Does MWI Adequately Explain Observer Branching in Quantum Mechanics?

[PeterDonis]

Does the spin up and spin down really can separately entangled to different photons in the environment

No, that's not how it works. Go back to post #4; remember we ended up with the state

$$
\vert \Psi' \rangle = a \vert + \rangle \vert U \rangle + b \vert - \rangle \vert D \rangle
$$

However, if we're now going to talk about the environment, this state is incomplete. We need to include the environment's state and how it's entangled with the state of the system and the measuring device (which, remember, is something microscopic like the first atom that interacts with the measured system). So we need something like this:

$$
\vert \Psi' \rangle \vert E \rangle = \left( a \vert + \rangle \vert U \rangle + b \vert - \rangle \vert D \rangle \right) \vert E \rangle
\rightarrow
a \vert + \rangle \vert U \rangle \vert E_U \rangle + b \vert - \rangle \vert D \rangle \vert E_D \rangle
$$

In other words, from the standpoint of unitary evolution/entanglement, the interaction with the environment is just more of the same: where in post #4, the state of the measuring device branched, here the state of the environment branches. But the difference is that the measuring device was microscopic, or at least it was such that we were able to keep track of its detailed microstate, so that we could in principle reverse the measurement. But the environment, by hypothesis, is something whose detailed microstate we cannot keep track of, so the $\rightarrow$ above is irreversible: once the entanglement with the environment happens, there is no way to reverse it.

So the states $E$, $E_U$, and $E_D$ are not microscopic quantum states like the states $+$, $-$, $U$, $D$; they are really huge subspaces of the environment's Hilbert space, and we can't keep track which individual microstate inside each subspace the environment is actually in. But in each subspace, the environment is composed of the same microscopic constituents--the same photons that bounced off the measured system/measuring device, the same atoms that interacted with it, the same molecules in your retinal cells that received the photons, etc., etc. The only difference is which subspace of the environment Hilbert space all those interactions ended up in.

 

[Jilang]

You're missing the point of what I said before. Measurement happens before decoherence. So asking what happens if you try to measure the particle after it's been subjected to decoherence doesn't make sense.

Peter, I'm having a real problem with this. I have always understood the opposite. Interactions (as a result of the setup for the measurement) cause the decoherence and the final measurement result is following the decoherence. If my understanding of measurement as a two stage process is flawed I would like to know as I would have been labouring under a serious misapprehension for many years!

 

[PeterDonis]

I have always understood the opposite.

Based on what sources?

Interactions (as a result of the setup for the measurement) cause the decoherence

If this were always true, there would be no such thing as reversible interactions. But in the quantum computing world they do reversible interactions on qubits all the time. They entangle them and then un-entangle them, swap entanglements from one qubit to another, etc., etc., all reversibly.

What makes all these interactions reversible, as I said, is that we can keep precise track of the exact microstates of all of the systems involved. (Actually, that's not quite true: some of the devices that are used to implement qubit operations, like beam splitters, are macroscopic devices and we can't keep track of their precise microstates. But the precise microstates of those devices--the ones that implement the operations on qubits, rather than the qubits themselves--turn out not to matter, because no information is stored in them; their states don't get entangled with the states of the qubits. A full discussion of that would take us way beyond the "B" level.)

So "interactions cause decoherence" can't be right, because we can do all kinds of interactions in the lab that don't cause decoherence. What causes decoherence is interactions with the environment, i.e., with stuff outside the lab, or at least outside the precisely designed, calibrated, etc. lab equipment that we know we can manipulated without causing decoherence (like the beam splitters that act on qubits). In other words, interactions with systems that (a) have a lot of degrees of freedom, that (b) we can't keep precise track of, that (c) store information, via entanglement, about the state of the system we are interested in.

