8.59: An Entanglement Swapping Protocol

In the field of quantum information, interference, superpositions and entangled states are essential resources. Entanglement, a non-factorable superposition, is routinely achieved when two photons are emitted from the same source, perhaps a parametric down converter (PDC). Entanglement swapping involves the transfer (teleportation) of entanglement to two photons that were produced independently and never previously interacted. The Bell states are the four maximally entangled two-qubit entangled basis for a four-dimensional Hilbert space and play an essential role in quantum information theory and technology, including teleportation and entanglement swapping. This analysis attempts to provide the essential matrix math needed to understand parts of "Entangled delayed-choice entanglement swapping" (arXiv1203.4834) and "Delayed-choice gedanken experiments and their realizations" (arXiv1407.2930). The following analysis deals exclusively with entanglement swapping and does not consider the delayed-choice aspect of the research presented in these papers.

Bell states and the identity operator:

\[ \begin{matrix} \Phi_p = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \\ 0 \\ 0 \\ 1 \end{pmatrix} & \Phi_m = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \\ 0 \\ 0 \\ -1 \end{pmatrix} & \Psi_p = \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \\ 1 \\ 1 \\ 0 \end{pmatrix} & \Psi_m = \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \\ 1 \\ -1 \\ 0 \end{pmatrix} & I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \end{matrix} \nonumber \]

A four-qubit state is prepared in which photons 1 and 2 are entangled in Bell state Ψ_m, and photons 3 and 4 are also entangled in Bell state Ψ_m. The state multiplication below is understood to be tensor vector multiplication.

\[ \begin{matrix} \Psi_{1234} = \Psi_{m12} \Psi_{m34} = \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \\ 1 \\ -1 \\ 0 \end{pmatrix} \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \\ 1 \\ -1 \\ 0 \end{pmatrix} & \Psi = \frac{1}{2} \begin{pmatrix} 0 & 0 & 0 & 0 & 0 & 1 & -1 & 0 & 0 & -1 & 1 & 0 & 0 & 0 & 0 & 0 \end{pmatrix}^T \end{matrix} \nonumber \]

The authors write this state as a superposition of Bell state products to suggest a way to transfer entanglement from 1&2-3&4 to 1&4-2&3: perform a Bell state measurement on photons 2 and 3.

\[ | \Psi \rangle_{1234} = \frac{1}{2} \left[ | \Psi_p \rangle_{14} \otimes | \Psi_P \rangle_{23} - | \Psi_m \rangle_{14} \otimes \Psi_m \rangle_{23} - | \Psi_m \rangle_{23} - | \Phi_p \rangle_{14} \otimes \Phi_p \rangle_{23} + | \Phi_m \rangle_{14} \otimes \Phi_m \rangle_{23} \right] \nonumber \]

The following calculations agree with this product of entangled photon pairs 1&4 and 2&3. Projection of photons 2 and 3 onto Ψ_p projects photons 1 and 4 onto Ψ_p.

Projection of photons 2 and 3 onto Φ_p projects photons 1 and 4 onto Ψ_m.

\[ \begin{matrix} \left( \text{kronecker} \left( \text{I, kronecker} \left( \Psi_p,~ \Psi_p^T,~ \text{I} \right) \right) \Psi \right)^T = \begin{pmatrix} 0 & 0 & 0 & 0.25 & 0 & 0.25 & 0 & 0 & 0 & 0 & 0.25 & 0 & 0.25 & 0 & 0 & 0 \end{pmatrix} \\ \frac{1}{ 2 \sqrt{2}} \left[ \begin{pmatrix} 1 \\ 0 \end{pmatrix} \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \\ 1 \\ 1 \\ 0 \end{pmatrix} \begin{pmatrix} 0 \\ 1 \end{pmatrix} + \begin{pmatrix} 0 \\ 1 \end{pmatrix} \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \\ 1 \\ 1 \\ 0 \end{pmatrix} \begin{pmatrix} 1 \\ 0 \end{pmatrix} \right]^T = \frac{1}{4} \begin{pmatrix} 0 & 0 & 0 & 1 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 \end{pmatrix} \end{matrix} \nonumber \]

Projection of photons 2 and 3 onto Ψ_m projects photons 1 and 4 onto -Ψ_m.

