# What if there are no Schrödinger's cats?

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1. Jan 5, 2016

### Petr Matas

Experiment description: Let us have a slight modification of the famous thought experiment, in which the cat is killed only if a particle decay is detected in a fixed time interval, say from 0 to 1 s. Let the box with the cat, the deadly device, and the radioactive sample with the detector be an isolated system, which at t = 0 s is in a pure quantum state (the initial state), which is known (although very complicated and practically impossible to determine). The quantum state of the system at t = 1000 s (the final state) is given by the time evolution operator, so it is a known pure state as well. Let us have an observable A indicating, whether the cat is alive or dead, but no measurement is going to be made.

Question: Is it true that for a majority of initial states meeting the experiment requirements (i.e. the radioactive substance amount is with high probability in the "interesting" range, etc.), the expectation value for A in the final state is very close to one of the two discrete values, getting even closer to that discrete value as time advances? Of course, it will be "alive" for some initial states and "dead" for others.

Motivation: I feel that the mutual interaction of the parts of the system will cause some sort of quantum decoherence, so no observer is needed: The isolated system will make the "almost-decision" itself. Am I right?

2. Jan 5, 2016

### StevieTNZ

At t = 1000s, the cat is going to be alive, only if there was no particle decay detected between t=0 and t=1 s.

Quantum decoherence does not turn a superposed system into one state or the other. All it is, is merely entanglement with the environment. It is often claimed due to decoherence the moon is always there, for example. But if that were the case there would be no measurement problem to resolve anymore. As far as I'm aware the measurement problem still exists.

3. Jan 5, 2016

### Petr Matas

That's what I thought, but the expectation value of an observable generally changes with time. For example, the expectation value of the position of an isolated particle moving to the right with v = 1.0 ± 0.1 km/s surely does, and no measurement is needed. The question is: Ignoring the duration of the cat's agony, does the expectation value of A change between t = 2 s and t = 3 s and how?

Maybe decoherence is not the right term to describe my idea. In a double-slit experiment, the photon incidence position probability distribution is influenced by interference. I expect the probability distribution of A at t = 1000 s to be influenced analogously, becoming 1:99 for one initial state, and 99:1 for another. Another analogy: The larger quantum computer you build, the lower the probability that it will work correctly, be the laboratory with the device an isolated system or not.

4. Jan 5, 2016

### Staff: Mentor

Whether the expectation value of an observable changes with time or not depends on whether that observable commutes with the Hamiltonian. If it does commute it will be a constant of the motion (for example, the magnitude of the angular momentum in a closed system); if it does not commute then its expectation value may change over time. Position does not commute with the Hamiltonian, and its expectation value does generally vary with time as you say.

5. Jan 5, 2016

### Staff: Mentor

The title of this thread is a bit misleading in that it asks a question for which the answer is already known: There are no Schrodinger's cats, and no one has ever seriously argued otherwise.

Schrodinger proposed his thought experiment to point out a problem in the then-current understanding of quantum mechanics. Common sense and millennia of observation say that there are no superposed dead/alive cats, so the box must contains either a dead cat or a living cat; there's no more quantum weirdness here than in having tossed a classical coin and knowing that it ended up either heads or tails. However, there was nothing in the 1920's-vintage understanding of QM that led to this outcome - on the contrary, although no one accepted the dead/alive superposition, it seemed to be an inescapable consequence of the theory.

As you suggest, decoherence goes a long ways towards addressing this problem. It doesn't solve the "measurement problem" (at least as mpst people think of it) but it does at least ensure that we will always get some sensible macroscopic outcome.

6. Jan 5, 2016

### Staff: Mentor

There's still a measurement problem even though the moon is always there and even though decoherence ensures that the cat is either dead or alive but not stuck in a superposition of the two. The problem is that although decoherence tells us that the state will evolve into a mixture (as opposed to a superposition) of dead and alive, it still doesn't tell us anything about how or why we eventually end up with exactly one of these outcomes. What happened to that unitarily evolving state to turn it into a particular definite outcome?

7. Jan 6, 2016

### Petr Matas

But in the isolated system it will be a superposition, right?

Is it possible to answer the Question from the first post?

8. Jan 6, 2016

### Petr Matas

The expectation value of A at t = 0 s is "alive". Later it is somewhere between "dead" and "alive". This shows that its expectation value may change, so it cannot commute with the Hamiltonian. But does it actually change between 2 s and 3 s?

9. Jan 6, 2016

### Staff: Mentor

But the cat is not isolated - its entangled with the radioactive nucleus.

Thanks
Bill

10. Jan 6, 2016

### Petr Matas

Yes, but I want to work with the quantum state of the entire composite system from the box to the radioactive nucleus, which is isolated by definition of the experiment. The operator A acts on the state of the composite system.

11. Jan 6, 2016

### Staff: Mentor

Then of course that evolves unitarily.

But I don't understand what such a view gains as far as understanding goes.

