Entanglement harvesting of three Unruh-DeWitt detectors

Abstract

We analyze a tripartite entanglement harvesting protocol with three Unruh-DeWitt detectors adiabatically interacting with a quantum scalar field in $(3+1)$-dimensional Minkowski spacetime. We consider linear, equilateral triangular, and scalene triangular configurations for the detectors, all of which remain static. We find that, under the same parameters, more entanglement can be extracted in the linear configuration than the equilateral one, consistent with single instantaneous switching results. No bipartite entanglement is required to harvest tripartite entanglement. Furthermore, we find that tripartite entanglement can be harvested even if one detector is at larger spacelike separations from the other two than in the corresponding bipartite case. We also find that for small detector separations bipartite correlations become larger than tripartite ones, leading to an apparent violation of the Coffman-Kundu-Wootters (CKW) inequality. We show that this is not a consequence of our perturbative expansion but that it instead occurs because the harvesting qubits are in a mixed state.

Appendices

Appendix A: The $\pi$-tangle for the scalene triangle configuration

Here we present the calculations for the $\pi$-tangle when the three detectors are in a scalene triangular configuration.

When the distance between each pair of detectors is arbitrary, the density matrix Eq. (6) becomes

$$\begin{aligned} {\rho }_{ABC} = \begin{pmatrix} 1-3P &{} 0 &{} 0 &{} 0 &{} X_{BC}^* &{} X_{AC}^* &{} X_{AB}^* &{} 0 \\ 0 &{} P &{} C_{BC}^* &{} C_{AC}^* &{} 0 &{} 0 &{} 0 &{} 0 \\ 0 &{} C_{BC} &{} P &{} C_{AB}^* &{} 0 &{} 0 &{} 0 &{} 0 \\ 0 &{} C_{AC} &{} C_{AB} &{} P &{} 0 &{} 0 &{} 0 &{} 0 \\ X_{BC} &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 \\ X_{AC} &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 \\ X_{AB} &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 \\ 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 \end{pmatrix}. \end{aligned}$$

(A1)

Following the same procedure for the equilateral triangular and linear configurations, we obtain

$$\begin{aligned} {\mathcal {N}}_{A(BC)}&= \max \left\{ 0,\frac{2}{\sqrt{3}}\sqrt{C_{BC}+|X_{AB}|^2+|X_{AC}|^2}\right. \nonumber \\&\quad \left. \cos \left[ \frac{\pi }{3}+\frac{1}{3}\arccos \left( \frac{3\sqrt{3}C_{BC}\left( X_{AB}X_{AC}^*+X_{AB}^*X_{AC}\right) }{2\big (C_{BC}^2+|X_{AB}|^2+|X_{AC}|^2\big )^{3/2}}\right) \right] -P\right\} \nonumber \\&\quad + \max \left\{ 0,\frac{2}{\sqrt{3}}\sqrt{C_{BC}+|X_{AB}|^2+|X_{AC}|^2}\right. \nonumber \\&\quad \left. \sin \left[ \frac{\pi }{6}+\frac{1}{3}\arccos \left( \frac{3\sqrt{3}C_{BC}\left( X_{AB}X_{AC}^*+X_{AB}^*X_{AC}\right) }{2\big (C_{BC}^2+|X_{AB}|^2+|X_{AC}|^2\big )^{3/2}}\right) \right] -P\right\} \end{aligned}$$

(A2)

