Wavefunction collapse: is that really an axiom

[lalbatros]

Can the wavefunction collapse not be derived or is it really an axiom?

How can the answer to this question (yes or no) be proven?

If it is an axiom, is it the best formulation, is it not a dangerous wording?

Let's enjoy this endless discussion !

 

[Demystifier]

There are several different interpretations of QM.
In some of them, there is no need for a collapse postulate.

 

[vanesch]

I think the thing we can sensibly say is that wavefunction collapse cannot follow from the unitary time evolution, which is easy to establish.

 

[lalbatros]

I think the thing we can sensibly say is that wavefunction collapse cannot follow from the unitary time evolution, which is easy to establish.

But it is if you include the interaction with the measuring device into the QM model.

 

[Hurkyl]

But it is if you include the interaction with the measuring device into the QM model.

Only if you choose some model other than unitary evolution for describing the measurement process.

 

[lalbatros]

Do you mean that the "measurement axiom" is contradictory to my/the postulate
that the (unitary) equation of evolution governs all interactions? (including measurement sytems)

 

[vanesch]

Do you mean that the "measurement axiom" is contradictory to my/the postulate that the (unitary) equation of evolution governs all interactions? (including measurement sytems)

yes, of course! That's the whole issue (or better, half the issue) in the "measurement problem". It is (to me at least) one of the reasons to consider seriously MWI.

There's no unitary evolution (no matter how complicated) that can result in a collapsed wavefunction. This can be shown in 5 lines of algebra.

 

[lalbatros]

yes, of course! That's the whole issue (or better, half the issue) in the "measurement problem". It is (to me at least) one of the reasons to consider seriously MWI.

There's no unitary evolution (no matter how complicated) that can result in a collapsed wavefunction. This can be shown in 5 lines of algebra.

The algebra is simple and true.
The problem is that the wavefunction collapse doesn't really exist.
Just like microreversibility doesn't contradict the second law: microreversibility doesn't necessarily imply the existence of a chaos demon.

 

[lalbatros]

yes, of course! That's the whole issue (or better, half the issue) in the "measurement problem". It is (to me at least) one of the reasons to consider seriously MWI.

There's no unitary evolution (no matter how complicated) that can result in a collapsed wavefunction. This can be shown in 5 lines of algebra.

The algebra is simple and true.
(even simple inspection of collapse algebra is enough for that, specially on the density matrix)
The problem is that the wavefunction collapse doesn't really exist.
Just like microreversibility doesn't contradict the second law: microreversibility doesn't necessarily imply the existence of a chaos demon.

In addition, I am quite sure that the collapse axiom can be derived from the Schrödinger equation. But the understanding is missing, to my knowledge.

 

[Demystifier]

In addition, I am quite sure that the collapse axiom can be derived from the Schrödinger equation. But the understanding is missing, to my knowledge.

There is actually a proof that it cannot. The proof is based on the fact that the Schrodinger equation involves only local interactions, while the collapse, including the cases with two or more entangled particles, requires nonlocal interactions.

There is, however, something that contains some elements of a collapse but can be obtained from the Schrodinger equation. This is the environment-induced decoherence. And it is closely related to the second law emerging from time-symmetric laws of a large number of degrees of freedom. See e.g.
http://xxx.lanl.gov/abs/quant-ph/0312059 (Rev. Mod. Phys. 76, 1267-1305 (2004))

 

[quetzalcoatl9]

This can be shown in 5 lines of algebra.

could you please show it then? i don't know about you guys, but i am so sick of qualitative arguments involving wavefunction collapse, etc.

to address the OP, it is my understanding is that if you postulate the Born interpretation then wavefunction collapse follows from that; in other words, physicists weren't just sitting around and postulated "wave collapse" as some popular books/shows would like one to believe.

