How do particles become entangled?

[Hans de Vries]

If I understand experts like Zurek and Omnes, things like the EPR experiment apparently involve a superposition of measurement outcomes and no faster-than-light effects.

There is definitely faster then light correlation in the teleporting theories.

The twisted argument is that it doesn't count because it can not
be checked immediately if an event is correlated to another distant
one. The check would have to go with the speed of light.

With some real bad luck I could get killed instantaneously by some
astronomical catastrophe some 100 lightyears away via entanglement.
It would however take 100 years to check that my dead was correlated
to the distant catastrophe.

So, am I not dead then as long as the correlation is not checked?
Doesn't it count?

Regards, Hans

 

[vanesch]

If I understand experts like Zurek and Omnes, things like the EPR experiment apparently involve a superposition of measurement outcomes and no faster-than-light effects.

Yes ! That's also what's my opinion. It is the observer who considers the correlation that does "the collapse thing", and not the two distant "observers" who just entangle their message with the part of the state of the received subsystem.
(but the price to pay for that opinion is a kind of solipsism...)

 

[alexepascual]

Caroline:
To make new thread, while looking at the posts on this thread, look at the top of the page. There should be a box that says:
Physics > Quantum Physics > How do particles become entangled ?
Click on "Quantum Physics", which will take you to the list of threads under this topic.
On the top left of that screen, just above the first thread, you'll see a button that says: "New Thread". Good luck!

Eye_in_the_sky:
Things came up and I am still reading your post. Some of it I understand, some of it I have questions. But I'll read it again and do a little more thinking before I respond.

 

[Caroline Thompson]

Yes ! That's also what's my opinion. It is the observer who considers the correlation that does "the collapse thing", and not the two distant "observers" who just entangle their message with the part of the state of the received subsystem.
(but the price to pay for that opinion is a kind of solipsism...)

Never mind the interpretation! What Bell showed was (as EPR had suspected), that the QM theory of entangled particles involves something in conflict with local realism. It is this that bothers me -- the failure of basic assumptions such as the ablity to predict probabilities of joint outcomes of independent events by multiplying the individual probabilities.

Before some people jump on me and say that the two sides are not independent, do look at the actual local realist calculation given in Appendix C of http://arXiv.org/abs/quant-ph/9903066! You will see then just what role the hidden variable, lambda, plays and at what stage the multiplication takes place. It is before integration over lambda.

Bell discusses the necessary assumptions in pp 36-37 of

Bell, John S, The Speakable and Unspeakable in Quantum Mechanics, Cambridge University Press 1987​

Caroline

 

[vanesch]

Never mind the interpretation! What Bell showed was (as EPR had suspected), that the QM theory of entangled particles involves something in conflict with local realism.

That's true if you think that the result of a measurement is classical and "graved in stone". However, if you consider that the "measurement" just transmits the entanglement, and that the result also comes in a superposed state, you just get, when you look at the correlations of the results from both distant photodetectors, a local interference of entangled systems. So this saves the "local" part. I'm not sure it saves the "realist" part
Imagine sending out Bob to the left photodetector, and Alice to the right photodetector. They come back in a superposed state, with all different possible results in a superposition, and it is when you look at both of them that the actual von Neuman measurement takes place, for you, as an observer. It's all in the mind

But ok, if you don't buy into photons, you won't buy into this. Although it is 100% in agreement with all of existing physics

 

[alexepascual]

Eye_in_the_Sky:
I figured that to get a good understanding of your posts I would have to study more about composite systems, tensor products, etc.
With respect to your expression:
|ψ>|φ> → Σk ak|ψk>|φk>
Now I understand what your intent was when you wrote that.
Even if that expression wasn't totally correct, it could be used to convey your idea. But I am not totally satisffied with that, because it appears that your expression is actually correct, and I suspect there is more to it than meets the eye.
I found that expression in an article on the web, but I don't know if the context is exactly the same. I could copy from that article but I'll try to make it shorter by just briefly telling you what it says.
The article says that it is always possible to find bases for H1 and H2 such that the composite state is represented by the right side of the equation above. It also says that the procedure to find such basis is known as "Schmidt decomposition" or "biorthogonal or polar expansion"
I looked up these on the web and there were many entries that came up which are related to entanglement.
The article also says that the dimmension of the index k is that of the smallest Hilbert space. In one of my previous responses I was assuming you meant that k was running from 1 to nxm, which would not be correct in the context of a "polar expansion".
So, I would like to know if a "polar expansion" is what you had in mind when you wrote that post.
Thanks Eye,

 

[chrismuktar]

I know that generally speaking entanglement of photons is done using 'parametric down conversion', where photons are collided in a material exhibiting non linear refractive indices. This process renders two entangled photons.

