Question: Is the transpose a valid quantum operation?

To make things a little more rigorous, let an operation $\Lambda$ on qubits be defined as $\Lambda(\rho)=\rho^T$, where $\rho^T$ denotes the transpose of $\rho$.

Consider the one qubit state $\ket{\psi_+}=\frac{1}{\sqrt{2}}(\ket{0}+i\ket{1})$ so that

\[ \begin{align*}

\ket{\psi_+}\bra{\psi_+}&=\frac{1}{2}(\ket{0}+i\ket{1})(\bra{0}-i\bra{1}) \\

&=\frac{1}{2}(\ket{0}\bra{0}-i\ket{0}\bra{1}+i\ket{1}\bra{0}+\ket{1}\bra{1}) \\

&=\frac{1}{2}\begin{pmatrix} 1&-i \\ i&1\end{pmatrix}.

\end{align*}\]

Then

\[ \begin{align*}

\Lambda(\ket{\psi_+}\bra{\psi_+})=\frac{1}{2}\begin{pmatrix} 1&-i \\ i&1\end{pmatrix}^T

&=\frac{1}{2}\begin{pmatrix} 1&i \\ -i&1\end{pmatrix} \\

&=\frac{1}{2}(\ket{0}\bra{0}+i\ket{0}\bra{1}-i\ket{1}\bra{0}+\ket{1}\bra{1}) \\

&=\frac{1}{2}(\ket{0}-i\ket{1})(\bra{0}+i\bra{1}) \\

&=\ket{\psi_-}\bra{\psi_-},

\end{align*}\]

where $\ket{\psi_-}=\frac{1}{\sqrt{2}}(\ket{0}-i\ket{1})$. Then the inner product of $\ket{\psi_-}$ and $\ket{\psi_+}$ is

\[ \begin{align*}

\ip{\psi_-}{\psi_+}&=\frac{1}{2}(\bra{0}+i\bra{1})(\ket{0}+i\ket{1}) \\

&=\frac{1}{2}(\ip{0}{0}+i\ip{0}{1}+i\ip{1}{0}-\ip{1}{1}) \\

&=\frac{1}{2}(1+0+0-1) \\

&=0.

\end{align*}\]

Thus, $\ket{\psi_+}$ and $\ket{\psi_-}$ are orthogonal pure states such that $\Lambda(\ket{\psi_+}\bra{\psi_+})=\ket{\psi_-}\bra{\psi_-}$.

Now, it will be proven that there does not exist a unitary operation $U$ such that $\Lambda(\rho)=U\rho U^\dagger$ for all $\rho$.

It suffices to show that such a unitary does not exist in the single qubit (two dimensional) case. For the sake of contradiction suppose that such a unitary $U$ does exist. Moreover, consider the two states $\ket{0}$ and $\ket{\psi_+}=\frac{1}{\sqrt{2}}(\ket{0}+\ket{1})$, whose density operators are given by

\[

\ket{0}\bra{0}=\begin{pmatrix} 1&0 \\0&0 \end{pmatrix} \ \ \ \text{and} \ \ \ \ket{\psi_+}\bra{\psi_+}=\frac{1}{2}\begin{pmatrix} 1&1\\1&1 \end{pmatrix}.

\]

Then

\[ \begin{align*}

\Lambda(\ket{0}\bra{0})&=\begin{pmatrix} 1&0 \\0&0 \end{pmatrix}^T=\begin{pmatrix} 1&0 \\0&0 \end{pmatrix}=\ket{0}\bra{0} \\

\Lambda(\ket{\psi_+}\bra{\psi_+})&=\frac{1}{2}\begin{pmatrix} 1&1\\1&1 \end{pmatrix}^T=\frac{1}{2}\begin{pmatrix} 1&1\\1&1 \end{pmatrix}=\ket{\psi_+}\bra{\psi_+},

