Interpreting a Density Matrix Resulting from Decoherence of a generalized GHZ State

Question

This questions extends on this question and on the given answer.

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What this question is about:

Decoherence model. Consider a simple decoherence process modelled by $$\mathcal{E}\left(\rho\right)=p_0\rho+(1-p_0)\sigma_z\rho\sigma_z$$ with $p_0=(1+e^{-\gamma t})/2$ that acts on every qubit of a system individually.

Case 1. The effect on a GHZ state $$|GHZ^n\rangle=\frac{1}{\sqrt{2}}\left(|0\rangle^{\otimes n}+|1\rangle^{\otimes n}\right)$$ can be looked at by calculating $$\mathcal{E}^{\otimes n}\left(\rho_{|\text{GHZ}^n}\rangle\right)= \frac{1}{2}\left(\left(|0\rangle \langle 0|\right)^{\otimes n}+r(t)\left(|0\rangle \langle 1|\right)^{\otimes n}+r(t)\left(|1\rangle \langle 0|\right)^{\otimes n}+\left(|\rangle 1\langle 1|\right)^{\otimes n}\right)$$ with $r(t)=e^{-\gamma n t}$.

It is straightforward to interpret this density matrix. The off diagonals show how "quantum" the state is - how many quantum features it has. All this is considered in regard to $|0\rangle^{\otimes n},|1\rangle^{\otimes n}$ or "alive" and "dead" (if you will), meaning, that the only option of interpretation is whether the density matrix shows a superposition of "dead" and "alive" or a classical mixture of the two (or, to what extend, a classical mixture of "dead", "alive" and "superposition of dead and alive"). The decoherence rate obviously can be directly observed: $r(t)$.

Case 2. Now consider the effect of said decoherence process on 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. Again, we calculate \begin{align}\tag{1} \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} with $c=\cos(\epsilon)$ and $s=\sin(\epsilon)$.

With this density matrix (1), it seems to be less intuitive to give an interpretation, because we are not able to write it in the form $$|0\rangle\langle 0|^{\otimes n}+|0\rangle\langle \epsilon|^{\otimes n}+|\epsilon\rangle\langle 0|^{\otimes n}+|\epsilon\rangle\langle \epsilon|^{\otimes n}$$ when using the definition of $|\epsilon\rangle$. It's not that easy to tell how (in respect to time t) one cannot observe superpositions of $|0\rangle$ and $|\epsilon\rangle$. Obviously, we could try $|\chi\rangle=c|0\rangle+s e^{-\gamma t}$ to get at least some shape $$|0\rangle\langle 0|^{\otimes n}+|0\rangle\langle \chi|^{\otimes n}+|\chi\rangle\langle 0|^{\otimes n}+|\chi\rangle\langle \chi|^{\otimes n}\tag{2}$$ but will fail, because we would need something like $s^2 e^{-2\gamma t}|1\rangle \langle 1|$ and not $ s^2 |1\rangle \langle 1|$ in the last bracket in (1). The second and third addend could be written as $|0\rangle\langle\chi|^{\otimes n}$ or $|\chi\rangle\langle0|^{\otimes n}$ though: $$\left[\mathcal{E}\left(|0\rangle\langle\epsilon|\right)\right]^{\otimes n}=\left(\sqrt{d}|0\rangle\langle\chi|\right)^{\otimes n}=d^{N/2}\left(|0\rangle\langle\chi|\right)^{\otimes n}$$ with $|\chi\rangle=\frac{1}{\sqrt{d}}\left(\cos(\epsilon)|0\rangle+\sin(\epsilon)e^{-\gamma t} |1\rangle\right)$ and $d=\cos(\epsilon)^2+\sin(\epsilon)^2\cdot e^{-2\gamma t}$.

In any case, the question is:

Question: Would you consider $||\left[\mathcal{E}(|0\rangle\langle \epsilon|)\right]^{\otimes n}||_1$ to calculate the decoherence rate? As seen in (1), there are also off-diagonals (whether its regarding the basis $|0\rangle,|1\rangle$ or something else) resulting from $(|\epsilon\rangle\langle\epsilon|)^{\otimes n}$, that should be considered too, no? If not, how would one interpret the result?

Answer 1

In the GHZ example of case 1, the relevant term that is decaying is $(|0\rangle\langle 1|)^{\otimes n}$, with coefficient $r(t)=e^{-\gamma nt}$. How can one observe this term and its decay?

