Twirling operation w.r.t. a group $U(d)$ is equivalent to a depolarizing channel

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