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

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!

• Since there are many answers now and it is a "textbook" question, let me ask, what did you try? Because all you need is to know the properties of the Pauli matrices and how to represent a qubit state in the Bloch representation. – keisuke.akira Jul 28 at 20:51

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.

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:

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

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 $$B$$.

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 $$\sum_A A_a \rho A_a^\dagger = \operatorname{unvec}\left(\sum_a (A_a\otimes A_a^*) \vec(\rho)\right) = \operatorname{unvec}(\lvert m\rangle ) = I_{\mathcal X}.$$

• It's not quite clear how it's equivalent to the question statement. – Danylo Y Jul 26 at 19:29
• @DanyloY I added some more detail. Makes more sense now? – glS Jul 26 at 20:00
• The idea is clear, though the details look a little bit messed up. The common vectorization formula is $\operatorname{vec}(ABC) = (C^T \otimes A)\operatorname{vec}(B)$. Also note that $\sigma_y^T = -\sigma_y$. – Danylo Y Jul 26 at 21:23
• @DanyloY indeed, going carefully over the steps there were a few more formal steps required. I guess one can state the result concisely as (1) thinking of the statement as a map acting on a state, the $\sigma_i$ are understood as Kraus ops (2) switching in the natural representation it becomes a question of what is $\sum_a A_a\otimes A_a^*$ (3) the Choi of $A\otimes A^*$ is $\mathrm{vec}(A)\mathrm{vec}(A)^\dagger$ (4) the latter is the identity if the vectorised ops are a basis. I have to say though, I think it's much easier to think componentwise here – glS Jul 27 at 7:33
• regarding the "vectorisation formula", I gues it depends on the convention? I'm simply using $\mathrm{vec}(A)_{12}=A_{12}$, thus $\mathrm{vec}(ABC)_{12}=(ABC)_{12}=A_{13}B_{34}C_{42}=[(A\otimes C^T)B]_{12}$. I'm not sure what is the point of your pointing out $\sigma_y^T=-\sigma_y$. – glS Jul 27 at 7:37

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.

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

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

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

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


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:

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