Applying CNOT in series to ancilla qubit

Question

$\renewcommand{ket}[1]{\left| #1 \right\rangle}$ $\renewcommand{bra}[1]{\left\langle #1 \right|}$Suppose we have to qubits both in the state $\ket{+ }= \frac{1}{\sqrt{2}}(\ket{0}+\ket{1})$, and we have an ancilla qubit in the state $\ket{0}$. What will the final state of the ancilla qubit be if we apply in series first a CNOT gate with qubit 1 as the control qubit and the ancilla qubit as the target, and then another CNOT gate this time with qubit 2 as the control qubit and the ancilla qubit as the target?

I tried computing:

$CNOT\ket{+0} = \frac{1}{\sqrt{2}}(CNOT\ket{00}+CNOT\ket{10}) = \frac{1}{\sqrt{2}}(\ket{00}+\ket{11})$

This was the first CNOT gate, followed by the next one. The ancilla qubit is now also in the state $\ket{+}$:

$CNOT\ket{+}\ket{+} = \frac{1}{2}(CNOT\ket{00}+CNOT\ket{01}+CNOT\ket{10}+CNOT\ket{11}) = \frac{1}{2}(\ket{00}+\ket{01}+\ket{11}+\ket{10})$

It seems to me the ancilla qubit is thus still in the state $\ket{+}$ and when measured in the standard basis, it gives $0$ or $1$ with equal chance. However, applying these gates is used in quantum error correction codes, and apparently you should get the outcome $0$ always when measuring the ancilla qubit. What am i doing wrong?

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Cross-posted on physics.SE

Answer 1

We have the initial state $|+\rangle|+\rangle|0\rangle$. The ordering I will be using is $q_2q_1a$. So, the first $\rm CNOT$ between $q_1$ and $a$ gives us the state

$$ \begin{align} |+\rangle\frac{1}{\sqrt{2}}(|00\rangle+|11\rangle) &= \frac{1}{2}(|0\rangle+|1\rangle)(|00\rangle+|11\rangle) \\ &= \frac{1}{2}(|000\rangle+|011\rangle+|100\rangle+|111\rangle) \end{align} $$

Now, it is easy to apply the second $\rm CNOT$ gate between $q_2$ and $a$. The state we end up with is

$$ \frac{1}{2}(|000\rangle+|011\rangle+|101\rangle+|110\rangle) $$

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Now, there is something wrong that you stated on your question.

This was the first $\rm CNOT$ gate, followed by the next one. The ancilla qubit is now also in the state $|+\rangle$.

This is not true. By expanding the tensor product, you can easily see that $|+\rangle|+\rangle$ is not equal to the Bell state $|00\rangle+|11\rangle$ (normalized).