Show that $I = \frac{\rho + \sigma_x\rho\sigma_x +\sigma_y\rho\sigma_y + \sigma_z\rho\sigma_z}{2}$ for all states $\rho$

Question

I am trying to show that for any qubit state p, the following holds:

$$I = \frac{\rho + \sigma_x\rho\sigma_x +\sigma_y\rho\sigma_y + \sigma_z\rho\sigma_z}{2}$$

I have tried different manipulations, but got stuck... Will be grateful for any help!

Answer 1

This method is largely similar to Davit's (this covers a slightly more general case where $\rho$ is any arbitrary matrix with trace 1, and you easily see how to adjust it without the trace 1 condition). Any $2\times 2$ matrix can be decomposed as $aI+\vec{n}\cdot\vec{\sigma}$ if we allow $a$ and $\vec{n}$ to take on arbitrary complex values. Moreover, two $2\times 2$ matrices are equal if an only if their values of $a$ and $\vec{n}$ are equal. So, let $$ \tau=\frac{\rho + \sigma_x\rho\sigma_x +\sigma_y\rho\sigma_y + \sigma_z\rho\sigma_z}{2}. $$ We want to show that $a=1$ and $\vec{n}=0$. Now, $$ a=\text{Tr}(\tau)/2,\qquad n_i=\text{Tr}(\sigma_i\tau)/2. $$ Remember that trace is invariant under cyclic permutations, so $$ a=\frac{1}{4}\text{Tr}(\rho + \sigma_x\rho\sigma_x +\sigma_y\rho\sigma_y + \sigma_z\rho\sigma_z)=\frac{1}{4}\text{Tr}(\rho + \rho\sigma_x^2 +\rho\sigma_y^2 + \rho\sigma_z^2)=\text{Tr}(\rho)=1. $$ Similarly, $$ n_x=\frac12\text{Tr}(\sigma_x\rho + \rho\sigma_x +\sigma_x\sigma_y\rho\sigma_y + \sigma_x\sigma_z\rho\sigma_z)=\frac12\text{Tr}(2\sigma_x\rho +\rho\sigma_y\sigma_x\sigma_y + \rho\sigma_z\sigma_x\sigma_z). $$ Now use the anti-commutation properties of the Pauli matrices to get $$ n_x=\frac12\text{Tr}(2\rho\sigma_x -\rho\sigma_x - \rho\sigma_x)=0. $$ The other two components are just the same.

Answer 2

For arbitrary one qubit density matrix we have:

$$\rho = \frac{I}{2} + \frac{r_x \sigma_x + r_y \sigma_y + r_z \sigma_z}{2}$$

where $|r| \le 1$. Here we should take into account that $\sigma_i \sigma_j \sigma_i = -\sigma_j$ where $i \ne j$ and $i, j \in \{x, y, z\}$, and, also $\sigma_i\sigma_i=I$. With this we will obtain the equality presented in the question. Let's see, for example what will be equal the $\sigma_x \rho \sigma_x$ term:

$$\sigma_x \rho \sigma_x = \frac{I}{2} + \frac{r_x \sigma_x - r_y \sigma_y - r_z \sigma_z}{2}$$

Similarly, we will obtain:

$$\frac{\rho + \sigma_x\rho\sigma_x +\sigma_y\rho\sigma_y + \sigma_z\rho\sigma_z}{2} = \\ =I + \frac{2r_x \sigma_x + 2r_y \sigma_y + 2r_z \sigma_z -2r_x \sigma_x - 2r_y \sigma_y - 2r_z \sigma_z}{2} = I $$

Answer 3

This is a special instance of a general linear algebra result.

Note that the identity matrix $\newcommand{\vec}{\operatorname{vec}}I$ can be decomposed as $I=\sum_k v_k\otimes v_k^*$ for any orthonormal basis $\{v_k\}_k$, and vice versa any such decomposition identifies the identity matrix.

Now notice that the Pauli matrices are an orthonormal basis in an enlarged Hilbert space, meaning that $$\operatorname{Tr}[(\sigma_i/\sqrt2)(\sigma_j/\sqrt2)]=\delta_{ij}.$$ More explicitly, this is saying that we can think of the matrices $\sigma_i$ as orthonormal vectors in some space, $i.e.$ we have $\langle \vec(\sigma_i/\sqrt2),\vec(\sigma_j/\sqrt2)\rangle=\delta_{ij}$, where $\vec(B)$ is the vectorisation of the operator $B$.