$$\begin{aligned} \nonumber \\ {\mathcal {N}}_{B(AC)}&= \max \left\{ 0,\frac{2}{\sqrt{3}}\sqrt{C_{AC}+|X_{AB}|^2+|X_{BC}|^2}\right. \nonumber \\&\quad \left. \cos \left[ \frac{\pi }{3} +\frac{1}{3}\arccos \left( \frac{3\sqrt{3}C_{AC}\left( X_{AB}X_{BC}^*+X_{AB}^*X_{BC}\right) }{2\big (C_{AC}^2+|X_{AB}|^2+|X_{BC}|^2\big )^{3/2}}\right) \right] -P\right\} \nonumber \\&\quad + \max \left\{ 0,\frac{2}{\sqrt{3}}\sqrt{C_{AC}+|X_{AB}|^2+|X_{BC}|^2}\right. \nonumber \\&\quad \left. \sin \left[ \frac{\pi }{6}+\frac{1}{3}\arccos \left( \frac{3\sqrt{3}C_{AC}\left( X_{AB}X_{BC}^*+X_{AB}^*X_{BC}\right) }{2\big (C_{AC}^2+|X_{AB}|^2+|X_{BC}|^2\big )^{3/2}}\right) \right] -P\right\} \end{aligned}$$

(A3)

$$\begin{aligned} \nonumber \\ {\mathcal {N}}_{C(AB)}&= \max \left\{ 0,\frac{2}{\sqrt{3}}\sqrt{C_{AB}+|X_{AC}|^2+|X_{BC}|^2}\right. \nonumber \\&\quad \left. \cos \left[ \frac{\pi }{3}+\frac{1}{3}\arccos \left( \frac{3\sqrt{3}C_{AB}\left( X_{AC}X_{BC}^*+X_{AC}^*X_{BC}\right) }{2\big (C_{AB}^2+|X_{AC}|^2+|X_{BC}|^2\big )^{3/2}}\right) \right] -P\right\} \nonumber \\&\quad + \max \left\{ 0,\frac{2}{\sqrt{3}}\sqrt{C_{AB}+|X_{AC}|^2+|X_{BC}|^2}\right. \nonumber \\&\quad \left. \sin \left[ \frac{\pi }{6}+\frac{1}{3}\arccos \left( \frac{3\sqrt{3}C_{AB}\left( X_{AC}X_{BC}^*+X_{AC}^*X_{BC}\right) }{2\big (C_{AB}^2+|X_{AC}|^2+|X_{BC}|^2\big )^{3/2}}\right) \right] -P\right\} \end{aligned}$$

(A4)

and

$$\begin{aligned} {\mathcal {N}}_{A(B)}&= {\mathcal {N}}_{B(A)} = \max \big [0,|X_{AB}|-P\big ]\,, \end{aligned}$$

(A5)

$$\begin{aligned} {\mathcal {N}}_{A(C)}&= {\mathcal {N}}_{C(A)} = \max \big [0,|X_{AC}|-P\big ]\,, \end{aligned}$$

(A6)

$$\begin{aligned} {\mathcal {N}}_{B(C)}&= {\mathcal {N}}_{C(B)} = \max \big [0,|X_{BC}|-P\big ]\,. \end{aligned}$$

(A7)

The $\pi$-tangle is then easily computed using Eqs. (12) and (11).

Appendix B: A toy model

The introduction of the $\pi$-tangle [92] has generally been thought to satisfy the CKW inequality. Specficially “For any pure $2\otimes 2\otimes 2$ states ${|{\phi }\rangle }_{ABC}$, the entanglement quantified by the negativity between A and B, between A and C, and between A and the single object BC satisfies the following CKW-inequality-like monogamy inequality:

$$\begin{aligned} {\mathcal {N}}^2_{AB}+{\mathcal {N}}^2_{AC}\le {\mathcal {N}}^2_{A(BC)}, \end{aligned}$$

where ${\mathcal {N}}_{AB}$ and ${\mathcal {N}}_{AC}$ are the negativities of the mixed states”. Although the preceding inequality holds only for pure tripartite systems, the form of the preceding statement suggests that violations of the CKW inequality are perhaps unexpected and somewhat counter-intuitive. To illustrate that such violations are not a consequence of the perturbative expression given in (6), we present a simple density matrix for a tripartite system of qubits that can have negative $\pi$-tangle over a wide range of parameters.

