Einstein synchronisation by quantum teleportation

In the seminal paper on the special relativity published in 1905 [1], Einstein considered a thought experiment for two spatially separated clocks to synchronize by exchanging light signals back and forth between them. The first light signal starts from one of the observers Alice at the time t_A0 as her clock indicates, reaches the other observer Bob at the time t_B1 measured by his clock and then the information of the numerical value t_B1 is sent back to Alice by the light signal, who receives the information of t_B1 at the time t_A2 by her clock. By this communication Alice obtains all of the necessary data ${t}_{A0},{t}_{B1}$ and t_A2 to check the synchronization criterion as shown in figure 1.

$$\begin{eqnarray}&&\displaystyle \frac{{t}_{A0}+{t}_{A2}}{2}={t}_{B1}.\end{eqnarray} \tag{ 1 }$$

Einstein then demanded that the clock synchronization above should hold also in an arbitrary inertial frame with a relative velocity from the rest frame assuming the frame independence of the light velocity and arrived at the Lorentz transformation between the coordinates of the inertial and the rest frames. He chose the criterion (1) on the basis of the equality of the light velocity in both ways.

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Nowadays, the Einstein clock synchronization is also practically important to implement the GPS system.

We notice that, although Einstein did not explicitly mention, Bob has to send back the information of t_B1 in some way to convince Alice that her clock is synchronized to Bob's by checking the criterion (1). Some people  (e.g., [2, 3])use the terminology 'Einstein synchronization' to broadly mean the date adjustment of the two spatially separated clocks by exchanging signals even without the criterion (1). In this paper the meaning of Einstein synchronization is closer to the original one in the sense that the criterion (1) is required as the validation. We take the criterion (1) as an operational one leaving aside its physical meaning. We stress that checking that the two classical clocks are synchronized is conceptually different from creating the two synchronized clocks, or quantum synchronization if they were quantum clocks. For the latter we will quickly review the recent development.

Jozsa et al [2] proposed a quantum mechanical synchronization using a prior entangled state. Let us briefly review the original protocol of the QCS to elucidates its drawbacks which are potentially harmful also to our own protocol of the validation of the clock synchronization. For elaborations and other quantum protocols, see e.g., [4, 5]. Suppose Alice and Bob have identical two-level quantum systems,i.e., a qubit consisting of the basis $| 0\rangle$ and $| 1\rangle$ with the energy levels 0 and ℏω. To make the model definite we choose a Hamiltonian $H={\hslash }\omega (1-{\sigma }_{z})/2$ and ${\sigma }_{z}| 0\rangle =| 0\rangle ,{\sigma }_{z}| 1\rangle =-| 1\rangle$. Here σ_x, σ_y and σ_z are the standard Pauli matrices.

Initially Alice and Bob share an EPR state:

$$\begin{eqnarray}&&| {EPR}\rangle =\displaystyle \frac{1}{\sqrt{2}}[| 1{\rangle }_{A}| 0{\rangle }_{B}-| 0{\rangle }_{A}| 1{\rangle }_{B}],\end{eqnarray} \tag{ 2 }$$

which remains the same up to an overal phase factor in the time evolution. This choice is important because the protocol is time-blind i.e., insensitive to the time when it starts. The singlet state $| {EPR}\rangle$ can also be written as

$$\begin{eqnarray}&&| {EPR}\rangle =\displaystyle \frac{1}{\sqrt{2}}[| -{\rangle }_{A}| +{\rangle }_{B}-| +{\rangle }_{A}| -{\rangle }_{B}],\end{eqnarray} \tag{ 3 }$$

where $| \pm \rangle =\tfrac{1}{\sqrt{2}}[| 0\rangle \pm | 1\rangle ]$ is the eigenstate of the observable σ_x with the eigenvalue ±1. Suppose Alice measures her σ_x to obtain $| \pm {\rangle }_{A}$. Then the state of Bob jumps to $| \mp {\rangle }_{B}$, correspondingly¹ . The state $| \mp {\rangle }_{B}=\tfrac{1}{\sqrt{2}}[| 0{\rangle }_{B}\mp | 1{\rangle }_{B}]$ evolves in time duration t as,

$$\begin{eqnarray}&&\displaystyle \frac{1}{\sqrt{2}}[| 0{\rangle }_{B}\mp {e}^{-i\omega t}| 1{\rangle }_{B}].\end{eqnarray} \tag{ 4 }$$

