# How can I represent the completely mixed state as $\frac I2=\frac14(\rho+X\rho X+Y\rho Y+Z\rho Z)$?

Consider the completely mixed state $$I/2$$. The equation comes from Eq.(8.101) of Nielsen's book:

$$\frac{I}{2}=\frac{\rho+X\rho X+Y\rho Y+Z\rho Z}{4}$$, How comes this equation?

• have you tried multiplying it out? Sep 19, 2022 at 9:00
• @Rammus What does "multiplying it out" mean? Sep 19, 2022 at 11:31
• – glS
Sep 20, 2022 at 8:27

There are two ways to see it: mathematically, and "intuitively".

## Mathematically

$$\rho$$ can be written as: $$\rho=\begin{pmatrix}a&b\\\overline{b}&1-a\end{pmatrix}$$ Thus, we have: $$X\rho X=\begin{pmatrix}0&1\\1&0\end{pmatrix}\begin{pmatrix}a&b\\\overline{b}&1-a\end{pmatrix}\begin{pmatrix}0&1\\1&0\end{pmatrix}=\begin{pmatrix}\overline{b}&1-a\\a&b\end{pmatrix}\begin{pmatrix}0&1\\1&0\end{pmatrix}=\begin{pmatrix}1-a&\overline{b}\\b&a\end{pmatrix}$$ $$Y\rho Y=\begin{pmatrix}0&-\mathrm{i}\\\mathrm{i}&0\end{pmatrix}\begin{pmatrix}a&b\\\overline{b}&1-a\end{pmatrix}\begin{pmatrix}0&-\mathrm{i}\\\mathrm{i}&0\end{pmatrix}=\begin{pmatrix}-\mathrm{i}\overline{b}&-\mathrm{i}(1-a)\\\mathrm{i}a&\mathrm{i}b\end{pmatrix}\begin{pmatrix}0&-\mathrm{i}\\\mathrm{i}&0\end{pmatrix}=\begin{pmatrix}1-a&-\overline{b}\\-b&a\end{pmatrix}$$ $$Z\rho Z=\begin{pmatrix}1&0\\0&-1\end{pmatrix}\begin{pmatrix}a&b\\\overline{b}&1-a\end{pmatrix}\begin{pmatrix}1&0\\0&-1\end{pmatrix}=\begin{pmatrix}a&b\\-\overline{b}&-(1-a)\end{pmatrix}\begin{pmatrix}1&0\\0&-1\end{pmatrix}=\begin{pmatrix}a&-b\\-\overline{b}&1-a\end{pmatrix}$$ Thus, we have: $$\rho+X\rho X + Y\rho Y + Z\rho Z=\begin{pmatrix}a+1-a+1-a+a&b+\overline{b}-\overline{b}-b\\\overline{b}+b-b-\overline{b}&1-a+a+a+1-a\end{pmatrix}=\begin{pmatrix}2&0\\0&2\end{pmatrix}=2I$$

## Intuitively

The state associated to this density matrix can be thought of as follow:

1. This is either, with probability $$\frac14$$, the original state $$\rho$$ with probability $$\frac14$$.
2. Or this is, with probability $$\frac14$$ the original state $$\rho$$ to which an $$X$$ gate has been applied.
3. Or this is, with probability $$\frac14$$ the original state $$\rho$$ to which a $$Y$$ gate has been applied.
4. Or this is, with probability $$\frac14$$ the original state $$\rho$$ to which a $$Z$$ gate has been applied.

Note that, up to an irrelevant global phase, the $$Y$$ gate is equivalent to a $$ZX$$ gate. In quantum computing, there are three types of errors: bit-flip errors (an $$X$$ gate has been applied), phase-flip errors (a $$Z$$ gate has been applied) and both (a $$ZX$$ gate has been applied).

Thus, the state $$\frac{\rho+X\rho X + Y\rho Y + Z\rho Z}{4}$$ is a state on which an unknown error (no error, bit-flip, phase-flip or both) has been applied (each case with the same probability). Intuitively, there's not much one knows about this state. In fact, as shown by this result, it is possible to show that we don't know anything about this state.

I'm no expert in the field and it is likely that the "intuitive" reasoning has flaws. Should someone spot such a flaw, please comment or edit the answer accordingly.