Phase Kickback

Quantum entanglement is one of the astonishing characteristics of quantum mechanics. Two entangled particles share a state of superposition no matter how far apart they are.

From a practical perspective, we can use entanglement to let one qubit control the state of another. For instance, the controlled NOT-gate (CNOT- or CX-gate) switches the amplitudes of a target qubit only if the control qubit is in state $|1\rangle$. Nothing happens if the control qubit is in state $|0\rangle$.

Such controlled quantum gates let us precisely manipulate a multi-qubit system. In the lesson Composing Quantum Computing Controls, we let certain states of the quantum system exhibit the measurement probabilities we want them to have. We use entanglement to create a fine-grained probabilistic system.

Another practical characteristic of controlled quantum gates is that they leave the control qubit untouched.

The following figure depicts the truth table of the CNOT-gate.

The control qubit values do not change, no matter which combination of qubit values we feed into the operation.

Having a practically applicable notion of quantum transformation gates is paramount when we work with qubits. However, every once in a while, we need to remember that a quantum operation is essentially a physical operation, as is the CNOT-gate.

For every action in physics, there is an opposite reaction. For this reason, we should be suspicious even of the apparent one-sidedness of the CNOT-gate.

Whenever we challenge our intuition in quantum computing, it’s good to consult the underlying math.

The CNOT-gate is a two-qubit gate. As a result, it transforms qubit states whose state we represent by a four-dimensional vector.

$$|\psi\rangle=\alpha|0\rangle|0\rangle+\beta|0\rangle|1\rangle+\gamma|1\rangle|0\rangle+\delta|1\rangle|1\rangle=\begin{bmatrix}\alpha \\ \beta \\ \gamma \\ \delta \end{bmatrix}$$

Accordingly, the CNOT-gate has a 4x4 transformation matrix.

$$CNOT=\begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&0&1\\0&0&1&0\end{bmatrix}$$

There is no effect if the control qubit, which isat the left-hand position in the Dirac notation, is in state $|0\rangle$.

$$CNOT\cdot|00\rangle=\begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&0&1\\0&0&1&0\end{bmatrix}\cdot \begin{bmatrix} 1\\ 0\\0 \\ 0\end{bmatrix}=\begin{bmatrix} 1\\ 0\\0 \\ 0\end{bmatrix}=|00\rangle$$

$$CNOT\cdot|01\rangle=\begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&0&1\\0&0&1&0\end{bmatrix}\cdot \begin{bmatrix} 0\\ 1\\0 \\ 0\end{bmatrix}=\begin{bmatrix} 0\\ 1\\0 \\ 0\end{bmatrix}=|01\rangle$$

However, if the control qubit is in state $|1\rangle$, then the controlled qubit switches from $|0\rangle$ to $|1\rangle$ and vice versa.

$$CNOT\cdot|10\rangle=\begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&0&1\\0&0&1&0\end{bmatrix}\cdot \begin{bmatrix} 0\\ 0\\1 \\ 0\end{bmatrix}=\begin{bmatrix} 0\\ 0\\0 \\ 1\end{bmatrix}=|11\rangle$$

$$CNOT\cdot|11\rangle=\begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&0&1\\0&0&1&0\end{bmatrix}\cdot \begin{bmatrix} 0\\ 0\\0 \\ 1\end{bmatrix}=\begin{bmatrix} 0\\ 0\\1 \\ 0\end{bmatrix}=|10\rangle$$

The math confirms our intuition.

When we describe the quantum states and operations in terms of mathematical formulae, we use the vectors $|0\rangle$ and $\mathrm{\mid }1$ ...