Assume that we have a density matrix of the form

$$\begin{aligned} \rho _{ABC} = \begin{pmatrix} 1-3P-3E-\Sigma &{} 0 &{} 0 &{} 0 &{} X^* &{} X^* &{} X^* &{} 0 \\ 0 &{} P &{} C &{} C &{} 0 &{} 0 &{} 0 &{} 0 \\ 0 &{} C &{} P &{} C &{} 0 &{} 0 &{} 0 &{} 0\\ 0 &{} C &{} C &{} P &{} 0 &{} 0 &{} 0 &{} 0\\ X &{} 0 &{} 0 &{} 0 &{} E &{} 0 &{} 0 &{} 0\\ X &{} 0 &{} 0 &{} 0 &{} 0 &{} E &{} 0 &{} 0\\ X &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} E &{} 0\\ 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} \Sigma \end{pmatrix}\,, \end{aligned}$$

(B1)

where $P,C,E,\Sigma \in {\mathbb {R}}$, which is clearly trace 1 and Hermitian. In order for this matrix to be a valid density matrix, it must have positive eigenvalues, which puts the following constraints on the elements of $\rho _{ABC}$:

$$\begin{aligned}&E \ge 0\,, \end{aligned}$$

(B2a)

$$\begin{aligned}&\Sigma \ge 0\,, \end{aligned}$$

(B2b)

$$\begin{aligned}&P \ge C\,, \end{aligned}$$

(B2c)

$$\begin{aligned}&P \ge -2C\,, \end{aligned}$$

(B2d)

$$\begin{aligned}&(1-3P-2E-\Sigma ) \nonumber \\&\quad \pm \sqrt{1-8E-16E^2-6P+24EP+9P^2+12|X|^2-2\Sigma +8E\Sigma +6P\Sigma +\Sigma ^2} \ge 0\,. \nonumber \\ \end{aligned}$$

(B2e)

From constraints Eqs. (B2c) and (B2d), we get the additional constraint:

$$\begin{aligned} P \ge 0. \end{aligned}$$

(B3)

If we assume that

$$\begin{aligned}&\xi :=1-3P-2E-\Sigma \ge 0\,, \end{aligned}$$

(B4)

which will be true if $P,E,\Sigma \ll 1$, then constraint (B2e) will be satisfied provided:

$$\begin{aligned} (1-3P-2E-\Sigma )^2&\ge \big (1-8E-16E^2-6P+24EP+9P^2+12|X|^2\nonumber \\&\quad -2\Sigma +8E\Sigma +6P\Sigma +\Sigma ^2\big ) \nonumber \\&\implies E\xi \ge E^2 + 3|X|^2\,, \end{aligned}$$

(B5)

which is consistent with the assumption (B4) since $E\ge 0$ and $|X|\ge 0$.

The partial transpose of $\rho _{ABC}$ with respect to A is

$$\begin{aligned} \rho _{ABC}^{T_A} = \begin{pmatrix} 1-3P-3E-\Sigma &{} 0 &{} 0 &{} 0 &{} X^* &{} C &{} C &{} 0 \\ 0 &{} P &{} C &{} X^* &{} 0 &{} 0 &{} 0 &{} 0 \\ 0 &{} C &{} P &{} X^* &{} 0 &{} 0 &{} 0 &{} 0\\ 0 &{} X &{} X &{} P &{} 0 &{} 0 &{} 0 &{} 0\\ X &{} 0 &{} 0 &{} 0 &{} E &{} 0 &{} 0 &{} 0\\ C &{} 0 &{} 0 &{} 0 &{} 0 &{} E &{} 0 &{} 0\\ C &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} E &{} 0\\ 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} \Sigma \end{pmatrix}\,, \end{aligned}$$

(B6)

which only has two eigenvalues that can be negative while maintaining non-negative eigenvalues for $\rho _{ABC}$:

$$\begin{aligned}&\frac{1}{2}\left( 2P+C - \sqrt{C^2+8|X|^2}\right) \,, \end{aligned}$$

