# Can Schrödinger's cat be revived?

According to Quantum Mechanics, Schrödinger’s cat is in a superposition state of $$\frac{1}{\sqrt{2}}(\left|A\right> + \left|D\right>)$$, where $$\left|A\right>$$ and $$\left|D\right>$$ correspond to alive and dead state respectively. An external observer can make a measurement $$M_1$$ to obtain a result of the cat being either dead or alive. Mathematically, $$M_1$$ is a self-adjoint operator $$M_1=\left|A\right>\left\left with eigen vector $$\left|A\right>$$ and $$\left|D\right>$$. After the measurement, the cat is in a state of either $$\left|A\right>$$ or $$\left|D\right>$$. It seems that theoretically we can perform a different kind of measurement $$M_2=\left|U\right>\left\left where $$\left|U\right>=\frac{1}{\sqrt{2}}(\left|D\right>+\left|A\right>)$$ and $$\left|U\right>=\frac{1}{\sqrt{2}}(\left|D\right>-\left|A\right>)$$. After $$M_2$$, the cat will be in either state $$\left|U\right>$$ or $$\left|V\right>$$. Now suppose we first do measurement $$M_1$$ and get the result that the cat is dead. Now we subsequently perform measurement $$M_2$$ and will get result either $$\left|U\right>$$ or $$\left|V\right>$$ with equal probability because $$\left|\left\right|=\left|\left\right| = \frac{1}{\sqrt{2}}$$. Until this step, everything seems plausible. But what if after measurement $$M_2$$, we perform measurement $$M_1$$ again. What will be the result? Because now the cat is now in state $$\left|U\right>$$ or $$\left|V\right>$$ and $$\left|\left\right|=\left|\left\right|=\left|\left\right|=\left|\left\right|$$, the result of this additional $$M_1$$ will be either dead or alive with equal probability. However, from the first measurement of $$M_1$$, we know the cat is already dead, how can we have equal probability of obtaining result of dead and alive from the second $$M_1$$? And actually, after the initial measurement $$M_1$$, we can repeat $$M_2$$ and $$M_1$$ for indefinite number of times until we see the cat is alive. So we can effectively revive the cat by measuring $$M_2$$ and $$M_1$$ many times. This seems to be implausible.

Edit:

I need to emphasize that here cat is only an example. And I admit that it's nearly impossible to get a cat in pure quantum state. But that does not make this question invalid. We can replace the cat with a microscopic object such as a molecule which can probably be isolated into a pure state.

My current thinking is that $$M_2$$ is the problem. Perhaps there is some theory about whether it's possible to measure $$M_2$$. I see several possibilities:

1. It's physically absolutely impossible to measure $$M_2$$. If it's this, I want to know the reason behind it.
2. It's nearly impossible to measure $$M_2$$. This might involves some thermodynamic argument. I want the argument to be concrete instead just saying something like the process is not reversible (in a thermodynamic sense). Basically, I want to know that given a mathematical construction of a measurement, how can we know whether it's thermodynamically possible or not.
3. It's actually possible to measure $$M_2$$ for some objects. This will be very interesting and I want to understand how we can physically measure $$M_2$$.
• I don't think this is implausible... assuming you can get the $M_2$ operator. The implausibility is because the operator needs to take a very complex macroscopic state and map it onto a very particular quantum state. The observation operator $M_1$ is much simpler. If you consider a particle spin instead the operation does not look too weird. – Anders Sandberg Apr 1 at 21:08
• To measure $M_2$ is to do something forbidden by thermodynamics, like spontaneously uncooking an egg. After all, measurements require interactions. This is not literally impossible, but quute implausible. – knzhou Apr 1 at 21:19
• Measuring $M_2$ on someone who is alive and well will kill that person with 50% probability. And you can also consider various observables for measuring the future states of a system. E.g. there exists an observable for tomorrow's stock market. If you measure it, the wave function will collapse to such a state that will cause it to evolve to yield the exact future state as you have measured the day before. – Count Iblis Apr 1 at 21:52
• Measuring $M_2$ is more difficult than being able to create a cat from scratch, because creating a cat from scratch can be done in a non-coherent way while for quantum measurements everything needs to stay coherent. Perhaps it is possible to measure $M_2$ for a virtual cat implemented by a quantum computer. – Count Iblis Apr 1 at 21:56
• In part, I think this is just the preferred basis problem. This is basically solved, AFAIK. See Joos and Zeh, "The emergence of classical properties through interaction with the environment," Z Phys B 59 (1985) 223. But the solution of the preferred basis problem may not be the full answer, since it just reduces the issue of pointer states to the properties of the detector and of the interaction with the environment that causes decoherence. Then we probably need a thermodynamic argument as to why we don't get detectors and interactions like $M_2$. – Ben Crowell Apr 2 at 1:04

## 2 Answers

You are following the misconception that there exists a wave function for a macroscopic object, the cat in this instance, that can be modeled quantum mechanically in the way you use logic to do.