\[ \begin{matrix} \left( \text{kronecker} \left( \text{I, kronecker} \left( \Psi_m,~ \Psi_m^T,~ \text{I} \right) \right) \Psi \right)^T = \begin{pmatrix} 0 & 0 & 0 & -0.25 & 0 & 0.25 & 0 & 0 & 0 & 0 & 0.25 & 0 & -0.25 & 0 & 0 & 0 \end{pmatrix} \\ \frac{-1}{ 2 \sqrt{2}} \left[ \begin{pmatrix} 1 \\ 0 \end{pmatrix} \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \\ 1 \\ -1 \\ 0 \end{pmatrix} \begin{pmatrix} 0 \\ 1 \end{pmatrix} - \begin{pmatrix} 0 \\ 1 \end{pmatrix} \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \\ 1 \\ -1 \\ 0 \end{pmatrix} \begin{pmatrix} 1 \\ 0 \end{pmatrix} \right]^T = \frac{1}{4} \begin{pmatrix} 0 & 0 & 0 & -1 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 1 & 0 & -1 & 0 & 0 & 0 \end{pmatrix} \end{matrix} \nonumber \]

Projection of photons 2 and 3 onto Φ_p projects photons 1 and 4 onto -Φ_p.

\[ \begin{matrix} \left( \text{kronecker} \left( \text{I, kronecker} \left( \Phi_p,~ \Phi_p^T,~ \text{I} \right) \right) \Psi \right)^T = \begin{pmatrix} -0.25 & 0 & 0 & 0 & 0 & 0 & -0.25 & 0 & 0 & 0 & 0 & 0 & -0.25 \end{pmatrix} \\ \frac{-1}{ 2 \sqrt{2}} \left[ \begin{pmatrix} 1 \\ 0 \end{pmatrix} \frac{-1}{ \sqrt{2}} \begin{pmatrix} 1 \\ 0 \\ 0 \\ 1 \end{pmatrix} \begin{pmatrix} 0 \\ 1 \end{pmatrix} + \begin{pmatrix} 0 \\ 1 \end{pmatrix} \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \\ 0 \\ 0 \\ 1 \end{pmatrix} \begin{pmatrix} 0 \\ 1 \end{pmatrix} \right]^T = \frac{1}{4} \begin{pmatrix} -1 & 0 & 0 & 0 & 0 & 0 & -1 & 0 & 0 & -1 & 0 & 0 & 0 & 0 & 0 & -1 \end{pmatrix} \end{matrix} \nonumber \]

Projection of photons 2 and 3 onto Φ_m projects photons 1 and 4 onto Φ_m.

\[ \begin{matrix} \left( \text{kronecker} \left( \text{I, kronecker} \left( \Phi_m,~ \Phi_m^T,~ \text{I} \right) \right) \Psi \right)^T = \begin{pmatrix} 0.25 & 0 & 0 & 0 & 0 & 0 & -0.25 & 0 & 0 & -0.25 & 0 & 0 & 0 & 0 & 0 & 0.25 \end{pmatrix} \\ \frac{1}{ 2 \sqrt{2}} \left[ \begin{pmatrix} 1 \\ 0 \end{pmatrix} \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \\ 0 \\ 0 \\ -1 \end{pmatrix} \begin{pmatrix} 1 \\ 0 \end{pmatrix} - \begin{pmatrix} 0 \\ 1 \end{pmatrix} \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \\ 0 \\ 0 \\ -1 \end{pmatrix} \begin{pmatrix} 0 \\ 1 \end{pmatrix} \right]^T = \frac{1}{4} \begin{pmatrix} 1 & 0 & 0 & 0 & 0 & 0 & -1 & 0 & 0 & -1 & 0 & 0 & 0 & 0 & 0 & 1 \end{pmatrix} \end{matrix} \nonumber \]

The initial four-particle state can be written in the H/V, A/D and R/L polarization bases. See the Appendix for details.

\[ \begin{matrix} \Psi_{1234} = \frac{1}{2} (H_1 V_2 - V_1 H_2) = (H_3 V_4 - V_3 H_4) = \frac{1}{2} (H_1 V_2 V_3 V_4 - H_1 V_2 V_3 H_4 - V_1 H_2 H_3 V_4 + V_1 H_2 V_3 H_4 + V_1 H_2 V_3 H_4) \\ \Psi_{1234} = \frac{1}{2} (A_1 D_2 - D_1 A_2) = (A_3 D_4 - D_3 A_4) = \frac{1}{2} (A_1 D_2 A_3 D_4 - A_1 D_2 A_3 D_4 - A_1 D_2 D_3 A_4 + D_1 A_2 A_3 D_4 + D_1 A_2 D_3 A_4) \\ \Psi_{1234} = \frac{1}{2} (L_1 R_2 - R_1 L_2) = (L_3 R_4 - R_3 L_4) = \frac{1}{2} (L_1 R_2 R_3 R_4 - L_1 R_2 R_3 L_4 - R_1 L_2 L_3 R_4 + R_1 L_2 R_3 L_4) \end{matrix} \nonumber \]

