Let's start with

$Tr_{\Omega}[|0,\Omega_{0}\rangle\langle0,\Omega_{0}|U^{\dagger}] = \sum_{\alpha}E_{\alpha}|0\rangle\langle0|E_{\alpha}^{\dagger}$ where $U$ be a unitary operator. The trace operator traces out $|\Omega\rangle$

$U|0,\Omega_{0}\rangle = \sqrt{p}|0,\Omega_{0}\rangle + \sqrt{\frac{1-p}{3}}\bigg(X|0,\Omega_{1}\rangle + Z|0,\Omega_{2}\rangle + Y|0,\Omega_{3}\rangle\bigg)$

In density operator, this is:

$\rho = p|0,\Omega_{0}\rangle\langle0,\Omega_{0}| + \frac{1-p}{3}\bigg(X|0,\Omega_{1}\rangle\langle0,\Omega_{1}| + Y|0,\Omega_{3}\rangle\langle0,\Omega_{3}| + Z|0,\Omega_{2}\rangle\langle0,\Omega_{2}|\bigg)$

Matching terms above one arrives at $E_{1} = \sqrt{\frac{1-p}{3}}X$

some questions:

$E_{1} = \sqrt\frac{1-p}{3}X$.

It is shown that $E_{1}|0\rangle\langle 0|E_{1}^{\dagger} = \frac{1-p}{3}X|0\rangle\langle0|$.

Here's my working:

$E_{1}|0\rangle\langle0|E_{1}^{\dagger} = \frac{1-p}{3}X|0\rangle\langle0|X$

In which $X = X^{\dagger}$.

I could have forgotten a key property that allows for the Pauli operator X to commute with $|0\rangle\langle0|$. Any help is appreciated.

  1. The physical description illustrating the action of unitary $U$ on $|0,\Omega_{0}\rangle\langle0,\Omega_{0}|$ is that there is a probability P that the unitary operator is invariant and probability $\frac{1-p}{3}$ that there is a bit - flip as shown by $\frac{1-p}{3}|0,\Omega_{1}\rangle\langle0,\Omega_{1}|$.

The subscript $1$ in the Omega is confusing. I would have thought that based on the physical description, it would be

$\frac{1-p}{3}|0,\Omega_{0}\rangle\langle0,\Omega_{0}|$ and not $\frac{1-p}{3}|0,\Omega_{1}\rangle\langle0,\Omega_{1}|$

Am I missing some details?

screen shot of material

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  • $\begingroup$ I believe there is some typo in the way you have defined your operator $E_1$. Also please do add some context, or snippet of the resource you are referring to. $\endgroup$
    – FDGod
    Oct 30, 2023 at 3:27
  • $\begingroup$ @FDGod please view the additional information I have provided in the OP $\endgroup$
    – Physkid
    Oct 30, 2023 at 3:41
  • $\begingroup$ the $X,Y,Z$ in the first equation acts on $|0\rangle$ or $|\Omega_n\rangle$? What are $|\Omega_n\rangle$? Can you just directly put a screenshot or link of the resource you are using? $\endgroup$
    – FDGod
    Oct 30, 2023 at 3:57
  • $\begingroup$ Also, I don't get the term matching, and the $E_1$ thing you are referring to is a scalar, not an operator. $\rho$ should be just $$\rho = U|0,\Omega_0\rangle \langle 0, \Omega_0 | U^{\dagger}$$ $\endgroup$
    – FDGod
    Oct 30, 2023 at 3:59
  • 1
    $\begingroup$ Okay, now everything makes sense. It's straightforward. I don't have time right now, but I will reply when I get free if no one else has answered by then. Also, keep only the questions related to algebra in this post. For shrinking the depolarizing channel, please create another post. $\endgroup$
    – FDGod
    Oct 30, 2023 at 4:38

1 Answer 1


You originally have the equation $(38)$. Then you calculate

$$\rho = U |0, \Omega_0 \rangle \langle 0,\Omega_0 | U^{\dagger}\,.$$

Taking the partial trace of $\rho$, $\text{Tr}_{\Omega}(\rho)$ will give you equation $(39)$.

Lets say for simplicity, $$q = \frac{1-p}{3}\,.$$

So you can rewrite equation $(39)$ as follows $$ \begin{align} \text{Tr}_{\Omega}(\rho) &= p|0\rangle\langle0| + \frac{1-p}{3}\bigg( X|0\rangle\langle0|X + Y|0\rangle\langle0|Y + Z|0\rangle\langle0|Z \bigg)\\ &= p|0\rangle\langle0| + q\big( X|0\rangle\langle0|X + Y|0\rangle\langle0|Y + Z|0\rangle\langle0|Z \big)\\ &= p|0\rangle\langle0| + q X|0\rangle\langle0|X + qY|0\rangle\langle0|Y + qZ|0\rangle\langle0|Z \\ &= \sqrt{p}I|0\rangle\langle0|\sqrt{p}I + \sqrt{q}X|0\rangle\langle0|\sqrt{q}X + \sqrt{q}Y|0\rangle\langle0|\sqrt{q}Y + \sqrt{q}Z|0\rangle\langle0|\sqrt{q}Z \\ &= E_0|0\rangle\langle0|E_0^{\dagger} + E_1|0\rangle\langle0|E_1^{\dagger} + E_2|0\rangle\langle0|E_2^{\dagger} + E_3|0\rangle\langle0|E_3^{\dagger} \\ \end{align} $$ where $\{E_\alpha\}$ are exactly the operators given in equation $(40)$. Now you can see that the equation $(37)$ holds true here.

$\text{Tr}_{\Omega}(\rho)$ is a mixed state, a density matrix. You cannot write it as a pure state the way you have described in the question.


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