Macroscopic objects quantum mechanically can be described in the density matrix formalism, in nanometer and micron situations successfully, because phases are retained in the off diagonal elements. For a macroscopic object there are only diagonal elements and the physics follows classical mechanics.

The quantum mechanics of the cat experiment , which lead to the simplification of "dead or alive" cat as two quantum states is a misuse of the cat as a detector, instead of a geiger counter, for detecting whether a nucleus has decayed. A decaying nucleus is the basic quantum state. If the cat is found dead, the nucleus has decayed, if it is found alive, the nucleus has not decayed. In the meantime in the closed box the cat is alive until the nucleus decays and releases the poison.

So analyzing the existence of the cat is analyzing the probability for bringing back the nucleus ( the source of quantum mechanics) to a non decayed state , which has zero probability if energy and new interactions are not brought into play. It has nothing to do with the cat.

If the nucleus has decayed, the Geiger counter counts one decay. When the cat is the counter, if the nucleus has decayed, the cat is dead. Any quantum analysis can only be done on the nucleus.

Please note that paradoxes appear in physics when one mixes the mathematics of the models for two different frameworks. In this analogue of the cat in a box, two mathematical frameworks are used the classical where energy, momentum follow the classical equations, for the cat, (which has different energy when dead or alive), and the quantum mechanical of a decaying nucleus ( which gives a probability for a decay at time t when the box is opened).

• Thanks for the answer. If you are not comfortable with describing a macroscopic object using a wave function, let’s focus on the nucleus. So it’s in a superposition of decay (D) and not decay (A). Will the question make sense now? – wei Apr 2 at 6:26
• In the standard interpretation of the predicted probability distributions from complex conjugate squaring the wave function, there is a probability of the nucleus non decaying even for infinity in time, it is an exponential, but with a very very small number. With the mathematics we use, no , you cannot bring back a decayed nucleus to its original wavefunction , because energy and momentum has been dispersed whi – anna v Apr 2 at 7:33
• ch, momentum, cannot be gathered back, the decaying parts go off to infinity. One could design an experiment to take the decayed nucleus add to it the gamma or the alpha or .. and brng back the original wavefunction for the undecayed nucleus, A specific nucleus of uraneum, for example is in an unknown state, decayed or not decayed, until measured with some chemical interaction, or maybe a second decay gives the signal that the original had also decayed. But the process is irreversible once a measurement has been made. – anna v Apr 2 at 7:33
• With a measurement the conditions have changed and a different wavefunction for the system of particles holds – anna v Apr 2 at 7:47
• It seems to me that the nucleus you are talking about is a statistical ensemble of $\left|D\right>$ and $\left|A\right>$. But that is not important. The important question is whether $\left|U\right> = \frac{1}{\sqrt{2}}(\left|D\right>+\left|A\right>)$ is a legitimate pure quantum state and whether it's an eigen vector of some legitimate observable. – wei Apr 2 at 16:10

Your question treats a macroscopic object (Schrödinger’s cat) as a particle spin. The Stern–Gerlach experiment (1922) evidenced the existence of incompatible operators (observables) in the microscopic world. Incompatibility means that if you change measurement basis you destroy the previous information; that is why you can get again a preceding outcome.

However in case of a macroscopic object, to destroy the previous information, would require an operator in condition to reverse the entropy of the object. Thermodynamics would exclude, with extremely high probability, the possibility of such an operator.

• This seems a little sketchy on the details. Thermodynamics would exclude, with extremely high probability, the possibility of such an operator. "The possibility of" seems vague to me here. Does it refer to mathematical existence? To embodiment in a measurement process? The OP has explicitly constructed such an operator, so it seems clear that it does exist mathematically. Although it seems likely that a live cat has lower entropy than a dead cat, the OP's question would seem to be equally interesting if the system was a molecule and the measurement was measuring its position. – Ben Crowell Apr 2 at 19:37
• @Ben Crowell. I do not question the mathematical construction of the operator. My arguing is that to reverse the entropy of a dead organism is by far extremely unlikely. If the system is an electron or a molecule I agree with the OP, it is what evidenced by the Stern–Gerlach experiment. – Michele Grosso Apr 3 at 15:38