more concretely,

$$\langle \Omega \rangle_\alpha = \sum_i \omega_i | \langle \omega_i | \alpha \rangle|^2$$

where the Born interpretation is that the quantity$$| \langle \omega_i | \alpha \rangle|^2$$ is to be interpreted as the probability amplitude of measuring a value $$\omega_i$$

$$\langle \Omega \rangle_\alpha = \sum_i \omega_i | \langle \omega_i | \alpha \rangle|^2$$
$$= \sum_i \omega_i \langle \omega_i | \alpha \rangle ^* \langle \omega_i | \alpha \rangle$$
$$= \sum_i \omega_i \langle \alpha | \omega_i \rangle \langle \omega_i | \alpha \rangle$$
$$= \sum_i \langle \alpha | \omega_i \rangle \omega_i \langle \omega_i | \alpha \rangle$$
$$= \sum_i \langle \alpha | \omega_i \rangle \langle \omega_i | \Omega | \omega_i \rangle \langle \omega_i | \alpha \rangle$$
$$= \langle \alpha |\left(\sum_i | \omega_i \rangle \langle \omega_i |\right) |\Omega |\left(\sum_i | \omega_i \rangle \langle \omega_i | \right)\alpha \rangle$$
$$=\langle \alpha | \Omega | \alpha \rangle$$

so for some state $$\phi = \sum_i c_i \psi_i$$ that is NOT an eigenket of the operator (but can always be formed from a linear combination of eigenkets), we have:

$$\langle \phi | \Omega | \phi \rangle = \langle \phi | \Omega | \sum_j c_j \psi_j \rangle$$
$$=\langle \phi | \sum_j c_j \omega_j \psi_j \rangle$$
$$=\sum_j \langle \sum_i c_i \psi_i | c_j \omega_j \psi_j \rangle$$
$$=\sum_i \sum_j c_i^* c_j \omega_j \langle \psi_i | \psi_j \rangle$$

and by orthogonality of states, we have:

$$=\sum_i |c_i|^2 \omega_i$$

which shows that the average value of our experiments will be a weighted average of the eigenstates, i.e. the "wavefunction collapse" s.t. any individual measurement will be a particular eigenvalue. in the classical limit, the spectra of eigenvalues is nearly continuous and so the effect is unnoticeable.

so what's the big deal?? can someone explain to me why this is, for some people, such a big damn mystery??

 

[meopemuk]

There's no unitary evolution (no matter how complicated) that can result in a collapsed wavefunction. This can be shown in 5 lines of algebra.

There is no need for any algebra to show that the wavefunction collapse is not described by a unitary evolution. This follows simply from the definition of the wavefunction. By definition, the wavefunction is a probability amplitude. This means that the measurements described by the wavefunction is a random probabilistic unpredictable process, which cannot be described by deterministic "unitary evolution". That's the whole point of quantum mechanics. In my opinion, looking for a unitary description of the collapse is equivalent to looking for "hidden variables".

Eugene.

 

[lalbatros]

I agree, meopemuk.
The collapse is not an unitary transformation,
and it is even not a transformation at all.

After the collapse, there is no wave function anymore, but a statistical mixture.

That's the axiom.

My view is that after the interaction of a small system with a measuring device,
the state of the small system loses its meaning,
and only the combined wavefunction has a meaning.

The problem that remains is how does the axiom emerge from the "complex" evolution.
This is a challenge, and I am confident that it will or can be explained trivially.
I am also sure that solving this problem is not really useful for the progress of QM,
that is it of the kind of problem that time and generations solves.

 

[vanesch]

Ok, here goes the "proof".

Axiom 1: every state of a system is a ray in Hilbert space.

Now, consider the system "measurement device + SUT" (SUT = system under test, say, an electron spin). This is quantum-mechanically described by a ray in hilbert space. As we have degrees of freedom belonging to the SUT and other degrees of freedom belonging to the measurement device, the hilbert space of the overall system is the tensor product of the hilbert spaces of the individual systems

H = H_m x H_sut

Now, consider that before the measurement, the SUT is in a certain state, say |a> + |b> and the measurement system is in a classically-looking state |M0>. As we have now individually assigned states for each of the subsystems, the overall state is given by the tensor product of both substates:

|psi0> = |M0> x ( |a> + |b> )

Now we do a measurement. That comes down to having an interaction hamiltonian between both subsystems, and from that interaction hamiltonian follows a unitary evolution operator over a certain time, say time T. We write this operator as U(0,T), it evolves the entire system from time 0 to time T.

Now, let us first consider that our SUT was in state |a> and our measurement system was in (classically looking) state |M0>, which is its state before a measurement was done.

"doing a measurement" would result in our measurement device get into a classically looking state |Ma> for sure, assuming that |a> was an eigenvector of the measurement. As such, our interaction between our system and our measurement apparatus, described by U(0,T) is given by:

U(0,T) { |M0> x |a> } = |Ma> x |a>

Indeed, the state of the measurement device is now for sure the classically-looking state Ma, and (property of a measurement on a system in an eigenstate) the SUT didn't change.

