# A Schrodinger's Cat and the thermal interpretation

#### Demystifier

2018 Award
[Moderator's note: spin-off from a previous thread since this discussion is a separate topic.]

If you are able to prepare a particular spin state (a superposition, say), it means that at the time of preparation, the state of the universe is such that the reduced density matrix of the spin is in this state. (See post #22 for full details.)
In this sense, the Schrodinger's famous thought experiment prepares the cat in a superposition of dead and alive. I still don't see how TI can possibly prevent it.

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#### A. Neumaier

In this sense, the Schrodinger's famous thought experiment prepares the cat in a superposition of dead and alive.
No; only the atom is prepared in a superposition of undecayed and decayed. Schrödinger arrives at your conclusion only by assuming that the subsystem consisting of a decaying atom and the cat is isolated, which is never the case.

Only the whole universe is isolated. The subsystem is open, interacting with the bottom of the box and with the air in it. Increasing the size of the subsystem does not help. This openness is sufficient to make the system dissipative; the quantum H-theorem (whose statistical mechanics derivation you conceded to be acceptable as proof) provides a proof of this. Thus the arguments of Section 5 of my Part III apply and produce a definite final state for the atom (undecayed or decayed) and the cat (alive or dead).

The validity of the quantum H-theorem also shows that increasing entropy does not prove that the system must end up in experimentally relevant times in an equilibrium state, thus making Valentini's and your conclusion of Bohmian mechanics soon being in quantum equilibrium as unwarranted.

#### stevendaryl

Staff Emeritus
No; only the atom is prepared in a superposition of undecayed and decayed. Schrödinger arrives at your conclusion only by assuming that the subsystem consisting of a decaying atom and the cat is isolated, which is never the case.

Only the whole universe is isolated.
Right, but I don't think that really changes anything, does it? Instead of a cat that is in a superposition of alive and dead, you have the whole universe in a superposition of a universe with a dead cat and a universe with a live cat.

#### stevendaryl

Staff Emeritus
Right, but I don't think that really changes anything, does it? Instead of a cat that is in a superposition of alive and dead, you have the whole universe in a superposition of a universe with a dead cat and a universe with a live cat.
In terms of density matrices, the evolution would produce something like this:

$\rho = p_1 \rho_{alive} + p_2 \rho_{dead} + \rho_{cross}$

where $\rho_{cross}$ represents the cross terms like $|dead\rangle\langle alive|$ and $|alive\rangle\langle dead|$ ("alive" and "dead" referring to states of the whole universe, not just the cat).

If you ignore the cross-terms, this can be interpreted as the cat having a probability of $p_1$ of being alive and a probability of $p_2$ of being dead. With the cross-terms, it's a little hard to say.

Regular quantum evolution is not going to change the probabilities $p_1$ and $p_2$. So I don't think that the universe will ever evolve into a state where the cat is definitely dead.

#### A. Neumaier

Right, but I don't think that really changes anything, does it? Instead of a cat that is in a superposition of alive and dead, you have the whole universe in a superposition of a universe with a dead cat and a universe with a live cat.
Superposition makes no sense for density operators.

We have a universe with an evolving density operator, and at the preparation time we have a binary subsystem whose reduced density operator is prepared as $\psi^S=\psi_S\psi_S^*$ with a superposition $\psi_S\in C^2$, and another subsystem describing a binary live/dead variable for a cat in local equilibrium, whose reduced density operator $\psi^C$ is prepared as $\psi_C\psi_C^*$ with $\psi_C={1\choose 0}$, where $A=\pmatrix{1 & 0\cr 0 & 0}$ is the degree of aliveness. The question is what happens to the reduced density matrix $\psi^C$ under the unitary evolution applied to the density operator of the universe. The claim is that only two states are stable under small perturbations.

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#### A. Neumaier

In terms of density matrices, the evolution would produce something like this:

$\rho = p_1 \rho_{alive} + p_2 \rho_{dead} + \rho_{cross}$

where $\rho_{cross}$ represents the cross terms like $|dead\rangle\langle alive|$ and $|alive\rangle\langle dead|$ ("alive" and "dead" referring to states of the whole universe, not just the cat).

If you ignore the cross-terms, this can be interpreted as the cat having a probability of $p_1$ of being alive and a probability of $p_2$ of being dead. With the cross-terms, it's a little hard to say.

