Is the math for collapse int different than for mwi?

[entropy1]

Because I understand that for unitary evolution, MWI is required, which suggests that for different interpretations, there may not be unitary evolution?

 

[vancouver_water]

All the interpretations have the same underlying mathematics (and therefore unitary evolution) and predict the same experimental results. It's more a question for philosophy rather than science. My personal favourite is the "shut up and calculate" interpretation.

 

[atyy]

Yes, the maths is different for standard quantum mechanics and for the MWI. It should be noted that there is not consensus that the MWI is conceptually coherent.

The standard interpretation postulates unitary evolution between measurements, but non-unitary evolution when a measurement is made. The MWI postulates unitary evolution at all times.

If MWI is correct, then the apparent non-unitary evolution of the standard interpretation can be derived, rather than needing to be postulated.

 

[entropy1]

If MWI is correct, then the apparent non-unitary evolution of the standard interpretation is can be derived, rather than needing to be postulated.

Is collapse a subset of MWI?

 

[Nugatory]

Is collapse a subset of MWI?

No.
They are two different ways of explaining the same thing, namely the apparent non-unitary evolution of the wave function after a measurement.

 

[entropy1]

They are two different ways of explaining the same thing

Is the math different in either case?

 

[martinbn]

The maths is the same. Otherwise the consequences would be different.

 

[entropy1]

The maths is the same. Otherwise the consequences would be different.

The consequence of collapse is selection of a single eigenstate, where the consequence of unitary evolution would be coexistence of all eigenstates, right? (Well, not really 'coexistence' I guess )

 

[PeterDonis]

for unitary evolution, MWI is required

Why do you think that?

 

[entropy1]

Why do you think that?

Is that not so?

 

[PeterDonis]

Is that not so?

No. Unitary evolution between measurements is part of the basic math of QM; it's the same for all interpretations.

 

[PeterDonis]

the maths is different for standard quantum mechanics and for the MWI

I think this needs to be clarified. The math that actually makes predictions that are compared with measurements is the same for all interpretations of QM. MWI, as you say, derives this math from the assumption of unitary evolution at all times, even through measurements; but that assumption does not lead to any different experimental predictions, it's just a different underlying set of assumptions.

 

[stevendaryl]

I would like to make a slight distinction: for all practical purposes, the mathematics of QM is the same with or without collapse. But it's not precisely the same.

They are the same for all practical purposes because of the practical impossibility of observing interference between macroscopically different possibilities.

 

[atyy]

I think this needs to be clarified. The math that actually makes predictions that are compared with measurements is the same for all interpretations of QM. MWI, as you say, derives this math from the assumption of unitary evolution at all times, even through measurements; but that assumption does not lead to any different experimental predictions, it's just a different underlying set of assumptions.

Yes. To clarify even more, I would say "tries to derive" since there is no consensus that the derivation is possible.

 

[PeterDonis]

To clarify even more, I would say "tries to derive" since there is no consensus that the derivation is possible.

Yes, agreed.

 

[PeterDonis]

They are the same for all practical purposes because of the practical impossibility of observing interference between macroscopically different possibilities.

I would view this somewhat differently. As I view it, in the standard math of QM, the definition of "macroscopic" is "observing interference is impossible". Or, to put it another way, the standard math of QM only says that a measurement has occurred when observing interference is impossible. (The decoherence program fleshes all this out with a lot more detail about how the point when observing interference is impossible is reached in practice.) If we had two alternatives that seemed "macroscopic" intuitively but between which interference was possible, a collapse interpretation would say no measurement had yet occurred.

 

[stevendaryl]

I would view this somewhat differently. As I view it, in the standard math of QM, the definition of "macroscopic" is "observing interference is impossible". Or, to put it another way, the standard math of QM only says that a measurement has occurred when observing interference is impossible. (The decoherence program fleshes all this out with a lot more detail about how the point when observing interference is impossible is reached in practice.) If we had two alternatives that seemed "macroscopic" intuitively but between which interference was possible, a collapse interpretation would say no measurement had yet occurred.

But there really is no cut-off where observing interference becomes completely impossible. Just as the systems get bigger, the interference term becomes smaller and smaller, compared with the non-interfering piece.

 

[martinbn]

If there were differences then it should be possible to point them out explicitly, say this equation or expression is different in this interpretation .

 

[stevendaryl]

If there were differences then it should be possible to point them out explicitly, say this equation or expression is different in this interpretation .

In a collapse interpretation, the wave function changes following a measurement. So the probabilities for the next measurement are different, depending on whether there was a previous collapse, or not. In principle, that's a difference, but in practice, it's not observable.