What I have been describing, in posts #4 and #36, is a highly idealized measurement where we can cleanly separate the initial "measurement interaction", where we entangle the state of the system with the state of some idealized "measuring device" whose state we can keep precise track of (so we can potentially reverse the measurement), from the subsequent decoherence, where the entanglement spreads to the environment, we can't keep precise track of it any more, and the whole process becomes irreversible. In most real situations, there is no such clean separation. When you see a bicycle (to use the example brought up earlier in this thread), there is no analogue to the initial idealized "measuring device", or even the initial "system" being measured. (I was hoping to get that point across by asking the question rhetorically.) The bicycle itself has a huge number of degrees of freedom, and those degrees of freedom are constantly interacting with each other and with an even huger number of degrees of freedom in the environment, like photons bouncing off the bicycle and then entering your eyes. There is no single "measurement" going on; it's just continuous interaction. And there isn't really decoherence going on in the bicycle case either, because there is no initial "coherence" (the analogue of the process I described in post #4) going on to be decohered from. The bicycle, and the photons bouncing off it, and your eyes, and your brain, body, etc., are never "coherent" in the first place (in the sense in which "decoherence" uses the term).

the final measurement result is following the decoherence

More precisely: knowledge of the final measurement result is following the decoherence. Or, to put it another way (the way an MWI theorist might put it): the final measurement result is only "final" after decoherence has occurred, because that is what makes that final measurement result constitute a separate "world". Once decoherence has taken place, that final measurement result is fixed, in that "world"--all observers will agree on it, further checks on what the result was will all agree, etc. The measurement can't be "undone" at that point. Whereas, as I said above, it might be that the measurement can be undone if all that has happened is an interaction, because the interaction might be between systems like qubits whose state we can keep precise track of.

 

[stevendaryl]

Peter, I'm having a real problem with this. I have always understood the opposite. Interactions (as a result of the setup for the measurement) cause the decoherence and the final measurement result is following the decoherence. If my understanding of measurement as a two stage process is flawed I would like to know as I would have been labouring under a serious misapprehension for many years!

Here's the way I understand measurement:

You have a macroscopic system that is in a metastable state. What that means is that it is stable, but it only takes a tiny push to send it into a drastically different state. For example, imagine a coin balanced on its edge. It's neither "heads" nor "tails". But give it a tiny push and it will fall into a more stable state: heads up or tails up.

Then you couple the metastable macroscopic system with the microscopic system that you are measuring. I don't know how you would do this, but suppose you make it so that if an electron is spin-up, it will give the coin a tiny push to make it fall heads-up, and if it is spin-down, it will make it fall tails-up.

So that's a crude measuring device for spin.

However, there is a hidden assumption here that is so deeply ingrained in our reasoning about cause and effect that we typically don't think about it: Why does the coin "prefer" to be on its side (heads or tails) instead of on its edge? The glib answer is that systems always try to minimize their energy, and the coin on its edge is a higher-energy state than the coin on its side. But that is not really a complete answer. Energy is conserved, so the total energy is the same, whether or not the coin is on its edge or on its side. The real story is that if the coin is on its edge, the energy is stored in the coin, while if the coin topples onto its side, the energy radiates away in the form of vibrations in the floor. So it's really not about energy, it's about entropy: having energy in the form of vibrations in the floor is a much higher entropy state than having energy in the form of a coin on its edge.

The above paragraph is speaking classically, but there is an quantum analogy: the coin falling onto its edge means coupling the state of the coin with the vibrational state of the floor; it means the coin's state becoming entangled with the states of the particles in the floor.

I think that classically, you haven't performed a measurement until an irreversible, entropy-increasing change to a more stable state. Quantum mechanically, you haven't performed a measurement until the measuring device becomes irreversibly entangled with the rest of the universe. So to me, measurement requires entanglement.

It's not true the other way around, though. You can have entanglement without anything that would count as a measurement.

