Many-Worlds Theory

ThomasT

I thought I was referring to a definite outcome.

Aren't all recorded results definite outcomes? Or is there a difference between results and outcomes?

What do you mean by 'definite outcomes'?

ThomasT

The boundary of a finite universe doesn't seem so strange if you think of that boundary as a wavefront expanding in a medium. Observations suggest that our universe is expanding, which suggests that it had a beginning, which implies that it's finite -- even if fapp it's infinite.

'Wavefunction collapse' refers to the registration of a detection, the appearance of a bit of data, the end of an interval that might define a trial where no detection is recorded, or the end of an experimental run. The term wasn't invented to eliminate "other worlds", at least afaik it wasn't.

"Other-worlds-elimination" isn't a problem for interpretations that don't produce them. The "other worlds" never appear. The simplest explanation for this is that they're interpretational fictions.

Hurkyl

We have a definite outcome if the wavefunction of the ________¹ is not in a combination² of states corresponding to different outcomes.

1: Fill the blank in with "universe" or "system" or whatever would be appropriate.
2: Either mixed or pure


e.g. if we have a qubit with basis states |0> and |1>, and we decide to call those states 'outcomes', then |0> and |1> are definite, whereas |0> + |1> is indefinite, as is a mixture 50% |0> and 50% |1>.


For a more elaborate example, suppose we had four qubits, three initialized to zero, and the other one an input to our experiment. We use a CNOT gate to add the input state to the first and second qubits (a toy 'measurement'), and then we used two more CNOT gates to add the first and second qubits to the third (a 'compare' 'measurement' -- set the third qubit to 1 if and only if the first and second qubits are different)

If the input qubit is |0> + |1>, then the end result if passing it through the device is the state: (the input qubit is on the fourth line)

A: Start with: |000>(|0> + |1>)
B: After first 'measurement' CNOT gate: |0000> + |1001>
C: After second 'measurement' CNOT gate: |0000> + |1101>
D: After the first 'compare' CNOT gate: |0000> + |1111>
E: After the second 'compare' CNOT gate: |0000> + |1101>

If we look at the (relative) state of the input qubit, it starts off in the pure state |0>+|1>, and after B, it decoheres into the mixture 50% |0> and 50% |1>, and stays there. The 'outcome' of the input qubit is indefinte.

If we look at the first qubit, it starts in the pure state |0>, and after B, it transisitions into the mixture 50% |0> and 50% |1>, and stays there. (It 'splits' into two 'copies': one 'measuring' a 0, and one 'measuring' a 1)

The third qubit, on the other hand, started in |0>, after D split into the mixture, and then at E it recombined into the pure state |0> -- we have the definite outcome that the first two qubits are measured to have the same value. (Despite the fact their actual value is indefinite!)

ThomasT

What I meant by definite outcome or definite result was a data bit or the physical state of a detector at a particular time. This is how the words, result, outcome, and definite are conventionally used, afaik.

Our discussion is about the physical meaning of quantum states. The physical meaning of the pure quantum bit state is that it will correspond to either 0 or 1, with 0 and 1 referring to, eg., two different voltage levels, two different polarizations, two different locations, etc. The data bits, 0 and 1, that the qubit superposition represents are mutually exclusive, definite outcomes.

CNOT gate demonstrations don't tell us that qubits refer to two, mutually exclusive, physical states simultaneously any more than any quantum superposition refers to the simultaneous existence of any number of mutually exclusive physical states.

Phrak

Classical logic levels are a recoding of qbits, in this case to two level saturable states. It doesn't have to be-one-to one does it? The process has physical meaning, as you've called it, within the constraints of the recoding mechanism.

Phrak

I tried to make some headway on this some time ago when this came up, as you may recall. I didn't get very far. It seemed, apriori unreasonable that two qbits would interact, with |C> effecting a change in |A>, without |A> effecting a change in |C>.

The best I obtained for an answer was an oblique hint that the phase of the control bit, |C> was effected by the state of |A>.

Is this correct?

Hurkyl

Alas, it has completely fallen out of my brain. Anyways, I don't follow your train of thought, but I can give my own.