Where,

\[ \begin{matrix} H = \begin{pmatrix} 1 \\ 0 \end{pmatrix} & V = \begin{pmatrix} 0 \\ 1 \end{pmatrix} & R = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \\ i \end{pmatrix} & L = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \\ 1 \end{pmatrix} & A = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \\ -1 \end{pmatrix} \end{matrix} \nonumber \]

\[ \begin{matrix} H = \frac{1}{ \sqrt{2}} (R+L) & V = \frac{i}{ \sqrt{2}} (L-R) & H = \frac{1}{ \sqrt{2}} (D+A) & V = \frac{1}{ \sqrt{2}} (D-A) \end{matrix} \nonumber \]

\[ \begin{matrix} R = \frac{1}{ \sqrt{2}} (H + iV) & L = \frac{1}{ \sqrt{2}} (H - iV) & R = \frac{1}{ \sqrt{2}} (D+A) \end{matrix} \nonumber \]

After production, photon 1 is sent to Alice and photon 4 is sent to Bob. Photons 2 and 3 are sent to Victor. Alice and Bob can measure their photons in either the H/V, R/L or A/D bases. Victor can measure his photons separately in the H/V basis or he can carry out one of the four Bell state measurements on his photon pairs. This later choice as shown earlier projects photons 1 and 4 into the corresponding entangled Bell state.

The following table shows the possible measurement results that Alice and Bob will obtain, depending on the type of measurement Victor makes on his photons. In experiments 1-4 Victor measures his photons separately in the H/V basis and Alice and Bob do the same. In the remaining experiments Victor does a Bell state measurement on his photons and Alice and Bob measure in any of the three bases.

\[ \begin{pmatrix} \text{Experiment} & 1 & 2 & 3 & 4 & ' & 5 & 6 & 7 & 8 & ' & 9 & 10 & 11 & 12 & ' & 13 & 14 & 15 & 16 \\ \text{Alice1} & H & H & V & V ' & H & V & H & V & ' & L & R & L & R & ' & D & A & A & D \\ \text{Bob4} & V & H & V & H & ' & V & H & H & V & ' & L & L & R & R & ' & D & D & A & A \\ \text{Victor23} & VH & VV & HH & HV & ' & \Psi_p & \Psi_m & \Phi_p & \Phi_m & ' & \Psi_p & \Psi_m & \Phi_p & \Phi_m ' & \Psi_p & \Psi_m & \Phi_p & \Phi_m \end{pmatrix} \nonumber \]

Results 1-4 are consistent with the original state expressed in the H/V basis.

\[ \Psi_{1234} = \frac{1}{2} (H_1 V_2 H_3 V_4 - H_1 V_2 V_3 H_4 - V_1 H_2 H_3 V_4 + V_1 H_2 V_3 H_4) \nonumber \]

In the remaining experiments Victor makes a Bell state measurement on photons 2 and 3, and projects photons 1 and 4 into the following Bell states. The table shows measurement results that Alice and Bob could make on their photons given these states. See the Appendix for more detail.

\[ \begin{matrix} \Psi_{p14} = \frac{1}{ \sqrt{2}} (H_1 V_4 + V_1 H_4) = \frac{i}{ \sqrt{2}} (L_1 L_4 - R_1 R_4) = \frac{1}{ \sqrt{2}} (D_1 D_4 - A_1 A_4) \\ \Psi_{m14} = \frac{1}{ \sqrt{2}} (H_1 V_4 - V_1 H_4) = \frac{i}{ \sqrt{2}} (L_1 L_4 - R_1 R_4) = \frac{1}{ \sqrt{2}} (A_1 D_4 - D_1 A_4) \\ \Phi_{p14} = \frac{1}{ \sqrt{2}} (H_1 H_4 + V_1 V_4) = \frac{i}{ \sqrt{2}} (L_1 R_4 + R_1 L_4) = \frac{1}{ \sqrt{2}} (A_1 A_4 - D_1 D_4) \\ \Phi_{m14} = \frac{1}{ \sqrt{2}} (H_1 H_4 - V_1 V_4) = \frac{i}{ \sqrt{2}} (L_1 L_4 + R_1 R_4) = \frac{1}{ \sqrt{2}} (A_1 D_4 + D_1 A_4) \end{matrix} \nonumber \]