(B7)

$$\begin{aligned}&\frac{1}{2} \Bigg ( (1-3P-2E-\Sigma ) \nonumber \\&\quad - \sqrt{(1-3P-2E-\Sigma )^2+4\left( 2C^2-E+3E^2+EP+|X|^2+E\Sigma \right) } \Bigg )\,. \end{aligned}$$

(B8)

In particular eigenvalue (B7) will only be non-negative if

$$\begin{aligned} (2P+C)^2 \ge \big (C^2+8|X|^2\big ) \implies P(P+C) \ge 2|X|^2 \end{aligned}$$

(B9)

and eigenvalue (B8) only be non-negative if

$$\begin{aligned} (1-3P-2E-\Sigma )^2&\ge \big [(1-3P-2E-\Sigma )^2\nonumber \\&\quad +4\left( 2C^2-E+3E^2+EP+|X|^2+E\Sigma \right) \big ] \nonumber \\&\implies E\xi \ge E^2 + |X|^2 + 2C^2. \end{aligned}$$

(B10)

Note that provided $C^2>|X|^2$, eigenvalue (B8) can be negative without contradicting condition (B5).

Finally, we calculate the negativity of $\rho _{AB}={{\,\mathrm{Tr}\,}}_C[\rho _{ABC}]$. The partial transpose of $\rho _{AB}$ is

$$\begin{aligned} \rho _{AB}^{T_A} = \begin{pmatrix} 1-2P-3E-\Sigma &{} 0 &{} 0 &{} C \\ 0 &{} P+E &{} X^* &{} 0 \\ 0 &{} X &{} P+E &{} 0 \\ C &{} 0 &{} 0 &{} E+\Sigma \end{pmatrix}\,, \end{aligned}$$

(B11)

which again only has two eigenvalues that can be negative while maintaining non-negative eigenvalues for $\rho _{AB}$:

$$\begin{aligned}&E+P-|X|\,, \end{aligned}$$

(B12)

$$\begin{aligned}&\frac{1}{2}\left( (1-2P-2E) - \sqrt{4C^2+(1-4E-2P-2\Sigma )^2}\right) \,. \end{aligned}$$

(B13)

Eigenvalue (B12) will be non-negative if

$$\begin{aligned} P+E > |X| \end{aligned}$$

(B14)

and eigenvalue (B13) will be non-negative if

$$\begin{aligned} (1-2P-2E)^2&\ge \big (4C^2+(-1+4E+2P+2\Sigma )^2\big ) \nonumber \\&\implies E\xi + \Sigma \xi -C^2 - E^2 - E\Sigma + PE + P\Sigma \ge 0.\nonumber \\ \end{aligned}$$

(B15)

Under the assumption that

$$\begin{aligned} P\ge \Sigma \,, \end{aligned}$$

(B16)

then condition (B15) can be related to condition (B10) by noticing that

$$\begin{aligned} E\xi + \Sigma \xi -C^2 - E^2 - E\Sigma + PE + P\Sigma&= \left( E\xi -E^2-|X|^2-2C^2\right) \nonumber \\&\quad + |X|^2 + C^2 + \Sigma \xi + P\Sigma + E(P-\Sigma ) \nonumber \\&\ge \left( E\xi -E^2-|X|^2-2C^2\right) \,, \end{aligned}$$

(B17)

meaning that if Eq. (B15) is not satisfied then neither will Eq. (B10); if Eq. (B10) is satisfied then Eq. (B15) will be too.

The only way that the CKW inequality is violated will be if ${\mathcal {N}}_{A(B)}$ is nonzero. In other words, at least one of the eigenvalues of the partial transpose of $\rho _{AB}$ must be negative. We will first consider the case where eigenvalue (B12) is negative, followed by the case where  (B13) is negative. We will also show that under the assumptions that (B4) and (B16) are both valid, it is not possible for both eigenvalues to be simultaneously negative.

