Suppose I have the density matrix: $$\rho = p|\psi^-\rangle\langle\psi^-| + (1-p)\times \frac14 \mathbb{I}_4 \,,$$where $p$ is some probability $<1$, $\mathbb{I}_4$ is the $4\times 4$ identity matrix, and $|\psi^-\rangle = \frac{1}{\sqrt{2}}(|01\rangle-|10\rangle)$ is one of the four Bell state basis. My goal here is to find the positive partial transpose (PPT) of $\rho$, and to do so in explicit matrix form.
My initial instinct was to do this in the Bell state basis, so that $$|\psi^-\rangle\langle\psi^-| = \begin{pmatrix} 0&0&0&0\\0&0&0&0\\ 0&0&0&0 \\ 0&0&0&1\end{pmatrix} \,.$$
(Is this even correct?)
My question is then: Would $\mathbb{I}_4$ still be written as 1's on the main diagonal, i.e. $$\mathbb{I}_4 = \begin{pmatrix} 1&0&0&0\\0&1&0&0\\ 0&0&1&0 \\ 0&0&0&1\end{pmatrix} \,,$$ or some other thing? (Perhaps by converting the standard basis {$|00\rangle$, etc.} to the Bell state basis?)
My instinct is the former case, then that would make this problem almost too trivial. (I don't have a problem with computing the PPT.)