Appendix

The Bell states are written in the H/V, R/L, and A/D bases.

\[ \begin{matrix} \Psi_p = \begin{pmatrix} 0 \\ 1 \\ 1 \\ 0 \end{pmatrix} & \frac{1}{ \sqrt{2}} (H_i V_j + V_i H_j ) & \begin{array}{|l} \text{substitute, } H_i = \frac{1}{ \sqrt{2}} (R_i + L_i) \\ \text{substitute, } V_j = \frac{i}{ \sqrt{2}} (R_j - L_j) \\ \text{substitute, } V_i = \frac{i}{ \sqrt{2}} (L_i - R_i) \\ \text{substitute, } H_j = \frac{1}{ \sqrt{2}} (R_j + L_j) \end{array} \rightarrow \sqrt{2} \left( - \frac{ R_i R_j i}{2} + \frac{L_i L_j i}{2} \right) \\ ~ & \frac{1}{ \sqrt{2}} (H_i V_j + V_i H_j ) & \begin{array}{|l} \text{substitute, } H_i = \frac{1}{ \sqrt{2}} (D_i + A_i) \\ \text{substitute, } V_j = \frac{i}{ \sqrt{2}} (D_j - A_j) \\ \text{substitute, } V_i = \frac{i}{ \sqrt{2}} (D_i - A_i) \\ \text{substitute, } H_j = \frac{1}{ \sqrt{2}} (D_j + A_j) \end{array} \rightarrow \sqrt{2} \left( \frac{ A_i A_j i}{2} - \frac{D_i D_j i}{2} \right) \\ \Psi_m = \begin{pmatrix} 0 \\ 1 \\ -1 \\ 0 \end{pmatrix} & \frac{1}{ \sqrt{2}} (H_i V_j - V_i H_j) & \begin{array}{|l} \text{substitute, } H_i = \frac{1}{ \sqrt{2}} (R_i + L_i) \\ \text{substitute, } V_j = \frac{i}{ \sqrt{2}} (L_j - R_j) \\ \text{substitute, } V_i = \frac{i}{ \sqrt{2}} (L_i - R_i) \\ \text{substitute, } H_j = \frac{1}{ \sqrt{2}} (R_j + L_j) \end{array} \rightarrow \sqrt{2} \left( - \frac{ L_i R_j i}{2} + \frac{L_i R_j i}{2} \right) \\ ~ & \frac{1}{ \sqrt{2}} (H_i V_j - V_i H_j ) & \begin{array}{|l} \text{substitute, } H_i = \frac{1}{ \sqrt{2}} (D_i + A_i) \\ \text{substitute, } V_j = \frac{i}{ \sqrt{2}} (D_j - A_j) \\ \text{substitute, } V_i = \frac{i}{ \sqrt{2}} (D_i - A_i) \\ \text{substitute, } H_j = \frac{1}{ \sqrt{2}} (D_j + A_j) \end{array} \rightarrow \sqrt{2} \left( \frac{ A_i D_j i}{2} - \frac{A_i D_j i}{2} \right) \\ \Phi_p = \begin{pmatrix} 1 \\ 0 \\ 0 \\ 1 \end{pmatrix} & \frac{1}{ \sqrt{2}} (H_i H_j + V_i V_j) & \begin{array}{|l} \text{substitute, } H_i = \frac{1}{ \sqrt{2}} (R_i + L_i) \\ \text{substitute, } V_j = \frac{i}{ \sqrt{2}} (L_j - R_j) \\ \text{substitute, } V_i = \frac{i}{ \sqrt{2}} (L_i - R_i) \\ \text{substitute, } H_j = \frac{1}{ \sqrt{2}} (R_j + L_j) \end{array} \rightarrow \sqrt{2} \left( \frac{ L_i R_j i}{2} + \frac{L_i R_j i}{2} \right) \\ ~ & \frac{1}{ \sqrt{2}} (H_i V_j + V_i H_j ) & \begin{array}{|l} \text{substitute, } H_i = \frac{1}{ \sqrt{2}} (D_i + A_i) \\ \text{substitute, } V_j = \frac{i}{ \sqrt{2}} (D_j - A_j) \\ \text{substitute, } V_i = \frac{i}{ \sqrt{2}} (D_i - A_i) \\ \text{substitute, } H_j = \frac{1}{ \sqrt{2}} (D_j + A_j) \end{array} \rightarrow \sqrt{2} \left( \frac{ A_i A_j i}{2} - \frac{D_i D_j i}{2} \right) \\ \Phi_m = \begin{pmatrix} 1 \\ 0 \\ 0 \\ -1 \end{pmatrix} & \frac{1}{ \sqrt{2}} (H_i H_j - V_i V_j) & \begin{array}{|l} \text{substitute, } H_i = \frac{1}{ \sqrt{2}} (R_i + L_i) \\ \text{substitute, } V_j = \frac{i}{ \sqrt{2}} (L_j - R_j) \\ \text{substitute, } V_i = \frac{i}{ \sqrt{2}} (L_i - R_i) \\ \text{substitute, } H_j = \frac{1}{ \sqrt{2}} (R_j + L_j) \end{array} \rightarrow \sqrt{2} \left( \frac{ L_i L_j i}{2} + \frac{R_i R_j i}{2} \right) \\ ~ & \frac{1}{ \sqrt{2}} (H_i V_j - V_i H_j ) & \begin{array}{|l} \text{substitute, } H_i = \frac{1}{ \sqrt{2}} (D_i + A_i) \\ \text{substitute, } V_j = \frac{i}{ \sqrt{2}} (D_j - A_j) \\ \text{substitute, } V_i = \frac{i}{ \sqrt{2}} (D_i - A_i) \\ \text{substitute, } H_j = \frac{1}{ \sqrt{2}} (D_j + A_j) \end{array} \rightarrow \sqrt{2} \left( \frac{ A_i D_j i}{2} - \frac{A_j D_i i}{2} \right) \end{matrix} \nonumber \]