We can tell now the same story if the system was in state b:

U(0,T) { |M0> x |b> } = |Mb> x |b>

where Mb is the classically looking state of the measurement apparatus with the pointer on "b".

Now from linearity of U follows that:

U(0,T) { |M0> x (|a> + |b>) } = |Ma> x |a> + |Mb> x |b>

We didn't find "sometimes |Ma> x |a> and sometimes |Mb> x |b>" ; the unitary evolution gave us an entangled superposition.

Now, you can say "yes, but |u> + |v> means: sometimes |v> and sometimes |u>"

but that's not true of course. Consider the other measurement apparatus N which does the following:

U(0,T) { |N0> x (|a> + |b>) } = |Nu> x (|a> + |b> )

U(0,T) { |N0> x (|a> - |b>) } = |Nd> x (|a> - |b>)

From this, we can deduce that

U(0,T) {|N0> x |a> } = 1/2 (|Nu> x { |a> + |b> } + |Nd> x { |a> - |b> })


Clearly if |u> + |v> means "sometimes u and sometimes v" then we could never have that |a> + |b> (which is then "sometimes a and sometimes b") always gives rise to Nu and never gives rise to Nd, because |a> gives rise to sometimes Nu and sometimes Nd.

 

[vanesch]

I agree, meopemuk.
The collapse is not an unitary transformation,
and it is even not a transformation at all.

After the collapse, there is no wave function anymore, but a statistical mixture.

That's the axiom.

The problem is that the transition for a density matrix to go from "superposition" to "statistical mixture" is the following transformation:

take the densitymatrix of the "superposition", and write it in the matrix form in the *correct basis*. Now put all non-diagonal elements to 0. You now have the statistical mixture.

But again, that is a point-wise state change (if you take the density matrix to define the state) which is not described by the normal evolution equation of the density matrix. In other words, the transformation "superposition" -> "mixture" for the density matrix is again a state change which is not described by a physical interaction (which is normally described by the usual evolution equation of the density matrix).

In other words, that's nothing else but another way of writing down a non-unitary evolution, which is not the result of a known physical interaction.

 

[meopemuk]

My guess is that 99% of all experiments involve a single measurement of the system's state. (The counterexample is the bubble chamber, where we repeatedly measure particle's position and obtain a continuous track) In these cases we do not care what is the state of the systems and its wavefunction after the measurement (collapse). It is important to realize that one needs to consider the abrupt change of the wavefunction after measurement only in (not very common) experiments with repeated measurements performed on the same system.

Eugene.

 

[lalbatros]

The problem is that the transition for a density matrix to go from "superposition" to "statistical mixture" is the following transformation:

take the densitymatrix of the "superposition", and write it in the matrix form in the *correct basis*. Now put all non-diagonal elements to 0. You now have the statistical mixture.

But again, that is a point-wise state change (if you take the density matrix to define the state) which is not described by the normal evolution equation of the density matrix. In other words, the transformation "superposition" -> "mixture" for the density matrix is again a state change which is not described by a physical interaction (which is normally described by the usual evolution equation of the density matrix).

In other words, that's nothing else but another way of writing down a non-unitary evolution, which is not the result of a known physical interaction.

I would rather say:

which is more conveniently described by a non-unitary transformation

Decoherence can already easily wipe off non-diagonal elements.

I also remember my master thesis 25+ years ago.
I worked on the Stark effect in beam-foil spectroscopy: the time-dependence with the quantum beats and the atomic decay.
(by the way the off-diagonal elements of the H-atoms exiting the foil were crucial in the simulation)

The hamiltonian had to simulate also the decays of the atomic levels.
Looks-like a non-unitary transformation too, isn't it?
I also didn't want to embarass myself with full QED stuff.
Guess how I modeled that:

- adding a non-hermitian term to the hamiltonian (related to the decay rates)
- and calculating the resulting non-unitary evolution operator
(the density matrix was therefore decaying, which is quite natural)

This is nothing strange or surprising and it shows clearly how non-unitary evolution can simply occur as a limit case of an unitary transformation.
With such a simple approach the Stark effect is very well calculated, for the energy levels, for the perturbed lifetimes and for the time-dependence and polarisations of light emission.