Alice sends a classical message which asks Bob to do the Hadamard transformation H_B. After that Bob will have a clock state:

$$\begin{eqnarray}&&\displaystyle \frac{1}{\sqrt{2}}[| +{\rangle }_{B}\mp {e}^{-i\omega {t}_{1}}| -{\rangle }_{B}]=\displaystyle \frac{1\mp {e}^{-i\omega {t}_{1}}}{2}| 0{\rangle }_{B}+\displaystyle \frac{1\pm {e}^{-i\omega {t}_{1}}}{2}| 1{\rangle }_{B}.\end{eqnarray} \tag{ 5 }$$

when Bob's clock indicates that the time is t = t₁.

Alice correspondingly has her clock state:

$$\begin{eqnarray}&&\displaystyle \frac{1\pm {e}^{-i\omega {t}_{1}}}{2}| 0{\rangle }_{A}+\displaystyle \frac{1\mp {e}^{-i\omega {t}_{1}}}{2}| 1{\rangle }_{A}\end{eqnarray} \tag{ 6 }$$

at t₁.Alice's state is identical to Bob's state if she flips her bit.

To implement the creation of the clock states of Alice and Bob, Alice sends the classical information ± of the outcome when she has performed the first measurement of σ_x to Bob, then Bob can see the time indicated by his own clock. Indeed the QCS proposal is not the check of the synchronization of the two clocks but rather an automatic synchronization of the two quantum clocks.

However, unfortunately there are at least two defects in the original QCS protocol.

(i) As remarked in their own paper [2] and briefly pointed out in the above, the spatial axis x at Bob cannot be definitely arranged relative to the x-axis of Alice so that the Hadamard transformation of Bob in the protocol is ambiguous and so is the resultant state by Alice's observation² .

(ii) The QCS has a drawback similar to Eddinton's synchronization [6]. Namely, the shared entangled state is contaminated by an uncontrollable phase δ in general,

$$\begin{eqnarray}&&\to \displaystyle \frac{1}{\sqrt{2}}[| 1{\rangle }_{A}| 0{\rangle }_{B}-{e}^{i\delta }| 0{\rangle }_{A}| 1{\rangle }_{B}],\end{eqnarray} \tag{ 7 }$$

because the environments of spatially separated Alice and Bob are different in general and therefore have different interactions [3]. This extra phase factor invalidates its isotropy, i.e.,the equation (3) and therefore the break-down of whole scheme. Recently, Ilo-Okeke et al [5] proposed a modification of QCS to resolve the frame ambiguity problem via entanglement purification but not the phase contamination problem.

The motivation of the present work is to propose a protocol for the validation of synchronization of classical clocks in the same sense of Einstein's original paper but a transmission of the time information from Bob to Alice's is done by using quantum teleportation, which does not have the two drawbacks of QCS mentioned above in spite of the use of the EPR pair state. The key to the resolution is the Bell measurements.

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Here we present a precise protocol of clock synchronization by the quantum teleportation.

• (I)  
  Prepare the initial state

  $$\begin{eqnarray}&&| \psi (0)\rangle =| {EPR}\rangle | 0{\rangle }_{B}^{{clock}},\end{eqnarray} \tag{ 8 }$$

  where the $| {EPR}\rangle$ is the state of EPR defined as before in equation (2).
• (II)  
  At an arbitrarily chosen and recorded time t_B0 Bob makes the Hadamard transformation H_B to the clock state $| 0{\rangle }_{B}^{{clock}}$ in equation (8). The state changes to

