Quantum Mechanics and Mind

[abashed]

@wolfpup

“Experts” in a scientific discipline are those who publish their research in peer-reviewed journals, not YouTube videos. Thomas Campbell is a mystic crackpot whose personal “theory of everything” is considered by the scientific community to be uninformed nonsense not really worth responding to, just like your posts on this subject. As already noted, you clearly have no understanding of quantum physics, yet continue to insult the knowledgeable posters here who do and who take the time to try to provide informed guidance. I’m not sure what you think you’re accomplishing here.

What was not nonsense was the result of the delayed eraser experiment he elucidated. His ideas about life in general, based on QM, is not the question here and I do not necessarily agree with them. The point is we know no more about what is going on now than they did 100 years ago so you do not know any more than I or anyone else about why these effects happen. I find your comments irrelevant and unhelp and franky, quite patronising. Just because you have a mathematical understand of QM does not mean you have cracked it since such mathematical models are merely ways that people try out to discover whether they are useful in predicting phenomena but nobody has any special access to why things happen. Mr. Campbell seems right about one thing, and that is the reluctance in the scientific community to question belief-based approaches that still cling on to a materialistic solution to QM phenomena and a denial that we cannot any longer use ‘mechanical’ interpretations that have long since been discredited.

[abashed]

@BeepKillBeep

I’ve thought about this a bit before. I think that it comes down to two things. First, non-experts like abashed, MantraPhilter et al. don’t understand how advanced and specialized science has become. They seem to think that it is much as it was like in the the first half of the 1900s (or late 1800s). Much of the basic understanding of most scientific disciplines is resolved and well-established. These days research is extremely specialized and generally very narrow. In some cases, they think that this has become dogmatic, but what they don’t understand is that it is well-established because so much has been built on top of it that verifies this understanding on a daily basis.

Second, they don’t understand the process of coming up with a scientific hypothesis and the process to investigate it. They seem to think that a hypothesis is just an idea and that such ideas can simply come from the ether. They don’t understand the process of reviewing literature and building on what has been firmly established, the process of looking for the little gaps in the literature that need to be filled in. This discussion could almost be an interesting thread itself, and if this goes anywhere I might start one because I am interested in what other people think.

The question is where do these ideas come from? Yes, from us, our consciousness! And when we try these hypotheses out we receive back information. Without first asking for some feedback no scientific knowledge is possible so we need both consciousness and information to make progress. You people either do not seem able to grasp this or refuse to acknowledge it, since it challenges your worldview.

[abashed]

@RaftPeople

I took the time to look at part of that video and his website.

The clarity of explanation was impressive:

Yes, it was wasn’t it? This is why I gave the link since he did put over the basic facts in a very conversational way and did not rush through it. I don’t necessarily agree with the rest of his philosophical interpretation about life in general based on his interpretation of QM, nevertheless, he does put his ideas over in a clear way.

[Mangetout]

@abashed

The question is where do these ideas come from? Yes, from us, our consciousness! And when we try these hypotheses out we receive back information. Without first asking for some feedback no scientific knowledge is possible so we need both consciousness and information to make progress. You people either do not seem able to grasp this or refuse to acknowledge it, since it challenges your worldview.

What people? We don’t exist until you think about us. This is your problem, not ours.

[abashed]

@Half_Man_Half_Wit

BeepKillBeep, I’d be very interested in such a thread, particularly if it were to focus on what could actually be done to help these people understand the defects of their reasoning. I mean, to me, it’s just such a great waste: millions of hours of intellectual work going to waste chasing mirages, simply because nobody ever managed to give them the right tools and expertise.

But most approaches seem to just lead to a doubling-down. You tell them they’re wrong, they’ll just consider you too stupid to appreciate their genius. You tell them why they’re wrong, and precisely because they’re lacking the right tools, it has no effect at all. So what’s a good strategy here?

You should just stick to the objective data gained from the double-slit and its variants and discuss that instead of thinking you have all the answers and other people have no clue. You and others have still not provided any convincing interpretation of the strange behaviour of QM phenomena. For example, how do you account for quantum entanglement? And if you are going to talk about hidden variables don’t bother because these have never been verified. It is not people like me who claim to be geniuses; it is people like you who think they know it all.

[abashed]

@Half_Man_Half_Wit

As I’ve told you, information has a very precise definition in technical literature. It’s given by the so-called Shannon entropy. As the wiki article says:
[

When physicists speak about something having ‘information’ or ‘knowledge’ about something else, what they mean is more specifically the [URL=“Mutual information - Wikipedia”]mutual information](Entropy (information theory) - Wikipedia)—a measure of how much the two systems are correlated. This does lead to some unfortunate misunderstandings for those not conversant with the terminology, but that’s a hard thing to avoid, without all the time inventing new words.
No. What you’re missing is the concept of decoherence: that, when quantum systems interact with the environment (including things like detectors), phenomena like interference become radically suppressed, and quickly unobservable.

It’s in fact pretty simple to demonstrate this. Assume we have a quantum object that can be in either of two states (say, to stay with the familiar example, going through slit one or two in an interference experiment). Label these states |1> and |2>.

(Note, don’t worry too much about the strange arrow-brackets. They’re a convention usually adopted in modern quantum mechanics; you basically should just read them as a reminder that what’s written between them—which could be any label whatsoever—refers to a quantum object.)

