# Kraus operators for two interacting harmonic oscillators: Problem with the calculation (Ex. 8.21 of Nielsen-Chuang)

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

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

where $$\hat{b}_E$$, $$\hat{b}^\dagger_E$$ are the creation/annihilation operators for the oscillator which models the environment. The evolution operator reads $$$$\hat{U}(\Delta t) = e^{-i \chi (\hat{a}^{\dagger}_S \otimes \hat{b}_E + \hat{a}_S \otimes \hat{b}^{\dagger}_E) \Delta t}$$$$ and represents a unitary transformation in SE which corresponds to the amplitude-damping channel of interest in S. This channel can be represented as an operator sum in S by tracing out the degrees of freedom of E, so the Kraus operators are $$\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 $$$$\hat{E_k} = \sum_n \sqrt{C^n_k (1-\gamma)^{n-k}\gamma^k} |n-k\rangle\langle n|,$$$$ where $$\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, $$$$\hat{E}_k = \frac{1}{\sqrt{k!}} \langle 0_E | (\hat{b}_E)^k \hat{U}(\Delta t) |0_E \rangle.$$$$ Expanding the exponent, $$$$\hat{E}_k = \frac{1}{\sqrt{k!}} \sum_n \frac{(-i \chi \Delta t)^n}{n!} \langle 0_E | (\hat{a}^\dagger_S \otimes (\hat{b}_E)^2 + \hat{a}_S \otimes \hat{b}_E \hat{b}^\dagger_E )^k (\hat{a}^\dagger_S \otimes \hat{b}_E + \hat{a}_S \otimes \hat{b}^\dagger_E )^{n-k} |0_E \rangle$$$$ At this point I am stuck since it is not clear for me how to calculate this matrix element. In particular, is there some way to simplify the expression $$(\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.

• Off hand, the square looks like the $kth$ term in the expansion of $(a + b)^n$, i.e. $C^n_k a^{n-k} b^k$? – jim Jul 3 '19 at 15:00
• Right, but I think it is just a consequence of completeness since the Kraus operators, in this case, should represent the expansion of unity, $\sum_k \hat{E}^\dagger_k \hat{E}_k = \hat{I}$. – Ciruzz Broncio Jul 4 '19 at 9:13
• I asked this question here and there's an answer for it now. – Bashir Jul 22 '19 at 18:09