  $$\begin{eqnarray}&&| \psi ^{\prime} (0)\rangle ={H}_{B}| \psi (0)\rangle =| {EPR}\rangle | +{\rangle }_{B}^{{clock}},\end{eqnarray} \tag{ 9 }$$

  and the state $| +{\rangle }_{B}^{{clock}}=\tfrac{1}{\sqrt{2}}[| 0{\rangle }_{B}^{{clock}}+| 1{\rangle }_{B}^{{clock}}]$ starts its time evolution, $| +{\rangle }_{B}^{{clock}}\mathop{\to }\limits^{t}\tfrac{1}{\sqrt{2}}[| 0{\rangle }_{B}^{{clock}}\,+{e}^{-i\omega (t-{t}_{B0})}| 1{\rangle }_{B}^{{clock}}]$ which plays a role of clock state.
• (III)  
  At an arbitrary time t_A0, Alice sends a classical signal to ask Bob to do another Hadamard transformation H_B, which Bob receives at time t_B1.After receiving it,Bob immediately makes the controlled unitary transformation by ${e}^{-i\omega {t}_{B0}}$ to cancel the t_B0 dependence of the state. Bob can do this transformation because he knows the value t_B0 from the classical record of his clock. The total state becomes at $t={t}_{B1}$

  $$\begin{eqnarray*}&&| \psi ^{\prime\prime} ({t}_{B1})\rangle =\displaystyle \frac{1}{\sqrt{2}}[| 1{\rangle }_{A}| 0{\rangle }_{B}-| 0{\rangle }_{A}| 1{\rangle }_{B}]\displaystyle \frac{1}{\sqrt{2}}[| 0{\rangle }_{B}^{{clock}}+{e}^{-i{\theta }_{1}}| 1{\rangle }_{B}^{{clock}}],\end{eqnarray*}$$

  where ${\theta }_{1}=\omega {t}_{B1}$. At this stage we notice that the resultant state $| \psi ^{\prime\prime} ({t}_{B1})\rangle$ can be rewritten as (7)

  $$\begin{eqnarray}\begin{array}{rcl}| \psi ^{\prime\prime} ({t}_{B1})\rangle & = & \displaystyle \frac{1}{2\sqrt{2}}(| 1{\rangle }_{A}+{e}^{-i{\theta }_{1}}| 0{\rangle }_{A})| {{\rm{\Psi }}}_{1}\rangle +(| 1{\rangle }_{A}-{e}^{-i{\theta }_{1}}| 0{\rangle }_{A})| {{\rm{\Psi }}}_{2}\rangle \\ & & +(-| 0{\rangle }_{A}-{e}^{-i{\theta }_{1}}| 1{\rangle }_{A})| {{\rm{\Psi }}}_{3}\rangle +(-| 0{\rangle }_{A}+{e}^{-i{\theta }_{1}}| 1{\rangle }_{A})| {{\rm{\Psi }}}_{4}\rangle ,\end{array}\end{eqnarray} \tag{ 10 }$$

  where ${{\rm{\Psi }}}_{1},{{\rm{\Psi }}}_{2},{{\rm{\Psi }}}_{3}$ and Ψ₄ are the Bell states defined by

  $$\begin{eqnarray}&&| {{\rm{\Psi }}}_{1}\rangle =\displaystyle \frac{1}{\sqrt{2}}(| 0{\rangle }_{B}| 0{\rangle }_{B}^{{clock}}-| 1{\rangle }_{B}| 1{\rangle }_{B}^{{clock}}),\end{eqnarray} \tag{ 11 }$$

  $$\begin{eqnarray}&&| {{\rm{\Psi }}}_{2}\rangle =\displaystyle \frac{1}{\sqrt{2}}(| 0{\rangle }_{B}| 0{\rangle }_{B}^{{clock}}+| 1{\rangle }_{B}| 1{\rangle }_{B}^{{clock}}),\end{eqnarray} \tag{ 12 }$$