Then, a generic state of the system has the form

|s> = a|1> + b|2>,

where a and b are complex numbers such that |a|[sup]2[/sup] + |b|[sup]2[/sup] = 1.

Now, say we want to know the probability of observing a certain state |o>, given that the system is in state |s>. That’s given by the Born rule:

P(o) = |<s|o>|[sup]2[/sup],

that is, the probability is given by the square of the overlap of the two vectors.

Now, in something like a two-slit experiment, we would naively expect the probability to land at a given point on the screen to be a sum of the probabilities for landing there via one slit, and landing there via the other—that is, in particular, a sum of two positive (or zero) numbers. That’s the classical story.

What we find, however, is different. The reason for this is as follows. The generic state |o> can be expanded in the basis used above, yielding

|o> = c|1> + d|2>.

Furthermore, we have that <1|1> = <2|2> = 1, and <1|2> = <2|1> = 0, which just says that these states are orthogonal—they have no overlap. Thus, the basis is an orthonormal basis.

Now, let’s calculate the probability P(o). We have that (if you’re wondering why the little arrows point sometimes in one, and sometimes in another direction, you can imagine these roughly like row- and column-vectors; the salient point is that <x|y> = <x|*|y> always yields a scalar, i.e. an ordinary (complex) number; the asterisk denotes complex conjugation, except where it denotes multiplication—sorry about that):

            P(o) =|<s|o>|[sup]2[/sup]
                 =|(a[sup]*[/sup]<1| + b[sup]*[/sup]<2|)*(c|1> + d|2>)|[sup]2[/sup]
                 =|a[sup]*[/sup]c + b[sup]*[/sup]d|[sup]2[/sup].

Here, I’ve merely used the above property of the orthonormal basis. If we then calculate the absolute square (recall that for a complex number z, |z|[sup]2[/sup] = zz[sup]*[/sup], i.e. the product of the number with its complex conjugate), we get:

            P(o) =|a[sup]*[/sup]c + b[sup]*[/sup]d|[sup]2[/sup]
                 =(a[sup]*[/sup]c + b[sup]*[/sup]d)(ac[sup]*[/sup] + bd[sup]*[/sup])
                 =|a[sup]*[/sup]c|[sup]2[/sup] + |b[sup]*[/sup]d|[sup]2[/sup] + a[sup]*[/sup]cbd[sup]*[/sup] + b[sup]*[/sup]dac[sup]*[/sup] 

Now, this is a very curious result! The first two terms are what we would have expected: two positive (or zero) numbers. They are, in fact, exactly what you’d get if you’d neglected the superposition—the sum of the probability of getting to the screen via slit 1, and getting there via slit 2.

However, the extra terms, due to involving products of complex numbers, need not be positive—they can become negative. So, this sum of probabilities is sometimes increased, and sometimes reduced, depending on the value of those extra terms. This is, of course, nothing but interference. So far, so good!

Now let’s assume that we have some which-path measuring detector. If the particle traverses slit 1, it enters into the state D[sub]1[/sub]; if it goes through slit 2, it enters the state D[sub]2[/sub]. Formulated in quantum-mechanical terms:

          |D>|1> ---> |D[sub]1[/sub]>|1> (= |D[sub]1[/sub],1>),
          |D>|2> ---> |D[sub]2[/sub]>|2> (= |D[sub]2[/sub],2>)

Here, the juxtaposition of two vectors formally denotes their tensor product; we’ll just use it as a shorthand for the combined state of two systems (particle and detector). (The part in parentheses is just a common notational simplification.)

Now, we want to calculate the probability of observing the particle in state |o>, given that the detector has made a detection as above. For this, we need two extra things: first, simply due to the fact that the two states of the detector are very high-dimensional (detectors consist of very many particles, typically) it can be shown that <D[sub]1[/sub]|D[sub]1[/sub]> and <D[sub]2[/sub]|D[sub]2[/sub]> are very, very close to one, and that <D[sub]1[/sub]|D[sub]2[/sub]> and <D[sub]2[/sub]|D[sub]1[/sub]> are very, very close to zero (the two states have near-vanishing overlap). In fact, for all practical purposes, their overlap is exactly zero.

Furthermore, we want to know the total probability to find the system in the state |o>, which is the sum of the probability of finding it in state |o> and the detector having made the detection 1, and the probability of finding it in state |o> and the detector having made the detection 2. In short:

P(o) = P(o,D[sub]1[/sub]) + P(o,D[sub]2[/sub]).