 

[martinbn]

In a collapse interpretation, the wave function changes following a measurement. So the probabilities for the next measurement are different, depending on whether there was a previous collapse, or not. In principle, that's a difference, but in practice, it's not observable.

Yes, but this is a difference in the interpretation not the mathematics. In a collapse interpretation one says that after a measurement the wave function collapses, say from ##\alpha|a\rangle+\beta|b\rangle## to ##|a\rangle##, and from then on you use ##|a\rangle##. In a different interpretation one says that after the measurement the universe splits (or some such thing), and since you observed ##\alpha## and things have decohered from then on you are in that branch and you use ##|a\rangle##.

 

[entropy1]

Should one study QM formalism to understand this? (which seems quite an endeavour)

Yes, but this is a difference in the interpretation not the mathematics. In a collapse interpretation one says that after a measurement the wave function collapses, say from ##\alpha|a\rangle+\beta|b\rangle## to ##|a\rangle##, and from then on you use ##|a\rangle##. In a different interpretation one says that after the measurement the universe splits (or some such thing), and since you observed ##\alpha## and things have decohered from then on you are in that branch and you use ##|a\rangle##.

So is 'outcome ##|a\rangle## given measurement outcome ##\alpha##' the same as ##|U_a\rangle|a\rangle##?

 

[stevendaryl]

Yes, but this is a difference in the interpretation not the mathematics. In a collapse interpretation one says that after a measurement the wave function collapses, say from ##\alpha|a\rangle+\beta|b\rangle## to ##|a\rangle##, and from then on you use ##|a\rangle##. In a different interpretation one says that after the measurement the universe splits (or some such thing), and since you observed ##\alpha## and things have decohered from then on you are in that branch and you use ##|a\rangle##.

Decoherence is not a black/white thing, though.

Suppose you have a system in state ##|A\rangle## and you're trying to calculate the probability that at a later time it will be in state ##|B\rangle##. For simplicity, there are two possible intermediate states, ##|C\rangle## and ##|D\rangle## (of course, in a real situation, there are infinitely many intermediate states).

The probability can be calculated as follows:

• Let ##\psi_{XY}## be the probability amplitude for going from state ##X## to state ##Y## (ignoring the time parameter for simplicity of notation)
• Then the amplitude for going from ##A## to ##B## is given by: ##\psi_{AB} = \psi_{AC} \psi_{CB} + \psi_{AD} \psi_{DB}##
• Let's write the probability amplitude for going from state ##X## to state ##Y## (ignoring the time parameter for simplicity of notation) this way: ##\psi_{XY} = \sqrt{P_{XY}} e^{i \phi_{XY}}##
• Then the probability of going from ##A## to ##B## is given by: ##P_{AB} = P_{AC} P_{CB} + P_{AD} P_{DB} + 2 cos(\alpha) \sqrt{P_{AC}P_{CB} P_{AD} P_{DB}}## where ##\alpha = \phi_{AC} + \phi_{CB} - \phi{AD} - \phi{DB}##.

So the term ##2 cos(\alpha) \sqrt{P_{AC}P_{CB} P_{AD}}## is the interference term.

If the difference between ##C## and ##D## is that it represents two alternative intermediate measurement results, then a collapse means that the interference term should be left out of the probability. No collapse means that the interference term should be included. That's what I mean by saying that there is a mathematical difference between the collapse and no-collapse interpretations.

Now, for practical purposes, if ##C## and ##D## are macroscopically distinguishable, then either ##P_{CB}## or ##P_{DB}## will be completely negligible. So the interference term will be effectively zero. But mathematically, it's not exactly zero.

 

[martinbn]

Now, for practical purposes, if ##C## and ##D## are macroscopically distinguishable, then either ##P_{CB}## or ##P_{DB}## will be completely negligible. So the interference term will be effectively zero. But mathematically, it's not exactly zero.

Yes, but whether you include it or not is a matter of interpretation, not mathematics. For instance you can say

If the difference between ##C## and ##D## is that it represents two alternative intermediate measurement results, then a collapse means that the interference term should be left out of the probability. No collapse means that the interference term should be included. That's what I mean by saying that there is a mathematical difference between the collapse and no-collapse interpretations.

If the difference between ##C## and ##D## is that it represents two alternative intermediate measurement results, then a split of the world means that the interference term should be left out of the probability. No split means that the interference term should be included. That's what I mean by saying that there is no a mathematical difference between interpretations.

 

[stevendaryl]

Yes, but whether you include it or not is a matter of interpretation, not mathematics.

Yes, whether you include it or not is a matter of interpretation. That's my point---the interpretation has (very tiny) mathematical consequences. Different interpretations don't make (precisely) the same predictions.