 

[Blue Scallop]

No, that's not how it works. Go back to post #4; remember we ended up with the state

$$
\vert \Psi' \rangle = a \vert + \rangle \vert U \rangle + b \vert - \rangle \vert D \rangle
$$

However, if we're now going to talk about the environment, this state is incomplete. We need to include the environment's state and how it's entangled with the state of the system and the measuring device (which, remember, is something microscopic like the first atom that interacts with the measured system). So we need something like this:

$$
\vert \Psi' \rangle \vert E \rangle = \left( a \vert + \rangle \vert U \rangle + b \vert - \rangle \vert D \rangle \right) \vert E \rangle
\rightarrow
a \vert + \rangle \vert U \rangle \vert E_U \rangle + b \vert - \rangle \vert D \rangle \vert E_D \rangle
$$

In other words, from the standpoint of unitary evolution/entanglement, the interaction with the environment is just more of the same: where in post #4, the state of the measuring device branched, here the state of the environment branches. But the difference is that the measuring device was microscopic, or at least it was such that we were able to keep track of its detailed microstate, so that we could in principle reverse the measurement. But the environment, by hypothesis, is something whose detailed microstate we cannot keep track of, so the $\rightarrow$ above is irreversible: once the entanglement with the environment happens, there is no way to reverse it.

So the states $E$, $E_U$, and $E_D$ are not microscopic quantum states like the states $+$, $-$, $U$, $D$; they are really huge subspaces of the environment's Hilbert space, and we can't keep track which individual microstate inside each subspace the environment is actually in. But in each subspace, the environment is composed of the same microscopic constituents--the same photons that bounced off the measured system/measuring device, the same atoms that interacted with it, the same molecules in your retinal cells that received the photons, etc., etc. The only difference is which subspace of the environment Hilbert space all those interactions ended up in.

What do you mean in each subspace, the environment is composed of the same microscopic constituents? Do you mean
$E_U$ = same subspace, same microscopic constituents as $E_D$
$E_D$ = same subspace, same microscopic constituents as $E_U$

If this is so. Then Spin Up is entangled with the $E_U$, Spin Down is entangled with the $E_D$, and since $E_U$=$E_D$, then spin up and spin down is still in superposition. This is because according to Wikipedia:

"Decoherence happens when different portions of the system's wavefunction become entangled in different ways with the measuring device. For two einselected elements of the entangled system's state to interfere, both the original system and the measuring in both elements device must significantly overlap, in the scalar product sense. If the measuring device has many degrees of freedom, it is very unlikely for this to happen."

But you are claiming that different portions of the system's wavefunction become entangled in SAME ways with the measuring device. Here "both the original system and the measuring in both elements device HAS significantly overlap, in the scalar product sense". Therefore there is still quantum coherence (between spin up and spin down) because you have the exact environment.. so what if we can't keep tract of the environment and can't reverse it.. the spin up and spin down has identical environment, therefore they should still interfere (or the left and right slit of the double slit should still interfere if they can exposed to the same huge subspace of the environment), ain't it?

(I wonder if spin up and spin down is good example of this being able to interfere.. in the spin 1/2 particle, can we say the spin up and spin down is interfering? maybe this is the case where the double slit experiment left and right slit path is better example or spin up and down is still the prime example of it?)

 

[PeterDonis]

Do you mean

$E_U$ = same subspace, same microscopic constituents as $E_D$

$E_D$ = same subspace, same microscopic constituents as $E_U$

Same microscopic constituents, yes. Same subspace, no; $E_U$ and $E_D$ are different, disjoint subspaces of the Hilbert space of the environment. The difference is not in the microscopic constituents, but in the microstates of those constituents.

But you are claiming that different portions of the system's wavefunction become entangled in SAME ways with the measuring device.

No, I'm not. See above.

 

[Blue Scallop]

Same microscopic constituents, yes. Same subspace, no; $E_U$ and $E_D$ are different, disjoint subspaces of the Hilbert space of the environment. The difference is not in the microscopic constituents, but in the microstates of those constituents.
No, I'm not. See above.