So first, we must know the state of our particle/detector system after the detection. The linearity of quantum mechanics dictates the following:

          |D>|s> = |D>(a|1> + b|2>)
                 = a|D>|1> + b|D>|2>
                 ---> a|D[sub]1[/sub],1> + b|D[sub]2[/sub],2>

Consequently, to evaluate the probability, we have to calculate the following:

         P(o,D[sub]1[/sub]) = |(a[sup]*[/sup]<D[sub]1[/sub],1| + b[sup]*[/sup]<D[sub]2[/sub],2|)*|o,D[sub]1[/sub]>|[sup]2[/sup]
                 = |a[sup]*[/sup]<D[sub]1[/sub],1|o,D[sub]1[/sub]> + b[sup]*[/sup]<D[sub]2[/sub],2|o,D[sub]1[/sub]>|[sup]2[/sup]

         P(o,D[sub]2[/sub]) = |(a[sup]*[/sup]<D[sub]1[/sub],1| + b[sup]*[/sup]<D[sub]2[/sub],2|)*|o,D[sub]2[/sub]>|[sup]2[/sup]
                 = |a[sup]*[/sup]<D[sub]1[/sub],1|o,D[sub]2[/sub]> + b[sup]*[/sup]<D[sub]2[/sub],2|o,D[sub]2[/sub]>|[sup]2[/sup]

Ultimately, thus, the total probability of observing the system in the state |o> works out to:

            P(o) = |a[sup]*[/sup]<D[sub]1[/sub],1|o,D[sub]1[/sub]> + b[sup]*[/sup]<D[sub]2[/sub],2|o,D[sub]1[/sub]>|[sup]2[/sup] + |a[sup]*[/sup]<D[sub]1[/sub],1|o,D[sub]2[/sub]> + b[sup]*[/sup]<D[sub]2[/sub],2|o,D[sub]2[/sub]>|[sup]2[/sup]

For brevity, we’ll only look at the first of these terms; the second one works exactly analogously.

         P(o,D[sub]1[/sub]) = |a[sup]*[/sup]<D[sub]1[/sub],1|o,D[sub]1[/sub]> + b[sup]*[/sup]<D[sub]2[/sub],2|o,D[sub]1[/sub]>|[sup]2[/sup]
                 = |a[sup]*[/sup]c<D[sub]1[/sub],1|1,D[sub]1[/sub]> + a[sup]*[/sup]d<D[sub]1[/sub],1|2,D[sub]1[/sub]> + b[sup]*[/sup]c<D[sub]2[/sub],2|1,D[sub]1[/sub]> + b[sup]*[/sup]d<D[sub]2[/sub],2|2,D[sub]1[/sub]>|[sup]2[/sup]

Here, I’ve done nothing but insert the decomposition of |o>, so far. But we can already read off the result: the second and third terms vanish immediately, because <1|2> = <2|1> = 0. But what about the fourth term? <2|2> = 1, so this looks like it might give a contribution.

Well, recall that we had already stipulated that, since |D[sub]1[/sub]> and |D[sub]2[/sub]> are macroscopic, we have that <D[sub]2[/sub]|D[sub]1[/sub]> = 0 to extremely good precision. So it turns out that this term vanishes as well. What are we left with?

The only nonvanishing term left is just |a[sup][/sup]c|[sup]2[/sup]. Things work out exactly the same for the probability P(o,D[sub]2[/sub]): there, the only term that’s left is |b[sup][/sup]d|[sup]2[/sup]. So we have, in total, for our probability now (drumroll please):

            P(o) = P(o,D[sub]1[/sub]) + P(o,D[sub]2[/sub])
                 = |a[sup]*[/sup]c|[sup]2[/sup] + |b[sup]*[/sup]d|[sup]2[/sup].

Now, if you’re still with me, you should recognize this as just the probability we calculated before—but without the terms responsible for the interference! Those have simply vanished.

However, at no point did we have to introduce any conscious observer whatsoever into the derivation. It’s enough for the system to couple to a system with very many degrees of freedom. In a sense, the coherence—which is responsible for the interference terms—gets ‘washed out’ across the total system, the combination of particle and detector, and thus, yields no observable effects anymore.

Indeed, after the interaction with the detector, the resulting state is entangled; thus, the detector and the particle are correlated (which is what’s typically called ‘having information about one another’). In a sense, all the quantum properties of the state are now distributed across all the degrees of freedom of the combined system of the particle and the detector, and thus, if we leave out the (which-path) detector, there’s just not enough ‘quantumness’ left in the rest, the single particle, to yield observable effects (at our detection on the ‘screen’, i.e. when we measure whether the system is in state |o>).

Consequently, such effects need no conscious observer.

The same goes for more complicated setups, such as the delayed choice experiment. The calculation is left as an exercise for the reader.

None of this stuff can read itself and no result of any experiment can have any meaning without an observer. How can a scientific experiment organize itself to put all the apparatus in place and do the experiment? You said; “The calculation is left as an exercise for the reader.” What if there is no reader? What don’t you understand about this?

Tell me, where exactly is the information contained within your post? Is it the pixels that make-up the symbols? NO. Is it the characters therein? NO. It is the interpretation encoded in the symbols made by an observer. YES!

[abashed]

@Half_Man_Half_Wit

As I’ve told you, information has a very precise definition in technical literature. It’s given by the so-called Shannon entropy. As the wiki article says:
[

When physicists speak about something having ‘information’ or ‘knowledge’ about something else, what they mean is more specifically the [URL=“Mutual information - Wikipedia”]mutual information](Entropy (information theory) - Wikipedia)—a measure of how much the two systems are correlated. This does lead to some unfortunate misunderstandings for those not conversant with the terminology, but that’s a hard thing to avoid, without all the time inventing new words.
No. What you’re missing is the concept of decoherence: that, when quantum systems interact with the environment (including things like detectors), phenomena like interference become radically suppressed, and quickly unobservable.

