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We know, Schrödinger's cat inside the box is in the equal superposition state of both alive and dead. We can express its state as $$|\text{cat}_\phi\rangle= \frac{|\text{alive}\rangle+e^{i\phi}|\text{dead}\rangle}{\sqrt{2}} \hspace{10mm} \text{where }\phi\text{ is relative phase}$$

If $\phi$ were $0$ or $\pi$ we could use Grover's algorithm to keep the cat alive.

But since we don't know $\phi$ and we don't want to measure the cat without being $100\%$ sure that the cat is now in $|\text{alive}⟩$ state, how can we proceed? Can we develop a more general version of Grover's algorithm?

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    $\begingroup$ This is similar to a question I posed a few weeks ago. I'm afraid the answer is pessimistic: quantumcomputing.stackexchange.com/questions/18591/… $\endgroup$
    – jecado
    Aug 29 at 16:22
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    $\begingroup$ Usually cat states are described as $\vert cat\rangle=\frac{1}{\sqrt{2}}(\vert 00\ldots 0\rangle + e^{i\phi} \vert 11\ldots 1\rangle)$. Aaronson has some public lectures on quantum necromancy; the punch line, I think translating it into your question, is that it's easy to measure $\phi$ to see the cat in superposition iff it's easy to swap and bring a dead cat back alive. $\endgroup$
    – Mark S
    Aug 29 at 17:13
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    $\begingroup$ Thank you very much. $\endgroup$ Aug 29 at 19:45
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    $\begingroup$ Can't this be done with a setup involving beam splitters? $\endgroup$
    – Michael
    Aug 30 at 0:48
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    $\begingroup$ Why do you say that you can use Grover's search to keep the cat alive? Grover's algorithm requires an operation that can repeatedly produce the state $|\text{cat}_{\phi}\rangle$. On the other hand, for any known $\phi$ there is a unitary that rotates the state to $|\text{alive}\rangle$. $\endgroup$
    – DaftWullie
    Aug 31 at 7:54
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TL;DR: This is probably going to be disappointing. If a cat enters a superposition and we lose track of the relative phase $\phi$ then there is only one deterministic operation that returns to the $|\text{alive}\rangle$ state: the state preparation channel. In other words, we have to get a new cat.


Let us represent the states of the cat on the Bloch sphere with $|\text{alive}\rangle$ at the North pole and $|\text{dead}\rangle$ at the South pole. The states $|\text{cat}_\phi\rangle$ are on the equator. Further, let us denote with $\mathcal{E}:L(\mathbb{C}^2)\to L(\mathbb{C}^2)$ the required quantum operation that saves the cat. In other words,

$$ \mathcal{E}(|\text{cat}_\phi\rangle\langle \text{cat}_\phi|) = |\text{alive}\rangle\langle \text{alive}|\quad\text{for all}\,\phi.\tag1 $$

Thus, $\mathcal{E}$ maps the equator of the Bloch sphere to the North pole. This immediately tells us that $\mathcal{E}$ is not bijective and hence not unitary.

Moreover, by linearity, $\mathcal{E}$ maps the entire equatorial plane of the Bloch sphere to the North pole. In particular, $\mathcal{E}$ maps the maximally mixed state $\frac{I}{2}$ to the North pole

$$ \mathcal{E}\left(\frac{I}{2}\right) = |\text{alive}\rangle\langle \text{alive}|.\tag2 $$

On the other hand,

$$ \mathcal{E}\left(\frac{I}{2}\right)=\mathcal{E}\left(\frac{|\text{alive}\rangle\langle \text{alive}|+|\text{dead}\rangle\langle \text{dead}|}{2}\right) = \frac12\rho_1+\frac12\rho_2\tag3 $$

where $\rho_1 = \mathcal{E}(|\text{alive}\rangle\langle \text{alive}|)$ and $\rho_2 = \mathcal{E}(|\text{dead}\rangle\langle \text{dead}|)$. Combining $(2)$ and $(3)$, we have

$$ |\text{alive}\rangle\langle \text{alive}| = \frac12\rho_1+\frac12\rho_2. $$

However, $|\text{alive}\rangle$ is an extreme point of the Bloch sphere and hence not a convex combination of states other than $|\text{alive}\rangle$. Therefore, $\rho_1=\rho_2=|\text{alive}\rangle\langle \text{alive}|$. Finally, since the set consisting of the equator and the poles contains a basis, we conclude that

$$ \mathcal{E}(\rho) = |\text{alive}\rangle\langle \text{alive}|\tag4 $$

for all states $\rho$. Thus, the only quantum operation satisfying $(1)$ is the state preparation channel $(4)$ for the $|\text{alive}\rangle$ state.

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    $\begingroup$ Thank you very much for the answer. Yeah, the result is disappointing though :( $\endgroup$ Aug 29 at 19:47
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    $\begingroup$ If you are willing to give up some certainty, is it possible to improve the chances of measuring the alive state above 50%? I suspect the answer is no, because you can't rotate the equator to all be part of the north hemisphere, but I'm out of practice on these calculations and I don't know if grover's algorithm changes the situation. $\endgroup$ Aug 30 at 2:20
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    $\begingroup$ ("we loose track""we lose track") $\endgroup$ Sep 2 at 11:03
  • $\begingroup$ Oops, thanks! Fixed. $\endgroup$ Sep 2 at 15:36

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