But in the double slit experiment, in the single at a time emission of a photon or electron which goes to left or right slit, it can't be the same photons (microscopic constituents) that hit each slit because there is a difference in position (of left and right slit). But in the spin-1/2 particle, it has same position (spin up and spin down) so it can get entangled with the same photons (but in different microstates or subspaces as you said). Yes?

 

[PeterDonis]

in the double slit experiment, in the single at a time emission of a photon or electron which goes to left or right slit, it can't be the same photons (microscopic constituents) that hit each slit

You don't detect the result of a double slit experiment by looking at the slits. You detect it by looking at the interference pattern on the screen that gets built up over time as each photon or electron goes through the experiment. The screen doesn't move.

 

[Blue Scallop]

You don't detect the result of a double slit experiment by looking at the slits. You detect it by looking at the interference pattern on the screen that gets built up over time as each photon or electron goes through the experiment. The screen doesn't move.

But decoherence occurs halfway.. in the C60 fullucume experiment, thermal baths can change the interference patterns by the path of the atoms being affected by thermal photons (which is halfway).

Anyway. In spin-1/2 particles. How do you know whether it is in superposition of spin up and spin down or it has already suffered decoherence.. how do you check? maybe if the measuring device sees spin up and spin down randomly.. it is in superposition.. if it always see spin up or always see spin down.. it is already decohered?

 

[PeterDonis]

in the C60 fullucume experiment, thermal baths can change the interference patterns by the path of the atoms being affected by thermal photons (which is halfway)

I don't understand what you mean by "halfway". Do you have a reference?

How do you know whether it is in superposition of spin up and spin down or it has already suffered decoherence.. how do you check?

Look again at what I wrote mathematically to describe the process. Superposition never disappears anywhere in the process. You start with a system, like an electron, in a superposition of eigenstates of some measurement operator. The measurement operator gets applied to the system and entangles it with a microscopic measuring device, which puts that device into a superposition (as I posted in post #4). Then the microscopic measuring device gets entangled with the environment (decoherence), which puts the environment into a superposition (as I posted in post #36). No superposition ever gets removed.

Now if we were talking about a collapse interpretation, then at some point the superposition would collapse onto one of its terms (and to be consistent with our observations, this would have to happen after decoherence). But we are talking about the MWI, where there is no collapse; the superposition just stays there (and keeps spreading, as parts of the environment interact and get entangled with other parts of the environment). The only difference decoherence makes is to make this process irreversible, because we can no longer keep track of the detailed microstate of the superposition, since it involves so many degrees of freedom in the environment. So if your observation is irreversible, then it occurs after decoherence.

 

[Blue Scallop]

I don't understand what you mean by "halfway". Do you have a reference?

It's very popular setup:

http://www.univie.ac.at/qfp/research/matterwave/c60/

the decoherence occurs halfway .see:

"It is intriguing that C60 can almost be considered to be a body obeying classical physics in view of its many excited internal degrees of freedom. Leaving the source, it has as much as 7 eV of internal energy stored in 174 vibrational modes, and highly excited rotational states with quantum numbers greater than 100. Fullerenes can emit and absorb blackbody radiation very much like a solid and they can no longer be treated as a simple few level system.
Quantum interference experiments with large molecules, of the kind first reported here, open up many novel possibilities among them decoherence studies and nanolithography experiments."

Look again at what I wrote mathematically to describe the process. Superposition never disappears anywhere in the process. You start with a system, like an electron, in a superposition of eigenstates of some measurement operator. The measurement operator gets applied to the system and entangles it with a microscopic measuring device, which puts that device into a superposition (as I posted in post #4). Then the microscopic measuring device gets entangled with the environment (decoherence), which puts the environment into a superposition (as I posted in post #36). No superposition ever gets removed.

