On Relativistic Quantum Information Properties of Entangled Wave Vectors of Massive Fermions

Abstract

We study special relativistic effects on the entanglement between either spins or momenta of composite quantum systems of two spin-$\frac{1}{2}$ massive particles, either indistinguishable or distinguishable, in inertial reference frames in relative motion. For the case of indistinguishable particles, we consider a balanced scenario where the momenta of the pair are well-defined but not maximally entangled in the rest frame while the spins of the pair are described by a one-parameter (η) family of entangled bipartite states. For the case of distinguishable particles, we consider an unbalanced scenario where the momenta of the pair are well-defined and maximally entangled in the rest frame while the spins of the pair are described by a one-parameter (ξ) family of non-maximally entangled bipartite states. In both cases, we show that neither the spin-spin (ss) nor the momentum-momentum (mm) entanglements quantified by means of Wootters’ concurrence are Lorentz invariant quantities: the total amount of entanglement regarded as the sum of these entanglements is not the same in different inertial moving frames. In particular, for any value of the entangling parameters, both ss and mm-entanglements are attenuated by Lorentz transformations and their parametric rates of change with respect to the entanglements observed in a rest frame have the same monotonic behavior. However, for indistinguishable (distinguishable) particles, the change in entanglement for the momenta is (is not) the same as the change in entanglement for spins. As a consequence, in both cases, no entanglement compensation between spin and momentum degrees of freedom occurs.

Appendices

Appendix A: Decomposition of Density Operators in Terms of Pauli Matrices

In this appendix, we derive (49). Consider a two-qubit system with Hilbert space $\mathcal{H}=\mathbb{C}^{2}\otimes \mathbb{C}^{2}$ and computational basis $\mathcal{B}_{\mathcal{H}}= \{ \vert 00 \rangle , \vert 01 \rangle , \vert 10 \rangle , \vert 11 \rangle \}$. It can be shown that a general two-qubit state $\hat{\rho}$ can always be written up to local unitary equivalence to a state of the following form [21],

$$\hat{\rho}=\frac{1}{4} \Biggl[ I_{4\times4}+\vec{a}\cdot\vec{\sigma}\otimes I_{2\times2}+I_{2\times2}\otimes\vec{b}\cdot\vec{\sigma}+\sum _{i=1}^{3}c_{i}\sigma_{i}\otimes\sigma_{i} \Biggr] , $$

(85)

that is, after some algebra,

$$\hat{\rho}=\frac{1}{4}\left ( \begin{array}[c]{c@{\quad }c@{\quad }c@{\quad }c}a_{3}+b_{3}+c_{3}+1 & b_{1}-ib_{2} & a_{1}-ia_{2} & c_{1}-c_{2}\\ \noalign {\vspace {2pt}}b_{1}+ib_{2} & a_{3}-b_{3}-c_{3}+1 & c_{1}+c_{2} & a_{1}-ia_{2}\\ \noalign {\vspace {2pt}}a_{1}+ia_{2} & c_{1}+c_{2} & b_{3}-a_{3}-c_{3}+1 & b_{1}-ib_{2}\\ \noalign {\vspace {2pt}}c_{1}-c_{2} & a_{1}+ia_{2} & b_{1}+ib_{2} & c_{3}-b_{3}-a_{3}+1\end{array} \right ) , $$

(86)

where $\vec{a}$ and $\vec{b}\in \mathbb{R}^{3}$ are given by $\vec{a}= ( a_{1}, a_{2}, a_{3} )$, $\vec{b}= ( b_{1}, b_{2}, b_{3} )$ and $\vec{\sigma}= ( \sigma_{x}, \sigma_{y}, \sigma_{z} )$ is the operator vector of Pauli matrices. Equating $\hat{\rho}_{\psi_{+}} ( \xi )$ in (48) to (86), we obtain

$$\begin{array}{c}a_{1}=a_{2}=b_{1}=b_{2}=0,\quad c_{1}=c_{2}=2\sqrt{\xi ( 1-\xi ) }, \\ \noalign {\vspace {6pt}}a_{3}=1-2 \xi, \quad b_{3}=- ( 1-2\xi ) , \quad c_{3}=-1.\end{array}$$