  $$\begin{eqnarray}&&| {{\rm{\Psi }}}_{3}\rangle =\displaystyle \frac{1}{\sqrt{2}}(| 1{\rangle }_{B}| 0{\rangle }_{B}^{{clock}}-| 0{\rangle }_{B}| 1{\rangle }_{B}^{{clock}}),\end{eqnarray} \tag{ 13 }$$

  $$\begin{eqnarray}&&| {{\rm{\Psi }}}_{4}\rangle =\displaystyle \frac{1}{\sqrt{2}}(| 0{\rangle }_{B}| 1{\rangle }_{B}^{{clock}}+| 1{\rangle }_{B}| 0{\rangle }_{B}^{{clock}}).\end{eqnarray} \tag{ 14 }$$
• (IV)  
  Bob performs the measurements of the Bell states. If he gets the state $| {{\rm{\Psi }}}_{1}\rangle$, the state of Alice at the time t_A2 would be $\tfrac{1}{\sqrt{2}}(| 1{\rangle }_{A}+{e}^{-i{\theta }_{1}}| 0{\rangle }_{A})$ and similar results for the cases $| {{\rm{\Psi }}}_{2}\rangle$ ,$| {{\rm{\Psi }}}_{3}\rangle$ and $| {{\rm{\Psi }}}_{4}\rangle$.
• (V)  
  Suppose that Alice gets the clock state

  $$\begin{eqnarray}&&\displaystyle \frac{1}{\sqrt{2}}[| 1{\rangle }_{A}+{e}^{-i{\theta }_{1}}| 0{\rangle }_{A}],\end{eqnarray} \tag{ 15 }$$

  teleported by Bob as well as the classical information that the result of the Bell measurement by Bob was $| {{\rm{\Psi }}}_{1}\rangle$. Then she waits for the tuned time duration $\tfrac{{t}_{A0}+{t}_{A2}}{2}$ knowing the dates t_A0 and t_A2 by her own clock. Then her clock state becomes

  $$\begin{eqnarray}&&| \psi {\rangle }_{A}=\displaystyle \frac{1}{\sqrt{2}}[{e}^{-i{\theta }_{2}}| 1{\rangle }_{A}+{e}^{-i{\theta }_{1}}| 0{\rangle }_{A}],\end{eqnarray} \tag{ 16 }$$

  where ${\theta }_{2}=\omega \tfrac{{t}_{A0}+{t}_{A2}}{2}$. Alice carries out the Hadamard transformation H_A

  $$\begin{eqnarray}&&| \psi ^{\prime} {\rangle }_{A}={H}_{A}| \psi {\rangle }_{A}=\displaystyle \frac{1}{\sqrt{2}}[{e}^{-i{\theta }_{2}}| -{\rangle }_{A}+{e}^{-i{\theta }_{1}}| +{\rangle }_{A}]={e}^{-i{\theta }_{1}}[\alpha | 0{\rangle }_{A}+\beta | 1{\rangle }_{A}],\end{eqnarray} \tag{ 17 }$$

  where

  $$\begin{eqnarray}&&\alpha =\displaystyle \frac{1+{e}^{-i({\theta }_{2}-{\theta }_{1})}}{2},\ \ \beta =\displaystyle \frac{1-{e}^{-i({\theta }_{2}-{\theta }_{1})}}{2}.\end{eqnarray} \tag{ 18 }$$

  Here we have tentatively assumed that the Hadamard transformation H_A of Alice is the same as the Hadamard transformation H_B of Bob for the simplicity of explanation. Later we will generalise it taking the frame ambiguity into account.

(VI) Finally Alice measures her qubit. The probability to obtain $| 0{\rangle }_{A}$ by the bit measurement under the condition that the Bell measurement resulted in $| {{\rm{\Psi }}}_{1}\rangle$ is

$$\begin{eqnarray}&&| \alpha {| }^{2}=\displaystyle \frac{1+\cos ({\theta }_{2}-{\theta }_{1})}{2},\end{eqnarray} \tag{ 19 }$$

which is equal to unity if the synchronization condition

$$\begin{eqnarray}&&{\theta }_{2}-{\theta }_{1}=\omega \left[\displaystyle \frac{{t}_{A0}+{t}_{A2}}{2}-{t}_{B1}\right]=0\end{eqnarray} \tag{ 20 }$$

is met. Namely, if Alice gets the state $| 0{\rangle }_{A}$ with a high probability she can convince herself that her clock is synchronized with Bob's. If the result of the Bell measurement was different from the $| {{\rm{\Psi }}}_{1}\rangle$, Alice can arrange different unitary transformation to check the synchronization criterion by the bit measurement $| 0{\rangle }_{A}$ and then check similarly as before.