Now if we were talking about a collapse interpretation, then at some point the superposition would collapse onto one of its terms (and to be consistent with our observations, this would have to happen after decoherence). But we are talking about the MWI, where there is no collapse; the superposition just stays there (and keeps spreading, as parts of the environment interact and get entangled with other parts of the environment). The only difference decoherence makes is to make this process irreversible, because we can no longer keep track of the detailed microstate of the superposition, since it involves so many degrees of freedom in the environment. So if your observation is irreversible, then it occurs after decoherence.

 

[PeterDonis]

the decoherence occurs halfway

I still don't understand what you mean by "halfway". Please explain in more detail.

 

[Blue Scallop]

I still don't understand what you mean by "halfway". Please explain in more detail.

Simple. As the C60 atom emits from source.. it can enter either the left or right slit.. but decoherence can make the path known before it even reaches the slit (destroying any interference in the screen).. the effect is like the left or right slit being covered. Why, how do you understand the c60 buckyball thermal experiment? where does decoherence occur there?

 

[PeterDonis]

decoherence can make the path known before it even reaches the slit

In which case you do not get interference fringes. But the page you linked to says interference fringes were detected. That means they did not do anything to make the path known in the middle of the experiment.

where does decoherence occur

It depends on what experiment you are running. If you run the experiment where you are detecting which slit the particle goes through, that is a different experiment from the experiment where you don't detect which slit the particle goes through. So you can't draw deductions about where decoherence occurs in the second experiment from the results of the first one. That's true whether the "particle" in the experiment is a photon, an electron, or a buckyball.

As for the general question of where decoherence occurs, I've already answered it, several times. Decoherence occurs when the process of interaction becomes irreversible because it has spread over a large enough number of degrees of freedom that we can't track the precise microstate any more.

 

[Blue Scallop]

Sure, just pick the basis in which the state $a \vert + \rangle + b \vert - \rangle$, which is the state that the system starts out in in post #4, is one of the basis states. The other basis state will then be $b \vert + \rangle - a \vert - \rangle$. It's easy to show that these two states are orthogonal, just as the $+$ and $-$ states are orthogonal.

$+$ and $-$ are orthogonal and yet they are in superposition.
$a \vert + \rangle + b \vert - \rangle$ and $b \vert + \rangle - a \vert - \rangle$ are orthogonal and yet they are not in superposition.
Why is the latter not in superposition? I think this is related to what Sabine wrote in http://backreaction.blogspot.com/2016/03/dear-dr-b-what-is-difference-between.html:

"All this is just to say that whether a particle is or isn’t in a superposition is ambiguous. You can always make its superposition go away by just wanting it to go away and changing the notation. Or, slightly more technical, you can always remove a superposition of basis states just by defining the superposition as a new basis state. It is for this reason somewhat unfortunate that superpositions – the cat being both dead and alive – often serve as examples for quantum-ness. You could equally well say the cat is in one state of dead-and-aliveness, not in a superposition of two states one of which is dead and one alive."

So let's say one basis is the cat is in state of dead-and-aliveness and another basis is in state of dead-minus-alive (or others). Why are these basis not in superposition?

 

[PeterDonis]

yet they are not in superposition

You're misstating it. The state $a \vert + \rangle + b \vert - \rangle$ is the superposition of $+$ and $-$ that we have been talking about. The state $b \vert + \rangle - a \vert - \rangle$ is also a superposition (a different superposition) of $+$ and $-$. And all of this depends on our having chosen $+$ and $-$ as a basis.

If, instead, we choose the states $a \vert + \rangle + b \vert - \rangle$ and $b \vert + \rangle - a \vert - \rangle$ as a basis (which we can since they are orthogonal and any state in the Hilbert space can be expressed as a linear combination of them), then the states $+$ and $-$ are now superpositions, whereas our new basis states are not. So yes, whether or not a state is a superposition is basis dependent, as Sabine says.