(87)

Therefore, in terms of Pauli matrices, $\hat{\rho}_{\psi_{+}} (\xi )$ becomes

$$\hat{\rho}_{\psi_{+}} ( \xi ) =\frac{1}{4}\left [ \begin{array}[c]{c}I_{2\times2}^{ ( 1 ) }\otimes I_{2\times2}^{ ( 2 )}+ ( 1-2\xi )\sigma_{z}^{ ( 1 ) }\otimes I_{2\times 2}^{ ( 2 ) }- ( 1-2\xi )I_{2\times2}^{ ( 1 )}\otimes\sigma_{z}^{ ( 2 ) }\\ \noalign {\vspace {5pt}}+2\sqrt{\xi ( 1-\xi )}\sigma_{x}^{ ( 1 ) }\otimes\sigma_{x}^{ ( 2 ) }+2\sqrt{\xi ( 1-\xi ) }\sigma_{y}^{ ( 1 ) }\otimes \sigma_{y}^{ ( 2 ) }-\sigma_{z}^{ ( 1 ) }\otimes \sigma_{z}^{ ( 2 ) }\end{array}\right ] . $$

(88)

Observe that for $\xi=\frac{1}{2}$ we obtain,

$$\hat{\rho}_{\psi_{+}} \biggl( \frac{1}{2} \biggr) =\frac{1}{4}\bigl[ I_{2\times2}^{ ( 1 ) }\otimes I_{2\times2}^{ ( 2 )}+\sigma_{x}^{ ( 1 ) }\otimes\sigma_{x}^{ ( 2 ) }+\sigma_{y}^{ ( 1 ) }\otimes\sigma_{y}^{ ( 2 ) }-\sigma_{z}^{ ( 1 ) }\otimes\sigma_{z}^{ ( 2 ) }\bigr] . $$

(89)

Similarly, equating $\hat{\rho}_{\psi_{-}} ( \xi )$ in (48) to (86), we get

$$\begin{array}{c}a_{1}=a_{2}=b_{1}=b_{2}=0,\quad c_{1}=c_{2}=-2\sqrt{\xi ( 1-\xi ) }, \\ \noalign {\vspace {6pt}}a_{3}=1-2 \xi, \quad b_{3}=- ( 1-2\xi ) ,\quad c_{3}=-1.\end{array}$$

(90)

In terms of Pauli matrices, $\hat{\rho}_{\psi_{-}} ( \xi )$ reads,

$$\hat{\rho}_{\psi_{-}} ( \xi ) =\frac{1}{4}\left [ \begin{array}[c]{c}I_{2\times2}^{ ( 1 ) }\otimes I_{2\times2}^{ ( 2 )}+ ( 1-2\xi ) \sigma_{z}^{ ( 1 ) }\otimes I_{2\times 2}^{ ( 2 ) }- ( 1-2\xi ) I_{2\times2}^{ ( 1 )}\otimes\sigma_{z}^{ ( 2 ) }\\ \noalign {\vspace {6pt}}-2\sqrt{\xi ( 1-\xi )}\sigma_{x}^{ ( 1 ) }\otimes\sigma_{x}^{ ( 2 ) }-2\sqrt{\xi ( 1-\xi ) }\sigma_{y}^{ ( 1 ) }\otimes \sigma_{y}^{ ( 2 ) }-\sigma_{z}^{ ( 1 ) }\otimes \sigma_{z}^{ ( 2 ) }\end{array}\right ] , $$

