1
$\begingroup$

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 $E_0$;

and

$\langle1_E|B|0_E\rangle = \beta \sin\theta|0\rangle$ 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

$\endgroup$
  • $\begingroup$ 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 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 at 11:52
  • $\begingroup$ @Mithrandir24601 Thanks for the edit. $\endgroup$ – Bashir Jul 20 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 at 17:32
  • $\begingroup$ Yes, you are right. There are non zero cross terms. $\endgroup$ – Bashir Jul 21 at 1:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.