标签：depolarizing langle group equivalent tilde right mathcal rangle left

I will give an extended explanation of Nielsen's proof, i.e. your first ref link.
The idea is that, \(\rho=\sum_ip_i|i\rangle\langle i|\), we can prove it's depolarizing channel for each \(|i\rangle\langle i|\) with same \(p\), then we are done.

I start after eq.(10):$$V \mathcal{E}_T(\rho) V^{\dagger}=\mathcal{E}_T\left(V \rho V^{\dagger}\right)\tag{1}$$
For one \(|i\rangle \langle i|\), we can choose \(V\) to be diagonal block with respect to \(|i\rangle \langle i|\) and \(I-|i\rangle \langle i|\), you can think it as written \(V\) in basis of \(|i\rangle\) as $\left( \begin{matrix}
a& 0\
0& B\
\end{matrix} \right) $ where \(a\) is a number and \(B\) is a matrix. Now by eq1 we have $V\mathcal{E} _T\left( |i\rangle \langle i| \right) V^{\dagger}=\mathcal{E} _T\left( |i\rangle \langle i| \right) $, hence we have \(\left[ V,\mathcal{E} _T\left( |i\rangle \langle i| \right) \right] =0\). I skip the proof that if \(\left[ V,\mathcal{E} _T\left( |i\rangle \langle i| \right) \right] =0\) for all block diagonal unitary of the form mentioned above, we can have $\mathcal{E} _T(|i\rangle \langle i|)=\alpha |i\rangle \langle i|+\beta \left( I-|i\rangle \langle i| \right) $. Notice that \(\mathcal{E} _T\) is trace preserving so we can rewrite it as \(\mathcal{E} _T(|i\rangle \langle i|)=pI/d+(1-p)|i\rangle \langle i|\) for some \(p\). Then we want to show that for different \(|i\rangle \langle i|\), the \(p\) is the same. To see this, we know that $|\tilde{i}\rangle $ and $|i\rangle $ can be connected with a \(U\) such that \(|\tilde{i}\rangle \langle \tilde{i}|=U|i\rangle \langle i|U^{\dagger}\). Then we will have

\[\mathcal{E} _T(|\tilde{i}\rangle \langle \tilde{i}|)=\mathcal{E} _T(U|i\rangle \langle i|U^{\dagger})=U\mathcal{E} _T(|i\rangle \langle i|)U^{\dagger} \\ =U\left( pI/d+(1-p)|i\rangle \langle i| \right) U^{\dagger} \\ =pI/d+(1-p)|\tilde{i}\rangle \langle \tilde{i}|\]

So for $|\tilde{i}\rangle $ we have the same \(p\).

Remark
Notice that twirling does not have to be done w.r.t. to unitary group \(U(d)\), any group \(G\) can have its corresponding twirling operation, but we can see from the end of the reasoning above that if we want to connect any two pure state \(|i\rangle\) and \(|\tilde i\rangle\), then we must have the twirling w.r.t. \(U(d)\).

标签：depolarizing,langle,group,equivalent,tilde,right,mathcal,rangle,left
来源： https://www.cnblogs.com/nana22/p/16683886.html

本站声明： 1. iCode9 技术分享网（下文简称本站）提供的所有内容，仅供技术学习、探讨和分享；
2. 关于本站的所有留言、评论、转载及引用，纯属内容发起人的个人观点，与本站观点和立场无关；
3. 关于本站的所有言论和文字，纯属内容发起人的个人观点，与本站观点和立场无关；
4. 本站文章均是网友提供，不完全保证技术分享内容的完整性、准确性、时效性、风险性和版权归属；如您发现该文章侵犯了您的权益，可联系我们第一时间进行删除；
5. 本站为非盈利性的个人网站，所有内容不会用来进行牟利，也不会利用任何形式的广告来间接获益，纯粹是为了广大技术爱好者提供技术内容和技术思想的分享性交流网站。