Regular quantum evolution is not going to change the probabilities $p_1$ and $p_2$. So I don't think that the universe will ever evolve into a state where the cat is definitely dead.
It is more complicated; only the q-expectation of $A$ of the cat needs to evolve such that a binary decision can be made. I'll provide some analysis in a separate thread, but not today; I need time to work out a reasonably intuitive form and to type the details.

#### Demystifier

2018 Award
It is more complicated; only the q-expectation of A of the cat needs to evolve such that a binary decision can be made. I'll provide some analysis in a separate thread, but not today; I need time to work out a reasonably intuitive form and to type the details.
I'm looking forward to see that. This might clarify some important things about TI.

#### A. Neumaier

It is more complicated; only the q-expectation of $A$ of the cat needs to evolve such that a binary decision can be made. I'll provide some analysis in a separate thread, but not today; I need time to work out a reasonably intuitive form and to type the details.
I'm looking forward to see that. This might clarify some important things about TI.
See Section 3 of my new paper here!

#### A. Neumaier

In terms of density matrices, the evolution would produce something like this: $\rho = p_1 \rho_{alive} + p_2 \rho_{dead} + \rho_{cross}$
where $\rho_{cross}$ represents the cross terms like $|dead\rangle\langle alive|$ and $|alive\rangle\langle dead|$ ("alive" and "dead" referring to states of the whole universe, not just the cat).
A detailed discussion of how Born's rule follows from the evolution of the state of the universe is given in the analysis in Section 3 of my Part IV. (Please discuss details in that thread.) The point is that one only needs to consider a binary pointer variable for the property ''atom decayed'' (or not), and that this decision is definitely made in a macroscopically noticeable way within a finite (macroscopically short) time, using the standard approximations used everywhere in statistical mechanics. Thus the cat is definitely dead or alive except during a short moment where the decay happens and nothing definite can be said.

#### Demystifier

2018 Award
The point is that one only needs to consider a binary pointer variable for the property ''atom decayed'' (or not), ... Thus the cat is definitely dead or alive except during a short moment where the decay happens and nothing definite can be said.
1. If the variable is binary, then it has only two possible values. Then how can there be a short moment where it does not have any of those two values? Did you actually mean that it has a continuum of values, but only two stable values?

2. What determines the time during which nothing definite can be said? Is it essentially the same as the decoherence time?

#### A. Neumaier

1. If the variable is binary, then it has only two possible values. Then how can there be a short moment where it does not have any of those two values? Did you actually mean that it has a continuum of values, but only two stable values?
Any actual binary display needs time for switching between 0 and 1, and cannot be meaningfully read during the switching time. Read Section 3 of Part IV for the details; everything is specified there.
2. What determines the time during which nothing definite can be said? Is it essentially the same as the decoherence time?
My argument is asymptotic and gives no information about the time needed. It surely depends on the response mechanism of the binary display and cannot be discussed in abstracto.

For a complicated binary decision, such as whether a cat is alive or dead, it might take minutes to make a solid decision. But if one doesn't use poison and a cat to register the decay then the decoherence time might be enough.

#### charters

For a complicated binary decision, such as whether a cat is alive or dead, it might take minutes to make a solid decision. But if one doesn't use poison and a cat to register the decay then the decoherence time might be enough.
Thus the cat is definitely dead or alive except during a short moment where the decay happens and nothing definite can be said.
But this is inconsistent with what you've said or agreed to elsewhere regarding the existence of deterministic hidden variables in the TI. In any deterministic HV interpretation, the real ontic state is always associated with *either* a cat that lives or a cat that dies, at all times. I think this is @Demystifier's point And here the Born rule is simply the result of respecting an equilibrium condition when assigning HVs. So you are creating unnecessary work for yourself that is obviated by concessions you make elsewhere.

Alternatively, in part IV section 3 you are saying the q-expectations/beables are just the reduced density matrices. This is actually pretty similar to Wallace and Timpson's Spacetime State Realism (https://arxiv.org/abs/0907.5294). But as they show, density operators as beables will not unitarily evolve to a single macroscopic world.

Put another way, you must commit once and for all as to whether the universal density operator represents a proper or improper mixture. Right now you seem to move between these two views. If the global mixture is proper, then the TI is an HV interpretation, and the q-expectations as density operators are not in fact the complete set of beables. The HVs are also beables. If the global mixture is improper, the density operators are the only beables, but it is many worlds under unitary time evolution.