1.1 Case #1: Assume that eigenvalue (B12) is negative

If eigenvalue (B12) is negative, then $|X|>P+E$ and we also have $|X|^2>C^2$ (since $P+E\ge P\ge C$). This means that in order that $\rho _{ABC}$ remain a valid density matrix, eigenvalue (B8) must be non-negative, implying that eigenvalue (B13) must be also non-negative. Additionally eigenvalue (B7) must also be negative since

$$\begin{aligned} |X|>P+E\ge P \implies 2|X|^2 > 2P^2 = P^2+P^2 \ge P^2+PC = P(P+C)\,.\nonumber \\ \end{aligned}$$

(B18)

In this case, the $\pi$-tangle is

$$\begin{aligned} \pi&= {\mathcal {N}}_{A(BC)}^2 - 2{\mathcal {N}}_{A(B)}^2 \nonumber \\&= \Bigg (\frac{1}{2}\left( \sqrt{C^2+8|X|^2}-2P-C\right) \Bigg )^2 - 2\big (|X|-P-E\big )^2\,, \end{aligned}$$

(B19)

which can be simplified by considering a Taylor expansion of the matrix elements of $\rho _{ABC}$ in powers of the coupling strength, $\lambda \ll 1$:

$$\begin{aligned} P&= \lambda ^2 P_2 + \lambda ^4 P_4 + \lambda ^6 P_6 + {\mathcal {O}}\left( \lambda ^8\right) , \nonumber \\ C&= \lambda ^2 C_2 + \lambda ^4 C_4 + \lambda ^6 C_6 + {\mathcal {O}}\left( \lambda ^8\right) , \nonumber \\ X&= \lambda ^2 X_2 + \lambda ^4 X_4 + \lambda ^6 X_6 + {\mathcal {O}}\left( \lambda ^8\right) , \nonumber \\ E&= \lambda ^4 E_4 + \lambda ^6 E_6 + {\mathcal {O}}\left( \lambda ^8\right) , \nonumber \\ \Sigma&= \lambda ^6 \Sigma _6 + {\mathcal {O}}\left( \lambda ^8\right) . \end{aligned}$$

(B20)

Under this expansion, the $\pi$-tangle becomes:

$$\begin{aligned} \pi&\approx \lambda ^4\Bigg (\frac{1}{4}\left( \sqrt{C_2^2+8|X_2|^2}-2P_2-C_2\right) ^2-2\big (|X_2|-P_2\big )^2\Bigg )\,, \end{aligned}$$

(B21)

which will be non-negative if

$$\begin{aligned}&\Big (8P_2|X_2|+C_2^2+2P_2C_2-2P_2^2\Big )^2 \nonumber \\&\quad \ge \left( (2P_2+C_2)\sqrt{C_2^2+8|X_2|^2}\right) ^2 \nonumber \\&\qquad \iff 4\Big [P_2^2(P_2-C_2)^2-2\big (|X_2|-P_2\big )\Big (C_2^2\big (2|X_2|-P_2\big )-|X_2|(2P_2-C_2)^2\Big )\Big ] \ge 0. \nonumber \\ \end{aligned}$$

(B22)

Fig. 10

A plot of the perturbative $\pi$-tangle (B21) where the value of $C_2$ is fixed to be $C_2=0.9P_2$ and $\lambda =0.1$. There are regions of the parameter space where the $\pi$-tangle is negative. The region where $P_2>|X_2|$ is excluded from the plot, since this corresponds to eigenvalue (B12) being non-negative

As the condition in (B22) is not particularly instructive, we plot the perturbative $\pi$-tangle (B21) in Fig. 10, where it is clear that there are regions of the parameter space where the $\pi$-tangle is negative, meaning the CKW inequality is not satisfied. For example, if

$$\begin{aligned} \lambda= & {} 0.1,\ P_2=1,\ C_2=0.9,\ |X_2|=4 \implies \pi \approx \frac{3261-29\sqrt{12\ 881}}{2\ 000\ 000} \\\approx & {} -1.52\times 10^{-5}. \end{aligned}$$