In somewhat pedantic words (I am not a mathematician): the limit of a series of unitary transformations may not be an unitary transformation, looks like that to me. And this can just mean that in some situations the unitary evolution may just be an academic vision: coarse graining makes it practical and non-unitary.

Why should I go for more science-fiction?

 

[Hurkyl]

Decoherence can already easily wipe off non-diagonal elements.

More precisely, decoherence can wipe off the non-diagonal elements of a relative state -- you have to use a partial trace to discard all of the information about the environment before you can get the matrix to look diagonal.

In somewhat pedantic words (I am not a mathematician): the limit of a series of unitary transformations may not be an unitary transformation, looks like that to me.

That is incorrect. If $T_i \to T$ then $T_i^* \to T^*$. Finally, because multiplication is continuous,

$$1 = \lim_i 1 = \lim_i T_i^* T_i = (\lim_i T_i^*) (\lim_i T_i) = T^* T.$$

 

[Hurkyl]

Why should I go for more science-fiction?

Nobody is asking you to go for science-fiction. They are asking you to go to MWI.

And, incidentally, your argument appears quite analogous to rejecting the kinetic theory of gases simply because temperature and pressure are good enough for his favorite applications.

 

[vanesch]

It is important to realize that one needs to consider the abrupt change of the wavefunction after measurement only in (not very common) experiments with repeated measurements performed on the same system.
.

Don't understand me wrong. Collapse is a very practical and good working "approximation", of course. The point is that in order for collapse to occur, you have to leave quantum mechanics. You have, as Bohr wanted it, to decide somehow about a transition to a classical world.

When you do classical mechanics, you can describe your measurement apparatus classically. That means, for instance, in a Lagrangian formulation, that you can give a generalized degree of freedom Qm to the pointer of your measurement apparatus if you want to. Say that you have an (oldfashioned) voltmeter, connected to a system (an electric network). You can solve simply for the behaviour of the network, calculate the voltage difference between two points, and "make your transition" to a measurement, that is, say that you stop there with the system physics, and the measurement apparatus, DEFINED to be a volt meter, will measure that difference.
But you can also include the voltmeter into your system, add the degree of freedom Qm (and all the other internal degrees of freedom of the apparatus) to your "system", and redo all the classical dynamics. You will now simply find that the end state of that variable Qm will simply indicate the result (the position of the pointer). So it is up to you to decide whether or not the internal dynamics of the apparatus and of Qm was worth the effort (taking into account the non-idealities of your measurement apparatus), but that doesn't change much the result.

But in quantum mechanics, if you REMAIN in quantum mechanics, and you do so, you find TWO TOTALLY DIFFERENT outcomes. Indeed, the state description of your pointer is now given by "pointer states", the classically-looking states |Qm>.
You might think, inspired by the classical example, that if you do the quantum-mechanical calculation completely, that if you include the apparatus, you would find sometimes a |Qm = 5V> and sometimes a |Qm=2V> and that the randomness is somehow given by tiny interactions with the environment or whatever. That this is the result of the apparent randomness of quantum mechanics, but that at the end of the day, you will find your apparatus in one or other pointer state, corresponding to what you observed.
In that case, one would have the quantum-mechanical equivalent of the above classical procedure, and it would be a matter of convenience whether or not we include the apparatus in the physical description.
We would have that:
|a> |Q0> would always evolve in |a> |Qa>
and that
(|a> + |b>) |Q0> would evolve half of the time in |a> |Qa> and half of the time in |b> |Qb>.

This would then be the "correspondence rule" and we would fully understand it. The dynamics of the interaction with the apparatus would somehow be responsible for the apparent random behaviour of quantum mechanics, and at the end of the day, we would see that the apparatus always ends up in one of its "pointer states". It would then be an approximation to go directly from (|a> + |b>) to |a> or to |b>, a very good one, exactly as in the case of the classical volt meter.

BUT THIS IS IMPOSSIBLE in quantum mechanics if the evolution is unitary, no matter how complicated the interaction will be. That's the little proof I provided. No matter how complicated the dynamics, if the time evolution is unitary, the above evolution is not possible. You DO NOT end up in a single pointer state.

