Phase kick-back

Takeaway

Phase kickback is a quantum mechanical effect where applying a controlled gate transfers the phase from a target qubit to a control qubit. This happens when the target qubit is in an eigenstate of the unitary operation being applied.

How It Works

The core principle of phase kickback relies on the setup of a controlled unitary operation (controlled-$U$ gate). This gate applies the unitary operator U to the target qubit only if the control qubit is in the state $|1\rangle$.

Let's represent the control qubit state as $|\psi\rangle = \alpha|0\rangle + \beta|1\rangle$ and the target (ancilla) qubit as $|t\rangle$. The combined initial state is:

$$|\psi_{\text{in}}\rangle = (\alpha|0\rangle + \beta|1\rangle) \otimes |t\rangle = \alpha|0\rangle|t\rangle + \beta|1\rangle |t\rangle$$

Applying the controlled-$U$ gate gives:

$$|\psi_{\text{out}}\rangle = \alpha|0\rangle|t\rangle + \beta|1\rangle(U|t\rangle)$$

Now, in order to perfrom the phase kick back, the target qubit $|t\rangle$ must be the eigenstate of the operator $U$. (An eigenstate is a state that, when acted upon by an operator, only changes by a scalar factor called the eigenvalue.) For a unitary operator, this eigenvalue is of the form $e^{i\phi}$.

So, let's assume $U|t\rangle = e^{i\phi}|t\rangle$. We substitute this into our equation:

$$|\psi_{\text{out}}\rangle = \alpha|0\rangle|t\rangle + \beta|1\rangle(e^{i\phi}|t\rangle)$$

We can now factor out the target qubit state $|t\rangle$, which is unchanged:

$$|\psi_{out}\rangle = (\alpha|0\rangle|t\rangle + e^{i\phi}\beta|1\rangle)|t\rangle$$

Notice that

• The target qubit $|t\rangle$ is completely unaffected.
• The control qubit state has changed from $\alpha|0\rangle|t\rangle + \beta|1\rangle$ to $\alpha|0\rangle|t\rangle + e^{i\phi}\beta|1\rangle$. A relative phase of $e^{i\phi}$ has been "kickback" from the target to the control qubit.

Example - CNOT gate

Let's consider a CNOT gate where the unitary operation is the Pauli-X gate X. We know that X has two eigenstates:

• $|+\rangle = \frac{|0\rangle + |1\rangle}{\sqrt{w}}$, with eigenvalue $e^{i0} = +1$.
• $|-\rangle = \frac{|0\rangle - |1\rangle}{\sqrt{w}}$, with eigenvalue $e^{i\pi} = -1$.

Let's set state $|-\rangle$ as our target qubit. We are expecting to get $X|-\rangle = -|-\rangle$ when we apply X on state $|-\rangle$ due to its eigenvalue. First we set our initial state as

$$|\psi_{\text{in}}\rangle = |+\rangle|-\rangle$$

the we apply the $CNOT$ gate,

$$\begin{array}{ll} CNOT(|+\rangle|-\rangle) & = CNOT \bigg(\frac{|0\rangle + |1\rangle}{\sqrt{2}} \otimes |-\rangle \bigg)\\ \ & = \frac{1}{\sqrt{2}}\bigg(CNOT|0\rangle|-\rangle + CNOT|1\rangle|-\rangle\bigg), \end{array}$$

remember that $CNOT$ clips the target if the control is $|1\rangle$. Thus,

• For the $|0\rangle|-\rangle$ part, the control is $|0\rangle$, so nothing happens.
• For the $|1\rangle|-\rangle$ part, the control is $|1\rangle$, so the X gate is applied to the target: $|1\rangle(X|-\rangle) = |1\rangle(-|-\rangle) = -|1\rangle|-\rangle$.

thus the final state will be

$$\begin{array}{ll} |\psi_{\text{out}}\rangle & = \frac{1}{\sqrt{2}}\bigg(|0\rangle|-\rangle-|1\rangle|-\rangle\bigg) = \bigg(\frac{1}{\sqrt{2}}(|0\rangle - |1\rangle)\bigg) \otimes |-\rangle\\ \ & = |-\rangle|-\rangle \end{array}$$

The phase eigenvalue of $-1$ from the target qubit was kicked back, flipping the control qubit from $|+\rangle$ to $|-\rangle$, while the target qubit remained in the $|-\rangle$ state.