To find operation elements for the Amplitude Damping channel, Nielsen and Chuang (in Section 8.3.5 of my copy) use the action of a beamsplitter on an initial state $ \alpha |0\rangle + \beta |1\rangle$. The output of the beamsplitter is $\alpha|0_E0\rangle + \beta \cos\theta |0_E1\rangle + \beta \sin\theta|1_E0\rangle$ where $E$ denotes environment. They say after tracing out the environment we get $E_0\rho E_0^\dagger + E_1\rho E_1^\dagger$ where

$E_0 =\begin{pmatrix}1 & 0 \\ 0 & \sqrt{1-\gamma}\end{pmatrix}$,

$E_1 =\begin{pmatrix} 0 & \sqrt{\gamma} \\ 0 & 0\end{pmatrix}$

with $\gamma = \sin^2\theta$.

Here is what I do:

\begin{align*}tr_E(\rho^\prime) &= tr_E(\alpha^2|0_E\rangle \langle0_E|\otimes|0\rangle \langle0| + \beta^2 \cos^2\theta |0_E\rangle \langle0_E|\otimes|1\rangle \langle1| + \beta^2 \sin^2\theta|1_E\rangle \langle1_E|\otimes|0\rangle \langle0|) \\&= \alpha^2|0\rangle \langle0| + \beta^2 \cos^2\theta|1\rangle \langle1| + \beta^2 \sin^2\theta|0\rangle \langle0|\end{align*}

Another approach I tried:

$\langle0_E|B|0_E\rangle = \alpha|0\rangle + \beta \cos\theta|1\rangle$

which gives

$\rho^\prime = \alpha^2|0\rangle \langle0| + \beta^2 \cos^2\theta|1\rangle \langle1|$

and $\begin{pmatrix}1 & 0 \\0 & \sqrt{1-\gamma}\end{pmatrix}$ for $E_0$;


$\langle1_E|B|0_E\rangle = \beta \sin\theta|0\rangle$ with $\rho^\prime = \beta^2 \sin^2\theta|0\rangle \langle0|$

which gives

$E_1 \rho E_1^\dagger= \begin{pmatrix}\beta^2\gamma & 0 \\0 & 0\end{pmatrix} = \begin{pmatrix}0 & \sqrt{\gamma} \\ 0 & 0 \end{pmatrix} \rho \begin{pmatrix}0 & 0 \\ \sqrt{\gamma} & 0 \end{pmatrix}$

Am I correct? If not, where am I wrong?

Here is a picture of the page for your reference: enter image description here

  • $\begingroup$ The density matrix from which you trace over are missing some terms. The will be terms like $\alpha^*\beta\cos(\theta)$ and so on... Some of these terms will go to zero when you take trace but some are left. $\endgroup$
    – Hemant
    Jul 20 '19 at 1:57
  • $\begingroup$ @Hemant Just so you're aware, you can Latex in comments! I've edited yours to add the Latex - hope you don't mind! (although it could be expanded into an answer...) $\endgroup$
    – Mithrandir24601
    Jul 20 '19 at 11:52
  • $\begingroup$ @Mithrandir24601 Thanks for the edit. $\endgroup$
    – Bashir
    Jul 20 '19 at 15:23
  • $\begingroup$ Mith, Thanks for the edit. @Bashir There are some cross terms that DO NOT go to zero as well. $\endgroup$
    – Hemant
    Jul 20 '19 at 17:32
  • $\begingroup$ Yes, you are right. There are non zero cross terms. $\endgroup$
    – Bashir
    Jul 21 '19 at 1:17

The approach I found the best: On page 380 of Nielsen and Chuang,

$B|0\rangle(a|0\rangle+b|1\rangle)=a|00\rangle+b \cos\theta|01\rangle + b \sinθ|10\rangle$.

So, to find the element, just use

$\langle 0|(B|0\rangle|\psi\rangle)=\langle 0(B|0\rangle(a|0\rangle+b|1\rangle))=a|0\rangle+b \cos\theta|1\rangle$ which is then equal to $E_0 |\psi\rangle$.


As a general rule, if a bipartite pure state $\Psi\in\mathbb C^n\otimes\mathbb C^m$ can be written as $$\Psi\equiv\sum_k (u_k\otimes v_k)$$ for some collection of orthogonal vectors $\{u_k\}_k$, then $$\operatorname{Tr}_1[\mathbb P(\Psi)] = \sum_k \|u_k\|^2 (v_k v_k^\dagger), \qquad \mathbb P(\Psi)\equiv \Psi\Psi^\dagger.$$ This is precisely the case at hand. Let $|\psi\rangle\equiv\alpha|0\rangle+\beta|1\rangle$ denote the initial state. The state after the beamsplitter is then $$U|\psi\rangle = |0\rangle\otimes(\alpha|0\rangle+\beta\cos\theta|1\rangle) + \beta\sin\theta |1\rangle\otimes|0\rangle,$$ which can equivalently be written as $$U|\psi\rangle = |0\rangle\otimes (A_0 |\psi\rangle) + |1\rangle\otimes (A_1 |\psi\rangle),$$ where $A_0|\psi\rangle\equiv \alpha|0\rangle+\beta\cos\theta|1\rangle$, and $A_1|\psi\rangle=\beta\sin\theta |0\rangle$. Consequently, $$A_0=\begin{pmatrix}1 & 0 \\ 0 & \cos\theta\end{pmatrix}, \qquad A_1=\begin{pmatrix}0 & \sin\theta \\ 0 & 0\end{pmatrix}.$$ It follows that $$\operatorname{Tr}_1[\mathbb P(|\Psi\rangle)] = A_0\mathbb P(|\psi\rangle) A_0^\dagger + A_1\mathbb P(|\psi\rangle) A_1^\dagger.$$

Note that the above recipe is completely general. What we are really using is the general relation between Stinespring isometry (equivalently, unitary representation) and Kraus representation of the channel. What I'm saying is that, if $$\Phi(X)=\sum_a A_a X A_a^\dagger,$$ then $\Phi(X)=\operatorname{Tr}_2[V X V^\dagger]$, where $$V|\psi\rangle = \sum_a (A_a|\psi\rangle)\otimes e_a.$$


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