# Define a traceless part of $\rho$ [closed]

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

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

• I’m voting to close this question because it is easy to solve if the asker is more careful Oct 23, 2023 at 13:08

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