It’s in fact pretty simple to demonstrate this. Assume we have a quantum object that can be in either of two states (say, to stay with the familiar example, going through slit one or two in an interference experiment). Label these states |1> and |2>.

(Note, don’t worry too much about the strange arrow-brackets. They’re a convention usually adopted in modern quantum mechanics; you basically should just read them as a reminder that what’s written between them—which could be any label whatsoever—refers to a quantum object.)

Then, a generic state of the system has the form

|s> = a|1> + b|2>,

where a and b are complex numbers such that |a|[sup]2[/sup] + |b|[sup]2[/sup] = 1.

Now, say we want to know the probability of observing a certain state |o>, given that the system is in state |s>. That’s given by the Born rule:

P(o) = |<s|o>|[sup]2[/sup],

that is, the probability is given by the square of the overlap of the two vectors.

Now, in something like a two-slit experiment, we would naively expect the probability to land at a given point on the screen to be a sum of the probabilities for landing there via one slit, and landing there via the other—that is, in particular, a sum of two positive (or zero) numbers. That’s the classical story.

What we find, however, is different. The reason for this is as follows. The generic state |o> can be expanded in the basis used above, yielding

|o> = c|1> + d|2>.

Furthermore, we have that <1|1> = <2|2> = 1, and <1|2> = <2|1> = 0, which just says that these states are orthogonal—they have no overlap. Thus, the basis is an orthonormal basis.

Now, let’s calculate the probability P(o). We have that (if you’re wondering why the little arrows point sometimes in one, and sometimes in another direction, you can imagine these roughly like row- and column-vectors; the salient point is that <x|y> = <x|*|y> always yields a scalar, i.e. an ordinary (complex) number; the asterisk denotes complex conjugation, except where it denotes multiplication—sorry about that):

            P(o) =|<s|o>|[sup]2[/sup]
                 =|(a[sup]*[/sup]<1| + b[sup]*[/sup]<2|)*(c|1> + d|2>)|[sup]2[/sup]
                 =|a[sup]*[/sup]c + b[sup]*[/sup]d|[sup]2[/sup].

Here, I’ve merely used the above property of the orthonormal basis. If we then calculate the absolute square (recall that for a complex number z, |z|[sup]2[/sup] = zz[sup]*[/sup], i.e. the product of the number with its complex conjugate), we get:

            P(o) =|a[sup]*[/sup]c + b[sup]*[/sup]d|[sup]2[/sup]
                 =(a[sup]*[/sup]c + b[sup]*[/sup]d)(ac[sup]*[/sup] + bd[sup]*[/sup])
                 =|a[sup]*[/sup]c|[sup]2[/sup] + |b[sup]*[/sup]d|[sup]2[/sup] + a[sup]*[/sup]cbd[sup]*[/sup] + b[sup]*[/sup]dac[sup]*[/sup] 

Now, this is a very curious result! The first two terms are what we would have expected: two positive (or zero) numbers. They are, in fact, exactly what you’d get if you’d neglected the superposition—the sum of the probability of getting to the screen via slit 1, and getting there via slit 2.

However, the extra terms, due to involving products of complex numbers, need not be positive—they can become negative. So, this sum of probabilities is sometimes increased, and sometimes reduced, depending on the value of those extra terms. This is, of course, nothing but interference. So far, so good!

Now let’s assume that we have some which-path measuring detector. If the particle traverses slit 1, it enters into the state D[sub]1[/sub]; if it goes through slit 2, it enters the state D[sub]2[/sub]. Formulated in quantum-mechanical terms:

          |D>|1> ---> |D[sub]1[/sub]>|1> (= |D[sub]1[/sub],1>),
          |D>|2> ---> |D[sub]2[/sub]>|2> (= |D[sub]2[/sub],2>)

Here, the juxtaposition of two vectors formally denotes their tensor product; we’ll just use it as a shorthand for the combined state of two systems (particle and detector). (The part in parentheses is just a common notational simplification.)

Now, we want to calculate the probability of observing the particle in state |o>, given that the detector has made a detection as above. For this, we need two extra things: first, simply due to the fact that the two states of the detector are very high-dimensional (detectors consist of very many particles, typically) it can be shown that <D[sub]1[/sub]|D[sub]1[/sub]> and <D[sub]2[/sub]|D[sub]2[/sub]> are very, very close to one, and that <D[sub]1[/sub]|D[sub]2[/sub]> and <D[sub]2[/sub]|D[sub]1[/sub]> are very, very close to zero (the two states have near-vanishing overlap). In fact, for all practical purposes, their overlap is exactly zero.

Furthermore, we want to know the total probability to find the system in the state |o>, which is the sum of the probability of finding it in state |o> and the detector having made the detection 1, and the probability of finding it in state |o> and the detector having made the detection 2. In short:

P(o) = P(o,D[sub]1[/sub]) + P(o,D[sub]2[/sub]).

