Probability in MWI and the Copenhagen interpretation

[entropy1]

Suppose we have an operator with three eigenvectors/eigenvalues $e_1$, $e_2$ and $e_3$. The operator measures wavefunction $\psi$. Could we say that we find outcome $e_x$ with probability $P(\psi,e_x)$, and could we extend this to an infinite dimensional operator as a spectrum of probability?

 

[PeterDonis]

Suppose we have an operator with three eigenvectors/eigenvalues $e_1$, $e_2$ and $e_3$. The operator measures wavefunction $\psi$. Could we say that we find outcome $e_x$ with probability $P(\psi,e_x)$, and could we extend this to an infinite dimensional operator as a spectrum of probability?

I assume you are asking how this question would be answered in the Copenhagen interpretation and the MWI, in order to compare and contrast the answers.

In the Copenhagen interpretation the answer is a simple "yes".

In the MWI the answer (unless you come up against a true hard-core MWI proponent who is willing to abandon the concept of "probability" altogether and just answer "no") will be a complicated explanation of how it's still ok to use the term "probability" even though all of the possible outcomes are realized.

 

[entropy1]

In the MWI the answer (unless you come up against a true hard-core MWI proponent who is willing to abandon the concept of "probability" altogether and just answer "no") will be a complicated explanation of how it's still ok to use the term "probability" even though all of the possible outcomes are realized.

I see it kind like this: In MWI, if we pair the (possible) outcomes with the probability of getting them, we practically end up with Copenhagen. The MWI in a way is the (preservation of the) probability distribution, while Copenhagen is the result of partitioning the measured wavefunction in orthogonal components given by the eigenvectors, causing that only one of those components can yield an outcome (because the possible outcomes are orthogonal). If I see it correctly, with Unitarity the probability distribution is preserved in the propagation of the wavefunction. We have entanglement between measured and measuring, no collapse yielding an outcome, but rather the process of decoherence that converges to a macroscopic result.

Perhaps is it also possible that the different macro-worlds of MWI are optional, and that choosing one of them retrocausally determines the microscopic outcome (like a collapse as a result of getting macroscopic by decoherence, a kind of "snap". Could that be possible?).

 

[PeterDonis]

In MWI, if we pair the (possible) outcomes with the probability of getting them, we practically end up with Copenhagen.

I have no idea what you mean here. MWI says all possible outcomes are realized. Copenhagen says only one is.

Copenhagen is the result of partitioning the measured wavefunction in orthogonal components given by the eigenvectors

This has nothing to do with Copenhagen; it's part of the basic math of QM, which applies to all interpretations. The same thing happens in MWI; the only difference is that all of the orthogonal components are realized, whereas in Copenhagen only one is.

causing that only one of those components can yield an outcome (because the possible outcomes are orthogonal)

This is not correct. The wave function is the same regardless of interpretation, including all of the components being orthogonal, but all of the components are realized in the MWI.

If I see it correctly, with Unitarity the probability distribution is preserved in the propagation of the wavefunction

I don't know what you mean by this.

We have entanglement between measured and measuring, no collapse yielding an outcome, but rather the process of decoherence that converges to a macroscopic result.

No, the process of decoherence realizes all of the possible results. It doesn't just realize one.

Perhaps is it also possible that the different macro-worlds of MWI are optional

This is personal speculation and is off limits here. It also makes no sense as a version of the MWI, since the whole point of the MWI is that nothing is "optional": unitary evolution all the time requires that all components of the wave function remain real always.

 

[Demystifier]

Suppose we have an operator with three eigenvectors/eigenvalues $e_1$, $e_2$ and $e_3$. The operator measures wavefunction $\psi$. Could we say that we find outcome $e_x$ with probability $P(\psi,e_x)$, and could we extend this to an infinite dimensional operator as a spectrum of probability?

Are you familiar with the von Neumann theory of quantum measurements? If not, I think you should first familiarize with it before asking such questions.

 

[entropy1]

@PeterDonis. If I understand correctly, under certain conditions, with measurement, the microscopic outcome entangles with the macroscopic outcome. So am I right saying that means that microscopic outcome $e_x$ corresponds to macroscopic outcome $M_x$? So would that mean that if we find ourselves reading macroscoping outcome $M_x$ off the measurement device, the microscopic outcome is (assumed to be) $e_x$?

 

[PeterDonis]

If I understand correctly, under certain conditions, with measurement, the microscopic outcome entangles with the macroscopic outcome.

This doesn't just happen under "certain" conditions; it always happens. The question is what happens next. In the MWI, all of the outcomes are realized: each entangled pair is a branch of the wave function and all branches are real. In Copenhagen, only one outcome is realized: one entangled pair gets picked according to the Born rule as the one that is realized.

would that mean that if we find ourselves reading macroscopic outcome $M_x$ off the measurement device, the microscopic outcome is (assumed to be) $e_x$?

Yes.

 

[entropy1]

In the MWI, all of the outcomes are realized: each entangled pair is a branch of the wave function and all branches are real. In Copenhagen, only one outcome is realized: one entangled pair gets picked according to the Born rule as the one that is realized.

Is the mechanism that determines which pair gets to be experienced completely unknown? Probably the answer for MWI is: all pairs get to be experienced.

 

[PeterDonis]

Is the mechanism that determines which pair gets to be experienced completely unknown?

Of course not. It's the interaction that entangles the measuring device, and the observer, with the measured system. Each branch of the entangled wave function that results from that interaction matches up a particular state of the measured system with the corresponding states of the measuring device and the observer.

Probably the answer for MWI is: all pairs get to be experienced.

Of course.