(91)

and for $\xi=\frac{1}{2}$ it becomes,

$$\hat{\rho}_{\psi_{-}} \biggl( \frac{1}{2} \biggr)=\frac{1}{4}\bigl[ I_{2\times2}^{ ( 1 ) }\otimes I_{2\times2}^{ ( 2 )}-\sigma_{x}^{ ( 1 ) }\otimes\sigma_{x}^{ ( 2 ) }-\sigma_{y}^{ ( 1 ) }\otimes\sigma_{y}^{ ( 2 ) }-\sigma_{z}^{ ( 1 ) }\otimes\sigma_{z}^{ ( 2 ) }\bigr] . $$

(92)

In summary, we have shown that

$$\everymath{\displaystyle }\begin{array}[b]{l}\hat{\rho}_{\psi\pm} ( \xi ) =\frac{1}{4}\left [ \begin{array}[c]{c}I_{2\times2}^{ ( 1 ) }\otimes I_{2\times2}^{ ( 2 )}+ ( 1-2\xi ) \sigma_{z}^{ ( 1 ) }\otimes I_{2\times 2}^{ ( 2 ) }- ( 1-2\xi ) I_{2\times2}^{ ( 1 )}\otimes\sigma_{z}^{ ( 2 ) } \\ \noalign {\vspace {5pt}}\pm2\sqrt{\xi ( 1-\xi )}\sigma_{x}^{ ( 1 ) }\otimes\sigma_{x}^{ ( 2 ) }\pm2\sqrt{\xi ( 1-\xi ) }\sigma_{y}^{ ( 1 ) }\otimes\sigma_{y}^{ ( 2 ) }-\sigma_{z}^{ ( 1 ) }\otimes\sigma_{z}^{ ( 2 ) }\end{array} \right ] , \\ \noalign {\vspace {8pt}}\end{array}$$

(93)

and in the limiting case of $\xi=\frac{1}{2}$ we recover the case of maximally entangled pure Bell states.

Appendix B: Lorentz Transformation of the Spin Density Operator

In this appendix, we derive (62). Substituting (21) and (22) into (25), we get

(94)

where $\hat{\rho}_{\mathrm{spin}}$ equals $\hat{\rho}_{\psi\pm} (\xi )$ in (49). Using (61), we observe that the following transformation rules hold

$$\everymath{\displaystyle }\begin{array}{rcl}\mathcal{D}_{A} ( p_{A_{1}} ) \sigma_{x}^{ (A)}\mathcal{D}_{A}^{\dagger} ( p_{A_{1}} ) & = &\biggl( \cos\frac{\varphi}{2}-i\sin\frac{\varphi}{2}\sigma_{z}^{ (A)}\biggr) \sigma_{x}^{(A)} \biggl( \cos\frac{\varphi}{2}+i\sin\frac{\varphi}{2}\sigma_{z}^{(A)} \biggr)\\ \noalign {\vspace {6pt}}& = &\cos \varphi\sigma_{x}^{(A)}+\sin\varphi\sigma_{y}^{ (A ) },\\ \noalign {\vspace {6pt}}\mathcal{D}_{B} ( p_{B_{1}} ) \sigma_{x}^{ (B)}\mathcal{D}_{B}^{\dagger} ( p_{B_{1}} ) & = &\biggl( \cos \frac{\varphi}{2}+i\sin\frac{\varphi}{2}\sigma_{z}^{ (B)}\biggr) \sigma_{x}^{(B)}\biggl( \cos\frac{\varphi}{2}-i\sin\frac{\varphi}{2}\sigma_{z}^{(B)} \biggr)\\ \noalign {\vspace {6pt}}& = &\cos \varphi\sigma_{x}^{(B)}-\sin\varphi\sigma_{y}^{ (B ) },\\ \noalign {\vspace {6pt}}\mathcal{D}_{A} ( p_{A_{1}} ) \sigma_{y}^{ (A)}\mathcal{D}_{A}^{\dagger} ( p_{A_{1}} ) & =&\biggl( \cos \frac{\varphi}{2}-i\sin\frac{\varphi}{2}\sigma_{z}^{ (A)}\biggr) \sigma_{y}^{(A)} \biggl( \cos\frac{\varphi}{2}+i\sin\frac{\varphi}{2}\sigma_{z}^{(A)} \biggr)\\ \noalign {\vspace {6pt}}& = &-\sin \varphi\sigma_{x}^{(A)}+\cos\varphi\sigma_{y}^{ (A ) },\\ \noalign {\vspace {6pt}}\mathcal{D}_{B} ( p_{B_{1}} ) \sigma_{y}^{ (B)}\mathcal{D}_{B}^{\dagger} ( p_{B_{1}} ) & =& \biggl( \cos \frac{\varphi}{2}+i\sin\frac{\varphi}{2}\sigma_{z}^{ (B)}\biggr) \sigma_{y}^{(B)} \biggl( \cos\frac{\varphi}{2}-i\sin\frac{\varphi}{2}\sigma_{z}^{(B)} \biggr)\\ \noalign {\vspace {6pt}}& = &\sin \varphi\sigma_{x}^{(B)}+\cos\varphi\sigma_{y}^{ (B ) },\end{array} $$

