes SL(2,\mathbb{C})$-invariant. What is not invariant is the normalized concurrence $\mathcal{C}(\psi)/\langle\psi|\psi\rangle$, because the spinor norm is not preserved by a boost: $\langle\tilde{\Lambda}\psi|\tilde{\Lambda}\psi\rangle = \langle\psi|\tilde{\Lambda}^{*}\tilde{\Lambda}|\psi\rangle \neq\langle\psi|\psi\rangle$ for a single boosted factor. This is the sharpest way to state the origin of the frame dependence in the framework. The invariant is a bilinear pairing of spinors, and a boost preserves it as a bilinear form; but the physically normalized state divides by a norm that the boost changes, and the ratio is frame-dependent. For a sharp momentum the normalization is common to both branches and cancels; for a momentum superposition the branch norms differ and the ratio does not.

The Wigner Angle and Its Momentum Dependence

The channel above is controlled by one number per branch, the Wigner angle, and the way that number depends on the momentum direction is the geometry of the frame dependence.

Definition. An element of $SU(2)$ is a unit real quaternion, and every such element is a rotation, $$ \tilde{W} = \cos\frac{\alpha}{2}\,e_0 + \sin\frac{\alpha}{2}\,\hat{\mathbf{n}}, \qquad |\hat{\mathbf{n}}| = 1 , $$ with $\alpha$ the Wigner angle and $\hat{\mathbf{n}}$ the rotation axis. For a boost of rapidity $\psi$ along $\hat{\mathbf{z}}$ and a particle of rapidity $\chi$ along a direction making angle $\theta$ with the boost axis, the axis is perpendicular to both, $$ \hat{\mathbf{n}} \propto \hat{\mathbf{z}}\times\hat{\mathbf{u}}, $$ and the angle depends on $\theta$ in a way that the following checks determine.

Vanishing at the endpoints. At $\theta = 0$ and $\theta = \pi$ the momentum is collinear with the boost, the two boosts are along one line and commute, and the Wigner rotation is the identity. This is verified directly: the rotor is $e_0$ up to the $10^{-6}$ of floating-point arithmetic at both endpoints.

Exact at perpendicular momentum. For $\theta = \pi/2$ the angle is given exactly by $$ \tan\alpha = \frac{\sinh\psi\,\sinh\chi}{\cosh\psi + \cosh\chi}, $$ which is verified to machine precision at several rapidities. The angle increases monotonically with both rapidities, tends to $\arctan(\sinh\psi)$ as $\chi\to\infty$, and is bounded by $\pi/2$.

Small rapidities. Expanding the same configuration for small $\psi,\chi$ gives $$ \alpha = \tfrac12\,\psi\chi\,\sin\theta + O(3), $$ with the coefficient $1/2$ confirmed by a numerical limit in which both rapidities are driven to zero at fixed ratio; since $\psi\approx v/c$ and $\chi\approx u/c$ at low speed, this is the leading Thomas–Wigner term $\alpha\approx vu\sin\theta/(2c^2)$, the rotation that accompanies the composition of two non-collinear boosts.

The maximum. For fixed rapidities the angle vanishes at the endpoints and has a single maximum in between. For a boost of rapidity $\psi = 1.5$ and a particle of speed $0.5c$, the angle rises from $0$ at $\theta = 0$ through $11.40^\circ$ at $30^\circ$ and $15.58^\circ$ at $45^\circ$ to a maximum of $19.60^\circ$ near $\theta = 80^\circ$, then returns to $0$ at $\theta = 180^\circ$. The maximum is therefore attained, to within a few degrees, at perpendicular momentum, but the exact location is somewhat below $90^\circ$ and drifts as the rapidities change.

What the angle controls. The angle enters every frame-dependent quantity through $\cos\alpha$: it is the damping factor of the transverse Bloch components, the value of the concurrence in the symmetric two-branch configuration, and the parameter of the spin entropy $h((1+\cos\alpha)/2)$ of a transverse single-particle state. At the maximum $\alpha = 19.60^\circ$ one has $\cos\alpha = 0.9421$, so a transverse spin is damped by about six per cent and the dephased single-particle entropy is $h(0.97103) = 0.1891$. The frame dependence is thus largest for a boost perpendicular to the particle's motion — where the Wigner rotation is largest — and absent for a boost along the motion.

