s two qubits as: |q)1=a|0)+B|1) and based on the definition of 1x) in Eq. (P1), we develop the 6-qubit tensor /D) that is input to the circuit, as follows (overlooking normalization factors): /D) = = (a0)1+B(1)1)0(0000) + |0111) + |1100) + |1011))2345 (a|00000) + x|00111) + x|01100) + a|01011) + + B|10111) + B|11100) + B|11011) 12345 (a|00000) + x|00111) + x|01100) + + (a'|0)6 B'|1)6) B|10000) + B|10111) + B|11100) + B|11011) 12345 aa'(|000000) + |001110) + |011000) + |010110)) + aB'(|000001) + |001111) + |011001) + |010111)) + + |101110) + |111000> + |110110)) + BB'(|100001) + B|101111) + |111001) + |110111)) 123456 We must then apply the action of the two CNOT gates C12, C56 in B, B', noted C12, C56 respectively, with the first and second indices representing the control and target qubits. Two-qubit teleportation For any binary combinations a, b, c, d, e, f = {0, 1} in a 6-qubit state |abcdef), it is clear that the gates have the corresponding action: C12|0abcde) = |0abcde) C12|1abcde) = |1abcde) C56|abcd0e) = |abcd0e) C56|abcd1e) = |abcd1e). Effecting the above rules in Eq. (P3), we obtain D' = C56C12/Q) as: aa'((000000)+001111)+011000)+010111)) + aB'(|000001) + |001110) + |011001) + |010110)) + x' (B(|110000) + |111111) + |101000> + |100111)) + BB'(|110001) + |111110) + |101001) + |100110)) 123456 Next, we must calculate the action of the subsequent Hadamard gates H1, H5 acting on qubits 1 and 5, respectively, recalling first that and, hence, for any binary combination a, b, C, d, e = {0, 1}: |labcde) H1|0abcde) = 1+abcde) = |labcde) H1|labcde) = |-abcde) = |abcd1e) H5\abcd0e) = \abcd + e) = % = - = Applying the two Hadamard gates H1, H5 onto I D') in Eq. (P5), and overlooking the normalization factor, we obtain (P) = H5H1 I'), as: aa'(1+000+0)+1+011-1)+1+110+0)+1+101-1))+ - aB(1+000+1)++011-0)+1+110 +1)+1+101 -0))+ - aB(-100+0)+1-111-1)+1-010+0)+1-001-1))+ - BB'( - + 1) + 111 - 0) + -010 + 1) + 1-001 - 0)) 123456 Appendix P where the states | are defined in Eq. (P6). Substituting these definitions in Eq. (P8