#### A. Neumaier

But this is inconsistent with what you've said or agreed to elsewhere regarding the existence of deterministic hidden variables in the TI.
No.
you must commit once and for all as to whether the universal density operator represents a proper or improper mixture. [...] it is many worlds under unitary time evolution.
Time evolution is unitary, and all density matrices represent improper mixtures; the notion of a proper mixture (and hence the distinction) makes no sense in the TI. But the meaning is very different from that in the MWI.

In MWI, only state decompositions in terms of a preferred basis have an interpretable meaning, with many terms implying many worlds.

In the TI, such decompositions are completely irrelevant. Instead, certain linear functionals of them (the q-expectations) have meaning, and these are unique at every time, so that there is only one world.
Wallace and Timpson's Spacetime State Realism (https://arxiv.org/abs/0907.5294). But as they show, density operators as beables will not unitarily evolve to a single macroscopic world.
Which theorem there do you refer to? There is no interpretation-independent concept of a world, and arguments about worlds don't mean anything in the TI.
In any deterministic HV interpretation, the real ontic state is always associated with *either* a cat that lives or a cat that dies, at all times.
Why??? Positing hidden variables does not imply any claim about what these mean, except that they exist independent of any observation or measurement.

The existence of the TI is a counterexample to your assertion. Its beables are ontic, hence are hidden variables in the conventional sense, and they evolve deterministically.

Its goal is to explain the real world, in which binary decisions cannot be made during a change of their value, and not a theoretical caricature where they are claimed to have binary values even when reality shows otherwise.
the Born rule is simply the result of respecting an equilibrium condition
The TI dispenses with such an equilibrium condition - this is one of its strength. Its probabilities are of the same nature as the probability of being in the left or right part of a Lorenz attractor. The probabilities are determined by the deterministic dynamics itself together with a limited temporal resolution, and not by any assumed equilibrium condition!

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#### charters

Time evolution is unitary, and all density matrices represent improper mixtures
Ok I'd like to focus on the first part of 3.4 in paper IV, up to eq (14). Tell me at which step you claim I go wrong.

1) A detector is completely described by q-expectation value beables for any relevant observable, which is given by the partial trace/reduced density matrix of the detector subsystem within the universal density matrix.

2) Like a standard EV in textbook QM, a q-expectation value that commutes with the Hamiltonian will be constant under unitary time evolution.

3) In section 3.4, you say to "consider an environmental operator $X^E$ that leads to a pointer variable $X_t$ which moves in a macroscopic time t > 0 a macroscopic distance to the left (in microscoic units, large negative) when p = 0 and to the right (large positive) when p = 1."

4) The q-expectation value of the detector pointer, prior to the measurement, is such that it is "pointing up" or 0.5, if we identify the left tilt with 0, and right tilt with 1.

5) The q-expectation value of the beam being measured can be prepared to also be 0.5 where the relevant observable commutes with the Hamiltonian. A concrete example of step 4 and 5 is just an n=1 beam prepared as $\sqrt 1/2 \left| up+down \right>$ that will undergo a spin measurement by a properly calibrated device.

6) By point 2, after the unitary interaction between beam and detector, the relevant q-expectation values for both beam and detector must still be 0.5.

7) A q-expectation value of 0.5 after measurement means the detector pointer did not even measure the beam, the pointer did not move. Alternatively, if the detector does move, and there is still only one world, its q-expectation is changing non-unitarily, ie it is not constant under time evolution even though the observable commutes with the Hamiltonian. In a sense MWI can be read as the explanation of how pointers can move while allowing the q-expectations to not change - because the left tilt in one world is offset by the right tilt in the other.

I think maybe you intend to say that somehow uncontrolled degrees of freedom in the environment solve this problem - that the deterministic outcomes are imprinted in these other degrees of freedom which are not considered above. But then, since the detector-environment split is arbitrary, this is really a way of saying the assumption of well calibrated detectors is flawed - in fact, when the pointer looks like it is "straight up" whether it is going to favor the left or right is already imprinted in the q-expectation of the nearby air, so the calibration is secretly false.

But if so, then to return to my original point, you never, not even briefly, have to worry about the state of Schrodinger's cat being alive *and* dead. To the extent it ever looks this way, its an artifact of the choice to ignore the degrees of freedom that already predict the outcome. It is not presenting the ontological issue that people are usually worried about when saying a cat is "alive AND dead".

