The Classical Limit of a Physical Theory and the Dimensionality of Space

Abstract

In the operational approach to general probabilistic theories one distinguishes two spaces, the state space of the “elementary systems” and the physical space in which “laboratory devices” are embedded. Each of those spaces has its own dimension—the minimal number of real parameters (coordinates) needed to specify the state of system or a point within the physical space. Within an operational framework to a physical theory, the two dimensions coincide in a natural way under the following “closeness” requirement: the dynamics of a single elementary system can be generated by the invariant interaction between the system and the “macroscopic transformation device” that itself is described from within the theory in the macroscopic (classical) limit. Quantum mechanics fulfils this requirement since an arbitrary unitary transformation of an elementary system (spin-1/2 or qubit) can be generated by the pairwise invariant interaction between the spin and the constituents of a large coherent state (“classical magnetic field”). Both the spin state space and the “classical field” are then embedded in the Euclidean three-dimensional space. Can we have a general probabilistic theory, other than quantum theory, in which the elementary system (“generalized spin”) and the “classical fields” generating its dynamics are embedded in a higher-dimensional physical space? We show that as long as the interaction is pairwise, this is impossible, and quantum mechanics and the three-dimensional space remain the only solution. However, having multi-particle interactions and a generalized notion of “classical field” may open up such a possibility.

Appendices

Appendix 1: Dynamics of Spin in Presence of Spin-Coherent State

Here we justify the approximation made in Sect. 4. Namely, we show that Eq. (22) can be realized within quantum mechanics. We follow the idea given in the work by Poulin [46]. Let the large system be a ferromagnet composed of N spin-1/2 particles with the Hamiltonian $H_0$. We assume that $H_0$ is rotationally invariant $U^{\otimes N}H_0 U^{\dagger \otimes N}=H_0$ for all single particle rotations $U\in \mathrm {SU}(2)$. One particular example of such a system is a Heisenberg ferromagnet with the Hamiltonian:

$$\begin{aligned} H_0=-\sum _{n,m=1}^NJ_{\textit{nm}}\vec {\sigma ^{(n)}}\vec {\sigma ^{(m)}}, \end{aligned}$$

(52)

with $J_{\textit{nm}}\ge 0$ are the coupling constants. The rotational invariance is an important assumption because there is no external reference direction. The large system itself can be used to define preferred direction in space. Referring to the well known result in solid state physics [63] such a system, although rotationally invariant, can still exhibit spontaneous magnetization bellow the critical temperature. At zero temperature all the spins are aligned along some direction, that we choose to be the $\mathbf {e}_z$-direction. Hence the ground state is $|\psi _0\rangle =|0\rangle ^{\otimes N}$ with the energy set to zero $E_0=0$ (this is always possible by changing the energy reference point). Let the small system be prepared in a state $|\phi \rangle =\alpha |0\rangle +\beta |1\rangle$ and assume $\sigma _3|0\rangle =|0\rangle$. The system interacts with the large system via Heisenberg interaction, therefore the total Hamiltonian reads

$$\begin{aligned} H=\sum _{n=1}^NJ_n\vec {\sigma ^{(0)}}\vec {\sigma ^{(n)}}+H_0, \end{aligned}$$

(53)

where $J_n$ is the coupling constant for the interaction between the small spin and nth spin of the large system. Our goal is to show that in macroscopic limit $N\rightarrow \infty$, the dynamics becomes separable:

$$\begin{aligned} e^{\textit{itH}}|\phi \rangle |\psi _0\rangle= & {} (e^{\textit{itH}_{\textit{eff}}}|\phi \rangle )|\psi _0\rangle , \end{aligned}$$

(54)

where $H_{\textit{eff}}$ is an effective Hamiltonian.