If we focus (relative) state of the control bit, then the typical result is that the CNOT gate forces decoherence: after passing through the CNOT gate, an initial pure state p|0> + q|1> would be transformed into a mixed state of |0> and |1> with probabilities |p|^2 and |q|^2 respectively. The same would happen with any input mixture. The only exception is if the control and target lines were entangled in a way that would cause interference.

So, as we would expect from a toy measurement, the object being 'measured' has decohered into a statistical mixture of the two outcomes.

And as long as there's not entanglement involved, the actual state of the target line is completely irrelevant. (At least, if I've done the calculation correctly)

Phrak

No problem, Hurkyl. Some 1000 threads back for both of us, you were explaining MWT and decoherence. The notion of a toy observer came up—I think for the first time—as a possible means to model decoherence in quantum mechanical observers.

---------------------------------

Maybe I should start over.

To make sense of MWT we need to understand an observer as a quantum mechanical system. To understand this, we could construct a toy observer. To do this we need to model the nature of human or machine information processing. To do this we need:

1) An underlying principle of classical information processing. We need this because it will tell us that classical information processing requires that information is discarded. The so-called reversible classical gates are reversible only in principle. Even a not gate discards information. This will invoke decoherence as a sufficient, though not necessary, element of classical information processing. (more on this later.)

and 2) A schema for building classical logic gates or neurons out of multiple quantum gates.


Now, if we are to model gates, or neurons, or the neurons of neuronetworks out of quantum gates, we had better darned well understand quantum gates.

I cannot believe that two quantum bits can interact, where qbit C changes qbit T, without qbit T changing qbit C. Say we have a c-not and C=1. What has happened to the information of the fomer state of T? If we are questioning the validity in the operation of a c-not gate, it’s not enough to say the value of T can be reacquired by acting a second c-not gate on T'. I think something has been left out of the popular description of a c-not gate.

I will have to go out on a limb in the following, because the fact of the matter is, I don’t know if the following is true or not. Let me know.

a) A quantum gate is reversible.
b)There is a relative phase between two qbits.
c) If the phase information is not preserved, the gate cannot be reversed.

For simplicity, assume the inputs to a c-not are both pure states; either |1> or |0>. Just as in boolean logic, there are 4 possible outcomes. If not, reversibility is violated; quantum mechanics would not obey time reverse symmetry.

d) I'm going to make a wild stab at the truth table of a c-not as follows.

[tex](c,t) \rightarrow (c',t')[/tex]
---------------------
[tex](0,0) e^{\delta} \rightarrow (0,0) e^{\delta}[/tex]
[tex](0,1) e^{\delta} \rightarrow (0,1) e^{\delta}[/tex]

[tex](1,0) e^{\delta} \rightarrow (1,1) e^{\delta + \phi}[/tex]
[tex](1,1) e^{\delta} \rightarrow (1,0) e^{\delta + \phi}[/tex]

[itex]\delta[/itex] is the relative phase of two qbits. [itex]\delta[/itex] is an unphysical gauge, that we could just as well set to zero. [itex]\phi[/itex] is the change in the relative phase of c and t. We should be free to attach the phase to either output bit, as long as we are consistant.

A second c-not gate acting on the primed qubits would have the truth table:

[tex](c',t') \rightarrow (c'',t'')[/tex]
---------------------
[tex](0,0) e^{\delta} \rightarrow (0,0) e^{\delta}[/tex]
[tex](0,1) e^{\delta} \rightarrow (0,1) e^{\delta}[/tex]

[tex](1,0) e^{\delta + \phi} \rightarrow (1,1) e^{\delta + 2\phi} [/tex]
[tex](1,1) e^{\delta + \phi} \rightarrow (1,0) e^{\delta + 2\phi} [/tex]

If [itex]\phi = i \pi[/itex], two c-not gates in series will restore the primed states to their original unprimed states.

So, have I told any fibs yet?

The above contention is testable with two electrons (C,T), entangled in a c-not, sent on separate paths encompasing a solenoid, then reentangled with a second c-not. Varying the strength of the solenoid field should cause the resultant spin states (C'',T'') to vary. Sending both electrons around the same side of the solenoid should obtain (C'',T'')=(C,T), independent of the strength of the solenoid field.

Hurkyl

What exactly do you mean by this? I understand what it would mean to talk about the overall phase of their joint state, but it's not clear how to make sense of the idea of a relative phase between the two qubits.

