Is There Only One True Outcome in the Many-Worlds Interpretation?

[entropy1]

Suppose we have a photon in superposition of reaching detector A or detector B. Then, in Everett-worlds, both outcomes (detection at A/detection at B) are true, but in different worlds (##|U_x\rangle|x\rangle##). But if we observe the law of conservation of energy and the quantisation of the detection, there can only be one detection. And this is what MWI tells us: in each world one outcome is true (##|x\rangle## part), and the rest is not (##|U_x\rangle## part)! This would be, in my eyes, the consequence of the quantisation of the measurement.

Would this be a valid observation?

 

[PeterDonis]

in Everett-worlds, both outcomes (detection at A/detection at B) are true, but in different worlds (##|U_x\rangle|x\rangle##)

This is not a good way to look at it if you are trying to evaluate conservation laws. In terms of the MWI, the whole system, which includes both the measuring device and the measured system, is in a single pure state. That state includes entanglement between the two subsystems (measuring device and measured system), but it's still just one state.

if we observe the law of conservation of energy and the quantisation of the detection, there can only be one detection

Unless the pure state of the whole system is an eigenstate of energy (i.e., of the Hamiltonian), then the system as a whole has no well-defined energy and there is no way to even ask the question whether energy is conserved. Conservation laws don't work the same in QM as they do in classical mechanics.

in each world one outcome is true (##|x\rangle## part), and the rest is not (##|U_x\rangle## part)!

This doesn't make sense. The pure state of the system as a whole evolves from ##U \left( |A\rangle + |B\rangle \right)## to ##|U_A\rangle |A\rangle + |U_B\rangle |B\rangle##. The two terms in the second state are the different "worlds" if you want to use that terminology, and each world has both a "U" factor and an "x" factor; there is no sense in which the "U" factor is in one world and the "x" factor is in another.

 

[entropy1]

This doesn't make sense. The pure state of the system as a whole evolves from ##U \left( |A\rangle + |B\rangle \right)## to ##|U_A\rangle |A\rangle + |U_B\rangle |B\rangle##. The two terms in the second state are the different "worlds" if you want to use that terminology, and each world has both a "U" factor and an "x" factor; there is no sense in which the "U" factor is in one world and the "x" factor is in another.

Thanks for answering. It is night here, I had a drink, I will read it over tomorrow.
Here I ment the state ##|U_x\rangle|x\rangle## in one of the worlds, ##|x\rangle## indicating outcome x, and ##|U_x\rangle## indicating none of the other outcomes (exclusion).

 

[PeterDonis]

Here I ment the state ##|U_x\rangle|x\rangle## in one world, ##|x\rangle## indicating outcome x, and ##|U_x\rangle## indicating none of the other outcomes (exclusion).

Ok. Yes, in one world, only one outcome occurs. However, in the MWI, all the other worlds also exist, so you cannot infer from the fact that only one outcome occurs in one world, that only one outcome occurs. (On a collapse interpretation, you can make that inference.)

As regards energy conservation, none of this really matters. The key thing about energy conservation, or any conservation law in QM, is that you can only assign definite values to anything if you measure it. So, for example, in the case of the photon in a superposition of reaching detector A or detector B (for example, by passing through a beam splitter), if you have a photon source that emits exactly one photon at a time and measures the photon's energy as it's emitted (and building such a source is a lot harder than you might think), and you have a setup that measures the energy delivered if a photon is detected at either detector, then you will find that, every time one photon is emitted from the source, either one photon will be detected at A, or one photon will be detected at B, never both; and whichever detector detects the photon will measure the same energy as the source measured when the photon was emitted. But all of those actual measurements have to be there in your experiment before you can say anything about energy conservation.

 

[entropy1]

Ok. Yes, in one world, only one outcome occurs. However, in the MWI, all the other worlds also exist, so you cannot infer from the fact that only one outcome occurs in one world, that only one outcome occurs. (On a collapse interpretation, you can make that inference.)

In each world, one outcome exists, which means that in world X, there exists outcome ##|A\rangle##, the other outcomes are excluded, while in world Y, there exists outcome ##|B\rangle##, while in that world ##|A\rangle## gets (explicitly!) excluded, so that in fact the worlds couldn't coexist, because they overlap yet exclude each other (world X and world Y together have both ##|A\rangle## included and excluded). It is like you have to make a choice, and that is what I mean by "there can be only one detection".

This is not a good way to look at it if you are trying to evaluate conservation laws. In terms of the MWI, the whole system, which includes both the measuring device and the measured system, is in a single pure state. That state includes entanglement between the two subsystems (measuring device and measured system), but it's still just one state.

If the probabilities of the worlds being real add up to 1, then each world has a probability of being real, right? Doesn't that mean that they are mutually exclusive in that way?

Unless the pure state of the whole system is an eigenstate of energy (i.e., of the Hamiltonian), then the system as a whole has no well-defined energy and there is no way to even ask the question whether energy is conserved. Conservation laws don't work the same in QM as they do in classical mechanics.

I don't get it: if I fire a photon with energy ##E_{ph}##, and it travels to a detector that measures its presence, could its energy have changed? I guess energy is probabilistic too?

 

[PeterDonis]

In each world, one outcome exists

No. In each world, one outcome "occurs". But you cannot say that only one outcome "exists" in the MWI. You can only say that in a collapse interpretation.

You are misleading yourself by trying to reason using ordinary language in a domain where ordinary language does not work. Physicists use math to avoid this kind of error.

If the probabilities of the worlds being real add up to 1

This is incorrect; in fact in the MWI it's meaningless. There is no such thing as "probabilities of worlds" in the MWI. All of the worlds exist; they are all contained in the wave function of the overall system.

then each world has a probability of being real, right?

No. In the MWI, all of the worlds are real.

I fire a photon with energy ##E_{ph}##, and it travels to a detector that measures its presence, could its energy have changed?

The photon is not the whole system. There is also the source and the detector. The photon is entangled with both of them. You can't think of the photon as a single system with a definite energy unless the whole system that includes the source, photon, and detector, has a definite energy. But that is only the case if the whole system is in an eigenstate of the overall Hamiltonian. In practice that is almost never the case.

I guess energy is probabilistic too?

If you measure the energy of a system that is not in an eigenstate of the Hamiltonian, then there are multiple possible energies that you could find as the result of the measurement. That's what not being in an eigenstate means.

Whether that is "probabilistic" depends on which interpretation you adopt. In the MWI, the system becomes entangled with the energy measuring device and multiple "worlds" exist, in each of which the system has a particular energy (one of the possible ones that could have been measured) and the measuring device records the result that shows that energy.

 

[entropy1]

@PeterDonis I will have to study that. Thanks for the answer.

On second sight, it is actually making sense to me what you are saying.