(95)

and,

$$\everymath{\displaystyle }\begin{array}{rcl}\mathcal{D}_{A} ( p_{A_{2}} ) \sigma_{x}^{ (A)}\mathcal{D}_{A}^{\dagger} ( p_{A_{2}} ) & = &\biggl( \cos \frac{\varphi}{2}+i\sin\frac{\varphi}{2}\sigma_{z}^{ (A)}\biggr) \sigma_{x}^{(A)} \biggl( \cos\frac{\varphi}{2}-i\sin\frac{\varphi}{2}\sigma_{z}^{(A)} \biggr)\\ \noalign {\vspace {6pt}}& =&\cos\varphi\sigma_{x}^{(A)}-\sin\varphi\sigma_{y}^{ (A ) },\\ \noalign {\vspace {6pt}}\mathcal{D}_{B} ( p_{B_{2}} ) \sigma_{x}^{ (B)}\mathcal{D}_{B}^{\dagger} ( p_{B_{2}} ) & = &\biggl( \cos\frac{\varphi}{2}-i\sin\frac{\varphi}{2}\sigma_{z}^{ (B)}\biggr) \sigma_{x}^{(B)} \biggl( \cos\frac{\varphi}{2}+i\sin\frac{\varphi}{2}\sigma_{z}^{(B)} \biggr)\\ \noalign {\vspace {6pt}}& =&\cos\varphi\sigma_{x}^{(B)}+\sin\varphi\sigma_{y}^{ (B ) },\\ \noalign {\vspace {6pt}}\mathcal{D}_{A} ( p_{A_{2}} ) \sigma_{x}^{ (A)}\mathcal{D}_{A}^{\dagger} ( p_{A_{2}} ) & =&\biggl( \cos\frac{\varphi}{2}+i\sin\frac{\varphi}{2}\sigma_{z}^{ (A)}\biggr) \sigma_{y}^{(A)} \biggl( \cos\frac{\varphi}{2}-i\sin\frac{\varphi}{2}\sigma_{z}^{(A)} \biggr)\\ \noalign {\vspace {6pt}}& =&\sin\varphi\sigma_{x}^{(A)}+\cos\varphi\sigma_{y}^{ (A ) },\\ \noalign {\vspace {6pt}}\mathcal{D}_{B} ( p_{B_{2}} ) \sigma_{x}^{ (B)}\mathcal{D}_{B}^{\dagger} ( p_{B_{2}} )& =&\biggl( \cos\frac{\varphi}{2}-i\sin\frac{\varphi}{2}\sigma_{z}^{ (B)}\biggr) \sigma_{y}^{(B)} \biggl( \cos\frac{\varphi}{2}+i\sin\frac{\varphi}{2}\sigma_{z}^{(B)} \biggr)\\ \noalign {\vspace {6pt}}& =&-\sin\varphi\sigma_{x}^{(B)}+\cos\varphi\sigma_{y}^{ (B ) }.\end{array} $$

