# 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$$, which are $$$$\tag{8.127} E_0 = \begin{bmatrix}1 & 0 \\ 0 & \sqrt{1-\lambda}\end{bmatrix}$$$$ $$$$\tag{8.128} E_1 = \left[\begin{matrix} 0 & 0 \\ 0 & \sqrt{\lambda}\end{matrix}\right],$$$$ where $$\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 am by no means an expert in this sort of calculation, but I think I (mostly) agree with you.

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.)

• I've gone through the question a couple of times and can't spot anything wrong either. I couldn't find anything in the errata either, so I assumed I was missing something but maybe not... Commented Dec 20, 2019 at 11:01

Let us use a more explicit notation $$H = \chi (a^\dagger \otimes a^\dagger a + a \otimes a^\dagger a)$$, where $$a$$ is a generic annihilation operator. (Notice that N&C mean $$b = a \otimes I$$.) As you said, we use the formula $$e^{A + B} = e^{-\frac{1}{2} [A, B]} e^A e^B$$ for operators $$A$$ and $$B$$ which commute with their commutator. In this case, we get $$$$U = e^{-\frac{1}{2} \chi^2 \Delta t^2 (I \otimes a^\dagger a)^2} e^{-i\chi \Delta t a^\dagger \otimes a^\dagger a} e^{-i\chi \Delta t a \otimes a^\dagger a}.$$$$

We want to compute the operation elements $$$$E_k = (\langle k| \otimes I) U (| 0 \rangle \otimes I) = \sum_{l, m = 0}^\infty |l\rangle (\langle k| \otimes \langle l|) U (|0 \rangle \otimes |m \rangle) \langle m|.$$$$

Clearly $$e^{-i\chi \Delta t a \otimes a^\dagger a} |0 \rangle \otimes |m \rangle = |0 \rangle \otimes |m \rangle$$, and also \begin{align*} e^{-i\chi \Delta t a^\dagger \otimes a^\dagger a} |0 \rangle \otimes |m \rangle &= \sum_{n = 0}^\infty \frac{(-i\chi \Delta t)^n}{n!} (a^\dagger)^n |0 \rangle \otimes (a^\dagger a)^n |m \rangle \\ &= \sum_{n = 0}^\infty \frac{(-i\chi \Delta t m)^n}{\sqrt{n!}} |n \rangle \otimes |m \rangle. \end{align*}

Notice that in this expression we need to interpret $$0^0 = 1$$. Moreover, $$|n \rangle \otimes |m \rangle$$ is an eigenstate of $$e^{-\frac{1}{2} \chi^2 \Delta t^2 (I \otimes a^\dagger a)^2}$$ with eigenvalue $$e^{-\frac{1}{2} \chi^2 \Delta t^2 m^2}$$, so we obtain $$$$(\langle k| \otimes \langle l|) U (|0 \rangle \otimes |m \rangle) = \delta_{lm} \frac{(-i\chi \Delta t m)^k}{\sqrt{k!}} e^{-\frac{1}{2} \chi^2 \Delta t^2 m^2}.$$$$ Therefore $$$$E_k = \sum_{m = 0}^\infty \frac{(-i\chi \Delta t m)^k}{\sqrt{k!}} e^{-\frac{1}{2} \chi^2 \Delta t^2 m^2} |m \rangle \langle m|,$$$$ where again $$0^0 = 1$$.

Indeed, there is an operation element for each $$k \in \mathbb{N}$$, but now our assumption is that $$\langle m| \rho |n \rangle$$ is non-zero only for $$m, n \in \{0, 1\}$$. For such a state $$\rho$$ we get \begin{align*} \sum_{k = 0}^\infty E_k \rho E_k^\dagger &= \sum_{k, m, n = 0}^\infty \frac{(\chi^2 \Delta t^2)^k m^k n^k}{k!} e^{-\frac{1}{2} \chi^2 \Delta t^2 (m^2 + n^2)} |m \rangle \langle m| \rho |n \rangle \langle n| \\ &= \langle 0| \rho |0 \rangle |0 \rangle \langle 0| + \sum_{k = 0}^\infty \frac{(\chi^2 \Delta t^2)^k}{k!} e^{-\chi^2 \Delta t^2} \langle 1| \rho |1 \rangle |1 \rangle \langle 1| \\ &\quad+ e^{-\frac{1}{2} \chi^2 \Delta t^2} \langle 0| \rho |1 \rangle |0 \rangle \langle 1| + e^{-\frac{1}{2} \chi^2 \Delta t^2} \langle 1| \rho |0 \rangle |1 \rangle \langle 0| \\ &= \rho_{00} |0 \rangle \langle 0| + \rho_{11} |1 \rangle \langle 1| + e^{-\frac{1}{2} \chi^2 \Delta t^2} ( \rho_{01} |0 \rangle \langle 1| + \rho_{10} |1 \rangle \langle 0| ). \end{align*} This shows the exponential damping of the off-diagonal terms, just like one expects for decoherence. In matrix form this is indeed the same as $$\left[ \begin{matrix} 1 & 0 \\ 0 & \sqrt{1 - \lambda} \end{matrix} \right] \left[ \begin{matrix} \rho_{00} & \rho_{01} \\ \rho_{10} & \rho_{11} \end{matrix} \right] \left[ \begin{matrix} 1 & 0 \\ 0 & \sqrt{1 - \lambda} \end{matrix} \right] + \left[ \begin{matrix} 0 & 0 \\ 0 & \sqrt{\lambda} \end{matrix} \right] \left[ \begin{matrix} \rho_{00} & \rho_{01} \\ \rho_{10} & \rho_{11} \end{matrix} \right] \left[ \begin{matrix} 0 & 0 \\ 0 & \sqrt{\lambda} \end{matrix} \right]$$ with $$\sqrt{1 - \lambda} = e^{-\frac{1}{2} \chi^2 \Delta t^2}$$. I think the cosine term is a mistake.