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If $A_a\in\mathrm{Lin}(\mathcal X,\mathcal Y)$ are a set of such orthonormal operators, we have $$\mathrm{tr}(A_a^\dagger A_b)=\delta_{ab} \Longleftrightarrow \langle\mathrm{vec}(A_a),\mathrm{vec}(A_b)\rangle=\delta_{ab},$$ where $\vec(A_a)\in\mathcal Y\otimes\mathcal X$ is the vectorisation of $A_a$. If the set is a basis, then we also have $$\sum_a (A_a)_{12} (A_a^*)_{34} = \delta_{13}\delta_{24} \Longleftrightarrow \sum_a\vec(A_a)\vec(A_a)^\dagger = I_{\mathcal Y\otimes\mathcal X} $$ Now, the statement we are interested in is of the form $\sum_a A_a \rho A_a^\dagger = I$. This amounts to $$ \sum_{a34} (A_a)_{13} (A_a^*)_{24} \rho_{34} = \delta_{12} \Longleftrightarrow \sum_a (A_a\otimes A_a^*)\vec(\rho) = \lvert m\rangle, $$ where $\lvert m\rangle\equiv\sum_k \lvert k,k\rangle$. The question is thus, what type of operator is $\sum_a A_a\otimes A_a^*$? Componentwise, the relation with $\sum_a \vec(A_a)\vec(A_a)^\dagger$ is clear: $$(A_a\otimes A_a^*)_{ij,nm} = (A_a)_{in} (A_a^*)_{jm} = (\vec(A_a)\vec(A_a)^\dagger)_{in,jm} = \delta_{ij}\delta_{nm},$$ that is, $A_a\otimes A_a^*$ is the Choi of $\vec(A_a)\vec(A_a)^\dagger$. Summing over $a$ this is the identity, which means that $\sum_a A_a\otimes A_a^*$ is the Choi of the identity, which is the projector over the maximally entangled state: $$\sum_a A_a\otimes A_a^*=\lvert m\rangle\!\langle m\rvert.$$ We conclude that $$\sum_A A_a \rho A_a^\dagger = \operatorname{unvec}\left(\sum_a (A_a\otimes A_a^*) \vec(\rho)\right) = \operatorname{unvec}(\lvert m\rangle ) = I_{\mathcal X}.$$

Answer 4

I know this is an old question, but I feel like giving, imo, the simplest answer following the question specified in N&C.

Firstly as specified in the question define:

$ \mathcal{E}(A) = \frac{A + XAX + YAY + ZAZ}{4}$ .

It it is easy to see that

$ \mathcal{E}(I) = \frac{I + XIX + YIY + ZIZ}{4} = \frac{I + XX + YY + ZZ}{4} = I$

For the other three quantities $\mathcal{E}(X),\mathcal{E}(Y),\mathcal{E}(Z)$ we can make use of the elementary identities:

$\sigma_i\sigma_i\sigma_i = \sigma_i$

$\sigma_i\sigma_j\sigma_i = -\sigma_j$

Plugging in these identities we can see $\mathcal{E}(X),\mathcal{E}(Y),\mathcal{E}(Z) = 0$. More generally, (after grouping the terms), we can see that:

$ \mathcal{E}(\sigma_i) = \frac{2\sigma_i - 2\sigma_i}{4} = 0$

Finally we know from e.q 2.175 that

$\rho = \frac{I + \vec{r} \cdot\vec{\sigma}}{2} = \frac{I + r_x\sigma_x + r_y\sigma_y + r_z\sigma_z}{2}$,

and plugging this into

$\mathcal{E}(\rho) = \frac{\mathcal{E}(I) + \mathcal{E}(r_x\sigma_x) + \mathcal{E}(r_y\sigma_y) + \mathcal{E}(r_z\sigma_z)}{2}$,

using the results from above we see all the $\mathcal{E}(r_i\sigma_i)=0$, leaving us just with