(96)

Therefore, using (95) and (96), after some tedious algebra it follows that $\hat{\rho}_{\mathrm{spin}}^{\prime}$ in (94) becomes

$$\hat{\rho}_{\mathrm{spin}}^{\prime}=\frac{1}{4}\left [ \begin{array}[c]{c}I_{2\times2}^{(A)}\otimes I_{2\times2}^{ (B)}+ ( 1-2\xi ) \sigma_{z}^{(A)}\otimes I_{2\times 2}^{(B)}- ( 1-2\xi ) I_{2\times2}^{ (A)}\otimes\sigma_{z}^{(B)}\\ \noalign {\vspace {7pt}}\pm2\sqrt{\xi ( 1-\xi )}\cos2\varphi\sigma_{x}^{(A)}\otimes\sigma_{x}^{ (B ) }\pm2\sqrt{\xi ( 1-\xi ) }\cos2\varphi\sigma_{y}^{ (A)}\otimes\sigma_{y}^{(B)}\\ \noalign {\vspace {7pt}}-\sigma_{z}^{(A)} \otimes\sigma_{z}^{(B)}.\end{array}\right ] , $$

(97)

that is, $\hat{\rho}_{\mathrm{spin}}^{\prime}$ equals $\hat{\rho}_{\psi\pm }^{ ( \Lambda ) } ( \xi, \varphi )$ in (62).

Appendix C: Lorentz Transformation of the Momentum Density Operator

In this appendix, we derive (68). Using (64), (65) and (67), it follows that

$$\everymath{\displaystyle }\begin{array}{l}\vert \Lambda\vec{p}_{A_{1}} \rangle \langle \Lambda\vec {p}_{A_{1}}\vert =\frac{I_{2\times2}^{(A)}+\tilde{\sigma}_{z}^{(A)}}{2},\quad \vert \Lambda\vec {p}_{B_{1}} \rangle \langle \Lambda\vec{p}_{B_{1}}\vert =\frac{I_{2\times2}^{(B)}+\tilde{\sigma}_{z}^{ (B)}}{2},\\ \noalign {\vspace {8pt}}\vert \Lambda\vec{p}_{A_{2}} \rangle \langle \Lambda\vec {p}_{A_{2}}\vert =\frac{I_{2\times2}^{(A)}-\tilde{\sigma}_{z}^{(A)}}{2},\quad \vert \Lambda\vec {p}_{B_{2}} \rangle \langle \Lambda\vec{p}_{B_{2}}\vert =\frac{I_{2\times2}^{(B)}-\tilde{\sigma}_{z}^{ (B)}}{2},\end{array} $$

(98)

and,

$$\everymath{\displaystyle }\begin{array}{l}\vert \Lambda\vec{p}_{A_{1}} \rangle \langle \Lambda\vec {p}_{A_{2}}\vert =\frac{\tilde{\sigma}_{x}^{ (A)}+i\tilde{\sigma}_{y}^{(A)}}{2},\quad \vert \Lambda \vec{p}_{A_{2}} \rangle \langle \Lambda\vec{p}_{A_{1}}\vert =\frac{\tilde{\sigma}_{x}^{(A)}-i\tilde{\sigma}_{y}^{ (A ) }}{2},\\ \noalign {\vspace {8pt}}\vert \Lambda\vec{p}_{B_{1}} \rangle \langle \Lambda\vec {p}_{B_{2}}\vert =\frac{\tilde{\sigma}_{x}^{ (B)}+i\tilde{\sigma}_{y}^{(B)}}{2},\quad \vert \Lambda \vec{p}_{B_{2}} \rangle \langle \Lambda\vec{p}_{B_{1}}\vert =\frac{\tilde{\sigma}_{x}^{(B)}-i\tilde{\sigma}_{y}^{ (B ) }}{2}.\end{array} $$

(99)