Firstly, let us compute the following

$$\begin{aligned} H|\phi \rangle |\psi _0\rangle= & {} \sum _{n=1}^N(J_n\vec {\sigma ^{(0)}}\vec {\sigma ^{(n)}}+H_0)|\phi \rangle |0\rangle ^{\otimes N}\\= & {} \sum _{n=1}^NJ_n\vec {\sigma ^{(0)}}\vec {\sigma ^{(n)}}|\phi \rangle |0\rangle ^{\otimes N}\nonumber \\= & {} \left( \sum _{n=1}^NJ_n\right) (\sigma _3|\phi \rangle )(\sigma _3^{(n)}|0\rangle ^{\otimes N})\nonumber \\+ & {} \sum _{n=1}^{N}J_n\sum _{i=1}^2\sigma _i^{(0)}\sigma _i^{(n)}|\psi \rangle |0\rangle ^{\otimes N}\nonumber \\= & {} \left( \sum _{n=1}^NJ_n\right) (\sigma _3|\phi \rangle )|0\rangle ^{\otimes N}+\sum _{n=1}^{N}J_n\sum _{i=1}^2\sigma _i^{(0)}\sigma _i^{(n)}|\psi \rangle |0\rangle ^{\otimes N}\nonumber \\= & {} |\chi \rangle +|\mu \rangle ,\nonumber \end{aligned}$$

(55)

where

$$\begin{aligned} |\chi \rangle= & {} \left( \sum _{n=1}^NJ_n\right) (\sigma _3|\phi \rangle )|0\rangle ^{\otimes N},\end{aligned}$$

(56)

$$\begin{aligned} |\mu \rangle= & {} \sum _{n=1}^{N}J_n\sum _{i=1}^2\sigma _i^{(0)}\sigma _i^{(n)}|\psi \rangle |0\rangle ^{\otimes N}. \end{aligned}$$

(57)

The norm of $|\chi \rangle$ is easy to compute $\langle \chi |\chi \rangle =(\sum _{n=1}^NJ_n)^2$. On the other hand, we have:

$$\begin{aligned} |\mu \rangle= & {} (\sigma _1|\psi \rangle )(J_1|1\rangle |0\rangle |0\rangle \dots +J_2|0\rangle |1\rangle |0\rangle \dots )\\\nonumber+ & {} (i\sigma _2|\psi \rangle )(J_1|1\rangle |0\rangle |0\rangle \dots +J_2|0\rangle |1\rangle |0\rangle \dots )\\\nonumber= & {} \left( (\sigma _1+i\sigma _2)|\psi \rangle \right) (J_1|1\rangle |0\rangle |0\rangle \dots +J_2|0\rangle |1\rangle |0\rangle \dots ). \end{aligned}$$

(58)

The norm of $|\mu \rangle$ is given by $\langle \mu |\mu \rangle =\sum _{n=1}^NJ^2_n$. Let us define the averages

$$\begin{aligned} \langle J\rangle _N= & {} \frac{1}{N}\sum _{n=1}^NJ_n,\end{aligned}$$

(59)

$$\begin{aligned} \langle J^2\rangle _N= & {} \frac{1}{N}\sum _{n=1}^NJ^2_n. \end{aligned}$$

(60)

We assume that $\langle J\rangle _N$ and $\langle J^2\rangle _N$ have finite values in macroscopic limit. Furthermore, we assume that $\lim _{N\rightarrow \infty }\langle J\rangle _N=\langle J\rangle \ne 0$. We can express the norms of $|\chi \rangle$ and $|\mu \rangle$ in terms of these quantities

$$\begin{aligned} \langle \chi |\chi \rangle= & {} N^2\langle J\rangle _N,\end{aligned}$$

(61)

$$\begin{aligned} \langle \mu |\mu \rangle= & {} N\langle J^2\rangle _N. \end{aligned}$$

(62)

Now, it is clear that $|\mu \rangle$ is a vector of short length as compared to $|\chi \rangle$ when N is large. Furthermore, in the macroscopic limit, we have $\lim _{N\rightarrow \infty }\frac{\langle \mu |\mu \rangle }{\langle \chi |\chi \rangle }=0$, therefore one can safely remove $|\mu \rangle$ from Eq. (55) when $N\rightarrow \infty$:

$$\begin{aligned} H|\phi \rangle |\psi _0\rangle =(H_{\textit{eff}}|\phi \rangle )|\psi _0\rangle , \end{aligned}$$

