Understanding phase kickback caused by the CNOT gate

Question

Applying the CNOT gate to the state |+-⟩ would result in the state |--⟩ as per:

What has occurred is a "phase kickback". The relative negative phase from the target qubit has transferred to the control qubit.

Let's look at the representation using vectors and matrices. The state |+-⟩ can be achieved by starting with the classical state |00⟩, flipping the second qubit using a NOT gate, e.g. (I⊕N)*|00⟩ = |01⟩ and then applying the hadamard gate to each qubit - (H⊕H)= |+-⟩. Using a vector column, this state would look like this:

\begin{bmatrix}1/2\\-1/2\\1/2\\-1/2\end{bmatrix}

If I apply the CNOT gate to this state, I would get:

My question is how am I supposed to interpret the resulting vector column? How can I convert from the computational basis after CNOT to the equivalent state in Hadamard basis to actually see that there's a negative phase infront of both qubits? I don't see that in the computational basis.

Answer 1

There are two ways to see this.

1. simply factor the sum you have:

$$\frac{1}{2}(|00\rangle - |01\rangle - |10\rangle + |11\rangle) \\ = \frac{1}{2}(|0\rangle - |1\rangle)(0\rangle - |1\rangle) \\ =|-\rangle|-\rangle $$

This of course may be more difficult to do when the states are less simple.

2. Use the fact that any quantum operation is a unitary, and a unitary is also a change of basis. Here you want to change from the computational basis to the X (phase) basis - so you need to apply a Hadamard on your state. You'll see that:

$$ \frac{1}{2}H_1H_2\begin{pmatrix}1\\ -1 \\ -1 \\ 1\end{pmatrix} = \begin{pmatrix}0\\ 0 \\ 0 \\ 1\end{pmatrix} $$ but here we need to understand what this means: that the state with the vector representation $$ \begin{pmatrix}1\\ -1 \\ -1 \\ 1\end{pmatrix} $$ in the computational basis is the same as the state $$ \begin{pmatrix}0\\ 0 \\ 0 \\ 1\end{pmatrix} $$ in the phase basis (and vice-versa).

This is much more powerful and allows you to switch between representations easily.