Using (98) and (99), $\hat{\rho}_{\mathrm{momentum}}^{\prime}$ in (68) becomes

$$\begin{array}[b]{rcl}\hat{\rho}_{\mathrm{momentum}}^{\prime} &=&\dfrac{1}{2}\left [ \begin{array}[c]{c}\langle 1|1 \rangle ( \frac{I_{2\times2}^{ (A)}+\tilde{\sigma}_{z}^{(A)}}{2} ) \otimes (\frac{I_{2\times2}^{(B)}+\tilde{\sigma}_{z}^{ (B)}}{2} ) + \langle 2|1 \rangle ( \frac{\tilde{\sigma}_{x}^{(A)}+i\tilde{\sigma}_{y}^{(A)}}{2} )\otimes ( \frac{\tilde{\sigma}_{x}^{(B)}+i\tilde{\sigma }_{y}^{(B)}}{2} )\\ \noalign {\vspace {6pt}}+ \langle 1|2 \rangle ( \frac{\tilde{\sigma}_{x}^{ (A ) }-i\tilde{\sigma}_{y}^{(A)}}{2} )\otimes ( \frac{\tilde{\sigma}_{x}^{(B)}-i\tilde{\sigma }_{y}^{(B)}}{2} ) + \langle 2|2 \rangle (\frac{I_{2\times2}^{(A)}-\tilde{\sigma}_{z}^{ (A)}}{2} ) \otimes ( \frac{I_{2\times2}^{(B)}-\tilde{\sigma}_{z}^{(B)}}{2} )\end{array} \right ] . \\ \noalign {\vspace {1pt}}\end{array} $$

(100)

Furthermore, substituting (47) into (21) and (22), it follows that 〈1|1〉=〈2|2〉=1 while 〈1|2〉 and 〈2|1〉 are given by,

(101)

and,

(102)

respectively. Thus, substituting (101) and (102) into (100), we obtain

$$\hat{\rho}_{\mathrm{momentum}}^{\prime}=\frac{1}{2}\left [ \begin{array}[c]{c}( \frac{I_{2\times2}^{(A)}+\tilde{\sigma}_{z}^{ (A ) }}{2} ) \otimes ( \frac{I_{2\times2}^{ (B)}+\tilde{\sigma}_{z}^{(B)}}{2} ) +[ \cos 2\varphi+i ( 2\xi-1 ) \sin2\varphi ]\\ \noalign {\vspace {6pt}}{}\times ( \frac {\tilde{\sigma}_{x}^{(A)}+i\tilde{\sigma}_{y}^{ (A ) }}{2} )\otimes ( \frac{\tilde{\sigma}_{x}^{ (B ) }+i\tilde{\sigma}_{y}^{(B)}}{2} )+ [ \cos2\varphi-i ( 2\xi-1 )\\ \noalign {\vspace {6pt}}{}\times \sin2\varphi ] (\frac{\tilde{\sigma}_{x}^{(A)}-i\tilde{\sigma}_{y}^{ (A ) }}{2} ) \otimes ( \frac{\tilde{\sigma}_{x}^{ (B ) }-i\tilde{\sigma}_{y}^{(B)}}{2} )\\ \noalign {\vspace {6pt}}+ ( \frac{I_{2\times2}^{(A)}-\tilde{\sigma}_{z}^{ (A)}}{2} )\otimes ( \frac{I_{2\times2}^{(B)} -\tilde{\sigma}_{z}^{(B)}}{2} )\end{array} \right ] . $$

(103)