So first, we must know the state of our particle/detector system after the detection. The linearity of quantum mechanics dictates the following:

          |D>|s> = |D>(a|1> + b|2>)
                 = a|D>|1> + b|D>|2>
                 ---> a|D[sub]1[/sub],1> + b|D[sub]2[/sub],2>

Consequently, to evaluate the probability, we have to calculate the following:

         P(o,D[sub]1[/sub]) = |(a[sup]*[/sup]<D[sub]1[/sub],1| + b[sup]*[/sup]<D[sub]2[/sub],2|)*|o,D[sub]1[/sub]>|[sup]2[/sup]
                 = |a[sup]*[/sup]<D[sub]1[/sub],1|o,D[sub]1[/sub]> + b[sup]*[/sup]<D[sub]2[/sub],2|o,D[sub]1[/sub]>|[sup]2[/sup]

         P(o,D[sub]2[/sub]) = |(a[sup]*[/sup]<D[sub]1[/sub],1| + b[sup]*[/sup]<D[sub]2[/sub],2|)*|o,D[sub]2[/sub]>|[sup]2[/sup]
                 = |a[sup]*[/sup]<D[sub]1[/sub],1|o,D[sub]2[/sub]> + b[sup]*[/sup]<D[sub]2[/sub],2|o,D[sub]2[/sub]>|[sup]2[/sup]

Ultimately, thus, the total probability of observing the system in the state |o> works out to:

            P(o) = |a[sup]*[/sup]<D[sub]1[/sub],1|o,D[sub]1[/sub]> + b[sup]*[/sup]<D[sub]2[/sub],2|o,D[sub]1[/sub]>|[sup]2[/sup] + |a[sup]*[/sup]<D[sub]1[/sub],1|o,D[sub]2[/sub]> + b[sup]*[/sup]<D[sub]2[/sub],2|o,D[sub]2[/sub]>|[sup]2[/sup]

For brevity, we’ll only look at the first of these terms; the second one works exactly analogously.

         P(o,D[sub]1[/sub]) = |a[sup]*[/sup]<D[sub]1[/sub],1|o,D[sub]1[/sub]> + b[sup]*[/sup]<D[sub]2[/sub],2|o,D[sub]1[/sub]>|[sup]2[/sup]
                 = |a[sup]*[/sup]c<D[sub]1[/sub],1|1,D[sub]1[/sub]> + a[sup]*[/sup]d<D[sub]1[/sub],1|2,D[sub]1[/sub]> + b[sup]*[/sup]c<D[sub]2[/sub],2|1,D[sub]1[/sub]> + b[sup]*[/sup]d<D[sub]2[/sub],2|2,D[sub]1[/sub]>|[sup]2[/sup]

Here, I’ve done nothing but insert the decomposition of |o>, so far. But we can already read off the result: the second and third terms vanish immediately, because <1|2> = <2|1> = 0. But what about the fourth term? <2|2> = 1, so this looks like it might give a contribution.

Well, recall that we had already stipulated that, since |D[sub]1[/sub]> and |D[sub]2[/sub]> are macroscopic, we have that <D[sub]2[/sub]|D[sub]1[/sub]> = 0 to extremely good precision. So it turns out that this term vanishes as well. What are we left with?

The only nonvanishing term left is just |a[sup][/sup]c|[sup]2[/sup]. Things work out exactly the same for the probability P(o,D[sub]2[/sub]): there, the only term that’s left is |b[sup][/sup]d|[sup]2[/sup]. So we have, in total, for our probability now (drumroll please):

            P(o) = P(o,D[sub]1[/sub]) + P(o,D[sub]2[/sub])
                 = |a[sup]*[/sup]c|[sup]2[/sup] + |b[sup]*[/sup]d|[sup]2[/sup].

Now, if you’re still with me, you should recognize this as just the probability we calculated before—but without the terms responsible for the interference! Those have simply vanished.

However, at no point did we have to introduce any conscious observer whatsoever into the derivation. It’s enough for the system to couple to a system with very many degrees of freedom. In a sense, the coherence—which is responsible for the interference terms—gets ‘washed out’ across the total system, the combination of particle and detector, and thus, yields no observable effects anymore.

Indeed, after the interaction with the detector, the resulting state is entangled; thus, the detector and the particle are correlated (which is what’s typically called ‘having information about one another’). In a sense, all the quantum properties of the state are now distributed across all the degrees of freedom of the combined system of the particle and the detector, and thus, if we leave out the (which-path) detector, there’s just not enough ‘quantumness’ left in the rest, the single particle, to yield observable effects (at our detection on the ‘screen’, i.e. when we measure whether the system is in state |o>).

Consequently, such effects need no conscious observer.

The same goes for more complicated setups, such as the delayed choice experiment. The calculation is left as an exercise for the reader.

None of this stuff can read itself and no result of any experiment can have any meaning without an observer. How can a scientific experiment organize itself to put all the apparatus in place and do the experiment? You said; “The calculation is left as an exercise for the reader.” What if there is no reader? What don’t you understand about this?

Tell me, where exactly is the information contained within your post? Is it the pixels that make-up the symbols? NO. Is it the characters therein? NO. It is the interpretation encoded in the symbols made by an observer. YES!

[abashed]

@Mangetout

What people? We don’t exist until you think about us. This is your problem, not ours.

Why don’t you exist? You need to explain this.

[abashed]

Originally Posted by Half Man Half Wit View Post
As I’ve told you, information has a very precise definition in technical literature. It’s given by the so-called Shannon entropy. As the wiki article says:
When physicists speak about something having ‘information’ or ‘knowledge’ about something else, what they mean is more specifically the mutual information—a measure of how much the two systems are correlated. This does lead to some unfortunate misunderstandings for those not conversant with the terminology, but that’s a hard thing to avoid, without all the time inventing new words.
No. What you’re missing is the concept of decoherence: that, when quantum systems interact with the environment (including things like detectors), phenomena like interference become radically suppressed, and quickly unobservable.

