7
$\begingroup$

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 = \left[\begin{matrix} 0 & 0 \\ 0 & \sqrt{\lambda}\end{matrix}\right], \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?

$\endgroup$
0

2 Answers 2

2
$\begingroup$

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.)

$\endgroup$
1
  • $\begingroup$ 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
1
$\begingroup$

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 \begin{equation} 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}. \end{equation}

We want to compute the operation elements \begin{equation} 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|. \end{equation}

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 \begin{equation} (\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}. \end{equation} Therefore \begin{equation} 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|, \end{equation} 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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.