What Is Frame-Dependent and What Is Not

Not frame-dependent. The entanglement of the full state, momentum included, is a property of a state under a unitary transformation and is invariant: the Lorentz group acts unitarily on the full one-particle Hilbert space, so the entanglement between the two particles including their momenta is a Lorentz invariant. The spin entanglement of two sharp-momentum particles is likewise invariant, by the theorem above. And no observable marginal can be changed by a local operation on the other particle, by the cyclicity of the partial trace — the framework's form of no-signaling, recorded in the entangled-subsystems companion.

Frame-dependent. The spin entanglement of a reduced state obtained by tracing an uncertain momentum is frame-dependent, because the boost entangles spin with momentum and the trace discards the momentum. The reduced spin entropy, the concurrence, and the spin correlation functions of a wave packet therefore all transform non-trivially. The effect is second order in the momentum spread in the sense that it vanishes as the distribution narrows to a delta function, and it vanishes as the boost is removed.

The physical location of the non-invariance. The invariance group of the spin factor is the little group $SU(2)$, and $SU(2)$ acts unitarily; the frame dependence is not a failure of unitarity but the consequence of an operation that does not factorise as spin $\times$ momentum. A boost is a unitary operation on the full Hilbert space which is not a product of an operation on the momentum factor and an operation on the spin factor; it correlates them. The spin state, which is a partial description, therefore changes. This is the same structure that produces the Unruh effect for accelerated observers and the thermal character of a restricted vacuum; the field-theoretic case belongs to the information theory of quantum fields, and the present article treats the single-particle kinematic mechanism only.

Relativistic Quantum Information: The Biquaternion Reading

The framework's contribution to this subject is a structural one, and it can be stated in three points. First, the two objects whose mismatch produces the frame dependence have a single algebraic locus. The Wigner rotation is the compact, material rotation generated by the non-compact, informational boosts: it is the little-group element $\tilde{W}\in SU(2)\subset\mathbb{B}$ extracted from a product of unit-norm biquaternions, and the momentum dependence of $\tilde{W}$ is the momentum dependence of a rotor, i.e. of an element of the same algebra that carries the spin. There is no second structure to be introduced: the boost and the spin rotation are elements of one algebra, and their composition is algebra multiplication.

Second, the entanglement measures are pairings. The correlation function, the concurrence, and the reduced state are all computed from the tensor-product trace pairing $\mathrm{Tr}(\tilde{\rho}\tilde{H})$ and the partial trace, and the relativistic action is a congruence on the factors. Read this way, the frame dependence of entanglement is a statement about which pairings the congruence preserves: the full pairing is preserved, the reduced pairing is not, and the difference is exactly the information the boost moves from the spin factor into the spin–momentum correlation.

Third, the invariant is a spinor bilinear form. The symplectic concurrence that survives the boost is the algebraic invariant of the spinor representation, and its non-invariance after normalization is the algebraic trace of the wave-packet character of the state. This is a genuinely biquaternionic statement, because it is the multiplication of spinors by biquaternions that produces both the invariant and its normalization; the framework makes the split between them explicit in a way that the operator formalism states only through the transformation of a density matrix.

Summary

For two particles of definite four-velocity, a Lorentz transformation acts on the spin state by the local unitary Wigner rotation $\tilde{W}_1\otimes\tilde{W}_2$ and therefore preserves the spin entanglement: the Schmidt coefficients, the concurrence, and the entanglement entropy are invariant. This is the sharp-momentum theorem, and it is exact.