For the purpose of quantum cryptography, people are trying to do this process inside an optical fibre. The process is called four wave mixing (FWM).

 

[Eye_in_the_Sky]

I figured that to get a good understanding ... I would have to study more about composite systems, tensor products, etc.

In that context, it would also be a good idea to look at "density operators" and the operation of "partial tracing" to obtain "reduced density operators".

With respect to your expression:
|ψ>|φ> → Σk ak|ψk>|φk>
Now I understand what your intent was when you wrote that.
Even if that expression wasn't totally correct, it could be used to convey your idea. But I am not totally satisffied with that, because it appears that your expression is actually correct, and I suspect there is more to it than meets the eye.

As I said, my intent was merely to indicate that, on account of the interaction, the initial joint product-state evolves into a final state which cannot be written as simple product-state.

The condition I gave was:

|ζ> = ∑_k a_k|ψ_k>|φ_k> ,

where each a_k ≠ 0, and there are at least two distinct values for k, and {|ψ_k>} and {|φ_k>} are each sets of linearly-independent vectors.[/color]

This condition is necessary and sufficient for the joint state |ζ> to have no expression as a simple product-state.

(Is there more to this than meets the eye? ... Well, maybe.

The sort of "interaction" I like to keep in mind as an example for an 'entanglement' scenario (although in this context, the word 'correlation' has a more appropriate connotation) is based upon the following type of unitary transformation. Consider two sets {|ψ_k>} and {|φ_k>} in H₁ and H₂ respectively, where the vectors in each of the sets are orthonormal (and not just linearly independent). Let |φ> be a specific state in H₂. Give a partial definition of a unitary operator U as follows:

U(|ψ_k>|φ>) = |ψ_k>|φ_k> .

Then, for any |ψ> ≡ ∑_k a_k|ψ_k>, we have by linearity

U(|ψ>|φ>) = ∑_k a_k|ψ_k>|φ_k> .

A unitary operator with this sort property is what is usually considered in elementary discussions of the so-called "Measurement Problem".)

... "Schmidt decomposition" or "biorthogonal or polar expansion" ... So, I would like to know if a "polar expansion" is what you had in mind when you wrote that post.

No, that is not what I had in mind.

"Schmidt Decomposition" goes like this:

For a given |ζ> Є H₁(x)H₂, there exist orthonormal bases {|α_i>} and {|β_j>} of H₁ and H₂ respectively, and nonnegative real numbers p_k, such that

|ζ> = ∑_k √p_k|α_k>|β_k>

and

∑_k p_k = 1 ,

where the summations are over k = 1, ..., min(dimH₁,dimH₂).[/color]

This condition is much stronger than mine. To know that two systems are entangled, you don't have to go to a Schmidt Decomposition. My weaker condition is enough.

 

[alexepascual]

Eye_in_the_Sky:
I am a little confused with respect to the process of obtaining an entangled state by applying a unitary operator.
Considering the case of two non-identical particles in which we look at spin in the z-direction, the composite Hilbert space would be 4-dimmensional. That is, in general we would have 4 components, each represented by a product state. Let's say we have two particles in a superposition of spin up and spin down. Let's also assume that, as you proposed in a previous post, the particles interact in a way that the final result is a state that contains |+>|-> and |->|+>. So the states |+>|+> and |->|-> now have a zero coefficient.
As I am writing this I realize that we could think of this transformation as a rotation of the state vector (originally spanning all four base vectors) into the subspace spanned by the resulting vectors. The thing is, I have seen something like this done with a projector instead of a unitary transformation. Actually, if we wanted to do it with a rotation in Hilbert space, the angle of rotation would depend on the original angle, which would preclude the use of the operator to represent such a process for an arbitrary vector.
Now, even if that unitary transformation was possible and correct, as far as I know, interactions are described in terms of Hamiltonian operators right?
Well, I don't see how I could use a Hamiltonian to get that entanglement. Would I have to apply the Hamiltonian for a very precise time interval?
I think it would perhaps be asking too much if I asked you to write the Hamiltonian Matrix, as this would be a 4x4 matrix, but maybe you can give me a few more hints.
Something that has me puzzled is that the process of entanglement would entail a reduction of the realm of possibilities, a "compression" of the Hilbert space, which has for me more the taste of a non-unitary process with a change in entropy.
I'll appreciate your comments Eye,
Alex