However, the choice of basis is not arbitrary once you specify a particular measurement. For example, in our spin example, we specified that we were measuring spin about the axis for which $+$ and $-$ are the eigenstates. That means the $+$ and $-$ basis is picked out by the physics of the measurement we are making; the choice of basis is no longer arbitrary. So the fact that the state $a \vert + \rangle + b \vert - \rangle$ is a superposition in this basis is now physically relevant, since it affects how the state of system + measuring device (+ environment once we bring that in) will evolve in time, in contrast to the way everything would evolve in time if the particle we were measuring were in the state $+$ (not a superposition in this basis).

Or, we could have chosen a different spin measurement, one for which the eigenstates were $a \vert + \rangle + b \vert - \rangle$ and $b \vert + \rangle - a \vert - \rangle$. Then a particle in the state $a \vert + \rangle + b \vert - \rangle$ would not be in a superposition, and everything would evolve in time the way you would expect if you measure a particle that is in an eigenstate of the measurement (i.e., no "branching" from the MWI point of view). We could still write things in the $+$ and $-$ basis, as I just did, but the measurement operator would look more complicated in this basis (the matrix would not be diagonal), and you would have to do a more complicated calculation to confirm what I just said about how things would evolve in time (whereas the calculation in the basis of eigenstates of the measurement is trivial).

We are rapidly approaching the point where I am going to close this thread as we have beaten this subject to death as much as we can in a "B" level thread. As I've said before, you really need to spend some time working through a basic QM textbook (and learning linear algebra, since all this that I've been saying about superpositions and basis and measurement operators is linear algebra 101).

 

[PeterDonis]

let's say one basis is the cat is in state of dead-and-aliveness and another basis is in state of dead-minus-alive (or others). Why are these basis not in superposition?

Because, as Sabine says, you can pick any pair of orthogonal states you like as a basis (note that we are treating the cat as if it were a spin-1/2 particle, which of course it isn't, but that complication is usually glossed over in discussions of this sort--although it really shouldn't be, see below). But, as I noted in my previous post, once you start talking about making measurements on the system, the choice of basis is no longer arbitrary--you can pick any basis you like, but only one basis will correspond to the basis of eigenstates of the measurement you are making. In the case of the cat, the alive/dead measurement is the one we know how to physically realize (or at least sort of realize--see below). We don't know how to physically realize a measurement which has "alive plus dead" and "alive minus dead" as its eigenstates.

(And, since the cat is not a spin-1/2 particle but a huge conglomeration of something like $10^{25}$ atoms, all of which are interacting, and which is interacting continuously with its environment, it's not clear that there even is a measurement that has "alive plus dead" and "alive minus dead" as eigenstates. That's because it's not clear that there even is a single quantum measurement that has "alive" and "dead" as eigenstates! The "alive state" of the cat is not a single quantum state; it's a huge subspace of the cat's Hilbert space, and the "dead state" is another huge subspace that we are assuming is disjoint from the first one. But we don't actually know that that's the case. We don't have a foolproof "alive/dead meter" that we can attach to a cat that goes one way if the cat is alive and another way if the cat is dead, the way we can unambiguously measure the spin of a spin-1/2 particle. When we say that a cat is "alive" or "dead", that is a judgment based on what amounts to a huge number of quantum measurements that has no useful relationship to the kind of clean, idealized measurement we make on the spin of a single spin-1/2 particle. So it's unfortunate that those two cases are so often conflated in discussions of this sort.)

 

[Denis]

Decoherence is necessary, together with locality, for solving the basis selection problem, which is crucial to "branching".

Which is one of the problems why MWI does not make sense.

Decoherence itself is a quite meaningful method to study various quantum systems, in particular greater ones which have some interaction with the environment which cannot be completely suppressed.

But this already presupposes some subdivision of the world into some parts, in particular the system to be studied and the environment. Without such a subdivision given, decoherence does not make sense. To apply decoherence, MWI has to presuppose something it officially does not presuppose. At least, I'm not aware what are these subdivisions into systems, environments and observers we need even before we can define the worlds - given that, as claimed, branching (and therefore the resulting branches) are not yet defined.