(63)

where $H_{\textit{eff}}=N\langle J\rangle \sigma _3$. Now we can prove (54):

$$\begin{aligned} e^{\textit{itH}}|\phi \rangle |\psi _0\rangle= & {} \sum _{k=0}^{+\infty }\frac{t^k}{k!}H^k|\phi \rangle |\psi _0\rangle \\\nonumber= & {} \sum _{k=0}^{+\infty }\frac{t^k}{k!}(H_{\textit{eff}}^k|\phi \rangle )|\psi _0\rangle \\\nonumber= & {} (e^{\textit{it}H_{\textit{eff}}}|\phi \rangle )|\psi _0\rangle . \end{aligned}$$

(64)

In general, if the large system exhibits the ground state $|\psi _0\rangle =|\vec {n}\rangle ^{\otimes N}$ (spin coherent state) magnetized along the direction $\vec {n}$, it will generate an effective Hamiltonian $H_{\textit{eff}}(\vec {n})=N\langle J\rangle \vec {n}\vec {\sigma }$.

Appendix 2: Groups Transitive on Spheres

The groups that are transitive on spheres are summarized in Table 1.

Table 1 Table taken from the Ref. [29]

For simplicity reasons, we shall study only the minimal group (therefore certainly within the set of physical transformations) that is transitive on a sphere $\mathcal {S}^{d-1}$. If d is odd, the minimal transitive group is the special orthogonal group $\mathrm {SO}(d)$ unless $d=7$. For $d=7$ the minimal group is the exceptional Lie group $\mathrm {G}_2$. If d is even, there are several options. We distinguish the cases whether the group contains the total inversion $E\mathbf {x}=-\mathbf {x}$ or not. The groups $\mathrm {U}(d/2), \mathrm {Sp}(d/4), \mathrm {Sp}(d/4)\times \mathrm {U}(1), \mathrm {Sp}(d/4)\times \mathrm {SU}(2), \mathrm {Spin}(7)$ and $\mathrm {Spin}(9)$ contain E as well as the group $\mathrm {SU}(d/2)$, if d is multiple of four $d=4k$ (Ref. [29], page 18). The only d-even groups that do not contain total inversion are $\mathrm {SU}(d/2)$ for $d=4k+2$, where $k=1,2,3,\dots$

Appendix 3: Kronecker Product of Irreducible Representations

Here we provide the proof of Lemma 1:

Lemma 1

CG series of the product $\Delta ^{(\mu )}\otimes \Delta ^{(\nu )}$, where $\Delta ^{(\mu )},\Delta ^{(\nu )}$ are real and irreducible, contains the trivial representation if and only if $\mu =\nu$ and then the trivial representation appears once, only.

Proof

Note that for a real, orthogonal representation D(g) we have $D(g^{-1})=D^{\mathrm {T}}(g)$, hence $\chi (g^{-1})=\mathrm {Tr}D(g^{-1})=\mathrm {Tr}D^{\mathrm {T}}(g)=\chi (g)$. We set $\mu =1$ with $\Delta ^{(1)}(g)=1$ (trivial representation) and $D(g)=\Delta ^{(\mu )}(g)\otimes \Delta ^{(\nu )}(g)$. We have the characters $\chi ^{(1)}(g)=1$ and $\chi (g)=\chi ^{(\mu )}(g)\chi ^{(\nu )}(g)$. The frequency is computed using Eq. (31)

$$\begin{aligned} a_{1}= & {} (\chi ^{(1)},\chi )\end{aligned}$$

(65)

$$\begin{aligned}= & {} \frac{1}{|\mathcal {G}|}\sum _{g\in \mathcal {G}}\chi ^{(1)}(g^{-1})\chi ^{(\mu )}(g)\chi ^{(\nu )}(g)\end{aligned}$$

(66)

$$\begin{aligned}= & {} \frac{1}{|\mathcal {G}|}\sum _{g\in \mathcal {G}}\chi ^{(\mu )}(g)\chi ^{(\nu )}(g)\end{aligned}$$