It’s not that simple I’m afraid.

If a detector is placed just behind one slit and photons are sent through one by one, after some time two bright areas build up where the photons behaved like particles. In this sense, yes, we get decoherence. *However, *if the detector stays in place but is switched off (i.e. unable to register particles) then an interference pattern is seen on the back screen! So it is not the fact that a detector (which is a physical object) causes decoherence but the fact that no information is made available about particles. When we examine the back screen after we switch off the detector we are causing reality to begin and at that moment no information about particles exist, therefore, an interference pattern is seen. So we are back to knowledge making a difference, not the environment. This was the point I was trying to make.

This is explained here: Double Slit Experiment explained! by Jim Al-Khalili - YouTube

[abashed]

“When physicists speak about something having ‘information’ or ‘knowledge’ about something else, what they mean is more specifically the mutual information—a measure of how much the two systems are correlated. This does lead to some unfortunate misunderstandings for those not conversant with the terminology, but that’s a hard thing to avoid, without all the time inventing new words.”

Actually, it could be that one system (a human observer) is correlated with another system via quantum entanglement, which would accommodate Wiki’s definition. But this is speculative.

[Mangetout]

@abashed

Why don’t you exist? You need to explain this.

I’m part of the universe - I don’t exist until you’ve worked out how to observe and describe me.

[BeepKillBeep]

@abashed

The question is where do these ideas come from? Yes, from us, our consciousness! And when we try these hypotheses out we receive back information. Without first asking for some feedback no scientific knowledge is possible so we need both consciousness and information to make progress. You people either do not seem able to grasp this or refuse to acknowledge it, since it challenges your worldview.

Meh. My worldview just resulted in a major discovery in the grammar theory research domain, solving a 45 year old research problem through the application of artificial intelligence. What is your worldview accomplishing?

[RaftPeople]

@abashed

Yes, it was wasn’t it? This is why I gave the link since he did put over the basic facts in a very conversational way and did not rush through it. I don’t necessarily agree with the rest of his philosophical interpretation about life in general based on his interpretation of QM, nevertheless, he does put his ideas over in a clear way.

I was joking when I wrote “the clarity of explanation was impressive.”

It was words without any substance. For example, when he says something like “entanglement is just an if-then statement”, I immediately have these types of questions:
1 - “if” what?
2 - “then” what?
3 - Is he trying to say that something is processing this “if-then” logic, like so:
If the particles are entangled then we need to make sure that their interactions operate using the entangled rules.

If so, what is keeping track of the entangled states of all the particles in the universe so it can perform the “if-then”?

etc.

[Half_Man_Half_Wit]

@abashed

None of this stuff can read itself and no result of any experiment can have any meaning without an observer. How can a scientific experiment organize itself to put all the apparatus in place and do the experiment?

It’s the other way around: experiments realize situations that occur in nature all the time within controlled circumstances, so that we can learn about what happens in nature. This learning is all that we do; what happens would have happened in just the same way if no early hominid (or wherever you want to draw the line) had ever had that first spark of consciousness.

Or at least, all that we know about the world, and quantum mechanics, is completely consistent with that picture. It’s true that other interpretations are possible; but they all suffer from extravagance. Furthermore, quantum theory has nothing to do with this: all that you’re saying can be said about classical mechanics just as well (and has, in fact, been said).

I know that it’s comforting to put consciousness, and by extension, humanity, back into the center of the universe, from where the Copernican and Darwinian revolutions have so thoroughly removed us, but nature is under no obligation to comfort us.

You said; “The calculation is left as an exercise for the reader.” What if there is no reader? What don’t you understand about this?

What you don’t understand is that nobody needs to do the calculations in order for the world to behave the way it does. Yes, there is no description of the world without somebody describing it; but the world does not depend on being described. Interference patterns vanished long before anybody came up with the math describing how they do so; they’ll continue to, long after the sun has burned this world to cinder.

As John Bell mockingly put it:

@John Bell

Was the wave function waiting to jump for thousands of millions of years until a single-celled living creature appeared? Or did it have to wait a little longer for some highly qualified measurer - with a PhD?

@abashed

If a detector is placed just behind one slit and photons are sent through one by one, after some time two bright areas build up where the photons behaved like particles. In this sense, yes, we get decoherence. *However, *if the detector stays in place but is switched off (i.e. unable to register particles) then an interference pattern is seen on the back screen! So it is not the fact that a detector (which is a physical object) causes decoherence but the fact that no information is made available about particles.

To the contrary, that’s exactly what follows from the math I presented above. If the detector is present, but does not make a detection, it simply falls out of the equation, and we’re left with what we had before.

In such a case, the evolution equations of a system going through either ‘slit’ are

          |D>|1> ---> |D>|1> (= |D,1>),
          |D>|2> ---> |D>|2> (= |D,2>).

This just means that the detector doesn’t change its state—it remains in the same state before and after having interacted with the particle (or, as the case may be, having not interacted with the particle). In other words, no ‘which-path information’ has been gathered.

