Never-ending experiments in MWI

[cube137]

I'd like to see an instance where experiments can end in mwi because it seems it never ends...

example.. if the radioactive source emits alpha, it triggers an explosion and Schrödinger is dead. If it didn't he is alive (we often use Schrödinger's Cat example with the consequence we never think what happens next because we think alive is the final state). so consider Schrödinger himself. When he entangled himself with the radioactive source, there are two terms in the superposition which is "Schrödinger dead when source emits" and "Schrödinger alive when source doesn't emit". When the experiment happens. In one term, he is alive, in one term he is dead.. but since the wave function never disappear and the unitary evolution goes on and on with more entanglements.. then the experiments never ends because in the Schrödinger dead branch, the police would be entangled with them (Schrödinger and the source) and in the Schrödinger alive branch, her lover would be entangled with them also (Schrödinger and the source). So is it correct that experiments never end in mwi? If you disagree. Please give example of one where experiment can end in mwi. Thank you.

 

[PeterDonis]

I'd like to see an instance where experiments can end in mwi

What do you mean by an experiment "ending"?

 

[cube137]

What do you mean by an experiment "ending"?

In the Copenhagen. an experiment ends when an outcome is chosen, for example Schrödinger alive when the radioactive source didn't emit alpha to trigger the explosive. But in MWI, there is never ending entanglements.. there is no way to cut the unitary evolution.. an experiment ends when the entanglement is cut. In the case of Schrödinger dead or alive...

quantum state = "Schrödinger alive when source didn't emit" + "Schrödinger dead when source emits"

then one includes the police...

quantum state = "Schrödinger alive when radioactive source didn't emit and police didn't arrive" + "Schrödinger dead when radioactive source emitted and police arrives"

one can include Schrödinger lover and the unitary evolution goes on and goes as the entanglement spreads.. so in mwi how do you stop or end an experiment?

 

[PeterDonis]

In the Copenhagen. an experiment ends when an outcome is chosen

So your definition of "the experiment ends" is "the wave function collapses". Then by your definition, there is no "experiment end" in the MWI, because MWI has no collapse.

However, that is a problem with your definition, not with the MWI. Actual experimental results are the same under both interpretations (they have to be, since both use the same underlying math).

 

[PeterDonis]

in mwi how do you stop or end an experiment?

When decoherence separates the different outcomes so that they can no longer interfere with each other.

 

[cube137]

When decoherence separates the different outcomes so that they can no longer interfere with each other.

I'm aware that Schrödinger is entangled with the radioactive source... so Schrödinger doesn't have a definite state... only both of them are... but even after decoherence, is it not the entanglement never breaks inside each mixed state (or branch) so Schrödinger is forever entangled with the radioactive source even when dead or alive and the entanglement can keep spreading within each mixed state or branch?

And in each branch, Schrödinger can have different numbers of children like 3 children versus none in other branch. So it's like new entanglements are formed to each mixed state, or in other words the branches would produce separate histories? We can easily imagine this where Schrödinger survived when the radioactive source didn't decay and explosive not set off and Schrödinger got out of the laboratory and went to dinner with his lover... this proves that even when it is in mixed state, the mixed state can produce more histories. Can the math produce this where mixed state can evolve on their own?

 

[PeterDonis]

in each branch, Schrödinger can have different numbers of children like 3 children versus none in other branch

Not as a result of the cat experiment. The only difference between the branches in the cat experiment is that in one, Schrödinger observes the cat to be alive, and in the other, Schrödinger observes the cat to be dead. Everything else about him is the same in both.

the branches would produce separate histories?

Basically, yes, that's what decoherence means.

this proves that even when it is in mixed state, the mixed state can produce more histories

No, it doesn't. There are no "mixed states" in the MWI. There are only superpositions and entanglements, and unitary evolution of pure states.

 

[cube137]

Not as a result of the cat experiment. The only difference between the branches in the cat experiment is that in one, Schrödinger observes the cat to be alive, and in the other, Schrödinger observes the cat to be dead. Everything else about him is the same in both.

How do you prepare an experimental setup where Schrödinger can have different children in different branches.. for example in one branch the radioactive source emits alpha which made him sterile and unable to bear children. In another branch he has many children. I didn't mention about the cat but you mentioned it.. so I thought you thought I was thinking about the cat.

