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 \begin{equation} \tag{8.126} H = \chi a^\dagger a\left(b+b^\dagger\right), \end{equation} 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 $E_k = \left<k_b|U|0_b\right>$, which are \begin{equation} \tag{8.127} E_0 = \begin{bmatrix}1 & 0 \\ 0 & \sqrt{1-\lambda}\end{bmatrix}\end{equation} \begin{equation} \tag{8.128} E_1 = \begin{bmatrix}1 & 0 \\ 0 & \sqrt{\lambda}\end{bmatrix}, \end{equation} 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?


1 Answer 1


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.)

  • $\begingroup$ 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... $\endgroup$
    – Mithrandir24601
    Dec 20, 2019 at 11:01

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