After some algebra, $\hat{\rho}_{\mathrm{momentum}}^{\prime}$ may be finally rewritten as $\hat{\rho}_{\mathrm{momentum}}^{\prime} ( \xi, \varphi )$ in (68),

$$\begin{array}[b]{rcl}\hat{\rho}_{\mathrm{momentum}}^{\prime}&\equiv&\hat{\rho}_{\mathrm{momentum}}^{\prime}( \xi, \varphi )\\ \noalign {\vspace {6pt}}&=&\dfrac{1}{4}\left [ \begin{array}[c]{c}I_{2\times2}^{(A)}\otimes I_{2\times2}^{ (B)}+\cos2\varphi ( \tilde{\sigma}_{x}^{(A)}\otimes \tilde{\sigma}_{x}^{(B)}-\tilde{\sigma}_{y}^{ (A)}\otimes\tilde{\sigma}_{y}^{(B)} ) +\tilde{\sigma}_{z}^{(A)}\otimes\tilde{\sigma}_{z}^{(B)}\\ \noalign {\vspace {6pt}}+ ( 1-2\xi ) \sin2\varphi ( \tilde{\sigma}_{x}^{ (A ) }\otimes\tilde{\sigma}_{y}^{(B)}+\tilde{\sigma}_{y}^{(A)}\otimes\tilde{\sigma}_{x}^{ (B) } )\end{array} \right ] .\\ \noalign {\vspace {1pt}}\end{array} $$

(104)

As a final remark, we stress that the density matrices (114) and (104) cannot be parametrized by means of the decomposition in (85). For them it is needed the most general parametrization for a two-qubit system with Hilbert space ℂ²⊗ℂ². It can be shown that any state $\hat{\rho}$ for such a system may be parametrized as [22],

$$\hat{\rho}=\frac{1}{4}\Biggl[ I_{4\times4}+\vec{c}\cdot\vec{\sigma}\otimes I_{2\times2}+I_{2\times2}\otimes\vec{d}\cdot\vec{\sigma}+\sum_{j, k=1}^{3}\gamma_{jk} \sigma_{j}\otimes\sigma_{k} \Biggr] ,$$

(105)

where $\vec{c}$, $\vec{d}\in \mathbb{R}^{3}$ and γ _jk are real numbers.

Appendix D: Additional Example for Distinguishable Particles

In this appendix, we consider the case of distinguishable particles with an unbalanced scenario where the momenta of the pair are well-defined and maximally entangled in the rest frame while the spins of the pair are described by a one-parameter (ξ) family of non-maximally entangled bipartite states. Unlike the distinguishable case considered in Sect. 4, we consider a composite quantum system described in the rest frame S by a total wave-vector that differs from (66) and characterized by a set of momentum constraints that differs from that in (50).

Consider the following set of working hypotheses on momentum states,

$$\vec{p}_{A_{1}}=-\vec{p}_{A_{2}},\quad \vec{p}_{B_{1}}=-\vec{p}_{B_{2}} \quad \hbox{and}\quad \vec{p}_{A_{1}}=a\vec{p}_{B_{2}}, $$

(106)

where $\alpha=\frac{m_{e}}{m_{\mu}}$. This expression for α is obtained by imposing that the two angles of Wigner’s rotations φ _A and φ _B be equal (for an explicit derivation of this specific expression for α, we refer to (59) in Sect. 5). The momentum states $\vert \vec{p}_{A_{1}} \rangle$ and $\vert \vec{p}_{A_{2}} \rangle$ are the eigenvectors of $\tilde{\sigma}_{z}^{ (A ) }$ with eigenvalues +1 and −1, respectively, while $\vert \vec{p}_{B_{1}} \rangle$ and $\vert \vec{p}_{B_{2}} \rangle$ are the eigenvectors of $\tilde{\sigma}_{z}^{(B)}$ with eigenvalues −1 and +1, respectively. Thus, we consider

$$\everymath{\displaystyle }\begin{array}{l}\vert \vec{p}_{A_{1}} \rangle =\vert 0 \rangle_{A}\equiv \binom{1}{0},\quad \vert \vec{p}_{A_{2}} \rangle =\vert 1 \rangle_{A}\equiv \binom{0}{1}, \\ \noalign {\vspace {6pt}}\vert \vec{p}_{B_{1}}\rangle =\vert 1 \rangle_{B}\equiv\binom{0}{1}\quad \hbox{and}\quad \vert \vec{p}_{B_{2}} \rangle =\vert 0 \rangle_{B}\equiv\binom{1}{0}. \end{array}$$