The initial state is written in the H/V, R/L, and A/D bases.

\[ \begin{matrix} \frac{1}{2} (H_1 V_2 - V_1 H_2) (H_3 V_4 - V_3 H_4 ) & \begin{array}{|l} \text{substitute, H}_1 = \frac{1}{ \sqrt{2}} (D_1 + A_1 ) \\ \text{substitute, V}_2 = \frac{1}{ \sqrt{2}} (D_2 - A_2 ) \\ \text{substitute, V}_1 = \frac{1}{ \sqrt{2}} (D_1 - A_1 ) \\ \text{substitute, H}_2 = \frac{1}{ \sqrt{2}} (D_2 + A_2 ) \\ \text{substitute, H}_3 = \frac{1}{ \sqrt{2}} (D_3 + A_3) \\ \text{substitute, V}_3 = \frac{1}{ \sqrt{2}} (D_3 - A_3) \\ \text{substitute, H}_4 = \frac{1}{ \sqrt{2}} (D_4 + A_4) \\ \text{substitute, V}_4 = \frac{1}{ \sqrt{2}} (D_4 - A_4) \end{array} \rightarrow \frac{A_1 D_2 - A_2 D_1)(A_3D_4-A_4D_3)}{2} \\ \frac{1}{2} (H_1 V_2 - V_1 H_2) (H_3 V_4 - V_3 H_4) & \begin{array}{|l} \text{substitute, H}_1 = \frac{1}{ \sqrt{2}} (R_1 + L_1 ) \\ \text{substitute, V}_2 = \frac{i}{ \sqrt{2}} (L_2 - R_2) \\ \text{substitute, V}_1 = \frac{i}{ \sqrt{2}} (L_1 - R_1) \\ \text{substitute, H}_2 = \frac{1}{ \sqrt{2}} (R_2 + L_2) \\ \text{substitute, H}_3 = \frac{1}{ \sqrt{2}} (R_3 + L_3) \\ \text{substitute, V}_3 = \frac{i}{ \sqrt{2}} (L_3 - R_3) \\ \text{substitute, H}_4 = \frac{1}{ \sqrt{2}} (R_4 + L_4) \\ \text{substitute, V}_4 = \frac{1}{ \sqrt{2}} (L_4 - R_4) \end{array} \rightarrow \frac{L_1 R_2 - L_2 R_1)(L_3R_4-L_4R_3)}{2} \end{matrix} \nonumber \]