(67)

$$\begin{aligned}= & {} \frac{1}{|\mathcal {G}|}\sum _{g\in \mathcal {G}}\chi ^{(\mu )}(g^{-1})\chi ^{(\nu )}(g)\end{aligned}$$

(68)

$$\begin{aligned}= & {} (\chi ^{(\mu )},\chi ^{(\nu )})\end{aligned}$$

(69)

$$\begin{aligned}= & {} \delta _{\mu \nu }. \end{aligned}$$

(70)

QED

Appendix 4: Irreducible Decomposition of the Two-Fold Tensor Representation of $\mathrm {SO}(d)$

Here we show that the decomposition (32)

$$\begin{aligned} \Delta ^{d}\otimes \Delta ^{d}=\Delta ^{1}\oplus \Delta ^{S}\oplus \Delta ^{AS}, \end{aligned}$$

(71)

is irreducible unless $d=4$.

Let the representation $D(\mathcal {G})$ of $\mathcal {G}$ acts on a vector space $\mathcal {V}$. By definition, $D(\mathcal {G})$ is irreducible on $\mathcal {V}$ if $\mathrm {span}\{D(g)\mathbf {x}~|~\forall g\in \mathcal {G}\}=\mathcal {V}$ for every non-zero vector $\mathbf {x}\in \mathcal {V}$.

Firstly, let us analyze the symmetric subspace of all $d\times d$ symmetric, traceless matrices

$$\begin{aligned} \mathcal {V}_S=\{H~|~H^{\mathrm {T}}=H~\wedge ~\mathrm {Tr}H=0\}. \end{aligned}$$

(72)

This is an invariant subspace under the action of $\mathrm {SO}(d)$, because $(\textit{RHR}^{\mathrm {T}})^{\mathrm {T}}=\textit{RHR}^{\mathrm {T}}$ for every $R\in \mathrm {SO}(d)$ and $H\in \mathcal {V}_S$. Our goal is to show that the action of $\mathrm {SO}(d)$ is irreducible on $\mathcal {V}_S$. Therefore, we have to prove that the set

$$\begin{aligned} \mathcal {W}(H)=\mathrm {span}\{\textit{RHR}^{\mathrm {T}}~|~R\in \mathrm {SO}(d)\}=\mathcal {V}_S, \end{aligned}$$

(73)

for every non-zero $H\in \mathcal {V}_S$. Let us write H in diagonal form $H=\sum _{i=1}^dh_i|i\rangle \langle i|$, where $H=\sum _{i=1}^dh_i=0$. Since $\mathrm {Tr}H=0$, the largest and lowest eigenvalue satisfy $h_{\max }>0$ and $h_{\min }<0$. For convenience we set $h_{\max }=h_1$ and $h_{\min }=h_2$. Consider the orthogonal matrix $F_{12}\in \mathrm {SO}(d)$ swapping the basis vectors $|1\rangle$ and $|2\rangle$ (swap-rotation in 12-subspace):

$$\begin{aligned} F_{12}=\mathrm {diag}\left[ \left( \begin{array}{cc} 0 &{} -1 \\ 1 &{} 0 \\ \end{array} \right) ,1,1,1,\dots \right] . \end{aligned}$$

(74)

We have $H'=\frac{1}{h_1-h_2}(H-F_{12}\textit{HF}_{12}^{\mathrm {T}})=|1\rangle \langle 1|-|2\rangle \langle 2|$, where $h_1-h_2>0$. If we further rotate in 12-subspace for $45^\circ$ we obtain

$$\begin{aligned} R_{45^\circ }H'R_{45^\circ }^{\mathrm {T}}=|1\rangle \langle 2|+|2\rangle \langle 1|=E_{12}. \end{aligned}$$

(75)

The matrix $E_{12}$ is the element of a standard basis in $\mathcal {V}_S$. Other basis elements $E_{\textit{ij}}$ can be obtained from $E_{12}$ by suitable rotations. Therefore we have completed the space $\mathcal {V}_S$ starting from an arbitrary element H, hence $\mathcal {W}_S(H)=\mathcal {V}_S$.

