# How can we keep Schrödinger's cat alive?

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?

• This is similar to a question I posed a few weeks ago. I'm afraid the answer is pessimistic: quantumcomputing.stackexchange.com/questions/18591/… Aug 29 at 16:22
• 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. Aug 29 at 17:13
• Thank you very much. Aug 29 at 19:45
• Can't this be done with a setup involving beam splitters? Aug 30 at 0:48
• 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 $|\text{alive}\rangle$. Aug 31 at 7:54

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.

• Thank you very much for the answer. Yeah, the result is disappointing though :( Aug 29 at 19:47
• 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. Aug 30 at 2:20
• ("we loose track""we lose track") Sep 2 at 11:03
• Oops, thanks! Fixed. Sep 2 at 15:36