Dephasing noise as a decoherence model - Effect on a generalized GHZ state

Question

Some input:

Consider a decoherence process modelled by $$\mathcal{E}\left(\rho\right)=p_0\rho+(1-p_0)\sigma_z\rho\sigma_z$$ with $p_0=(1+e^{-\gamma t})/2$. One can readily find that this leads to $\mathcal{E}\left(|i\rangle\langle i|\right)=|i\rangle\langle i|$ for $i=0,1$ and $\mathcal{E}\left(|i\rangle\langle j|\right)=e^{-\gamma t}|i\rangle\langle j|$ for $i\neq j;i,j=0,1$. With the linearity of the quantum operation $\mathcal{E}$, properties of tensor products and $\mathcal{E}^{\otimes n}(\sigma^{\otimes n})=[\mathcal{E}(\sigma)]^{\otimes n}$ one readily finds that the effect of $\mathcal{E}^{\otimes n}$ on a GHZ state $$|GHZ\rangle=\frac{1}{\sqrt{2}}\left(|0\rangle^{\otimes n}+|1\rangle^{\otimes n}\right)$$ can be evaulated by only looking at the "off diagonal terms" in the density matrix (regarding the computational basis $|0\rangle,|1\rangle$) $$\rho=\frac{1}{2}\left(|0\rangle\langle 0|^{\otimes n}+|1\rangle\langle 0|^{\otimes n}+|0\rangle\langle 1|^{\otimes n}+|1\rangle\langle 1|^{\otimes n}\right),$$ since the others don't change.

We have $\mathcal{E}^{\otimes n}\left(|0\rangle\langle 1|^{\otimes n}\right)=\left[e^{-\gamma t}|0\rangle\langle 1|\right]^{\otimes n}=e^{-n\gamma t}|0\rangle\langle 1|$. The exponential factor is considered as the decoherence rate of the ghz state, since it determines how quickly the state becomes maximally mixed (in the computational basis $|0\rangle,|1\rangle$ I assume). So far so good.

Now consider the generalized ghz state $$|GHZ\rangle=\frac{1}{\sqrt{K}}\left(|0\rangle^{\otimes n}+|\epsilon\rangle^{\otimes n}\right)$$ with $|\epsilon\rangle=\cos\epsilon|0\rangle+\sin\epsilon|1\rangle$ and some $K$ to normalize the state.

Question:

Is it true, that one can (here) also consider only the "off diagonal terms" $|0\rangle\langle\epsilon|$ to compute the decoherence rate?

My thoughts:

No, because here we cannot use that $|\epsilon\rangle\langle\epsilon|$ or $|\epsilon\rangle\langle\epsilon|^{\otimes n}$ is unchanged, since $\mathcal{E}(|\epsilon\rangle\langle\epsilon|)\neq |\epsilon\rangle\langle\epsilon|$! We would have to consider all parts of the density matrix and write it respective to the basis $|0\rangle,|\epsilon\rangle$ (or $|0\rangle,|1\rangle$ ??) to compare.

Answer 1

The only fact one needs to answer any question of this sort really comes from a definition: $$\mathcal{E}^{\otimes n}(\sigma_1\otimes \sigma_2\otimes\cdots\otimes\sigma_n)=\mathcal{E}(\sigma_1)\otimes\mathcal{E}( \sigma_2)\otimes\cdots\otimes\mathcal{E}(\sigma_n).$$ You can use this to compute the effect of the overall channel on any basis state, simply affecting the off-diagonal elements of each $\sigma_i$, then build up the action on any state by taking linear combinations of basis states.