The different environments of Alice and Bob induces the uncontrollable phase δ to the EPR state to give

$$\begin{eqnarray}&&| {EPR}\rangle \to | {ENT}\rangle =\displaystyle \frac{1}{\sqrt{2}}[| 1{\rangle }_{A}| 0{\rangle }_{B}-{e}^{i\delta }| 0{\rangle }_{A}| 1{\rangle }_{B}].\end{eqnarray} \tag{ 21 }$$

Correspondingly the total state becomes

$$\begin{eqnarray}\begin{array}{rcl}| \psi ^{\prime\prime} ({t}_{B1})\rangle & = & (| 1{\rangle }_{A}+{e}^{-i{\theta }_{1}+i\delta }| 0{\rangle }_{A})| {{\rm{\Psi }}}_{1}\rangle +(| 1{\rangle }_{A}-{e}^{-i{\theta }_{1}+i\delta }| 0{\rangle }_{A})| {{\rm{\Psi }}}_{2}\rangle \\ & & +(-{e}^{i\delta }| 0{\rangle }_{A}-{e}^{-i{\theta }_{1}}| 1{\rangle }_{A})| {{\rm{\Psi }}}_{3}\rangle +(-{e}^{i\delta }| 0{\rangle }_{A}+{e}^{-i{\theta }_{1}}| 1{\rangle }_{A})| {{\rm{\Psi }}}_{4}\rangle ,\end{array}\end{eqnarray} \tag{ 22 }$$

If the Bell measurement gives $| {{\rm{\Psi }}}_{1}\rangle$ state,the state of Alice would be $\tfrac{1}{\sqrt{2}}(| 1{\rangle }_{A}+{e}^{-i{\theta }_{1}+i\delta }| 0{\rangle }_{A})$.After waiting for the tuned time duration $\tfrac{{t}_{A1}+{t}_{A2}}{2}$ she performs the Hadamard transformation and then make the bit observation just as explained before in the case that $\delta =0$. The probability for Alice to obtain $| 0{\rangle }_{A}$ is now

$$\begin{eqnarray}&&{P}_{1}=| \alpha {| }^{2}=\displaystyle \frac{1+\cos ({\theta }_{2}-{\theta }_{1}+\delta )}{2},\end{eqnarray} \tag{ 23 }$$

instead of (19), where the δ is the unknown phase contamination.The case of the Bell state $| {{\rm{\Psi }}}_{2}\rangle$ gives the same probability so that the subsequent bit measurement does not add new information. However, for the case of $| {{\rm{\Psi }}}_{3}\rangle$ the probability for Alice to obtain $| 0{\rangle }_{A}$ is

$$\begin{eqnarray}&&{P}_{3}=\displaystyle \frac{1+\cos ({\theta }_{2}-{\theta }_{1}-\delta )}{2}.\end{eqnarray} \tag{ 24 }$$

Note the sign difference in front of δ from the case of P₁. For $| {{\rm{\Psi }}}_{4}\rangle$, the result P₄ is the same as P₃.

Eliminating the unwanted δ in Eqs.(23, 24), we obtain an expression for ${\theta }_{2}-{\theta }_{1}$ as

$$\begin{eqnarray}&&{\theta }_{2}-{\theta }_{1}=\displaystyle \frac{1}{2}[{{Cos}}^{-1}(2{P}_{1}-1)+{{Cos}}^{-1}(2{P}_{3}-1)].\end{eqnarray} \tag{ 25 }$$

The right hand side is experimentally accessible. If its value is close to unity, the synchronization is validated. Therefore, the phase contamination problem has been overcome by the combination of the more than two Bell measurements.