Basically, yes, that's what decoherence means.
No, it doesn't. There are no "mixed states" in the MWI. There are only superpositions and entanglements, and unitary evolution of pure states.

 

[PeterDonis]

How do you prepare an experimental setup where Schrödinger can have different children in different branches..

You would have to make some aspect of his having children dependent on quantum indeterminacy.

 

[cube137]

You would have to make some aspect of his having children dependent on quantum indeterminacy.

like this?
quantum state = "Schrödinger healthy when radioactive source didn't emit" + "Schrödinger sick (and infertile) when radioactive source emits which activates a box in the lab that emits more gamma radiation"

so when decoherence separates the different outcomes.. then the outcomes would produce different histories...
but even without decoherence, the outcomes would evolve separately and produce separate histories too.. is it not?

 

[PeterDonis]

like this?

Basically, yes.

so when decoherence separates the different outcomes.. then the outcomes would produce different histories...

Yes.

even without decoherence, the outcomes would evolve separately and produce separate histories too.. is it not?

No, because without decoherence, the outcomes can interfere with each other.

 

[cube137]

Basically, yes.
Yes.
No, because without decoherence, the outcomes can interfere with each other.

If the outcomes interfere with each other.. why can't it produce separate histories?

 

[PeterDonis]

If the outcomes interfere with each other.. why can't it produce separate histories?

Because "separate histories" means that each branch can treat itself as self-contained, not influenced by any other branches. That is how the MWI explains our everyday observation that experiments have outcomes where, for example, Schrödinger is in a definite state. But that isn't true if the outcomes can interfere with each other.

 

[cube137]

Because "separate histories" means that each branch can treat itself as self-contained, not influenced by any other branches. That is how the MWI explains our everyday observation that experiments have outcomes where, for example, Schrödinger is in a definite state. But that isn't true if the outcomes can interfere with each other.

Ok.. unitary conserves information.. that is why after decoherence occurs, there is no more communication between the different branches... but if we can pump negative entropy much like how the ATP in the mitochondria can produce negative entropy to power our cells... can't another source of negative entropy from outside pump the wave function such that it can produce a link between branches? Why is this not possible?

 

[PeterDonis]

if we can pump negative entropy much like how the ATP in the mitochondria can produce negative entropy to power our cells... can't another source of negative entropy from outside pump the wave function such that it can produce a link between branches? Why is this not possible?

Where are you getting this stuff about negative entropy?

 

[cube137]

Where are you getting this stuff about negative entropy?

If unitary conserves information.. then we can simply supply more information from outside.. in thermodynamics.. it's negative entropy. In the wave function.. I don't know the equivalent but just illustrating the idea... what's the equivalent of negative entropy (from outside) in the unitary evolution that can connect two branches together?

 

[PeterDonis]

If unitary conserves information.. then we can simply supply more information from outside

No, you can't. There is no "outside" if you are doing unitary evolution correctly, because to do it correctly, the state vector has to already include everything that can possibly interact with whatever you're interested in.

in thermodynamics

In thermodynamics you don't model the system using a state vector and you don't use unitary evolution; you throw away information in your model. So thermodynamics is irrelevant to what we're discussing here.

 

[cube137]

Because "separate histories" means that each branch can treat itself as self-contained, not influenced by any other branches. That is how the MWI explains our everyday observation that experiments have outcomes where, for example, Schrödinger is in a definite state. But that isn't true if the outcomes can interfere with each other.

So if asked what happened after the experiment, whether Schrödinger would be healthy or sick with radiation, the answer is "both, each history occurs in different branches".. we can't say Schrödinger doesn't have definite state after the experiment.. we can't say only Schrödinger and radioactive sources have a definite state.. because in MWI, the experiment technically never end even after decoherence, because the branches evolve separately on their own.. and so Schrödinger is forever entangled with the radioactive sources for his entire lifetimes in each branch.. so we can really say Schrödinger and the radioactive sources are forever entangled right?