(107)

In such case, the wave-vector |Ψ〉 _S in the rest frame S reads,

$$\begin{array}[b]{rcl}\bigl \vert \Psi_{e\mu}^{ ( \mathrm{new} ) } \bigr\rangle & =&\vert \Psi \rangle_{S}\stackrel{\mathrm {def}}{=}\dfrac{1}{\sqrt{2}}\bigl[ \vert \vec{p}_{A_{1}}, \vec{p}_{B_{1}} \rangle \otimes \vert 0\rangle_{S}+\vert \vec{p}_{A_{2}},\vec{p}_{B_{2}} \rangle \otimes \vert 0 \rangle_{S} \bigr]\\ \noalign {\vspace {6pt}}& =&\dfrac{1}{\sqrt{2}}\vert 10 \rangle \bigl[ \sqrt{1-\xi }\vert \uparrow \downarrow \rangle \pm\sqrt{\xi} \vert \downarrow\uparrow \rangle \bigr] \mp \dfrac{1}{\sqrt{2}}\vert 01 \rangle \bigl[ \sqrt{1-\xi} \vert \uparrow\downarrow \rangle \pm\sqrt{\xi} \vert \downarrow\uparrow \rangle \bigr] .\end{array} $$

(108)

Note that for $\xi=\frac{1}{2}$, |Ψ〉 _S in (108) becomes the wave-vector studied in [15],

$$\vert \Psi_{\pm} \rangle \stackrel{\mathrm {def}}{=}\frac{1}{\sqrt{2}}\vert 10 \rangle \vert \phi_{\pm} \rangle \mp\frac {1}{\sqrt{2}}\vert 01 \rangle \vert \phi_{\pm} \rangle , $$

(109)

with |ϕ _±〉 defined in (28). Focusing on |Ψ₊〉 (the same results are obtained using |Ψ₋〉 and following the same line of reasoning presented in Sect. 5, we obtain that the spin and momentum reduced density operators in the rest frame S become,

(110)

and,

(111)

respectively. The concurrence of $\rho_{\mathrm{spin}}^{ (\text{rest-frame} ) } ( \xi )$ and $\rho_{\mathrm{momentum}}^{ ( \text{rest-frame} ) }$ are,

$$\mathcal{C}_{\mathrm{spin}}^{ ( S ) } ( \xi ) =\sqrt {4\xi ( 1-\xi ) }\quad \hbox{and}\quad \mathcal{C}_{\mathrm{momentum}}^{ (S ) }=1, $$

(112)

respectively. The Lorentz-transformed spin density operator in (25) reads,

(113)

Similarly, the Lorentz-transformed momentum density operator in (26) is given by

(114)

Furthermore, omitting technical details, it can be shown that the concurrences of the Lorentz-transformed spin and momentum density matrices in the moving reference frame S′ are given by,

$$\mathcal{C}_{\mathrm{spin}}^{ ( S^{\prime} ) } ( \xi, \varphi ) =\sqrt{4\xi ( 1-\xi )}\vert \cos 2\varphi \vert , $$

(115)

and,

$$\mathcal{C}_{\mathrm{momentum}}^{ ( S^{\prime} ) } ( \xi, \varphi ) =\sqrt{1-4\xi ( 1-\xi ) \sin^{2}2\varphi}, $$

(116)

respectively.

We remark that although (108) and (106) differ from (66) and (50), respectively, we have obtained the same final conclusions in both distinguishable cases considered: (115) and (116) are equal to (73) and (77), respectively. We also point out that when $\xi=\frac{1}{2}$ and m _e =m _μ (two electrons), we recover the main result appeared in [15]. Finally, we emphasize that for distinguishable particles, the change in entanglement for the momenta is not the same as the change in entanglement for spins (see (115) and (116) in addition to Figs. 3, 4, 5, 6).