# Is the math for collapse int different than for mwi?

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

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.

• Michael Price, entropy1 and Mentz114
Gold Member
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
Mentor
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.

• bhobba and entropy1
Gold Member
They are two different ways of explaining the same thing
Is the math different in either case?

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

Gold Member
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
Mentor
2020 Award
for unitary evolution, MWI is required

Why do you think that?

Gold Member
Why do you think that?
Is that not so?

PeterDonis
Mentor
2020 Award
Is that not so?

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

• bhobba, Michael Price and entropy1
PeterDonis
Mentor
2020 Award
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.

• bhobba, martinbn and entropy1
stevendaryl
Staff Emeritus
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
Mentor
2020 Award
To clarify even more, I would say "tries to derive" since there is no consensus that the derivation is possible.

Yes, agreed.

• bhobba
PeterDonis
Mentor
2020 Award
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
Staff Emeritus
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
Staff Emeritus
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
Gold Member
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
Staff Emeritus
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.

• entropy1
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.

• entropy1
stevendaryl
Staff Emeritus
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.

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