For $n=1$, the standard procedure is to perform an interference experiment to measure such "coherences" (off-diagonal elements of a density matrix). An interference experiment projects your state $\rho$ onto a coherent superposition of the two states in question, $$|\psi(\phi)\rangle=\frac{|0\rangle+e^{i\phi}|1\rangle}{\sqrt{2}},$$ which will oscillate in probability as \begin{align} p(\phi)\equiv\langle \psi(\phi)|\mathcal{E}(\rho)|\psi(\phi)\rangle=\frac{1+r(t)\cos\phi}{2}. \end{align} The amplitude of the oscillations decay with $t$, showing that the state is becoming less and less coherent in the $\{|0\rangle,|1\rangle\}$-basis as time progresses.

If we generalize this to any $n$, there are lots of off-diagonal elements to inspect for coherence. Do we want to compare to the state $|\psi(\phi_1)\rangle\otimes \cdots\otimes |\psi(\phi_n)\rangle$? This would give us insight into the coherences present in each individual qubit through a complicated interference pattern $[1+r(t)\cos(\phi_1+\cdots+\phi_n)]/2^n$. But we know that the GHZ case should only give us one off-diagonal matrix element on each side of the diagonal, so we might as well ask about the most generic interference experiment comparing to the state $$|\psi(\phi)^n\rangle=\frac{|0\rangle^{\otimes n}+e^{i\phi}|1\rangle^{\otimes n}}{\sqrt{2}},$$ which gives the exact same interference pattern as for $n=1$ but with $r(t)$ being defined for any $n$. Notably, the amplitude of the interference decays with $t$ and with $n$.

Notably, in interferometry one only cares about the interference visibility $$V=\frac{\max_\phi p(\phi)-\min_\phi p(\phi)}{\max_\phi p(\phi)+\min_\phi p(\phi)},$$ which means we end up only caring about the ratio of the coefficients of the constant term and of the cosine term. So either way, for the GHZ case, we expect an interference pattern where the amplitude of the oscillation decays as $V=r(t)$.

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Now, can we do the same thing using $|0\rangle$ and $|\epsilon\rangle$ as our basis of choice? If we start with $n=1$ and compare our evolved state to some superposition $$|\psi(\phi)\rangle=\frac{(1+e^{i\phi}\cos\epsilon )|0\rangle+e^{i\phi}\sin\epsilon|1\rangle}{\sqrt{2+2\cos\epsilon\cos\phi}}\propto |0\rangle+e^{i\phi}|\epsilon\rangle,$$ we find $$p(\theta)=\frac{2 r \sin ^2(\epsilon ) (\cos (\epsilon )+1) (\cos (\epsilon )+\cos (\phi ))+(\cos (\epsilon )+1)^2 \left(2 \cos (\epsilon ) \cos (\phi )+\cos ^2(\epsilon )+1\right)+\sin ^4(\epsilon )}{4 (\cos (\epsilon )+1) (\cos (\epsilon ) \cos (\phi )+1)}.$$ The limit of $\cos\epsilon=0$ matches the GHZ case and the limit of $\sin\epsilon=0$ has a uniform probability of $1$ (no interference, the state is always $|0\rangle$). Now, to find the visibility, we need to compute the maximum and minimum with respect to $\phi$. An easy way to find those is by taking a derivative with respect to $\phi$ and realizing that the zeros all have $\sin\phi=0$. We might not know which is the maximum and which is the minimum, depending on the value of $\epsilon$, so we take the absolute values of the differences to yield $$V=\frac{|p(\pi)-p(0)|}{p(\pi)+p(0)}=\cos^2\epsilon+r(t)\sin^2\epsilon.$$ The visibility seems to be bigger wheneverthe state $|\epsilon\rangle$ acquires more of a $|0\rangle$ component! That's weird! But it is probably because we don't know what we want to look for: the state $|0\rangle+|\epsilon\rangle$ changes less than a state $|0\rangle+|1\rangle$ does under the map $\mathcal{E}$, so maybe we should have expected the interference visibility to remain higher for the former. But that is not what I would have expected.

This can be continued for arbitrary $n$ along either of the generalizations for GHZ states above. It is important to ask a more precise question (what exact physical properties are we looking for) in order to get a unique answer.