Thanks
Bill

12. Jan 6, 2016

### Petr Matas

1. Let us have a composite isolated system H1 × H2. At the start, both subsystems are in a pure state. Then an interaction between them occurs, so they become entangled, and H1 is in a mixed state. Then another interaction between them occurs and I thought that it is not possible to model the process precisely without taking the correlation between the subsystems into account, i.e. to keep track of the inseparable composite state.
2. I am not very familiar with mixed states.

13. Jan 6, 2016

### Staff: Mentor

14. Jan 7, 2016

### Petr Matas

Thanks for the link, I have made it to the end of section 1.2.3, so I think that I have an idea what information I can get from a mixed state. So am I right that precise simulation of an isolated system consisting of repeatedly interacting subsystems is not possible by tracking just each subsystem's density matrix?

15. Jan 7, 2016

### StevieTNZ

If I understand correctly, decoherence only occurs because we cannot fully specify all the information about the systems involved (leading to a density matrix?)? But in principle, if we specify all the information then we have a superposition.

16. Jan 7, 2016

### Petr Matas

Although I don't fully understand what @Nugatory said here, I think that I can answer: Imagine you are doing an experiment on a quantum system H1 and you want it to evolve unitarily in order to be in a superposition just before making the measurement on H1 (otherwise the quantum effects would not be observed). But the system interacts with another system H2 (the environment), so H1 will be in a mixture. Knowing the composite-system state, which is really a superposition, does not help, because you are making your measurement on H1.
I would even say that H1 being in a mixture does not matter; what matters is that its state has been altered by an unwanted interaction.

Last edited: Jan 7, 2016
17. Jan 7, 2016

### StevieTNZ

18. Jan 7, 2016

### Staff: Mentor

Of course not since those subsystems are likely entangled with other subsystems.

Thanks
Bill

19. Jan 8, 2016

### Petr Matas

Did you mean "Of course you are not right" or "Of course it is not possible"?

20. Jan 8, 2016

### Staff: Mentor

Of course the state of the sub-sytems cant tell you the state of the system because they can be entangled.

Thanks
Bill

21. Jan 8, 2016

### Petr Matas

Now it should be clear why I want to work with the quantum state of the entire composite system, so I hope that we can return to the initial question: Is it true that for a majority of (pure) initial states, the expectation value for A in the (pure) final state is very close (but not equal) to one of the two discrete values ("alive" for some initial states and "dead" for others)?

22. Jan 8, 2016

### Staff: Mentor

Its not quite clear cut.

See post 22:

Now look what happens to the composite system when the cat is actually observed. That is the issue I think really concerns you.

Once you get that we can discuss this important issue that doesn't get discussed a lot.

Thanks
Bill

23. Jan 12, 2016

### Petr Matas

Ok, so it is perfectly sufficient to make the measurement on a subsystem A consisting of a single qubit, whose base is { |ai> }, i ∈ {0, 1}. The rest of the composite system is a subsystem B with a base { |bj> }, j = 1..n.
The final (unnormalized) composite pure state is $|\psi\rangle = \sum_{i,j} c_{i,j} |a_i\rangle |b_j\rangle$. Due to (unspecified) properties of the system the coefficients $c_{i,j}$ are obtained using some linear map L from coefficients $\{ d_k \}$, so the possible final states form a subspace of the space of all $|\psi\rangle$ of the mentioned form.
The probability of obtaining result i from measurement on A in the final state is $$p(i) = { \sum_j |c_{i,j}|^2 \over \sum_{i',j} |c_{i',j}|^2 }.$$ Question from post #1 suggests that for e.g. 39% of possible final states is p(1) ≤ 0.01, for another 59% is p(1) ≥ 0.99, and for the remaining 2% is 0.01 < p(1) < 0.99. Thanks to the map L being linear, it seems that the suggestion can't be true, although I have not figured out yet how to prove it. It actually seems that for the majority of states, p(1) will be very near to some constant value in between, like 0.6, given by the radioactive sample activity.

So if my analysis is correct, the answer to the question is "no". Prior to a measurement the box really contains a Schrödinger's cat in an improper mixture of states. What do you think? Of course, after obtaining that single classical bit of information from the measurement, we know pretty much about the cat: Whether it is breathing, its heart beating, etc...

Last edited: Jan 12, 2016
24. Jan 12, 2016

### Staff: Mentor

I cant follow what you are trying to say.

My point was simple. The system is in a mixed state as explained in my link. However if you observe the system there is still 'collapse' because you are observing outside the entangled system.

I think I will have to leave it there.

Thanks
Bill

25. Jan 12, 2016

### Petr Matas

I see the collapse in this: In post #23, after the bit is obtained from the measurement on subsystem A, the subsystem B is no longer in a mixture, but in a pure state $\sum_j c_{i,j} |b_j\rangle$, where $i$ is the result of the measurement on A. The Question from post #1 is basically: What if, due to the mutual interaction of subsystems A and B, the result $i$ is known almost certainly (depending on the initial state of the system) even before the measurement is made. In post #23 I am trying to disprove the answer being "yes".

Last edited: Jan 12, 2016