The end state, namely |a> |Qa> + |b> |Qb> is not a good approximation of "sometimes |a> |Qa> and sometimes |b> |Qb>". We obtain, when we carefully work out the dynamics of the interaction of the measurement apparatus with the system, a totally different state, which is NOT a pointer state. THIS is the difficulty in principle.
There is no description, in terms of quantum mechanics, which corresponds to the classical end state and which follows out of the dynamics. It is not even a close approximation. You have to LEAVE quantum mechanics in order to be able to say that the "apparatus is now in pointer state Qb". Quantum-mechanically, you can't claim that, because out of the calculation does NOT follow that the apparatus is in state |Qb>.

 

[vanesch]

Decoherence can already easily wipe off non-diagonal elements.

Only in the REDUCED density matrix, by taking the partial trace. In fact, that comes down to stating that the overall wavefunction takes on the form:

u|a> |env1> + v|b> |env2> + ...

and that |env1> and |env2> will remain orthogonal throughout evolution. The interpretation of the reduced density matrix as a full density matrix with probabilistic meaning is nothing else but another formulation of collapse.
It means that we take a state such as above to mean that we have |u|^2 chance to have |a> and v^2 chance to have |b>. It's just the same. Just with different clothes.

Of course, you can model, in different ways, the application of collapse, or its equivalents such as tracing out the reduced density matrix and so on. This all works very well. It is the way to do practical computations in quantum mechanics. But you have to realize that it means that at a certain point, you've decided to leave quantum dynamics proper.

It is as if, in classical mechanics, it were impossible to calculate the movement of the pointer of the voltmeter through classical mechanics, and that if you did so, you found nonsensical results. So then you should stop doing mechanics and introduce an ad hoc procedure "I apply the voltmeter", and you'd get out the right result.

 

[lalbatros]

More precisely, decoherence can wipe off the non-diagonal elements of a relative state -- you have to use a partial trace to discard all of the information about the environment before you can get the matrix to look diagonal.

That is incorrect. If $T_i \to T$ then $T_i^* \to T^*$. Finally, because multiplication is continuous,
$$1 = \lim_i 1 = \lim_i T_i^* T_i = (\lim_i T_i^*) (\lim_i T_i) = T^* T.$$

Agree on all that, but by dispersing the density matrix on a larger and larger number of degrees of freedom, you can reach a point where the (very small and dispersed) coherence terms are practically without effects. Reminds me the Poincaré recurrence theorem: recurrence times larger than the age of the universe just make the recurrence meaningless.

Is it really meaningless to assume that the "wavefunction collapse" occurs for similar reasons.

And you know well that there are numerous other arguments: can we define a wavefunction for the whole universe.
In classical mechanics we can -in principle- define a phase space for the whole universe.
Why not a wavefunction for the whole universe. And what about the WF collapse then ?

 

[Anonym]

More precisely, decoherence can wipe off the non-diagonal elements of a relative state -- you have to use a partial trace to discard all of the information about the environment before you can get the matrix to look diagonal.

Consider single particle in some potential well: QM system with only three levels. Please explain how you define a relative state(s) for that system.

Regards, Dany.

 

[vanesch]

Consider single particle in some potential well: QM system with only three levels. Please explain how you define a relative state(s) for that system.

In such a universe (with 3 states) there wouldn't be anything like "observation".

 

[vanesch]

Agree on all that, but by dispersing the density matrix on a larger and larger number of degrees of freedom, you can reach a point where the (very small and dispersed) coherence terms are practically without effects. Reminds me the Poincaré recurrence theorem: recurrence times larger than the age of the universe just make the recurrence meaningless.

The Poincare recurrence theorem is only applicable to bound phase space systems, and indeed, in such a system, the second law of thermodynamics is only valid in certain areas of the dynamics, with "special" starting conditions. In an infinite universe (Newtonian flat space with infinite extend for instance) the Poincare theorem doesn't hold.
As such, the second law is only an approximation because we are far from equilibrium, and evolving in the "right" direction of time.

And you know well that there are numerous other arguments: can we define a wavefunction for the whole universe.
In classical mechanics we can -in principle- define a phase space for the whole universe.
Why not a wavefunction for the whole universe. And what about the WF collapse then ?

Indeed, that's the whole argument. If you define a wavefunction for the whole universe, with a unitary time evolution, then there's no way to have any genuine collapse. But you CAN have decoherence, of course, in the sense that the wavefunction takes on the aspect of a sum of components which don't mix anymore. Nevertheless, there's no way to make these different components disappear except for one. It's the whole idea behind MWI.