However, I separately worry this environmental degrees of freedom story is afoul of Von Neumann's HV theorem (see https://arxiv.org/abs/1006.0499) as the environment seems to be simultaneously, deterministically encoding outcomes of measurements on all possible bases. I don't think VN's theorem allows you to just use other quantum subsystems as HVs, I think it requires an additional layer of beables like in Bohmian mech.

#### A. Neumaier

Ok I'd like to focus on the first part of 3.4 in paper IV, up to eq (14). Tell me at which step you claim I go wrong. [...]

5) The q-expectation value of the beam being measured can be prepared to also be 0.5 where the relevant observable commutes with the Hamiltonian. A concrete example of step 4 and 5 is just an n=1 beam prepared as $\sqrt 1/2 \left| up+down \right>$ that will undergo a spin measurement by a properly calibrated device.
In step 5 you add the unwarranted assumption that relevant quantities commute with the Hamiltonian. But such quantities cannot measure anything, as you correctly argue. Thus pointer variables don't commute with the Hamiltonian.

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#### charters

Thus pointer variables don't commute with the Hamiltonian.
Then measurement records won't even be stable in time/under decoherence. Zurek even defines the pointer variables by the fact they commute with the Hamiltonian: https://arxiv.org/abs/quant-ph/9805065

But such quantities cannot measure anything, as you correctly argue.
It works fine in the Everettian/decoherence story.

#### A. Neumaier

Then measurement records won't even be stable in time/under decoherence. Zurek even defines the pointer variables by the fact they commute with the Hamiltonian: https://arxiv.org/abs/quant-ph/9805065
My definition is different from Zureks: my pointer variables are not operators but special q-expectations that have the property defined before (14). This is possible only if $X^E$ does not commute with the Hamiltonian.

Note that I didn't model the amplification process that goes into a real measurement ending up with a true pointer on a scale, but only the initial step where something microscopic (e.g., an electron in a photodetector) is moved by a macroscopic distance (where it would trigger an amplifying cascade). On this level, a temporary microscopic position would be the pointer. The sign of the motion decides already the final amplified binary result - the amplification only makes it macroscopically visible and irreversible. This sign is dynamically stable in the situation analyzed.
- that the deterministic outcomes are imprinted in these other degrees of freedom which are not considered above. But then, since the detector-environment split is arbitrary,
Note that there is no arbitrary detector-environment split, only the (obviously necessary) distinction of the variable that produces the binary decision! My argument in Section 3 encodes the effects of all degrees of freedom of the universe. It involves no approximation at all, apart from the replacement of finite times in Subsection 3.4 by an infinite time limit, as usual in arguments about microscopic scattering processes, and the final discussion at the end of Subsection 3.4.

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#### charters

Ok, I'll try one more approach, the prior wasn't effective.

You claim the TI is unitary and has deterministic outcomes of measurements. You also claim macroscopic cats/measuring devices evolve from a state of ontic uncertainty during the measurement, into a fixed state of the measurement basis, specifically:

Thus the cat is definitely dead or alive except during a short moment where the decay happens and nothing definite can be said
But unitary transformations cannot take a state of ontic uncertainty (a superposition or improper mixture) on some basis into a state of certainty on that basis, or vice versa. T'Hooft (who I generally disagree with on foundations, but on this he is certainly right) has actually just put out a paper calling this a "conservation of ontology" principle: https://arxiv.org/abs/1904.12364

So, either the TI is not deterministic, or the uncertainty associated with the cat/detector is just an epistemic uncertainty, due to ignoring the external/environmental subsystems or (more reasonably) HVs whose state deterministically predicts the outcome. In a deterministic interpretation, you don't need to, shouldn't want to, and can't talk about "dead AND alive" states, even ephemerally.

#### A. Neumaier

You claim the TI is unitary and has deterministic outcomes of measurements.
yes.
You also claim macroscopic cats/measuring devices evolve from a state of ontic uncertainty during the measurement, into a fixed state of the measurement basis
No.
A basis never figures in my argument. I only claim that there is a stable sign of expectations, enough to get a decision based on the TI beables. Also, there is no uncertainty at all in my arguments, except in the details about the probabiliy with which a given sign is obtained - which depends on the state of the unverse at preparation time and on the precise dynamics.
In a deterministic interpretation, you don't need to, shouldn't want to, and can't talk about "dead AND alive" states, even ephemerally.
I never talked about that. I only talk about the microscopic anlogue of an apparently dying cat - during the time where the sign cannot yet be read off with certainty.