 

[Gigaz]

The line of thought is known as the quantum suicide:
https://en.wikipedia.org/wiki/Quantum_suicide_and_immortality

The thing is that it is on the edge of scientific method and quantum terminology. I mean, does dying constitute a measurement? Can a result be considered "scientific" if the experiment is subjective and can only be confirmed by each of us in person and not by proxy? How do we determine if the experiment failed and we are in fact dead?

 

[kith]

My experiment ends, when I decide that I have taken enough data. An experiment always involves a person who does the experiment, so there's always an inside view. QM enables me to make predictions about what I will find, if I do a certain experiment.

In MWI, the outside view -which involves my own quantum state- can be taken and there may be many terms in the quantum state of the universe which correspond to totally different events. But if it is a correct description, it needs to involve a term where I experience the end of my experiment.

 

[PeterDonis]

so we can really say Schrödinger and the radioactive sources are forever entangled right?

In the MWI, yes.

 

[PeterDonis]

if it is a correct description, it needs to involve a term where I experience the end of my experiment.

According to the MWI, the "you" in each branch, does experience the end of the experiment. That occurs, as I said before, when decoherence has separated all the branches so they can't interfere with each other.

 

[kith]

According to the MWI, the "you" in each branch, does experience the end of the experiment. That occurs, as I said before, when decoherence has separated all the branches so they can't interfere with each other.

Although I was talking about a more complex situation, I agree in principle. My intention with the statement you quoted wasn't to challenge the MWI but to show that experiments have to "end" in the MWI.

 

[cube137]

According to the MWI, the "you" in each branch, does experience the end of the experiment. That occurs, as I said before, when decoherence has separated all the branches so they can't interfere with each other.

The take home lessons in MWI is once entanglement forms, it keeps branching.. that means we must avoid meeting bad people or avoid certain situations because we can be forever entangled with them.. for example you meet jack the ripper.. you can form branches with him where you are his assistance or you are his victim. Remember there are quantum choices right in our body atomic processes.

What concerns me right now is the energy. Sometime I feel fatique and wondering if my other selves in other branches are using my energy. Let's go back to Schrödinger's.

quantum state = "Schrödinger healthy when radioactive source didn't emit" + "Schrödinger sick when radioactive source emits which activates a box in the lab that emits more gamma radiation"

Let's say the energy or the Hamiltonian operator has 10 Gev defined for the entire quantum state. Does it mean each term (or branches) would have 5 Gev? How do the branches partition the energy? And if there are more branches, there are less energy for each branch? And what is the right unit for energy (instead of Gev) in the Hamiltonian operator?

 

[Nugatory]

The take home lessons in MWI is once entanglement forms, it keeps branching.. that means we must avoid meeting bad people or avoid certain situations because we can be forever entangled with them.

The take-home lesson from MWI is that as you're thinking about the future evolution of whatever quantum system you've just measured... You are expected to calculate the value of a whole bunch of terms that you will never care about and then ignore them, whereas with collapse interpretations you can just set these terms to zero so there's nothing to ignore.

 

[PeterDonis]

Let's say the energy or the Hamiltonian operator has 10 Gev defined for the entire quantum state. Does it mean each term (or branches) would have 5 Gev?

No.

How do the branches partition the energy?

They don't. You apply an operator to the quantum state, not to parts of it. So it's meaningless to ask what the energy of just one branch is.

what is the right unit for energy (instead of Gev)

You can use any energy units you like. GeV is a perfectly valid energy unit.

 

[cube137]

No.
They don't. You apply an operator to the quantum state, not to parts of it. So it's meaningless to ask what the energy of just one branch is.
You can use any energy units you like. GeV is a perfectly valid energy unit.

How do you determine or compute the energy of the whole quantum state? can it be done? for example... Schrödinger being entangled to the radioactive source? do you weigh Schrödinger and the radioactive source?

 

[PeterDonis]

How do you determine or compute the energy of the whole quantum state? can it be done?

Of course, you just apply the Hamiltonian operator to the state.

for example... Schrödinger being entangled to the radioactive source? do you weigh Schrödinger and the radioactive source?

In non-relativistic QM, mass and energy are not the same, so weighing things doesn't tell you anything about their energy.

I strongly suggest that you spend some time working through a QM textbook such as Ballentine. We are getting to the point where a "B" level thread simply isn't sufficient.