 

[Anonym]

In such a universe (with 3 states) there wouldn't be anything like "observation".

My question is not related to any universe or the "observation". I am interesting to know what the physical meaning of Hilbert-Schmidt decomposition is. It is valid in my example also.

Regards, Dany.

 

[vanesch]

My question is not related to any universe or the "observation". I am interesting to know what the physical meaning of Hilbert-Schmidt decomposition is. It is valid in my example also.

Eh, you have only one single system, so what's the decomposition in tensor products ?

 

[Anonym]

Eh, you have only one single system, so what's the decomposition in tensor products ?

That is my question. Whether I need to introduce additional body to define ref frame? For this reason I took example with three levels. I may construct tensor products with let say ground state and still have non-trivial two or three level system. Why not?

Regards, Dany.

 

[Hurkyl]

And what about the WF collapse then ?

Facts:

(1) We have a well-tested theory of unitary evolution that fundamental particles obey.
(2) We have an ad-hoc application of non-unitary evolution to produce measurements.
(3) We have evidence that (2) can be a result of the thermodynamic properties of (1).

So which is the appropriate conclusion?

(A) The evolution of the universe is unitary, and (2) is just a simplifying approximation
(B) The evolution of the universe is non-unitary, because (2) is exactly correct.

 

[vanesch]

That is my question. Whether I need to introduce additional body to define ref frame? For this reason I took example with three levels. I may construct tensor products with let say ground state

I see. It is not clear to me whether the dynamics generates a natural "subsystem decomposition", or whether one has to introduce that by hand. It could be that the dynamics is such, that certain (local ?) subsystems "factor naturally off" in that they give rise to the only stable Schmidt decomposition. Or it could be that one has to put in by hand what is "an observer subsystem", in the same way as one has somehow to define by hand what is "your body" and what part of the physical universe is "not your body".
But once a certain subsystem (that is, a certain set of degrees of freedom) is defined to be "an observer", then one can hope that decoherence sets in, and brings naturally this "observer state" in a kind of Schmidt decomposition wrt the rest of the universe.

 

[f95toli]

Wojciech Zurek is one of the most influencal theorists in the fields of open quantum system and measurement theory (he is actually quite well known in the whole physics community) and he has written a few reviews that I believe are relevant to the discussion in this thread; some of which are freely available on the arXiv

This short review, "Decoherence and the transition from quantum to classical ", from 2003 is an updated version of an article from Physics Today (meaning it is not a "real" paper so it is quite easy to read)

You can download it from
http://arxiv.org/abs/quant-ph/0306072

Higly recommended! I think it gives a good idea of how these problem are handled i nowadays and how we have at least in part solved the measurement problem.
It is also quite interesting to see how much progress was made between 1991 and 2003.

Zurek has also written a more "technical" review that appeared in Reviews of Modern Physics a few years ago. It can also be found on the arXiv.

I have already recommended "The Theory of Open Quantum System" by Breuer and Petruccione. It is a very good book but you need to be familiar with "ordinary" QM in order to understand it (with ordinary I mean the contents of e.g Sakurai, no QFT or relativistic QM needed)

 

[Anonym]

It is not clear to me whether the dynamics generates a natural "subsystem decomposition", or whether one has to introduce that by hand. It could be that the dynamics is such, that certain (local ?) subsystems "factor naturally off" in that they give rise to the only stable Schmidt decomposition. Or it could be that one has to put in by hand what is "an observer subsystem", in the same way as one has somehow to define by hand what is "your body".

I am not certain yet how the relative state should be defined. Your conditions are the natural requirements. I try to obtain that without introducing anything by hand but locally and for the closed systems (just as in CM).

Regards, Dany.

 

[lalbatros]

Facts:

(1) We have a well-tested theory of unitary evolution that fundamental particles obey.
(2) We have an ad-hoc application of non-unitary evolution to produce measurements.
(3) We have evidence that (2) can be a result of the thermodynamic properties of (1).

So which is the appropriate conclusion?

(A) The evolution of the universe is unitary, and (2) is just a simplifying approximation
(B) The evolution of the universe is non-unitary, because (2) is exactly correct.

(A) is my best choice

But an excellent approximation, and not a simplification at all.