\end{align*}\]

which shows that these two states remain the same under the transposition operation $\Lambda$. Therefore, these two states must also be left unchanged by the action of the unitary $U$. That is, it must be the case that

\[\begin{align*}

U\ket{0}\bra{0}U^\dagger&=\ket{0}\bra{0},\\

U\ket{\psi_+}\bra{\psi_+}U^\dagger&=\ket{\psi_+}\bra{\psi_+},

\end{align*}\]

or equivalently that $U\ket{0}=\ket{0}$ and $U\ket{\psi_+}=\ket{\psi_+}$. By thinking of these states and the action of $U$ as a rotation on the Block sphere, this implies that both of these states remain fixed and must therefore lie on the axis of rotation of $U$. More explicitly, recall that a one-qubit unitary can be represented as

\[

U=e^{i\alpha}\cos(\theta/2)I-ie^{i\alpha}\sin(\theta/2)\left(c_x X +c_Y Y + c_Z Z\right),

\]

where $\alpha$ is just a phase factor, $\theta$ gives the angle of rotation, and $(c_X,c_Y,c_Z)$ is a unit vector.

Then since the state $\ket{0}$ lies on the $z$-axis of the Bloch sphere, the rotation of $U$ must have its axis of rotation as the $z-$ axis in order to keep $\ket{0}$ fixed, which implies that $c_X,c_Y=0$. Furthermore, since the state $\ket{\psi_+}$ lies along the $y$-axis, this also implies that $U$ must rotate about the $y$-axis in order to ket $\ket{\psi_+}$ fixed, but this implies that $c_X,C_Z=)$. These implications together imply that $c_X,c_Y,c_Z=0$,so that $U=I$. However, as seen above the transpose operation $U$ does not act as the identity operation all states. This is a contradiction, and therefore there cannot exist a unitary $U$ satisfying the desired conditions.

Another, perhaps more general, way to see why such a unitary cannot exist is too note that any operation of the form $\Phi(\rho)=U\rho U^\dagger$ defined with some unitary $U$ always yields a valid quantum operation. This means that $\Phi(\rho)$ is both trace preserving and a completely positive map by definition. On the other hand, although the transpose operation $\Lambda$ is trace preserving it is not completely positive in general, and thus cannot define a valid quantum operation by definition.

To make things a little more rigorous, let an operation $\Lambda$ on qubits be defined as $\Lambda(\rho)=\rho^T$, where $\rho^T$ denotes the transpose of $\rho$.

Consider the one qubit state $\ket{\psi_+}=\frac{1}{\sqrt{2}}(\ket{0}+i\ket{1})$ so that

\[ \begin{align*}

\ket{\psi_+}\bra{\psi_+}&=\frac{1}{2}(\ket{0}+i\ket{1})(\bra{0}-i\bra{1}) \\

&=\frac{1}{2}(\ket{0}\bra{0}-i\ket{0}\bra{1}+i\ket{1}\bra{0}+\ket{1}\bra{1}) \\

&=\frac{1}{2}\begin{pmatrix} 1&-i \\ i&1\end{pmatrix}.

\end{align*}\]

Then

\[ \begin{align*}

\Lambda(\ket{\psi_+}\bra{\psi_+})=\frac{1}{2}\begin{pmatrix} 1&-i \\ i&1\end{pmatrix}^T

&=\frac{1}{2}\begin{pmatrix} 1&i \\ -i&1\end{pmatrix} \\

&=\frac{1}{2}(\ket{0}\bra{0}+i\ket{0}\bra{1}-i\ket{1}\bra{0}+\ket{1}\bra{1}) \\

&=\frac{1}{2}(\ket{0}-i\ket{1})(\bra{0}+i\bra{1}) \\

&=\ket{\psi_-}\bra{\psi_-},

\end{align*}\]

where $\ket{\psi_-}=\frac{1}{\sqrt{2}}(\ket{0}-i\ket{1})$. Then the inner product of $\ket{\psi_-}$ and $\ket{\psi_+}$ is

\[ \begin{align*}

\ip{\psi_-}{\psi_+}&=\frac{1}{2}(\bra{0}+i\bra{1})(\ket{0}+i\ket{1}) \\

&=\frac{1}{2}(\ip{0}{0}+i\ip{0}{1}+i\ip{1}{0}-\ip{1}{1}) \\

&=\frac{1}{2}(1+0+0-1) \\

&=0.

