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I saw in a paper: $|\bar{\rho}\rangle\rangle=|\rho\rangle\rangle-|\hat{I}\rangle\rangle / 2^{n / 2}$ for the $4^n$-dimensional vector representing the traceless part of $\rho$. https://arxiv.org/abs/2308.15648v1

For example, $n=1$, $ |\rho\rangle\rangle = \begin{pmatrix} a \\ b \\ c \\ d \end{pmatrix}, $ $a+d=1$. \begin{equation} |\hat{I}\rangle\rangle = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ 0 \\ 0 \\ 1 \end{pmatrix}, \end{equation} \begin{equation} |\bar{\rho}\rangle\rangle = \begin{pmatrix} a - \frac{1}{\sqrt{2}} \\ b \\ c \\ d - \frac{1}{\sqrt{2}} \end{pmatrix}, \end{equation} $tr(\bar{\rho})=a+d-\frac{2}{\sqrt{2}}\ne0$.

Here, why not $|\bar{\rho}\rangle\rangle=|\rho\rangle\rangle-|\hat{I}\rangle\rangle / 2^{n}$?

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    $\begingroup$ I’m voting to close this question because it is easy to solve if the asker is more careful $\endgroup$
    – karry
    Oct 23, 2023 at 13:08

2 Answers 2

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You've not been careful enough with the factors of $2^{n/2}$. There are two of them: one in making $\hat I$ instead of $I$ and another when subtracting $|\hat I\rangle\rangle$. In other words, your $|\bar\rho\rangle\rangle$ is incorrect and the terms really are $a-\frac12$ etc. as you expect.

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  • $\begingroup$ I see, there are two factors $2^{n/2}$, omg, thank you $\endgroup$
    – karry
    Oct 23, 2023 at 13:04
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It is right, be careful! $n=1$, $ |\rho\rangle\rangle = \begin{pmatrix} a \\ b \\ c \\ d \end{pmatrix}, $ $a+d=1$. \begin{equation} |\hat{I}\rangle\rangle = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ 0 \\ 0 \\ 1 \end{pmatrix}, \end{equation} \begin{equation} |\bar{\rho}\rangle\rangle = \begin{pmatrix} a - \frac{1}{2} \\ b \\ c \\ d - \frac{1}{2} \end{pmatrix}, \end{equation} $tr(\bar{\rho})=a+d-1=0$.

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