And it’s important that you realize here that that’s all that which-path information (as scientists use the term) is: if a state-change has been induced in the detector (say, a light has come on, indicating ‘slit 1’, or ‘slit 2’), then which-path information has been collected; if nothing happens, we have no such ‘information’.

This means in particular that the general state of our particle-detector system afterwards is factorizable—i.e. you can write it as a product of two states of the two subsystems. This is the formal definition of not being entangled, i.e. separable.

So, whereas in the case where we did collect which-path information, we had the entangled state

          |D,s> = a|D[sub]1[/sub],1> + b|D[sub]2[/sub],2>,

after the particle has passed the slit, we now have the separable state

         |D',s'> = a|D,1> + b|D,2>
                 = |D>(a|1> + b|2>).

This makes all the difference: since there is no correlation between the detector and the particle, the quantum coherence does not get ‘smeared out’ over the whole system, and consequently, the particle system itself stays coherent; hence, we will recover an interference pattern.

We now have:

            P(o) = P(o,D)
                 = |(a[sup]*[/sup]<D,1| + b[sup]*[/sup]<D,2|)*|o,D>|[sup]2[/sup]
                 = |a[sup]*[/sup]<D,1|o,D> + b[sup]*[/sup]<D,2|o,D>|[sup]2[/sup]
                 = |a[sup]*[/sup]<1|o> + b[sup]*[/sup]<2|o>|[sup]2[/sup]
                 = |a[sup]*[/sup]<1|(c|1> + d|2>) + b[sup]*[/sup]<2|(c|1> + d|2>)|[sup]2[/sup]
                 = |a[sup]*[/sup]c<1|1> + a[sup]*[/sup]d<1|2> + b[sup]*[/sup]c<2|1> + b[sup]*[/sup]d<2|2>|[sup]2[/sup]
                 = |a[sup]*[/sup]c + b[sup]*[/sup]d|[sup]2[/sup]
                 = (a[sup]*[/sup]c + b[sup]*[/sup]d)(ac[sup]*[/sup] + bd[sup]*[/sup])
                 = |a[sup]*[/sup]c|[sup]2[/sup] + |b[sup]*[/sup]d|[sup]2[/sup] + a[sup]*[/sup]cbd[sup]*[/sup] + b[sup]*[/sup]dac[sup]*[/sup]

Here, in the third line, I used that <D|D> = 1, in the fourth line, I’ve introduced the general form of |o>, the fifth line is just a rearrangement, in the sixth line, I’ve used that <1|1> = <2|2> = 1 and <1|2> = <2|1> = 0, and then I just repeated the steps from our original calculation to once again arrive at a result showing prominent interference terms.

So, you see: it’s not whether anybody reads the information, or interprets it, or holds it somehow in conscious experience; rather, the vanishing of interference solely depends on whether the detector makes a detection, indicated by changing its state, or not, standing just idly by.

[Trinopus]

@abashed

. . . Tell me, where exactly is the information contained within your post? Is it the pixels that make-up the symbols? NO. Is it the characters therein? NO. It is the interpretation encoded in the symbols made by an observer. YES!

Computers can communicate meaningfully with each other, without any benefit of “mind.”

[Half_Man_Half_Wit]

@Trinopus

Computers can communicate meaningfully with each other, without any benefit of “mind.”

Hmm. Not on the usual meaning of ‘meaningfully’, I’d say—one of the most important characteristics of computers is that they key on a symbol’s syntactic, as opposed to semantic, properties. So, a letter, a word, what have you, to a computer is really just a series of switches being flipped or not. When you type the word ‘dog’ into a computer, and it shows you a picture of a dog, then there’s no understanding of what you asked it; rather, ‘dog’ is shorthand for pushing a couple of switches that light up a certain array of pixels—dogs don’t enter into it.

[abashed]

@Half_Man_Half_Wit

It’s the other way around: experiments realize situations that occur in nature all the time within controlled circumstances, so that we can learn about what happens in nature. This learning is all that we do; what happens would have happened in just the same way if no early hominid (or wherever you want to draw the line) had ever had that first spark of consciousness.

Or at least, all that we know about the world, and quantum mechanics, is completely consistent with that picture. It’s true that other interpretations are possible; but they all suffer from extravagance. Furthermore, quantum theory has nothing to do with this: all that you’re saying can be said about classical mechanics just as well (and has, in fact, been said).

I know that it’s comforting to put consciousness, and by extension, humanity, back into the center of the universe, from where the Copernican and Darwinian revolutions have so thoroughly removed us, but nature is under no obligation to comfort us.
What you don’t understand is that nobody needs to do the calculations in order for the world to behave the way it does. Yes, there is no description of the world without somebody describing it; but the world does not depend on being described. Interference patterns vanished long before anybody came up with the math describing how they do so; they’ll continue to, long after the sun has burned this world to cinder.

As John Bell mockingly put it:

To the contrary, that’s exactly what follows from the math I presented above. If the detector is present, but does not make a detection, it simply falls out of the equation, and we’re left with what we had before.

In such a case, the evolution equations of a system going through either ‘slit’ are

          |D>|1> ---> |D>|1> (= |D,1>),
          |D>|2> ---> |D>|2> (= |D,2>).