$\mathcal{E}(\rho) = \frac{\mathcal{E}\left({I}\right)}{2} = \frac{I}{2}$,

finally to complete the proof

$2\mathcal{E}(\rho) = I$

Answer 5

Given that you're only working with a single qubit state it is possible to also show this by direct calculation on a parametrised state. That is, we can write any single qubit $\rho$ as $$ \rho = \begin{pmatrix} a & \beta \\ \overline{\beta} & 1-a \end{pmatrix} $$ with $a\in[0,1]$ and $\beta \in \mathbb{C}$ such that $(1-2a)^2 + 4 |\beta|^2 \leq 1$. Then we can directly compute the action of the Pauli conjugation $$ \sigma_x \rho \sigma_x = \begin{pmatrix} 1-a & \overline{\beta} \\ \beta & a \end{pmatrix} $$ $$ \sigma_y \rho \sigma_y = \begin{pmatrix} 1-a & -\overline{\beta} \\ -\beta & a \end{pmatrix} $$ $$ \sigma_z \rho \sigma_z = \begin{pmatrix} a & -\beta \\ -\overline{\beta} & 1-a \end{pmatrix}. $$ Summing these up with $\rho$ and dividing through by $2$ we get the desired result.

Answer 6

Chapter VII. E. in Daniel Lidar's notes. Use $\rho = \frac{1}{2}(I + \vec{v}\cdot\vec{\sigma})$ and products of Pauli matrices:

Check for each pair that: $\sigma_i \sigma_j = \delta_{ij} I + i \epsilon_{ijk}\sigma_k$

Use it to show: $ \sigma_i \sigma_j \sigma_k = \delta_{ij} \sigma_k - \delta_{ik} \sigma_j + \delta_{jk} \sigma_i + i \epsilon_{ijk} I $

One more step $ \sigma_i \sigma_j \sigma_i = 2\delta_{ij} \sigma_i - \sigma_j = \begin{cases} +\sigma_j &, i = j\\ -\sigma_j &, i \neq j \end{cases} $

with this go to eq. 189 from Daniel Lidar: $$ \sigma_x(I + \vec{v}\cdot \vec{\sigma}) \sigma_x = I + v_x \sigma_x - v_y \sigma_y - v_z \sigma_z $$ $$ \sigma_y(I + \vec{v}\cdot \vec{\sigma}) \sigma_y = I - v_x \sigma_x + v_y \sigma_y - v_z \sigma_z $$ $$ \sigma_z(I + \vec{v}\cdot \vec{\sigma}) \sigma_z = I - v_x \sigma_x - v_y \sigma_y + v_z \sigma_z $$

add it together with $$ I(I + \vec{v}\cdot \vec{\sigma}) I = I + v_x \sigma_x + v_y \sigma_y + v_z \sigma_z $$

to get $$ 2(\rho + \sigma_x \rho \sigma_x + \sigma_y \rho \sigma_y + \sigma_z \rho \sigma_z) = 4I $$

Answer 7

Assuming $\rho$ is a pure state, these are the explicit calculations. You can easily generalize to mixed states.

$\newcommand{\ket}[1]{|{#1}\rangle}$ $\newcommand{\bra}[1]{\langle{#1}|}$ Let $\ket{\psi} = \alpha\ket{0}+ \beta\ket{1}$, where $\alpha,\beta \in \mathbb{C}^2$ and $|\alpha|^2 + |\beta|^2 = 1$.

Thus $\rho = \ket{\psi}\bra{\psi} = \begin{pmatrix} \alpha\alpha^* & \alpha\beta^* \\ \beta\alpha^* & \beta\beta^*\end{pmatrix}$.

Now $\sigma_x = \begin{pmatrix} 0 & 1 \\ 1 & 0\end{pmatrix}$, so $ \sigma_x \rho \sigma_x = \begin{pmatrix} \beta\beta^* & \beta\alpha^* \\ \alpha\beta^* & \alpha\alpha^*\end{pmatrix}$.

Similarly, you can compute $\sigma_y\rho\sigma_y = \begin{pmatrix} \beta\beta^* & -\beta\alpha^* \\ -\alpha\beta^* & \alpha\alpha^*\end{pmatrix}$ and $\sigma_z\rho\sigma_z = \begin{pmatrix} \alpha\alpha* & -\alpha\beta* \\ -\beta\alpha* & \beta\beta*\end{pmatrix}$.

Finally, summing up:

$\frac{1}{2} (\rho + \sigma_x \rho \sigma_x + \sigma_y \rho \sigma_y + \sigma_z \rho \sigma_z) = \frac{1}{2}\begin{pmatrix} 2(\alpha\alpha^* + \beta\beta^*) & 0 \\ 0 & 2(\alpha\alpha^* + \beta\beta^*) \end{pmatrix} = I$.