 

[Eye_in_the_Sky]

I am a little confused with respect to the process of obtaining an entangled state by applying a unitary operator.
Considering the case of two non-identical particles in which we look at spin in the z-direction, the composite Hilbert space would be 4-dimmensional. That is, in general we would have 4 components, each represented by a product state. Let's say we have to particles in a superposition of spin up and spin down. Let's also assume that, as you proposed in a previous post, the particles interact in a way that the final result is a state that contains |+>|-> and |->|+>. So the states |+>|+> and |->|-> now have a zero coefficient.

Actually, when I spoke of an interaction leading to entanglement, I did not have in mind a spin-spin system. (The spin-spin entangled "singlet" state which I mentioned earlier is something which, so to speak, just 'pops out' on account of conservation of angular momentum, where the original system was already in a state of zero total angular momentum.)

The example you are suggesting is not so straight forward. Presumably, the proposed interaction would respect conservation of angular momentum. But, as you can see, in your example the total spin for the joint system will not be conserved. So, we would then need to consider something more than just the joint spin-space of the two-particle system. Perhaps then, we should take into account the position parts of the joint system as well, and then think about if (and how) we can get a 'trade-off' between orbital angular momentum and the spins. Or perhaps we should have to consider a third entity which interacts with these two, which can thereby 'carry off' or 'put in' some angular momentum. ... This is too difficult.

... However, what you are really looking for is a clear, simple, and physically real example of a process which leads to "entanglement". So, here is one.

Consider a single electron with corresponding Hilbert space spanned by the "spin-position" basis {|z+>|r >, |z->|r > | r Є R³}. Suppose that the electron is initially in the state

(a|z+> + b|z->)|φ_o> ,

and it serves as input to a Stern-Gerlach device which is set to measure the spin component along the z-axis. Let the initial spatial wavefunction
φ_o(r) be a packet centered at the origin and moving in the +x direction towards the Stern-Gerlach device.

What will you get when you solve the Schrödinger equation for this situation? ... You will find that the initial spatial wave packet splits into two parts, |φ₊> and |φ_–>, one part correlated with |z+>, and the other with |z->. That is,

(a|z+> + b|z->)|φ_o> → a|z+>|φ₊> + b|z->|φ_–> ,

where the "spin-up" packet φ₊(r) will be centered at a point with a positive value of z, and the "spin-down" packet φ_–(r) will be centered at a point with a negative value of z.

In this example, it is with regard to the state vector for a single electron that the spin and position parts become "entangled" (or (perhaps, more appropriately) "correlated") with one another. This is accomplished by means of the unitary (Schrödinger) evolution of an electron in the magnetic field of a Stern-Gerlach device.

 

[alexepascual]

The example you are suggesting is not so straight forward. Presumably, the proposed interaction would respect conservation of angular momentum. But, as you can see, in your example the total spin for the joint system will not be conserved.

I can't see how momentum is not conserved. It was zero to start and is still zero after the interaction.
With respect to your example, I understand how that works physically, and I thank you for trying to find an example that is experimentally feasible. I think I understand your description. But the part where the wave packet splits is a little too complex for me. What I mean is that even I can understand that this is exactly what would happen in a Stern-Gerlach apparatus, working out the math in detail might be a little over my head. I am also affraid that we might need to consider the coupling between the field and the electron, and I don't know anything about quantum field theory.
So, while your example helps, I am still looking for a simpler case. Maybe that's not possible and things are more complex than I would like them to be.
Also, what I was originally looking for was a case where two separate particles get entangled. I wonder if, after sacrificing the advantage of a physically reallistic example, we could find a mathematical description of the entanglement process for two two-state systems.
Let's say that we have these two two-state systems, and that we ignore the physical nature of the states. These systems are not entangled. Then we come up with some matrix U, expressed in the same basis as the states, which transforms the compound system into an entangled system. I wonder if this is possible. I understand that it migh be hard to come up with a physical process that will achieve this, but I was wondering if at least in the abstract it is possible (using some unitary matrix).

 

[Eye_in_the_Sky]

I can't see how momentum is not conserved.