In the case of antisymmetric subspace we define

$$\begin{aligned} \mathcal {V}_{\textit{AS}}=\mathrm {span}\{H~|~H^{\mathrm {T}}=-H\}. \end{aligned}$$

(76)

Our goal is to show $\mathcal {W}(A)=\mathcal {V}_{\textit{AS}}$ for arbitrary $A\in \mathcal {V}_{\textit{AS}}$. Let $A_{\textit{ij}}=|i\rangle \langle j|-|j\rangle \langle i|$, for $j>i$ be the standard basis in $\mathcal {V}_{\textit{AS}}$. It is sufficient to show that $A_{12}\in \mathcal {W}(A)$, and the other basis elements can be obtained from $A_{12}$ by suitable rotations. For an arbitrary antisymmetric matrix $A\in \mathcal {V}_{\textit{AS}}$ we can find the canonical form by applying suitable rotation $T\in \mathrm {SO}(d)$:

$$\begin{aligned} \nonumber A'=\textit{TAT}^{\mathrm {T}}= & {} \mathrm {diag}\left[ \left( \begin{array}{cc} 0 &{} -a_1 \\ a_1 &{} 0 \\ \end{array} \right) ,\left( \begin{array}{cc} 0 &{} -a_2 \\ a_2 &{} 0 \\ \end{array} \right) ,\dots ,0,0,\dots \right] \\= & {} a_1A_{12}+a_2A_{34}+\dots . \end{aligned}$$

(77)

If only $a_1\ne 0$, than $A=a_1A_{12}$ and we can generate the full basis $\{A_{\textit{ij}}\}$ in $\mathcal {V}_{AS}$ by applying suitable rotations. Otherwise, we assume that at least two elements $a_i$ are non-zero, and for convenience we set $a_{1}\ne 0$ and $a_2\ne 0$. Let $R_{\textit{ij}}$ be the rotation that flips ith and jth coordinate only, i.e. $R_{\textit{ij}}|k\rangle =s|k\rangle$, where $s=-1$ if $k=i$ or $k=j$, otherwise $s=1$. We get the following

$$\begin{aligned} A''=A'-R_{13}A'R_{13}^{\mathrm {T}}=2a_1A_{12}+2a_2A_{34}. \end{aligned}$$

(78)

Now if $d>4$ we further apply $R_{15}$ to $A''$ and obtain the following

$$\begin{aligned} A''-R_{15}A''R_{15}^{\mathrm {T}}=4a_1A_{12}. \end{aligned}$$

(79)

From here we can generate the full basis $A_{\textit{ij}}$, hence $\mathcal {W}(H)=\mathcal {V}_{\textit{AS}}$. If $d=4$ the construction above is no longer possible ($R_{15}$ does not exist). In this case the antisymmetric space is reduced to two three-dimensional irreducible subspaces as follows

$$\begin{aligned} \Delta ^{4}\otimes \Delta ^{4}=\Delta ^{1}\oplus \Delta ^{9}\oplus \Delta ^{3}_{+}\oplus \Delta ^{3}_{-}. \end{aligned}$$

(80)

We leave the proof to the curious reader.

Appendix 5: $d=3$ Solution

We begin with analyzing the fourth tensor power of $\Delta ^{d}$ representation of $\mathrm {SO}(d)$ group, as defined in the main text. We have

$$\begin{aligned} \Delta ^{d}\otimes \Delta ^{d}\otimes \Delta ^{d}\otimes \Delta ^{d}= & {} (\Delta ^{d}\otimes \Delta ^{d})\otimes (\Delta ^{d}\otimes \Delta ^{d})\\\nonumber= & {} \left( \Delta ^{1}\oplus \Delta ^{\textit{AS}}\oplus \Delta ^{S}\right) \otimes \left( \Delta ^{1}\oplus \Delta ^{\textit{AS}}\oplus \Delta ^{S}\right) . \end{aligned}$$

(81)