Additionally, if the non-perturbative expression for the $\pi$-tangle (B19) is considered, it is still possible to find regions of the parameter space where the CKW inequality is not satisfied. For example,

$$\begin{aligned}&P=0.01,\ C_2=0.009,\ |X_2|=0.04,\ E=0.00011 \\&\quad \implies \pi = \frac{8\ 218\ 379 - 72\ 500\ \sqrt{12\ 881}}{5\ 000\ 000\ 000} \approx -1.99\times 10^{-6}. \end{aligned}$$

1.2 Case #2: Assume that eigenvalue (B13) is negative

If eigenvalue (B13) is negative, then $\sqrt{4C^2+(1-4E-2P-2\Sigma )^2}>(1-2P-2E)$ and eigenvalue (B8) will also be negative. Recall that in order to ensure that $\rho _{ABC}$ is a valid density matrix, we also require $C^2>|X|^2$. This also means that eigenvalue (B12) is non-negative since $|X|^2 < C^2 \le P^2 \le (P+E)^2$. Also notice that if $C\ge 0$, then

$$\begin{aligned} P(P+C) \ge C(C+C) = 2C^2 > 2|X|^2\,, \end{aligned}$$

and if $C<0$, then Eq. (B2d) becomes $P\ge -2C=2|C|$, and

$$\begin{aligned} P(P+C) = P\big (P-|C|\big ) \ge 2|C|\big (2|C|-|C|\big ) = 2|C|^2 = 2C^2 > 2|X|^2\,,\nonumber \\ \end{aligned}$$

(B23)

so, eigenvalue (B7) is also non-negative regardless of the sign of C. Therefore, the $\pi$-tangle is

$$\begin{aligned} \pi&= {\mathcal {N}}_{A(BC)}^2 - 2{\mathcal {N}}_{A(B)}^2 \nonumber \\&= \left[ \frac{1}{2}\Big (\sqrt{(1-3P-2E-\Sigma )^2+4\left( 2C^2-E+3E^2+EP+|X|^2+E\Sigma \right) }\right. \nonumber \\&\quad \left. - (1-3P-2E-\Sigma )\Big )\right] ^2 \nonumber \\&\quad -2 \Bigg [\frac{1}{2}\left( \sqrt{4C^2+(1-4E-2P-2\Sigma )^2} - (1-2P-2E)\right) \Bigg ]^2 \end{aligned}$$

(B24)

$$\begin{aligned}&\approx \lambda ^8\Big (2\big (C_2^2+|X_2|^2\big )^2-\big (E_4+|X_2|^2\big )^2\Big ), \end{aligned}$$

(B25)

where the last line is from the perturbative expansion of $\rho _{ABC}$ [Eq. (B20)]. In this case, the $\pi$-tangle will be non-negative if

$$\begin{aligned} E_4 \le \big (\sqrt{2}-1\big )|X_2|^2 + \sqrt{2}C_2^2\,. \end{aligned}$$

(B26)

Once again, we can find regions of the parameter space where the $\pi$-tangle is negative, meaning the CKW inequality is not satisfied. For example if

$$\begin{aligned}&\lambda =0.1,\ P_2=1,\ C_2=0.9,\ |X_2|=0.8,\ E_4=1.5 \\&\quad \implies \pi = -\frac{1873}{500\ 000\ 000\ 000} \approx -3.75\times 10^{-9}. \end{aligned}$$

Additionally, if the non-perturbative expression for the $\pi$-tangle (B24) is considered, it is still possible to find regions of the parameter space where the CKW inequality is not satisfied. For example if

$$\begin{aligned}&P=0.01,\ C=0.009,\ |X|=0.008,\ E=0.00015,\ \Sigma = 10^{-6}\\&\quad \implies \pi = \frac{-978\ 879\ 246\ 503+35\ 269\ 200\ \sqrt{2\ 961\ 556\ 921}-969\ 699\ \sqrt{940\ 638\ 421\ 201}}{2\ 000\ 000\ 000\ 000} \\&\quad \approx -2.43\times 10^{-9}. \end{aligned}$$