I now retract my earlier statement in which I wrote:

... in your example the total spin for the joint system will not be conserved.

The most general unitary transformation consistent with conservation of total spin for the composite system is given by:

|+>|+> → e^iα|+>|+>

|->|-> → e^iα|->|->

|+>|-> → [(e^iα + e^iβ)/2]|+>|-> + [(e^iα - e^iβ)/2]|->|+>

|->|+> → [(e^iα - e^iβ)/2]|+>|-> + [(e^iα + e^iβ)/2]|->|+>

where α and β are arbitrary real parameters.

[Sorry about my earlier confusion.]

Thus, for example, we can take α = 0 and β = π/2 ... and this would give:

|+>|+> → |+>|+>

|->|-> → |->|->

|+>|-> → a|+>|-> + a^*|->|+>

|->|+> → a^*|+>|-> + a|->|+>

where a = (1+i)/2 .

----------------------------
There is, however, one important point to note here in relation to what you said about the total spin:

It was zero to start and is still zero after the interaction.

Note that in a {|J,m_J> | J = 0,1 ; m_J = -J,...,J} basis, we can write

|+>|-> = (1/√2) [ |1,0> + |0,0> ] → (1/√2) [ e^iα|1,0> + e^iβ|0,0> ] ,

|->|+> = (1/√2) [ |1,0> - |0,0> ] → (1/√2) [ e^iα|1,0> - e^iβ|0,0> ] .

So, for these states, it is incorrect to say that the total spin (before or after) is zero; rather, these states are each a superposition of a J=1 state with the J=0 state.
----------------------------

... Okay, fine. So, we can use the example above, or we can consider your alternative suggestion:

Let's say that we have these two two-state systems, and that we ignore the physical nature of the states. These systems are not entangled. Then we come up with some matrix U, expressed in the same basis as the states, which transforms the compound system into an entangled system. I wonder if this is possible. I understand that it migh be hard to come up with a physical process that will achieve this, but I was wondering if at least in the abstract it is possible (using some unitary matrix).

Ah ... this will make it easier. First, consider the following:

A unitary transformation is equivalent to the linear extension of a mapping of one orthonormal basis to another.

So, suppose we have two orthonormal bases {|u_k>} and {|v_k>}. We then define a mapping by

|u_k> → |v_k> , k = 1,2,... ,

and extend this mapping to the whole Hilbert space by linearity. This mapping is then a unitary transformation. (The converse is obviously also true.)

We are now equipped with all we need to build our own "unitary-entangling" operator.

Here, I'll build one that I like. Let system A have a basis {|0>, |1>} and let system B have a basis {|↑>, |↓>}. Suppose that when system A is in the |0> state, the interaction invokes no change ... but, when system A is in the |1> state, the interaction invokes a "flip-flop" in B. That is:

|0>|↑> → |0>|↑>

|0>|↓> → |0>|↓> ... i.e. no change ,

and

|1>|↑> → |1>|↓>

|1>|↓> → |1>|↑> ... i.e. "flip-flop" in B .


"But where is the entanglement?" you ask. Here is one spot (in the linear extension):

(a|0> + b|1>)|↓> → a|0>|↓> + b|1>|↑> .

------

Now, it seems to me you may want to ask:

What is a necessary and sufficient condition for a unitary operator to be able to induce entanglement?

Well ... a trivial response is:

There do not exist unitary operators V and W (on H_A and H_B , respectively) such that U = V (x) W.

Quite generally, whenever the Hamiltonian of the joint system involves terms which indicate interaction between the two subsystems, then we will not have U = V (x) W, and therefore, entanglement will occur for (at least) some initial states.
----------------------------------
----------------------------------

With respect to your example, I understand how that works physically, and I thank you for trying to find an example that is experimentally feasible. I think I understand your description. But the part where the wave packet splits is a little too complex for me. What I mean is that even I can understand that this is exactly what would happen in a Stern-Gerlach apparatus, working out the math in detail might be a little over my head. I am also affraid that we might need to consider the coupling between the field and the electron, and I don't know anything about quantum field theory.

The "splitting" of the wave packet should not be too mind boggling. I will just sketch out the basic idea.

Fist of all, I should have picked a neutral particle (with a magnetic moment), rather than a charged one, like the electron. That way, there will be no Lorentz force on the particle to consider.

Okay. So let's use a hydrogen atom in the ground state (treating it as an electrically neutral particle with magnetic moment equal that of the electron (the magnetic moment of the proton can be neglected, and in the ground state L=0)).

