How can we keep Schrödinger's cat alive?

Question

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?

Answer 1

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.

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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.