For the state in question, we collect terms and write \begin{align} |GHZ\rangle\langle GHZ|&=\frac{1}{K}\left[\left(|0\rangle\langle 0|\right)^{\otimes n}+\left(|0\rangle\langle \epsilon|\right)^{\otimes n}+\left(|\epsilon\rangle\langle 0|\right)^{\otimes n}+\left(|\epsilon\rangle\langle \epsilon|\right)^{\otimes n}\right]\\ &=\frac{1}{K}\left[\left(|0\rangle\langle 0|\right)^{\otimes n}+\left(c|0\rangle\langle 0|+s|0\rangle\langle 1|\right)^{\otimes n}+\left(c|0\rangle\langle 0|+s|1\rangle\langle 0|\right)^{\otimes n}+\left(c^2|0\rangle\langle 0|+s^2|1\rangle\langle 1|+cs|1\rangle\langle 0|+cs|0\rangle\langle 1|\right)^{\otimes n}\right]. \end{align} Using the definition of the channel acting on a set of states, we find (using $c$ and $s$ as shorthand for the cosine and sine terms) \begin{align} \mathcal{E}^{\otimes n}(|GHZ\rangle\langle GHZ|)=\frac{1}{K}\left\{\mathcal{E}\left[\left(|0\rangle\langle 0|\right)\right]^{\otimes n}+\mathcal{E}\left[\left(c|0\rangle\langle 0|+s|0\rangle\langle 1|\right)\right]^{\otimes n}+\mathcal{E}\left[\left(c|0\rangle\langle 0|+s|1\rangle\langle 0|\right)\right]^{\otimes n}+\mathcal{E}\left[\left(c^2|0\rangle\langle 0|+s^2|1\rangle\langle 1|+cs|1\rangle\langle 0|+cs|0\rangle\langle 1|\right)\right]^{\otimes n}\right\}. \end{align}
As before, we only need to worry about terms $|0\rangle\langle 1|$ and $|1\rangle\langle 0|$, so we find \begin{align} \mathcal{E}^{\otimes n}(|GHZ\rangle\langle GHZ|)=\frac{1}{K}\left[\left(|0\rangle\langle 0|\right)^{\otimes n}+\left(c|0\rangle\langle 0|+se^{-\gamma t}|0\rangle\langle 1|\right)^{\otimes n}+\left(c|0\rangle\langle 0|+se^{-\gamma t}|1\rangle\langle 0|\right)^{\otimes n}+\left(c^2|0\rangle\langle 0|+s^2|1\rangle\langle 1|+cse^{-\gamma t}|1\rangle\langle 0|+cse^{-\gamma t}|0\rangle\langle 1|\right)^{\otimes n}\right]. \end{align}

The properties of this final state can be inspected in any manner you like. One straightforward thing to compute is the reduced state of any of the qubits, which will all be identical and equal to $$[(1+2c^n+c^2)|0\rangle\langle 0|+s^2|1\rangle\langle 1|+se^{-\gamma t}(c^{n-1}+c)(|1\rangle\langle 0|+|0\rangle\langle 1|)]/(K),$$ which for $n=1$ is $$[(1+c)^2|0\rangle\langle 0|+s^2|1\rangle\langle 1|+se^{-\gamma t}(1+c)(|1\rangle\langle 0|+|0\rangle\langle 1|)]/(2+2c).$$

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We can compare this final state to the putative state one would get from simply enacting $|0\rangle\langle \epsilon|\to e^{-\gamma t}|0\rangle\langle \epsilon|$. We only need to consider the $n=1$ case to show that the results differ. For this supposed transformation, we would get, up to normalization \begin{align} |0\rangle\langle 0|+|\epsilon\rangle\langle \epsilon|+|0\rangle\langle \epsilon|+|\epsilon\rangle\langle 0|\to &|0\rangle\langle 0|+|\epsilon\rangle\langle \epsilon|+e^{-\gamma t}|0\rangle\langle \epsilon|+e^{-\gamma t}|\epsilon\rangle\langle 0|\\ &=|0\rangle\langle 0|+c^2|0\rangle\langle 0|+s^2|1\rangle\langle 1|+cs(|1\rangle\langle 0|+|0\rangle\langle 1|)+ce^{-\gamma t}|0\rangle\langle 0|+se^{-\gamma t}|0\rangle\langle 1|+ce^{-\gamma t}|0\rangle\langle 0|+se^{-\gamma t}|1\rangle\langle 0|\\ &=(1+c^2+2ce^{-\gamma t})|0\rangle\langle 0|+s^2 |1\rangle\langle 1|+s(e^{-\gamma t}+c)(|1\rangle\langle 0|+|0\rangle\langle 1|). \end{align} One only difference (for $n=1$) is the $2ce^{-\gamma t}|0\rangle\langle 0|$ term, which is enough to show the two strategies to be different and this latter strategy to be incorrect, but you may still argue that in the $t\to\infty$ limit this difference would disappear. The more important difference is comparing the coefficient $se^{-\gamma t}(1+c)$ from the former strategy to $s(e^{-\gamma t}+c)$ from the latter. In the $t\to\infty$ limit these give different results: the channel I defined at the beginning will have not off-diagonal terms $|0\rangle\langle 1|+|1\rangle\langle 0|$, while the channel defined in this section would retain those off-diagonals with a coefficient $cs$.