In PT-symmetric quantum mechanics, the inner product is given by

\begin{equation}

\label{eq:pt-product}

\langle v_1|P|v_2\rangle

\end{equation}

where $P=\begin{bmatrix}

0 & 1\\

1 & 0

\end{bmatrix}$.

A *PT-symmetric operator* $H$ is a $2\times 2$ symmetric matrix that satisfies

$$\langle Hv_1|P|v_2\rangle=\langle v_1|P|Hv_2\rangle$$

which is equivalent to

$$\bar HP=PH$$

So, PT-symmetric operators are self-adjoint operators with respect to the inner product \eqref{eq:pt-product}. PT-symmetric operators take the form

$$\begin{bmatrix}

a & b\\

b & \bar a

\end{bmatrix}$$

In PT-symmetric quantum mechanics, $\mathrm{NOT}$ gate is given by

$$\mathrm{NOT}=\begin{bmatrix}

1 & 0\\

0 &-1

\end{bmatrix}$$

The reason for this is that in PT-symmetric quantum mechanics the qubits are

$$|0\rangle=\frac{1}{\sqrt{2}}\begin{bmatrix}

1\\

1

\end{bmatrix},\ |1\rangle=\frac{1}{\sqrt{2}}\begin{bmatrix}

1\\

-1

\end{bmatrix}$$

The eigenvectors of $\mathrm{NOT}$ corresponding to the eigenvalues $\pm 1$ are, respectively, $\begin{bmatrix}

1\\

0

\end{bmatrix}$ and $\begin{bmatrix}

0\\

1

\end{bmatrix}$ and their norms are zero with respect to the inner product \eqref{eq:pt-product}. This means that the spectral decomposition of $\mathrm{NOT}$ in terms of its eigenvectors does not exist in PT-symmetric quantum mechanics. This also hints us that $\sqrt{\mathrm{NOT}}$ does not likely exist. In PT-symmetric quantum mechanics, an operator $U$ is called *$P$-unitary* if $U^\dagger PU=P$ and *$P$-antiunitary* if $U^\dagger PU=-P$.

For the same physical reason, in PT-symmetric quantum mechanics, gates are required to be $P$-unitary or $P$-antiunitary. $\mathrm{NOT}$ is $P$-antiunitary. It can be easily seen that $\sqrt{\mathrm{NOT}}$ does not exist. If $\sqrt{\mathrm{NOT}}$ exists, then it is either $P$-unitary or $P$-antiunitary. If $\sqrt{\mathrm{NOT}}$ is $P$-unitary, then

\begin{align*} \mathrm{NOT}^\dagger P\mathrm{NOT}&=\sqrt{\mathrm{NOT}}^\dagger(\sqrt{\mathrm{NOT}}^\dagger P\sqrt{\mathrm{NOT}})\sqrt{\mathrm{NOT}}\\ &=\sqrt{\mathrm{NOT}}^\dagger P\sqrt{\mathrm{NOT}}\\ &=P \end{align*}

This contradicts to the fact that $\mathrm{NOT}$ is $P$-antiunitary. If $\sqrt{\mathrm{NOT}}$ is $P$-antiunitary, then

\begin{align*} \mathrm{NOT}^\dagger P\mathrm{NOT}&=\sqrt{\mathrm{NOT}}^\dagger(\sqrt{\mathrm{NOT}}^\dagger P\sqrt{\mathrm{NOT}})\sqrt{\mathrm{NOT}}\\ &=-\sqrt{\mathrm{NOT}}^\dagger P\sqrt{\mathrm{NOT}}\\ &=P \end{align*}

Again, this is a contradiction. Therefore, $\sqrt{\mathrm{NOT}}$ does not exist. As seen here, the standard hermitian quantum mechanics admits $\sqrt{\mathrm{NOT}}$ gate. This implies that the standard hermitian quantum mechanics and PT-symmetric quantum mechanics cannot be consistent with each other. While no one has yet come up with a physical implementation of $\sqrt{\mathrm{NOT}}$ gate, it is interesting to see that the physical viability of a quantum theory hangs on the physical realizability of a quantum gate.