This just means that the detector doesn’t change its state—it remains in the same state before and after having interacted with the particle (or, as the case may be, having not interacted with the particle). In other words, no ‘which-path information’ has been gathered.

And it’s important that you realize here that that’s all that which-path information (as scientists use the term) is: if a state-change has been induced in the detector (say, a light has come on, indicating ‘slit 1’, or ‘slit 2’), then which-path information has been collected; if nothing happens, we have no such ‘information’.

This means in particular that the general state of our particle-detector system afterwards is factorizable—i.e. you can write it as a product of two states of the two subsystems. This is the formal definition of not being entangled, i.e. separable.

So, whereas in the case where we did collect which-path information, we had the entangled state

          |D,s> = a|D[sub]1[/sub],1> + b|D[sub]2[/sub],2>,

after the particle has passed the slit, we now have the separable state

         |D',s'> = a|D,1> + b|D,2>
                 = |D>(a|1> + b|2>).

This makes all the difference: since there is no correlation between the detector and the particle, the quantum coherence does not get ‘smeared out’ over the whole system, and consequently, the particle system itself stays coherent; hence, we will recover an interference pattern.

We now have:

            P(o) = P(o,D)
                 = |(a[sup]*[/sup]<D,1| + b[sup]*[/sup]<D,2|)*|o,D>|[sup]2[/sup]
                 = |a[sup]*[/sup]<D,1|o,D> + b[sup]*[/sup]<D,2|o,D>|[sup]2[/sup]
                 = |a[sup]*[/sup]<1|o> + b[sup]*[/sup]<2|o>|[sup]2[/sup]
                 = |a[sup]*[/sup]<1|(c|1> + d|2>) + b[sup]*[/sup]<2|(c|1> + d|2>)|[sup]2[/sup]
                 = |a[sup]*[/sup]c<1|1> + a[sup]*[/sup]d<1|2> + b[sup]*[/sup]c<2|1> + b[sup]*[/sup]d<2|2>|[sup]2[/sup]
                 = |a[sup]*[/sup]c + b[sup]*[/sup]d|[sup]2[/sup]
                 = (a[sup]*[/sup]c + b[sup]*[/sup]d)(ac[sup]*[/sup] + bd[sup]*[/sup])
                 = |a[sup]*[/sup]c|[sup]2[/sup] + |b[sup]*[/sup]d|[sup]2[/sup] + a[sup]*[/sup]cbd[sup]*[/sup] + b[sup]*[/sup]dac[sup]*[/sup]

Here, in the third line, I used that <D|D> = 1, in the fourth line, I’ve introduced the general form of |o>, the fifth line is just a rearrangement, in the sixth line, I’ve used that <1|1> = <2|2> = 1 and <1|2> = <2|1> = 0, and then I just repeated the steps from our original calculation to once again arrive at a result showing prominent interference terms.

So, you see: it’s not whether anybody reads the information, or interprets it, or holds it somehow in conscious experience; rather, the vanishing of interference solely depends on whether the detector makes a detection, indicated by changing its state, or not, standing just idly by.

To be honest, I have no clue about the calculations you give here, but that does not change anything.

You say nature is ‘there’, regardless of whether we look at it or not, but what you keep overlooking is the fact that the models human beings create can never exist without human beings! Regardless of what is* really * there (although this doesn’t make much sense), all one can do is invent representations of nature which are really very subjective and never existed before we evolved. Language and mathematics never existed before they were invented so how can something that never existed exist and not exist at the same time, which is effectively what you are saying when you assert nature is always there even if we aren’t? Now, something *did * exist before we did, undoubtedly, but whatever that was only existed as a probability which combined with our consciousness to appear to us as what we call ‘reality’. Your mathematical calculations are mere models that your mind has constructed in order to ‘prove’ something or other but what does it all really prove? It proves mathematics is creative imagination that tries to ‘mimic’ reality and does this successfully up to a point but sooner or later fails to give accurate predictions.

[abashed]

@Half_Man_Half_Wit

Hmm. Not on the usual meaning of ‘meaningfully’, I’d say—one of the most important characteristics of computers is that they key on a symbol’s syntactic, as opposed to semantic, properties. So, a letter, a word, what have you, to a computer is really just a series of switches being flipped or not. When you type the word ‘dog’ into a computer, and it shows you a picture of a dog, then there’s no understanding of what you asked it; rather, ‘dog’ is shorthand for pushing a couple of switches that light up a certain array of pixels—dogs don’t enter into it.

Now here, we are in agreement.

[abashed]

@RaftPeople

I was joking when I wrote “the clarity of explanation was impressive.”

It was words without any substance. For example, when he says something like “entanglement is just an if-then statement”, I immediately have these types of questions:
1 - “if” what?
2 - “then” what?
3 - Is he trying to say that something is processing this “if-then” logic, like so:
If the particles are entangled then we need to make sure that their interactions operate using the entangled rules.

If so, what is keeping track of the entangled states of all the particles in the universe so it can perform the “if-then”?

etc.

I can’t say I recall him saying this. Perhaps you misunderstood.

[abashed]

@BeepKillBeep

Meh. My worldview just resulted in a major discovery in the grammar theory research domain, solving a 45 year old research problem through the application of artificial intelligence. What is your worldview accomplishing?

Well done. So what’s your point?