Given that ... then, making now a simplification to 1 spatial dimension, the Hamiltonian can be written as

H = -(h_bar²/2m)(∂/∂z)² + (eh_bar/2m)σ_zB(z) ,

where σ_z is the 2x2 matrix

1 0
0 -1 .

It does not take much effort to see that the time-dependent Schrödinger equation will (quite trivially) split into two decoupled equations, one for |z+> and one for |z->, the only difference between them being a "+" or a
"-" in front of the "interaction" part of the Hamiltonian. ... I hope this clears up some of the 'fog'.

... Also, there is no need here to quantize the magnetic field (QFT is not required in this approximation).

 

[gptejms]

>Hans wrote :
>There is definitely faster then light correlation in the teleporting theories.

>The twisted argument is that it doesn't count because it can not
>be checked immediately if an event is correlated to another distant
>one. The check would have to go with the speed of light.

>With some real bad luck I could get killed instantaneously by some
>astronomical catastrophe some 100 lightyears away via entanglement.
>It would however take 100 years to check that my dead was correlated
>to the distant catastrophe.

>So, am I not dead then as long as the correlation is not checked?
>Doesn't it count?

>Regards, Hans

The above post by Hans went unanswered.What do you guys say to this--I mean how do you answer it?

 

[vanesch]

>Hans wrote :
>There is definitely faster then light correlation in the teleporting theories.

>The twisted argument is that it doesn't count because it can not
>be checked immediately if an event is correlated to another distant
>one. The check would have to go with the speed of light.

>With some real bad luck I could get killed instantaneously by some
>astronomical catastrophe some 100 lightyears away via entanglement.
>It would however take 100 years to check that my dead was correlated
>to the distant catastrophe.

>So, am I not dead then as long as the correlation is not checked?
>Doesn't it count?

>Regards, Hans

The above post by Hans went unanswered.What do you guys say to this--I mean how do you answer it?

Ok, there are 2 aspects. The first one is that you cannot "die by entanglement". You die locally, by something that came over to you at less than lightspeed, say a meteor. The probabilities of what happens locally are of course independent of whatever an entangled system far away might undergo or do. So the chance of Hans dying by the meteor is locally fixed.
However, indeed, in order to find out that this meteor was entangled with something else far away, we will indeed have to wait at least 100 years before we could tell about the correlation (and it wouldn't make a lot of sense since there has been only 1 event :-)

Again and again and again: entanglement does NOT imply "action at a distance". I've seen some stupid proposal of having entangled ions in a spacecraft , which could then be excited by exciting their partners on earth, but that is totally wrong of course. Whatever happens to the ions in the spacecraft has probability distributions which are independent of what you do to their partners on earth. You could at best do measurements on them, and find out later that they were correlated.

cheers,
Patrick.

 

[alexepascual]

Eye_in_the_sky:
I am thinking about your post.
I'll let you know as soon as I have digested it.

With respect to the last two posts:
I would argue that it is possible to die by entanglement. You could device a kind of Schrodinger's cat experiment with two boxes that contain atoms with entangled spins and a person in each box. You take the boxes far appart. Then, within each of the boxes there is a mechanism that kills the person depending on the result of a spin measurement. The death of one person would be correlated to the death of the other person. Well, actually you don't need to put the person in a box.
Now, this does not mean that you can do anything at one end to provoke the death of the person at the other end.
With respect to the "faster than light" issue, yes the events could be space-like related, but as there is nothing going from one end to the other (at the time between events) , it is missleading to talk about "faster than light".
I don't agree that the fact that you may have to wait 100 years to check the correlation really matters. you can confirm the correlation 100 years later and thus show that an experiment that was done 100 years ago proved or disproved whaterver is was designed to prove.
The fact that it is only one event is not relevant. You could design your experiment with 1000 events. You could send 1000 people in a spaceship, each of which carries an atom whose spin is entangled to that of an atom on earth. Their death depends on their finding the spin to be up or down. Of course there might be practical difficulties that might prevent an experiment like this being done. The main problem might be finding the money. Sending 1000 people to a possible death might not be a problem from a political point of view. If that were a problem you can always do it with people from a country you don't like, which automatically makes them less human. (as they would be in outer space, their death would not be under the jurisdiction of any eartly court). Well, enough about this, let's leave politics to the politicians and get back to the science. Yes, I started it. Sorry for the digression.