Since $S\ne \textit{AS}$ for $d>1$ and $d\ne 4$ (see Appendix 4), according to Lemma 1 the only contributing terms to the trivial representation are $\Delta ^{1}\otimes \Delta ^{1}$, $\Delta ^{\textit{AS}}\otimes \Delta ^{\textit{AS}}$ and $\Delta ^{S}\otimes \Delta ^{S}$, each of which appears once. Therefore, the tensor $K_{\textit{ijk}l}$ that is invariant under $\mathrm {SO}(d)$ belongs to the three dimensional IR subspace. We can form a basis in it by combining Kronecker delta tensors $\delta _{\textit{ij}}$. There are three different ways to combine them into a four-fold tensor, therefore:

$$\begin{aligned} K_{\textit{ijk}l}=\alpha \delta _{\textit{ij}}\delta _{kl}+\beta \delta _{ik}\delta _{jl}+\gamma \delta _{il}\delta _{jk}. \end{aligned}$$

(82)

From the analysis given in the main text, only $d=3$ case exhibits non-trivial invariant dynamics. The most general dynamical law for the global state $\psi =(\mathbf {x},\mathbf {y},T,\Lambda )$ is given by:

$$\begin{aligned} \frac{\textit{dx}_i}{\textit{dt}}= & {} a\epsilon _{\textit{ijk}}T_{\textit{jk}}+L^{(1)}_{\textit{in}}\lambda _n,\end{aligned}$$

(83)

$$\begin{aligned} \frac{\textit{dy}_i}{\textit{dt}}= & {} b\epsilon _{\textit{ijk}}T_{jk}+L^{(2)}_{\textit{in}}\lambda _n,\end{aligned}$$

(84)

$$\begin{aligned} \frac{dT_{\textit{ij}}}{dt}= & {} -a\epsilon _{\textit{ijk}}x_k-b\epsilon _{\textit{ijk}}y_k+L^{(12)}_{\textit{ijn}}\lambda _n+K_{\textit{ijk}l}T_{kl},\end{aligned}$$

(85)

$$\begin{aligned} \frac{d\lambda _n}{\textit{dt}}= & {} Q_{\textit{nm}}\lambda _m-L^{(1)}_{\textit{in}}x_{i}-L^{(2)}_{\textit{in}}y_{i}-L^{(12)}_{\textit{ijn}}T_{\textit{ij}}. \end{aligned}$$

(86)

Note that the reversibility requires $K_{\textit{ijk}l}=-K_{\textit{klij}}$. If we apply this constraint to the Eq. (82), we obtain $K=0$.

Next we will find the consistent values for the constants $a,b,L^{(1)}_{\textit{in}},L^{(2)}_{\textit{in}},L^{(12)}_{\textit{ijn}}$ such that the solutions to the dynamical Eqs. (83)–(86) above always lead to non-negative probabilities in Eq. (13). We look at the simplest case where all the couplings to global parameters are zero $L^{(1)}_{\textit{in}}=L^{(2)}_{\textit{in}}=L^{(12)}_{\textit{ijn}}=0$. If our initial state is a product state, than the global parameters remain zero during the evolution and we can safely neglect them from the analysis. In other words, the solution to the dynamical equations admits local tomography ($\Lambda =0$) and it can be found by solving the following set of equations:

$$\begin{aligned} \frac{\textit{dx}_i}{\textit{dt}}= & {} a\epsilon _{\textit{ijk}}T_{\textit{jk}},\end{aligned}$$

(87)

$$\begin{aligned} \frac{\textit{dy}_i}{\textit{dt}}= & {} b\epsilon _{\textit{ijk}}T_{\textit{jk}},\end{aligned}$$

(88)

$$\begin{aligned} \frac{\textit{dT}_{\textit{ij}}}{\textit{dt}}= & {} -a\epsilon _{\textit{ijk}}x_k-b\epsilon _{\textit{ijk}}y_k. \end{aligned}$$

(89)

