# Derive phase damping quantum operation

I am reading about the phase damping quantum operation on page 384 of Nielsen & Chuang's Quantum Computation and Quantum Information (10th Anniversary Edition).

Nielsen & Chuang derived the operation elements from an interaction model of two harmonic oscillators where only the first two levels $$|0\rangle$$ and $$|1\rangle$$ are considered. Here's a clipping of the corresponding contents in the book:

Another way to derive the phase damping operation is to consider an interaction between two harmonic oscillators, in a manner similar to how amplitude damping was derived in the last section, but this time with the interaction Hamiltonian $$$$\tag{8.126} H = \chi a^\dagger a\left(b+b^\dagger\right),$$$$ Letting $$U = \exp\left(-iH\Delta t\right)$$, considering only the $$\left|0\right>$$ and $$\left|1\right>$$ states of the $$a$$ oscillator as our system, and taking the environment oscillator to initially be $$\left|0\right>$$, we find that tracing over the environment gives the operation elements $$E_k = \left$$, which are $$$$\tag{8.127} E_0 = \begin{bmatrix}1 & 0 \\ 0 & \sqrt{1-\lambda}\end{bmatrix}$$$$ $$$$\tag{8.128} E_1 = \begin{bmatrix}1 & 0 \\ 0 & \sqrt{\lambda}\end{bmatrix},$$$$ where $$\lambda = 1-\cos^2\left(\chi\Delta t\right)$$

I just could not work out the calculations. Anybody can help me with the $$\sqrt{1-\lambda}$$ and $$\sqrt{\lambda}$$ terms?

Actually, when I attempted to derive the operation elements along this way, I got the very different answer:

Firstly, we know that if $$[A,[A,B]]=[B,[A,B]]=0$$ then $$e^{A+B}=e^A e^B e^{-[A,B]/2}$$. So we have $$E_0=\langle 0_b| e^{-i\chi\Delta t a^\dagger a(b+b^\dagger)} |0_b\rangle =\langle 0_b| e^{-i\chi\Delta t a^\dagger a b} e^{-i\chi\Delta t a^\dagger a b^\dagger} |0_b\rangle e^{(\chi\Delta t a^\dagger a)^2/2}$$Now using $$e^{-i\chi\Delta t a^\dagger a b^\dagger} |0_b\rangle = \sum_{n=0}^{\infty} \dfrac{(-i\chi\Delta t a^\dagger a)^n}{n!} (b^\dagger)^n |0_b\rangle = \sum_{n=0}^{\infty} \dfrac{(-i\chi\Delta t a^\dagger a)^n}{\sqrt{n!}} |n_b\rangle$$ and $$\langle 0_b| e^{-i\chi\Delta t a^\dagger a b} = \sum_{n=0}^{\infty} \langle 0_b| b^n \dfrac{(-i\chi\Delta t a^\dagger a)^n}{n!} = \sum_{n=0}^{\infty} \langle n_b| \dfrac{(-i\chi\Delta t a^\dagger a)^n}{\sqrt{n!}}$$ we are able to get $$E_0 = \sum_{n=0}^{\infty} \dfrac{(-i\chi\Delta t a^\dagger a)^{2n}}{n!} e^{(\chi\Delta t a^\dagger a)^2/2} = e^{-(\chi\Delta t a^\dagger a)^2/2}$$ Following the same line, using $$\langle 1_b| e^{-i\chi\Delta t a^\dagger a b} = \sum_{n=0}^{\infty} \langle 1_b| b^n \dfrac{(-i\chi\Delta t a^\dagger a)^n}{n!} = \sum_{n=1}^{\infty} \langle n_b| \dfrac{(-i\chi\Delta t a^\dagger a)^{n-1}}{\sqrt{n!}} n$$ we are to obtain $$E_1 = \sum_{n=0}^{\infty} \dfrac{(-i\chi\Delta t a^\dagger a)^{2n+1}}{n!} e^{(\chi\Delta t a^\dagger a)^2/2} = (-i\chi\Delta t a^\dagger a) e^{-(\chi\Delta t a^\dagger a)^2/2}$$

Therefore, my answer will be $$E_{0}=\left[\begin{array}{cc}{1} & {0} \\ {0} & {e^{-(\chi\Delta t)^2/2}}\end{array}\right]$$ and $$E_{1}=\left[\begin{array}{cc}{0} & {0} \\ {0} & {-i\chi\Delta t e^{-(\chi\Delta t)^2/2}}\end{array}\right]$$. What is the problem?

I divided the calculation up slightly differently, which simplified things notationally. Firstly, I considered the input $$|0\rangle_A|0\rangle_B$$. Clearly, $$H$$ acting on this is just 0, so this state doesn't evolve. So, the lop-left element of $$E_0$$ is 1, and that of $$E_1$$ is 0.
Then I considered the input $$|1\rangle_A|0\rangle_B$$. We know that $$a^\dagger a$$ will always just return $$|1\rangle_A$$, so we only need to consider the evolution of the second system, i.e. $$e^{-i\chi\Delta t(b+b^\dagger)}|0\rangle_B.$$ We can then follow your strategy for the calculation (I've only gone through the $$E_0$$ case) to find the bottom-right element is $$e^{-\Delta t^2\chi^2/2}$$. (Digression: if I assume the b operators are fermionic, then $$b+b^\dagger$$ is basically just the Pauli $$X$$ matrix on a qubit. Then you recover the formula that's given.)
What at first glance seems confusing is why you should only consider $$E_0$$ and $$E_1$$. Surely, there are also $$E_k$$ for all natural numbers $$k$$? Of course, they will all be of the same form as $$E_1$$ up to some constant of proportionality. Let's assume $$E_k=\alpha_k|1\rangle\langle 1|$$ for $$k\geq 1$$. Then the relevant terms of the Master equation look like $$\sum_k\frac12 E_k^\dagger E_k\rho+\frac12\rho E_k^\dagger E_k-E_k\rho E_k^\dagger=\frac12 \beta|1\rangle\langle 1|\rho+\frac12 \beta\rho|1\rangle\langle 1|-\beta|1\rangle\langle 1|\rho|1\rangle\langle 1|.$$ This is entirely equivalent to the action of a single operator $$E_1'=\sqrt{\beta}|1\rangle\langle 1|$$ with $$\beta=\sum_k\alpha_k^2$$. Moreover, by the fact that the map will be trace preserving, I don't need to bother actually calculating $$\beta$$. I know that $$\beta+\langle 1|E_0|1\rangle^2=1.$$ (At least this part is consistent with what N&C is telling us.)