Let us turn to the frame adjustment problem. The Hadamard transformation H_A by Alice can in general be different from H_B of Bob by unadjustable rotation around z-axis as noted below equation (18). Let the unitary transformation be $U(\phi )={e}^{i\zeta {\sigma }_{z}/2}$ and the Hadamard transformation be $\tilde{{H}_{A}}$ explicitly given by

$$\begin{eqnarray}&&\tilde{{H}_{A}}=U(\zeta ){H}_{A}{U}^{\dagger }(\zeta ),\end{eqnarray} \tag{ 26 }$$

so that

$$\begin{eqnarray*}\begin{array}{rcl} & & \tilde{{H}_{A}}| {0}_{A}\rangle =\displaystyle \frac{1}{\sqrt{2}}[| 0{\rangle }_{A}+{e}^{-i\zeta }| 1{\rangle }_{A}]\\ & & \tilde{{H}_{A}}| 1{\rangle }_{A}=\displaystyle \frac{1}{\sqrt{2}}[{e}^{i\zeta }| 0{\rangle }_{A}-| 1{\rangle }_{A}].\end{array}\end{eqnarray*}$$

The teleported state $\tfrac{1}{\sqrt{2}}[(| 1{\rangle }_{A}+{e}^{-i{\theta }_{1}+i\delta }| 0{\rangle }_{A})]$ in (22) for the Bell result Ψ₁ is Hadamard transformed to

$$\begin{eqnarray}&&\tilde{{H}_{A}}\displaystyle \frac{1}{\sqrt{2}}\left[(| 1{\rangle }_{A}+{e}^{-i{\theta }_{1}+i\delta }| 0{\rangle }_{A})\right]=\left[\displaystyle \frac{1}{2}({e}^{i\zeta }+{e}^{-i{\theta }_{1}+i\delta })| 0{\rangle }_{A}+\displaystyle \frac{1}{2}(-1+{e}^{-i{\theta }_{1}-i\zeta +i\delta })| 1{\rangle }_{A}\right].\end{eqnarray} \tag{ 27 }$$

The waiting procedure explained around (16) replaces θ₁ by θ₁ − θ₂.

To sum up,the probability to get ${{\rm{\Psi }}}_{1}$ then $| 0{\rangle }_{A}$ is

$$\begin{eqnarray}&&{P}^{1}(0)=\displaystyle \frac{1}{2}[1+\cos ({\theta }_{2}-{\theta }_{1}+\delta -\zeta )].\end{eqnarray} \tag{ 28 }$$

Similarly the probability to get Ψ₃ then $| 0{\rangle }_{A}$ is

$$\begin{eqnarray}&&{P}^{3}(0)=\displaystyle \frac{1}{2}[1+\cos ({\theta }_{2}-{\theta }_{1}-\delta +\zeta )].\end{eqnarray} \tag{ 29 }$$

Note that the uncontrollable phases ζ and δ appear only in the combination of δ − ζ. Therefore, the formula (25) remains to work for the check of the Einstein synchronisation. This is not a coincidence, because the selective spin rotation of Alice only effectively induces the phase contamination δ.

We admit that the Bell measurement is idealized in the sense that it can be done instantaneously. More practically, it needs many repeated measurements, during which the uncontrollable phase δ and ζ might change. We have not estimated the inaccuracy of the synchronization due to this effect.

To summarize, we have presented a quantum protocol to transmit the time information by teleportation for the Einstein synchronization of two distant clocks, which is free from the frame adjustment and the phase contamination problems. However,there remain points to improve the QCS further. We add a comment on the Eddington synchronisation [6],an alternative of the Einstein synchronization. A quantum version of the Eddinton synchronization in which a clock state is physically exchanged has also been proposed [7].

The author is preparing a paper which shows that the present protocol of synchronisation is consistent with relativity.