0
$\begingroup$

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

enter image description here

$\endgroup$
8
  • $\begingroup$ 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

1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.