Let us find the solution for the initial conditions $\vec {\psi }^{\pm }(0)=\{\mathbf {e}_3,\pm \mathbf {e}_3,\pm \mathbf {e}_3\mathbf {e}_3^{\mathrm {T}}\}$, where $\mathbf {e}_3=(0,0,1)^{\mathrm {T}}$. The only components that evolve in time are $x_3(t)$, $y_3(t)$ and $T_{12}(t)=-T_{12}(t)$, hence the solution has the form:

$$\begin{aligned} \psi ^{\pm }(t)=\left\{ \left( \begin{array}{c} 0 \\ 0 \\ x^{\pm }(t) \\ \end{array} \right) ,\left( \begin{array}{c} 0 \\ 0 \\ y^{\pm }(t) \\ \end{array} \right) ,\left( \begin{array}{ccc} 0 &{} \tau (t) &{} 0 \\ -\tau (t) &{} 0 &{} 0 \\ 0 &{} 0 &{} \pm 1 \\ \end{array} \right) \right\} , \end{aligned}$$

(90)

where $x^{\pm }(t),y^{\pm }(t)$ and $\tau (t)$ are the solutions to:

$$\begin{aligned} \frac{\textit{dx}^{\pm }}{\textit{dt}}= & {} 2a\tau ,\end{aligned}$$

(91)

$$\begin{aligned} \frac{\textit{dy}^{\pm }}{\textit{dt}}= & {} 2b\tau ,\end{aligned}$$

(92)

$$\begin{aligned} \frac{d\tau }{\textit{dt}}= & {} -a x^{\pm }-b y^{\pm }. \end{aligned}$$

(93)

Note that the state $\vec {\psi }^{\pm }(t)$ has to be physical state, that is, probability of Eq. (13) is non-negative $P_{12}(\vec {\psi }|~\mathbf {a},\mathbf {b})\ge 0$ for arbitrary choice of local measurements $\mathbf {a}$ and $\mathbf {b}$. If we set $\mathbf {a}=\mathbf {e}_3$ and $\mathbf {b}=-\mathbf {e}_3$, the positivity condition reads $\frac{1}{4}(x^{\pm }(t)-y^{\pm }(t))\ge 0$. Similarly for $\mathbf {a}=-\mathbf {e}_3$ and $\mathbf {b}=\mathbf {e}_3$ we have $\frac{1}{4}(-x^{\pm }(t)+y^{\pm }(t))\ge 0$. This is possible only if $x^{\pm }(t)=y^{\pm }(t)$.

In order to eliminate $\tau (t)$ from the dynamical equations we find the second derivatives in time of $x^{\pm }$ and $y^{\pm }$. We obtain:

$$\begin{aligned} \frac{d^2x^{\pm }}{\textit{dt}^2}= & {} -2a^2x^{\pm }-2\textit{ab}y^{\pm },\end{aligned}$$

(94)

$$\begin{aligned} \frac{d^2y^{\pm }}{\textit{dt}^2}= & {} -2\textit{ab}x_i-2b^2y^{\pm }. \end{aligned}$$

(95)

This set of equation leads to the symmetric solution $x^{\pm }(t)=y^{\pm }(t)$ only if $a^2=b^2$ or equivalently $b=\pm a$. Note that $a=-b$ case brings new symmetry to the set of dynamical equations, the invariance under particle swap. If one requires such a symmetry, the case $a=b$ can be safely eliminated. However, we will use another argument that has been used in the work of Ref.  [26]. We distinguish two cases, and label different solution as $\vec {\psi }_{\mathrm {MQM}}^{\pm }(t)$ and $\vec {\psi }_{\mathrm {QM}}^{\pm }(t)$, for $a=b$ and $a=-b$ respectively. The label QM and MQM stands for quantum mechanics and mirror quantum mechanics and the meaning of notation we explain shortly.

