# Identity Matrix in other bases

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.)

• The matrix for the identity operator is $\operatorname{diag}(1,1,\dots)$ in any basis. – Luke Pritchett Jul 13 '17 at 18:06
• You are aware that the (partial) transpose is a basis dependent concept? – Norbert Schuch Jul 13 '17 at 21:06
• @NorbertSchuch Yes, I am aware. Is thst a problem here? – Troy Jul 14 '17 at 1:21
• Not if you know what you are doing. But partial transposition only makes sense w.r.t. a product basis. What are you going to do with the matrix in the Bell basis? – Norbert Schuch Jul 14 '17 at 13:59
• @NorbertSchuch I didn't think about it because it was a diagonal matrix, (and so I assumed it would remain the same) but I think I was going for transposing the off-diagonal 2x2 blocks. Which, now thinking about it, doesn't really make sense in the Bell basis since I can't identify a particle A and particle B (and therefore the 2x2 blocks) to do a partial transpose on. – Troy Jul 14 '17 at 15:27

By definition the unit (or identity) matrix $\mathbb{I}_n$ is the matrix with $1$'s on the diagonal and $0$'s elsewhere. This is so because, if $A$ is any $n\times n$ square matrix, one must have $$\mathbb{I}_n\cdot A = A\cdot \mathbb{I}_n = A\, .$$