If the difference between ##C## and ##D## is that it represents two alternative intermediate measurement results, then a split of the world means that the interference term should be left out of the probability. No split means that the interference term should be included. That's what I mean by saying that there is no a mathematical difference between interpretations.

Many Worlds doesn't actually have any splits. The wave function evolves unitarily, and the interference terms are there.

 

[entropy1]

So if no-collapse is asymptotically the same as collapse, is no-collapse the more general formulation?

 

[martinbn]

Yes, whether you include it or not is a matter of interpretation. That's my point---the interpretation has (very tiny) mathematical consequences. Different interpretations don't make (precisely) the same predictions.

Sorry, what I meant was that everyone includes it or not, they just use different words to justify it.

 

[stevendaryl]

Sorry, what I meant was that everyone includes it or not, they just use different words to justify it.

Many Worlds as I understand it would always include the interference terms.

There is a popularization of Many Worlds that posits that every time you perform a measurement, the world splits into as many copies as there are different possible outcomes. That really is a collapse interpretation grafted onto an ensemble view of probability. It's not the original Everett idea of a universal wavefunction that evolves unitarily.

 

[Agrippa]

Because I understand that for unitary evolution, MWI is required

Perhaps it's more correct to say: If unitary Schroedinger evolution is the only evolution and realism is true, then MWI is required.
(Bohmian mechanics keeps both realism and unitary Schroedinger evolution, but adds additional variables and dynamics in the form of the guiding equation).
(QBism keeps unitary Schroedinger evolution as the only evolution, but denies realism.)

which suggests that for different interpretations, there may not be unitary evolution?

Certainly for the collapse interpretations, the math and empirical predictions are significantly different. These differences arise from modifying the Schroedinger equation by adding non-linear terms that describe the collapse process.

 

[DarMM]

There would be a difference I think. Without the other branches being "pruned", they will eventually begin to cohere again due to the cyclic nature of the interference term coefficients. This far future interference is not something predicted by collapse theories.

(QBism keeps unitary Schroedinger evolution as the only evolution, but denies realism.)

For anybody not familiar with these terms, different authors use "realism" to mean slightly different things. Under Cabello's distinction (https://arxiv.org/abs/1509.04711), QBism is anti-realist in the sense that our measurement results are not purely properties/results of the subatomic systems themselves, but created by the act of measurement, a new property of the device/atom system. However this applies in general in the interpretation, not just to laboratory measurements, i.e. one atom can do it to another. This feature of the results being created by interactions is called instead by Cabello (after conversation with Christopher Fuchs I believe) "participatory realism", rather than anti-realism.

 

[Michael Price]

Many Worlds as I understand it would always include the interference terms.

There is a popularization of Many Worlds that posits that every time you perform a measurement, the world splits into as many copies as there are different possible outcomes. That really is a collapse interpretation grafted onto an ensemble view of probability. It's not the original Everett idea of a universal wavefunction that evolves unitarily.

No. The original Everett article talks about branchings and splittings AND maintains the universal wavefunction obeys unitary evolution at all times.

 

[stevendaryl]

No. The original Everett article talks about branchings and splittings AND maintains the universal wavefunction obeys unitary evolution at all times.

I think that in Everett's original paper, the emphasis was slightly different than the world splitting. Instead, the state of the world is relative to the observer's state. Each observer experiences a slightly different world.

 

[Michael Price]

I think that in Everett's original paper, the emphasis was slightly different than the world splitting. Instead, the state of the world is relative to the observer's state. Each observer experiences a slightly different world.

Seeing a different world is the same as saying the world has split. Everett never called the state relative to an observer a "world", but he used the language of splitting and branching everywhere. It is just a terminology thing.

 

[stevendaryl]

Seeing a different world is the same as saying the world has split. Everett never called the state relative to an observer a "world", but he used the language of splitting and branching everywhere. It is just a terminology thing.

This might be splitting hairs (no pun intended) but to me there is a difference between saying "The world splits every time a measurement is made" and "The state of the universe is relative to the observer". The latter has the consequence that no two observers share exactly the same world.

 

[Michael Price]

Surely the former implies this as well? When the world or timelines splits, each branch is slightly different, since each contains an observer having recorded a different result and the object system in a different state.

 

[stevendaryl]

Surely the former implies this as well? When the world or timelines splits, each branch is slightly different, since each contains an observer having recorded a different result and the object system in a different state.

Let's take a Schrodinger's cat example. With the "branching worlds" interpretation (which isn't Everett), there's a world where the cat is alive afterward, and there is a world in which the cat is dead. But in a relative state interpretation, the cat might be alive for some observers, dead for other observers, in a superposition for yet other observers. (Well, the cat by himself can't be in a superposition, because of decoherence, but the cat + environment can be in a superposition).