 

[RandallB]

entanglement does NOT imply "action at a distance". Patrick.

Maybe you can help me be sure I understand entangement correctly.Lets say we generate two balls A & B that are entangled going left and right.
We know we have a ball coming to us at the left we may even be fairly sure that it is Ball A, but there is some probability that it is ball B. The QM superposition point is that it isn't ether ball A or ball B it can be either one until a commitment is required by some interaction or test. Hit it with a BAT and see it. Now if both A and B are completely identical as many are we cannot tell if we actually got ball B when we expected A.
But if there is a characteristic that is different, A is RED and B is GREEN then our testing will show if we hit the unexpected GREEN ball. AND we know anyone in the right area is going to "or already has" seen a RED Ball. But they a free to catch it or hit it as they local choose.

Now if it is true that most likely the RED Ball is the most likely result in the left area; nothing in QM allows us to be tipped to what that most likely one is. Thus it could also be true that the chance is the GREEN Ball instead can be very high say .49 at first and reduce to a smaller chance as the pair separate. Even down to such a low chance that entanglement could be considered to have been lost. (i.e. You can't stick them in separate jars for later use somehow) . Since it cannot be shown if RED really was most likely to start with, it should be very hard to prove or disprove the "decay" of entanglement.
The question - - is the idea of entanglement decay viable within the math of QM?

Personally I'd think it should be expected.
Randall B

 

[alexepascual]

I don't know much about the possibility of a gradual decline in the degree of entanglement. I'll let other members answer to that.
But I would like to tell you that in my opinion, it would be more advantageous for you to think about superpositions not in terms of a superposition of different "balls" but a superposition of different properties in the same ball.
It migh not be a good idea either to give "balls" as an example as this suggests that you are dealing with macroscopic objects, which have never been found in a state of superposition.
I also think that while your question about a "decay" in entanglement might be a valid one, you might not be able to understand this issue without studying QM in depth, including all the math and jargon.

 

[DrChinese]

The question - - is the idea of entanglement decay viable within the math of QM?

Personally I'd think it should be expected.
Randall B

There is no decay of entanglement due to distance in the ideal case, but... In the real world, a lot of things can cause the entanglement to end. A measurement does, and a lot of things can act as if they are measurements.

http://arxiv.org/PS_cache/quant-ph/pdf/9810/9810080.pdf performed testing over a distance of 1 kilometer and violated Bell's Inequality. I think Tittel et al did it across 10 kilometers and saw high correlations, but less than Weihs.

 

[vanesch]

Eye_in_the_sky:
I am thinking about your post.
I'll let you know as soon as I have digested it.

With respect to the last two posts:
I would argue that it is possible to die by entanglement. You could device a kind of Schrodinger's cat experiment with two boxes that contain atoms with entangled spins and a person in each box. You take the boxes far appart. Then, within each of the boxes there is a mechanism that kills the person depending on the result of a spin measurement. The death of one person would be correlated to the death of the other person.

Yes, the deaths would be correlated. But not "caused by". If the local person only looked locally at his spin, he knows that he has 50% chance of dying. And locally, whether this local atom is entangled or not with a faraway atom doesn't change anything for the probabilities of dying.
So, yes, entanglement can make it such that people die in correlated ways. But they do not die more or less because there is entanglement. We just observe that there are correlations in their deaths, when we know the result of both.
I had understood (maybe wrongly) that "to die by entanglement" would mean that you somehow die because the particle was entangled, and if it weren't, you wouldn't die. Maybe this was not what the original question was about, in which case my answer was next to the point.

In fact, in an MWI-like interpretation, everything is about entangled with everything else, and this is the basis of decoherence, which makes us avoid seeing superpositions of classical states of macroscopic subsystems, except in rare circumstances

cheers,
patrick.

 

[alexepascual]

I agree, Patrick
Thanks for clarifying your point of view.
By the way, I kind of like the MWI. For the moment it appears to be the most elegant, and elegance is something I appreciate as much in physics as in women.
-Alex-

 

[RandallB]

MWI.

Actually I think using a “Ball” is an ok analogy as long as it isn’t performing any impossible tests on itself, like trying to decide if it’s really dead or not! Better than “Schrodinger's cat” anyway.

Looked in other parts of thread but couldn’t find MWI defined. What is MWI-like?
EDIT: Think I found it --- from MWT to "Immortality"
Not so we'd ever know.