For a momentum superposition, the boost acts branch by branch and is a momentum-controlled rotation; tracing the momentum leaves the spin state acted on by the mixed-unitary channel $$ \tilde{\rho}\ \longmapsto\ \sum_i q_i\,\bigl(\tilde{W}_{1,i}\otimes\tilde{W}_{2,i}\bigr)\,\tilde{\rho}\,\bigl(\tilde{W}_{1,i}\otimes\tilde{W}_{2,i}\bigr)^\dagger . $$ The channel is unitary exactly when the Wigner rotation is branch-independent, and otherwise it changes the entanglement. In the symmetric two-branch case it is the dephasing about the Wigner axis, with the parallel Bloch component preserved and the transverse components damped by $\cos\alpha$; the single-particle entropy produced is $h((1+r')/2)$, and the concurrence of a maximally entangled pair decreases monotonically with the boost rapidity, equalling $\cos\alpha$ for the symmetric configuration. For a state with $|\chi\rangle = (|\!\uparrow\uparrow\rangle+|\!\downarrow\downarrow\rangle)/\sqrt2$, $A$ in a symmetric two-branch superposition with $\theta = 0.8$ and speed $0.5c$, and $B$ at sharp momentum along the boost axis, the concurrence falls from $1$ at $\psi = 0$ to $0.98127650$ at $\psi = 1$ and $0.92765115$ at $\psi = 2.5$, with the von Neumann entropy of the two-qubit spin state rising from $0$ to $0.076532$ and $0.224468$.

The invariant behind the computation is the spinor symplectic concurrence $\mathcal{C}(\psi) = |\psi^{\mathsf T}(\epsilon\otimes\epsilon)\psi|$, which is exactly $SL(2,\mathbb{C})\otimes SL(2,\mathbb{C})$-invariant because $\tilde{\Lambda}^{\mathsf T}\epsilon\tilde{\Lambda} = \det(\tilde{\Lambda})\epsilon = \epsilon$. The frame dependence appears only in the normalized ratio, because a boost preserves the bilinear pairing but not the spinor norm; a sharp momentum normalizes both branches identically and the effect cancels. The entanglement of the full state, momentum included, is frame-independent, and no local operation changes the marginal statistics of the other particle; what is frame-dependent is the partial description that discards the momentum, and what produces it is the momentum dependence of the little-group rotor.

Summary of Notation

Further Reading

• Asher Peres, Petra F. Scudo, and Daniel R. Terno, "Quantum entropy and special relativity," Physical Review Letters 88 (2002) 230402, for the frame dependence of the reduced entropy of a spin.
• Robert M. Gingrich and Christoph Adami, "Quantum entanglement of moving bodies," Physical Review Letters 89 (2002) 270402, for the frame dependence of spin entanglement under Wigner rotations.
• Paul M. Alsing and Gerard J. Milburn, "On entanglement and Lorentz transformations," Quantum Information and Computation 2 (2002) 487–512, for the Lorentz transformation of entangled spin states.
• Marek Czachor, "Einstein–Podolsky–Rosen–Bohm experiment with relativistic massive particles," Physical Review A 55 (1997) 72–77, for the relativistic spin-state transformation and its nonlinearity.
• Asher Peres and Daniel R. Terno, "Quantum information and relativity theory," Reviews of Modern Physics 76 (2004) 93–123, for the review of relativistic quantum information.
• William K. Wootters, "Entanglement of formation of an arbitrary state of two qubits," Physical Review Letters 80 (1998) 2245–2248, for the concurrence and its mixed-state form.
• Scott Hill and William K. Wootters, "Entanglement of a pair of quantum bits," Physical Review Letters 78 (1997) 5022–5025, for the entanglement of formation of two qubits.
• Michael A. Nielsen and Isaac L. Chuang, Quantum Computation and Quantum Information (Cambridge, 2000), for the Schmidt decomposition, the partial trace, and the entanglement measures.
• E. P. Wigner, "On unitary representations of the inhomogeneous Lorentz group," Annals of Mathematics 40 (1939) 149–204, for the little group and the Wigner rotation.
• S. Weinberg, The Quantum Theory of Fields, Vol. 1 (Cambridge, 1995), for the unitary action of the Lorentz group on one-particle states.