Q-Sphere representation of Bell States

Question

I'm currently going through Lab1 of the Qiskit Quantum Computing Course, where one task is to create the Bell State $$\frac{1}{\sqrt{2}}\left(\vert01\rangle + \vert10\rangle\right)$$ Technically it should be achived by applying a Hadamard, followed by a CNOT gate on the initial state $\vert01\rangle$, if I understand it correctly. My Qiskit implementation is the following:

sv = Statevector.from_label('01')
mycircuit = QuantumCircuit(2)
mycircuit.h(0)
mycircuit.cx(0,1)
new_sv = sv.evolve(mycircuit)
plot_state_qsphere(new_sv.data)

I'm only confused about the resulting Q-Sphere image:

The way I understood the Q-Sphere, this image should correspond to the state $$\frac{1}{\sqrt{2}}\left(\vert11\rangle - \vert00\rangle\right)$$ and not to the requested one. Was my creation of the asked Bell State wrong, or my interpretation of the Q-Sphere? Thanks for any help!

Answer 1

If you want to create $\frac{|01\rangle + |01\rangle }{\sqrt{2}} $, you actually need to start from the state $|10\rangle$, not $|01\rangle$. Quickly written, the calculation is:

$$ |10\rangle \xrightarrow{I\otimes H} \frac{1}{\sqrt{2}}|1\rangle (|0\rangle + |1\rangle) \xrightarrow{CX} \frac{1}{\sqrt{2}}(|10\rangle + |01\rangle) $$

and

$$ |01\rangle \xrightarrow{I\otimes H} \frac{1}{\sqrt{2}}|0\rangle (|0\rangle - |1\rangle) \xrightarrow{CX} \frac{1}{\sqrt{2}}(|00\rangle - |11\rangle) $$

So change this

sv = Statevector.from_label('10')

should be enough to solve your issue, I tried and it worked! ;)

I hope this is clear enough, please tell me if not :)

Answer 2

The circuit you shown is to create the state $ \dfrac{|00\rangle + |11\rangle}{\sqrt{2}}$ have you started from the state $|00\rangle = |0\rangle \otimes |0\rangle$. Now to get this state to the state $|\psi \rangle = \dfrac{|01\rangle + |10\rangle}{\sqrt{2}} $ you need to apply another $X$ gate to any of the two qubits in addition to what you had before. The circuit is as follows:

         ┌───┐          
q_0:|0>  ┤ H ├──■───────
         └───┘┌─┴─┐┌───┐
q_1:|0>  ─────┤ X ├┤ X ├
              └───┘└───┘

If you want to generate this circuit with qiskit you can do it as follows:

from qiskit import QuantumRegister, ClassicalRegister, QuantumCircuit

qreg_q = QuantumRegister(2, 'q')
creg_c = ClassicalRegister(2, 'c')
circuit = QuantumCircuit(qreg_q, creg_c)

circuit.h(qreg_q[0])
circuit.cx(qreg_q[0], qreg_q[1])
circuit.x(qreg_q[1])

The reason for this is because upon applying the operator $X$ to the first qubit and do nothing to the second is the same as applying the operator $X \otimes I$ (note I am writing the tensor product in term of qiskit's standard where they used little endian) to the state $\dfrac{|00\rangle + |11\rangle}{\sqrt{2}}$. Explicitly, we have:

$$ (X \otimes I)\dfrac{|00\rangle + |11\rangle}{\sqrt{2}} = \dfrac{X|0\rangle \otimes I |0\rangle + X|1\rangle \otimes I|1\rangle}{\sqrt{2}} = \dfrac{|1\rangle \otimes |0\rangle + |0\rangle \otimes |1\rangle }{\sqrt{2} } = \dfrac{|1 0\rangle + |01\rangle }{\sqrt{2} }= \dfrac{|01\rangle + |10 \rangle }{\sqrt{2} } $$