#### charters

Ok, I don't think we're going to make progress. I just don't agree it is possible in principle to have unitary, determinsitic, single world quantum mechanics in which the subsystems are exhaustively described by improper mixtures, ie states that display ontic uncertainty. One needs additional hidden variables to underwrite this single world determinism at all times, effectively rendering the mixtures proper. I see all this as following from the definitions of these terms, and as such the point is too general to be sensitive to any idiosyncracies of the TI or any particular interpretation. But I guess I'm not able to convey the argument effectively.

#### PeterDonis

Mentor
states that display ontic uncertainty
As I understand it, there is no ontic uncertainty in the TI. The ontic state is the q-expectation, and q-expectations are not uncertain. The uncertainty is epistemic--we can't make infinitely precise measurements of the q-expectations.

#### charters

As I understand it, there is no ontic uncertainty in the TI. The ontic state is the q-expectation, and q-expectations are not uncertain. The uncertainty is epistemic--we can't make infinitely precise measurements of the q-expectations.
But the q-expectations are described by reduced density matrices/partial traces (see 3.3 in paper IV; axiom A5 in paper I). The uncertainty of these mathematical objects is ontic by definition (this fact is the core of the measurement problem). Equivalently, the uncertainty of improper mixtures is ontic, whereas proper mixtures encode epistemic uncertainty. And, at least in #71 above, Arnold says:

all density matrices represent improper mixtures; the notion of a proper mixture (and hence the distinction) makes no sense in the TI

#### PeterDonis

Mentor
The uncertainty of these mathematical objects is ontic by definition
Not if you don't define it that way. I don't see how you can declare by fiat that a particular mathematical object can only have one ontic meaning. Mathematical objects don't have any ontic meaning at all apart from definitions we choose.

#### charters

It's a requirement of the uncertainty principle/noncommuting observables that some of the uncertainty in quantum theory (without hidden variables) is ontic. The (state vector or) density matrix can be in an eigenstate when basis A is diagonal, but not when we diagonalize for conjugate observable B. To say the uncertainty is epistemic is to say the system is *really* in an eigenstate of two conjugate bases at once, and we are just ignorant of which they are. But due to interference effects and the mathematical framework, we know the uncertainty of quantum systems has no such simple ignorance interpretation.

I stress the truth of this is interpretation independent. In fact, interpretations exist only because we aren't sure how to handle ontic uncertainty and the Born rule at the same time.

Sometimes a subsystem is not in *any* local eigenstate, so it has this ontic uncertainty on every basis, rather than merely having ontic uncertainty on most bases. An example is one of the qubits when the state is a Bell state. In this case, we say the qubit's state is an improper mixture, rather than a pure state.

One can choose different definitions for terms, but doing so would not change the reality of the situation, which is that quantum mechanics requires additional (hidden) variables to admit a purely epistemic uncertainty interpretation similar to classical mechanics.

Also, just think about the idea of an expectation value being a beable (this btw is also how MWI works). If the uncertainty that necessitated the use of an EV was merely epistemic, just a matter of our ignorance about the state, it would make no sense to ever suggest the EV was a beable. The beable would obviously just be the underlying ontic state. Classical mechanics works exactly like this, where the beables are represented by a point in phase space, which is embedded in a probability distribution over the same phase space, to represent our lack of exact knowledge of the point. In this case, we would call a random sampling of possible underlying points a proper mixture.

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#### lukesfn

My understanding is only very rudimentary, but I come up with similar issues as expressed in this thread with trying to figure TI.

A basis never figures in my argument. I only claim that there is a stable sign of expectations, enough to get a decision based on the TI beables. Also, there is no uncertainty at all in my arguments, except in the details about the probabiliy with which a given sign is obtained - which depends on the state of the unverse at preparation time and on the precise dynamics.
This kind of point sounds like something starting conditions could be considered as implicit hidden variables.

You could look at TI from the point of view of it being a generalization of Bohmian Mechanics, or other non-local hidden variable interpretations, but with out all the explicit details that would bring any additional hidden variables or poincare invariance, or super determinism into the light.

"Schrodinger's Cat and the thermal interpretation"

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