I'm working with Exercise 8.21 of the Nielsen-Chuang book on quantum information. It illustrates the amplitude-damping quantum channel by the interaction between two harmonic oscillators (the first one models the system S, the second one models the environment E) described by the Hamiltonian

\begin{equation} \hat{H} = \chi (\hat{a}^{\dagger}_S \otimes \hat{b}_E + \hat{a}_S \otimes \hat{b}^{\dagger}_E), \end{equation}

where $\hat{b}_E$, $\hat{b}^\dagger_E$ are the creation/annihilation operators for the oscillator which models the environment. The evolution operator reads \begin{equation} \hat{U}(\Delta t) = e^{-i \chi (\hat{a}^{\dagger}_S \otimes \hat{b}_E + \hat{a}_S \otimes \hat{b}^{\dagger}_E) \Delta t} \end{equation} and represents a unitary transformation in SE which corresponds to the amplitude-damping channel of interest in S. This channel can be represented as an operator sum in S by tracing out the degrees of freedom of E, so the Kraus operators are $\hat{E}_k = \langle k_E | \hat{U}(t) |0_E\rangle $, where $| k_E \rangle$ are the Fock states of E with $k$ quanta.

Nielsen and Chuang say that the Kraus operators read \begin{equation} \hat{E_k} = \sum_n \sqrt{C^n_k (1-\gamma)^{n-k}\gamma^k} |n-k\rangle\langle n|, \end{equation} where $\gamma = 1 - \cos^2(\chi \Delta t)$ is the probability of emitting a quantum of energy. Physically the form of the Kraus operators is clear: we finish with a statistical mixture of all the possible scenarios, namely, emitting from $0$ to $n$ quanta, where $n$ is the number of quanta which has the system oscillator, with the corresponding probabilities expressed by the binomial distribution in the square root.

However, I get a problem in a precise derivation of these Kraus operators. Rewriting the Fock state $|k_E\rangle = \frac{(\hat{b}^\dagger_E)^k}{\sqrt{k!}} |0_E\rangle$ we have to find the diagonal matrix element of the following operator corresponding to the vacuum state, \begin{equation} \hat{E}_k = \frac{1}{\sqrt{k!}} \langle 0_E | (\hat{b}_E)^k \hat{U}(\Delta t) |0_E \rangle. \end{equation} Expanding the exponent, \begin{equation} \hat{E}_k = \frac{1}{\sqrt{k!}} \sum_n \frac{(-i \chi \Delta t)^n}{n!} \langle 0_E | (\hat{a}^\dagger_S \otimes (\hat{b}_E)^2 + \hat{a}_S \otimes \hat{b}_E \hat{b}^\dagger_E )^k (\hat{a}^\dagger_S \otimes \hat{b}_E + \hat{a}_S \otimes \hat{b}^\dagger_E )^{n-k} |0_E \rangle \end{equation} At this point I am stuck since it is not clear for me how to calculate this matrix element. In particular, is there some way to simplify the expression $(\hat{A} + \hat{A}^\dagger)^n$?

I see that the Kraus operators due to Nielsen and Chuang are a kind of powers of the annihilation operator with the changed normalization due to the binomial distribution. However, I do not see where the coefficient $\sqrt{C^n_k (1-\gamma)^{n-k}\gamma^k}$ comes from.

  • $\begingroup$ Off hand, the square looks like the $kth$ term in the expansion of $(a + b)^n$, i.e. $C^n_k a^{n-k} b^k$? $\endgroup$ – jim Jul 3 at 15:00
  • $\begingroup$ Right, but I think it is just a consequence of completeness since the Kraus operators, in this case, should represent the expansion of unity, $\sum_k \hat{E}^\dagger_k \hat{E}_k = \hat{I}$. $\endgroup$ – Ciruzz Broncio Jul 4 at 9:13
  • $\begingroup$ I asked this question here and there's an answer for it now. $\endgroup$ – Bashir Jul 22 at 18:09

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.