Symbolically, the joint state, expressed as a vector in Hilbert space in the 0-1 basis, is of the form

a |00> + b |01> + c |10> + d |11>

If b=c=d=0, for example, then we can interpret a as being the (unphysical) overall phase of the state vector, but there is nothing here expressing the idea of a relative phase between the two bits.

We could use the partial trace to extract the single particle states out of this joint state -- but the calculation for that I know involves using density matrices, and thus completely obliterates all information about the overall phase. (as well as destroying information about entanglement)

Phrak

But you do understand the problem, don't you? If we can't properly account for quantum mechanical information, it's problematic that a translation to classical information will be correct.

Two particles of the same species have a relative phase. Either that, or I've misunderstood something. I'm referencing Mark P. Silverman, on the correlations of two electron interferometry, if that helps.

But things are worse than I though. The quantum gates described in most web sites, including Wikipedia, ignore the gate apparatus. The unitary operations are so schematical that entanglement with the apparatus enabling the operation is ignored. I have managed to discover that in an NMR (Nuc. Mag. Resonance) Hadamard gate, for instance, the qubit is entangled with a radio frequency source. Nowhere does this show up in the truth-table.

Hurkyl

It's not clear what exact question you're asking here. If you're referring to the loss of information when decomposing a 2-particle state into two 1-particle states, then yes, that would be a problem if we insisted that the two 1-particle states told us everything about the 2-particle state. Fortunately, we know better than to insist such a thing!

If you're referring to the fact I don't know what you mean by 'relative phase' here, then basically, one of the following is true:

A. You are talking about something that isn't a 2-qubit system
B. There is relative phase in the state I described, and I simply don't know what you mean by the term

I guess I should point out that the place I am familiar seeing 'relative phase' is when talking about the components of a quantum state relative to some basis. e.g. for the single qubit state a|0>+b|1>, both the magnitude and the phase of the value (b/a) are physically significant.


If the action of the gate apparatus on the qubits is not completely described by the truth table, then it's not a quantum gate.

I'm somewhat skeptical of your description though: I don't think the action of the gate on the qubit state can be unitary if it results in an entanglement with the gate apparatus. However, there is no problem introducing a temporary entanglement while the apparatus is operating, as long as it gets wiped out by the end of the operation.

I confess, however, I don't really know anything about the engineering of quantum gates.

Phrak

Re: engineering of quantum gates.

This is what it comes down to. I really won't be convinced of many things without a description of the experiental apparatus that manages a universal gate set.

ccrummer

Many Worlds, like any other interpretation is not forced on us like, e.g. the Schroedinger equation is. In fact none of the (carefully thought out) interpretations is. Consider the following experiment. A fat coin is tossed. (It's fat so that the probability of it landing on its edge is not very small.) Spinning in the air, it is in a superposition of three states: heads, tails and edge. (There are hidden variables here of course, namely the initial conditions of the flip.) In a universe the coin lands on its edge. The people in this universe have never seen such a thing before. In the same universe the coin is flipped again. In another universe the coin lands on its edge and the people in this universe have the memory of two edges in a row. This goes on through many universes and the people in the "edge" universes begin to suspect that the laws of classical physics are just not applicable to this experiment for them. As time goes on and many flips have happened, in the "edge" universes the results of the flips are seen as just plain miraculous. If you extend this reasoning, you get an infinite number of universes where the laws of physics are wantonly broken. To me, this is unacceptable and a definite bye-bye to MWI. Comments?

Hurkyl

Classical physics permits the laws of physics to be "wantonly broken" too, y'know.

ccrummer

Hurkyl How does "Classical physics permit[s] the laws of physics to be 'wantonly broken'[?]"

Hurkyl

There's nothing in classical physics that forbids a flipped coin from landing on edge a million times in a row, for example. And if you treated the situation statistically, such events are definitely part of the set of outcomes, and with nonzero weight too.

ThomasT

The 'worlds' of MWI shouldn't be taken literally. MWI has no particular physical meaning. Just as standard quantum theory has no particular physical meaning.

So, with that in mind we can explore your question regarding, if there's an infinite number of universes, then why hasn't at least one of these universes collided with our universe? Well, maybe one has, and we just haven't noticed the effects of it in our sector of our universe yet.

Do you see how silly this sort of speculation can get?

Forget it. Don't be a fan. Just study physics.