# The analogous IBM circuit representation for the quantum scheme

I am trying to implement a quantum scheme of sharing a secret via EPR states. Here is my initial state of 6 qubits $$$$|\psi\rangle=\left(\dfrac{|0\rangle_1|0\rangle_2+ |1\rangle_1|1\rangle_2}{\sqrt{2}}\right) \left(\dfrac{|0\rangle_3|0\rangle_4+ |1\rangle_3|1\rangle_4}{\sqrt{2}}\right)\left(\dfrac{|0\rangle_5|0\rangle_6+ |1\rangle_5|1\rangle_6}{\sqrt{2}}\right)$$$$,

1. The qubits $$(1,4)$$ belong to $$P_1$$, $$(2,6)$$ to $$P_2$$, $$(3,5)$$ to $$P_3$$.
2. An operator is applied to qubit $$1$$ say the Pauli $$Z$$ gate, to get $$$$|\psi\rangle \left(\dfrac{|0\rangle_1|0\rangle_2-|1\rangle_1|1\rangle_2}{\sqrt{2}}\right) \left(\dfrac{|0\rangle_3|0\rangle_4+ |1\rangle_3|1\rangle_4}{\sqrt{2}}\right)\left(\dfrac{|0\rangle_5|0\rangle_6+ |1\rangle_5|1\rangle_6}{\sqrt{2}}\right)$$$$. Now after each party measures his qubits the operator is revealed. For instance say if $$P_1$$ measures $$|\alpha^+\rangle_{14}$$, and $$P_2$$ and $$P_3$$ measure $$|\beta^-\rangle_{26}$$ and $$|\beta^+\rangle_{35}$$ respectively, they write their state as after swapping as $$$$|\alpha^-\rangle_{12}|\alpha^+\rangle_{34} |\alpha^+\rangle_{56}$$$$, now they know the initial state of the qubits $$(1,2)$$, hence they deduce the operator as $$Z$$.

I have tried hard, but how to make the circuit for this scheme on an IBM-QE or for that matter any hardware.

The circuit provided is my try.

1. The part before the first barrier is where the initial state is prepared with the operator $$\mathbb{Z}$$, (this the operator that the parties have to find out by combing their measurements).

2. $$(2,6)$$ undergoes a Bell Basis measurement with the C-NOT and hadamard.

3. This makes $$(1,5)$$ entangled, represented by the C-NOT between (1,5) (a kind of reverse C-NOT- I don't know why i did this reverse, just out of intuition).

4. $$(3,5)$$ undergoes a Bell Basis measurement with the C-NOT and Hadamard.

5. This makes $$(1,4)$$ entangled, represented by the reverse C-NOT between (1,4) (again intuition).

6. Then the Bell basis measurement on $$(1,4)$$ with the C-NOT and Hadamard.

But, i am still not knowing how to proceed, the theoretical calculations say that the final state should be $$$$|\alpha^-\rangle_{12}|\alpha^+\rangle_{34}|\alpha^+\rangle_{56}$$$$ so that they apply the $$Z$$ on the first qubit to get $$$$|\psi\rangle=\left(\dfrac{|0\rangle_1|0\rangle_2+ |1\rangle_1|1\rangle_2}{\sqrt{2}}\right) \left(\dfrac{|0\rangle_3|0\rangle_4+ |1\rangle_3|1\rangle_4}{\sqrt{2}}\right)\left(\dfrac{|0\rangle_5|0\rangle_6+ |1\rangle_5|1\rangle_6}{\sqrt{2}}\right)$$$$ Can't somebody atleast hint at where to start from? Do i have to use conditional statements , because that is what the theory says, but how does one use these statements? I know for instance that the state 0000 on the 2,6 ,3,5 qubits will correspond to an identity operator on the 1st qubit and similarly we do these for others but does one code this on a simulator

• Why do you introduce additional entanglement in steps (3) and (5)? The separable state over all of the qubits (1, 2), (3, 4), and (5, 6) will become entangled with a bell basis measurement between (2, 6) without the need for your adding additional gates. – forky40 Jul 28 at 6:13
• Okay so there are just two measurements on (2,6) and (3,5). I got this, but how to do the conditioning and coding for this – Upstart Jul 28 at 7:15