Confusion regarding the interpretation of no-pancake theorem

Question

I am reading about phase damping channel from Preskill's notes on quantum information. It is shown that the channel causes decay of the $x$ and $y$ components of the spin polarization of the density matrix representing a single qubit, resulting the Block ball to shrink to a prolate spheroid along $z$ axis. Then he discusses whether there can be a channel which shrinks only one component of the polarization, making the Bloch ball an oblate spheroid instead. In this context, he says (page $29$ of the notes)

You might wonder whether there is a quantum channel which causes just one component of the polarization to decay, mapping the Bloch ball to an oblate spheroid which touches the Bloch sphere along its equator. In fact no such map can be completely positive (the no-pancake theorem). Up to a unitary change of basis, a pancake map shrinks the value of $P_2$ while preserving $P_1$ and $P_3$, which is equivalent to taking the transpose of the density operator with nonzero probability, and we have seen in $\S3.2.6$ that the transpose map is not completely positive.

However,I am unable to make any connection of the transpose of the said density matrix with the pancake map mentioned in the notes. I do not understand how we can write this map as the transpose of the density operator. Can anyone please explain how the calculation can be done?1

Answer 1

Consider the map $\Phi(\rho)=p\rho+(1-p) \rho^T$. For example if $\rho$ is a state in $\mathbb{C}^2$ then this reads explicitly $$\Phi(\rho)=\begin{pmatrix}\rho_{11} & p\rho_{12}+(1-p)\rho_{21}\\ p\rho_{21}+(1-p)\rho_{12} & \rho_{22} \end{pmatrix},$$ and you can directly verify that $$\operatorname{tr}[X\Phi(\rho)]=2\operatorname{Re}(\Phi(\rho)_{12}) = 2\operatorname{Re}(\rho_{12})=\operatorname{tr}[X\rho],$$ and similarly that $$\operatorname{tr}[Z\Phi(\rho)]=\Phi(\rho)_{11}-\Phi(\rho)_{22}=\rho_{11}-\rho_{22}=\operatorname{tr}[Z\rho],$$ hence the shrinking in only one direction.

These mean that the expectation values of $X$ and $Z$ are left untouched by this operation. On the other hand, $$\operatorname{tr}[Y\Phi(\rho)]= 2\operatorname{Im}(\Phi(\rho)_{21}) = 2[p\operatorname{Im}(\rho_{21})+(1-p)\operatorname{Im}(\rho_{12})] =(2p-1)\operatorname{tr}[Y\rho].$$