It is straightforward to evaluate the solution of dynamical equations:

$$\begin{aligned}&\psi _{\mathrm {MQM}}^{+}(t)=\left\{ \left( \begin{array}{c} 0 \\ 0 \\ \cos 2\textit{at} \\ \end{array} \right) ,\left( \begin{array}{c} 0 \\ 0 \\ \cos 2\textit{at} \\ \end{array} \right) ,\left( \begin{array}{ccc} 0 &{} -\sin 2\textit{at} &{} 0 \\ \sin 2\textit{at} &{} 0 &{} 0 \\ 0 &{} 0 &{} 1 \\ \end{array} \right) \right\} ,~~~~\nonumber \\&\psi _{\mathrm {MQM}}^{-}(t)=\left\{ \left( \begin{array}{c} 0 \\ 0 \\ 1 \\ \end{array} \right) ,\left( \begin{array}{c} 0 \\ 0 \\ -1 \\ \end{array} \right) ,\left( \begin{array}{ccc} 0 &{} 0 &{} 0 \\ 0 &{} 0 &{} 0 \\ 0 &{} 0 &{} -1 \\ \end{array} \right) \right\} ,\\&\psi _{\mathrm {QM}}^{-}(t)=\left\{ \left( \begin{array}{c} 0 \\ 0 \\ \cos 2\textit{at} \\ \end{array} \right) ,\left( \begin{array}{c} 0 \\ 0 \\ -\cos 2\textit{at} \\ \end{array} \right) ,\left( \begin{array}{ccc} 0 &{} \sin 2\textit{at} &{} 0 \\ -\sin 2\textit{at} &{} 0 &{} 0 \\ 0 &{} 0 &{} -1 \\ \end{array} \right) \right\} ,~~~\nonumber \\&\psi _{\mathrm {QM}}^{+}(t)=\left\{ \left( \begin{array}{c} 0 \\ 0 \\ 1 \\ \end{array} \right) ,\left( \begin{array}{c} 0 \\ 0 \\ -1 \\ \end{array} \right) ,\left( \begin{array}{ccc} 0 &{} 0 &{} 0 \\ 0 &{} 0 &{} 0 \\ 0 &{} 0 &{} -1 \\ \end{array} \right) \right\} . \end{aligned}$$

(96)

Our goal is to show that $\psi _{\mathrm {QM}}$ and the associate dynamics corresponds to quantum mechanics for two qubits, whereas $\psi _{\mathrm {MQM}}$ belongs to so called mirror quantum mechanics [26]. The later case has the set of states obtained by partial transpose of two-qubit states. We introduce the matrix representation of $\vec {\psi }=(\mathbf {x},\mathbf {y},T)$:

$$\begin{aligned} \rho (\vec {\psi })=\frac{1}{4}(\mathbb {1}\otimes \mathbb {1}+x_i\sigma _i\otimes \mathbb {1}+y_i\mathbb {1}\otimes \sigma _i+T_{\textit{ij}}\sigma _i\otimes \sigma _j), \end{aligned}$$

(97)

where $\sigma _i,~i=1,2,3$ are the Pauli matrices. Straightforward calculation shows that $\rho (\psi _{\mathrm {QM}}^{-}(t))=|\psi (t)\rangle \langle \psi (t)|$ is a density matrix, furthermore, it is a pure quantum state, where $|\psi (t)\rangle =\cos \textit{at}|0\rangle |1\rangle +i\sin \textit{at}|1\rangle |0\rangle$. Similarly, one can show that the matrix representation of mirror state $\psi _{\mathrm {MQM}}^{+}(t)$ is a non-quantum state (unless $\psi _{\mathrm {MQM}}^{+}(t)$ is product state) that can be obtained from $\psi _{\mathrm {QM}}^{-}(t)$ by applying total inversion $\mathbf {y}\mapsto -\mathbf {y}$ on the second spin. Note, that is a non-quantum operation. Mirror quantum mechanics is shown to be mathematically inconsistent theory for the tripartite case [26]. Therefore we will adopt only quantum solution.

The set of dynamical Eq. (87) has the corresponding matrix form:

$$\begin{aligned} \frac{d\rho (\vec {\psi })}{\textit{dt}}=i[H_{12},\rho (\vec {\psi })], \end{aligned}$$

(98)

where $H_{12}$ is the Heisenberg spin-spin interaction $H_{12}=\frac{a}{2}\vec {\sigma _1}\vec {\sigma _2}=\frac{a}{2}\sum _{i=1}^{3}\sigma _i\otimes \sigma _i$.