# The statevector of three-qubit bit-flip encoding circuit for entangled state

I was trying to implement a three-qubit bit-flip code for shared entangled state. I was curious to analyze this mathematically, but the problem is I can't calculate the statevector after encoding. Here is the circuit:

I used Qiskit to calculate the statevector, but it turns out to be 0's for every qubit. I am very confused because I don't know if this is valid or not!

Could you please guide me a bit? Also, should this circuit have higher fidelity? Please comment on this as well.

• This is certainly not the output that you should be getting for the circuit you've displayed. (It should be $(|000000\rangle+|111111\rangle)/\sqrt{2}$.) I'm not a qiskit expert, but I'm sure the experts on this site would find it helpful to see the whole code that you've written to define bit_flip_qc. Apr 25 at 6:36
• I have found the solution. Thanks everyone! Apr 25 at 9:40

The method Statevector.evolve()[1] returns the evolution result. It does not change the instance it called in. So all what you need is to change your code to become:

av = Statevector.from_label('000000')
result = av.evolve(bit_flip_qc)
result.data

• Wow, it's working now. Thanks a lot!! Apr 25 at 9:39

I suggest you make 2 repetition logical qubits first and entangle them with logical CNOT operator. you can check every single steps separately.

1. Make $$|+_L\rangle_{Alice} = {1\over\sqrt{2}} (|000 \rangle + |111 \rangle )$$ and $$|0_L\rangle_{bob} = |000 \rangle$$

2. Apply three CNOT operators transversally

3. Final state will be $${1\over\sqrt{2}} (|000_{Alice}000_{bob} \rangle + |111_{Alice}111_{bob} \rangle )$$