\end{align*}\]

Thus, $\ket{\psi_+}$ and $\ket{\psi_-}$ are orthogonal pure states such that $\Lambda(\ket{\psi_+}\bra{\psi_+})=\ket{\psi_-}\bra{\psi_-}$.

Now, it will be proven that there does not exist a unitary operation $U$ such that $\Lambda(\rho)=U\rho U^\dagger$ for all $\rho$.

It suffices to show that such a unitary does not exist in the single qubit (two dimensional) case. For the sake of contradiction suppose that such a unitary $U$ does exist. Moreover, consider the two states $\ket{0}$ and $\ket{\psi_+}=\frac{1}{\sqrt{2}}(\ket{0}+\ket{1})$, whose density operators are given by

\[

\ket{0}\bra{0}=\begin{pmatrix} 1&0 \\0&0 \end{pmatrix} \ \ \ \text{and} \ \ \ \ket{\psi_+}\bra{\psi_+}=\frac{1}{2}\begin{pmatrix} 1&1\\1&1 \end{pmatrix}.

\]

Then

\[ \begin{align*}

\Lambda(\ket{0}\bra{0})&=\begin{pmatrix} 1&0 \\0&0 \end{pmatrix}^T=\begin{pmatrix} 1&0 \\0&0 \end{pmatrix}=\ket{0}\bra{0} \\

\Lambda(\ket{\psi_+}\bra{\psi_+})&=\frac{1}{2}\begin{pmatrix} 1&1\\1&1 \end{pmatrix}^T=\frac{1}{2}\begin{pmatrix} 1&1\\1&1 \end{pmatrix}=\ket{\psi_+}\bra{\psi_+},

\end{align*}\]

which shows that these two states remain the same under the transposition operation $\Lambda$. Therefore, these two states must also be left unchanged by the action of the unitary $U$. That is, it must be the case that

\[\begin{align*}

U\ket{0}\bra{0}U^\dagger&=\ket{0}\bra{0},\\

U\ket{\psi_+}\bra{\psi_+}U^\dagger&=\ket{\psi_+}\bra{\psi_+},

\end{align*}\]

or equivalently that $U\ket{0}=\ket{0}$ and $U\ket{\psi_+}=\ket{\psi_+}$. By thinking of these states and the action of $U$ as a rotation on the Block sphere, this implies that both of these states remain fixed and must therefore lie on the axis of rotation of $U$. More explicitly, recall that a one-qubit unitary can be represented as

\[

U=e^{i\alpha}\cos(\theta/2)I-ie^{i\alpha}\sin(\theta/2)\left(c_x X +c_Y Y + c_Z Z\right),

\]

where $\alpha$ is just a phase factor, $\theta$ gives the angle of rotation, and $(c_X,c_Y,c_Z)$ is a unit vector.

Then since the state $\ket{0}$ lies on the $z$-axis of the Bloch sphere, the rotation of $U$ must have its axis of rotation as the $z-$ axis in order to keep $\ket{0}$ fixed, which implies that $c_X,c_Y=0$. Furthermore, since the state $\ket{\psi_+}$ lies along the $y$-axis, this also implies that $U$ must rotate about the $y$-axis in order to ket $\ket{\psi_+}$ fixed, but this implies that $c_X,C_Z=)$. These implications together imply that $c_X,c_Y,c_Z=0$,so that $U=I$. However, as seen above the transpose operation $U$ does not act as the identity operation all states. This is a contradiction, and therefore there cannot exist a unitary $U$ satisfying the desired conditions.

Another, perhaps more general, way to see why such a unitary cannot exist is too note that any operation of the form $\Phi(\rho)=U\rho U^\dagger$ defined with some unitary $U$ always yields a valid quantum operation. This means that $\Phi(\rho)$ is both trace preserving and a completely positive map by definition. On the other hand, although the transpose operation $\Lambda$ is trace preserving it is not completely positive in general, and thus cannot define a valid quantum operation by definition.

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