Einstein and "spooky actions"

[LegionOnomaMoi]

This one:

I can represent each card as an entry in a single vector in 52-dimensional space. I can do this in a number of ways. I can make all number cards an entry in the vector that has a value corresponding to that number, make aces have values of 1, and make face cards either 11, 12, & 13 (jack, queen, & king respectively) or choose one value for all face cards as in a number of card games.

I actually like the other one better, but let's go with the first, and lets say I'm numbering the cards as such that aces are 1's and face cards go from 11to 13.

There are a couple ways to do this. One would be just to partition the vector into four subvectors 1-13, 14-27, etc., and then designate which partition corresponds to which suit.

This is actually not what partitioning would look like. It should just be one vector with horizontal bar separating the entries into 4 partitions, but I don't know how to do that, so we'll pretend that instead of what looks like a vector in a vector is really just one set of () partitioned with something like this (--) only with 4 bars.'

That would be one way. Another way would simply to have a entries 1 through 52 represent variables and one of 4 scalars in front of each variable (card, whether represented numerically or by letters or whatever), like d3 for the 3 of diamonds or sk for the king of spades. Or I could have a vector of ordered pairs:

such that each entry gives the suit and then a number from 1-13 representing ace to king. All of those would be valid vectors in 52-dimensional space.

I did not understand that at all. Can you rephrase it?

Those dots (...) indicate a series (the summation of n terms of a sequence) which ends in a generic formula. It's not a polynomial. It a formula for curve fitting a polynomial function at a = 0 for an arbitrary degree. It's worse than just generic:

This is using Maclaurin to derive the obvious forumula (or theorem), but it shows you how the iteration process of taking derivatives does this. We can start evaluating at a=0 (hence Maclaurin) and plugging in the formula into our function f. For a generic x, the left side shows us what we'd get, while for the actual value 0 we get 1. We take the derivative using the power rule and we and again get both the generic x and the derivative of f at 0. We take the 2nd derivative and the same thing happens. And thanks to the fact that we have the formula already (and, according to legend from a little boy named Gauss who summed all the numbers between 1 and 100 in a few seconds), we can prove the theorem.
You asked if I'd seen a polynomial written like that. The answer is trivially no, because that isn't a polynomial. Notice that we have a formula in my example that allows us to evaluate each term. In your case, we'd have a polynomial to evaluate in a similar way. We plug it in starting at a = 0 and start taking derivatives up to n. That last part makes the entire thing unnecessary because it gives you everything you need to begin with. Whatever the degree of the polynomial, that's the nth derivative.

The reason it was useful was because we were dealing with polynomials fit to data, where we did have dp/dx(0). (Or, Δp(0) )

You have to plug the polynomial into your regression (interpolation, curve fitting/sketching, or whatever you are doing) formula. All you have is the series showing the general procedure we'd follow up to any n. But as we have no formula, the entire thing is like opening a model car kit and finding only the instructions, no pieces. You can't use polynomial approximations (Legendre, Hermite, whatever), without a polynomial.

I don't see how the two parts of this sentence connect

You start your evaluation at a = 0. a of what? That is, if I'm using R, and just starting a simple polynomial regression, I need something more than reading into a variable polyN or something and the using the plot() and model() functions. I need x and y values and my data. In MATLAB, for simple fits all I need to do is enter these in and use the polyfit function. Whether I am trying to using regression, fitting a curve using derived paramaters, or just finding roots, you've given nothing with which to do that. You don't have a polynomial but maybe a procedure for using one.

I am pointing out that defining an object in terms of its own values is, while slightly confusing on first glance, a perfectly true statement.

You didn't do that. At all. You gave some series with no actual polynomial to use, and even then the terms were marked differently and clearly according to a notation used for at least a century. In other words, you supplied not only the wrong context

You've never seen a polynomial written like this?

by saying "a polynomial" when that clearly isn't one, yet I was still able to read the notation well enough to realize that
1) it was a series
2) you were either providing indices weirdly or an argument of 0
3) that it was for an approximation

All without actually even having a polynomial and being told I did.

Just because it is convention for A to represent a matrix doesn't mean using it as a vector is wrong, if that's what its defined as.

Wrong? No. Imprecise and misleading? Yes. Look what you compared it too. I had less info in your example which was a written out as only part of a standard way of describing a series and I knew anyway. Why? Notation.

But matrices represent something apart from the systems of equations. This is most obvious because the operations (row exchange, row scaling) that preserve system identity do not preserve matrix identity.

??? Are you confusing the identity matrix with something else?

That's because the matrix in the example is not a system of equations at all, but instead a transform acting on the general vector [x,y,z]^T.

A transform is a matrix.

its the definition of a vector.

I am asking about the space, and this:

The space as a whole is defined by how many distinct linear combinations it is possible to build.)

is wrong. You can built infinitely many linear combinations with two vectors (actually you can do it with one by some definition). Nothing about that definition is accurate or close to accurate. A space is spanned by linear combinations and these concepts combine in ways that have to meet particular conditions.

So which linear combination of [1,1,2] and [1,2,1] does not lie on the plane -3x+y+z=0?

It's not that I can span a space, but that I can and I cannot get out of that space by the 2 operations. Also, to answer your question:

Row reduce all you want, but you have 2 free variables to take you out of any plane you wish to span.

I pointed that out earlier.

You said 52D multiple times. If that's the space we're in, then we aren't in infinite dimensional space.

You appear to be claiming that F and V<F> are the same object

No. One is simply a list of conditions, and the other is the application ofthese to form an algebraic structure.

or that V<F> is a field

from Linear Algebraic Groups (2nd ed.) by T. A. Springer.

Are you having a problem imagining an operator A such that

As I said in my other latest reply to you here, I'm having a problem understanding why you are saying the things you are about vector spaces, fields, etc.

 

[LegionOnomaMoi]

Math and stuff.

opcorn:

Someone said something about some poker. In QM poker what is the probability of getting 5 ace of spades. Is it really just as likely as any other combo?

That depends on whether or not it's my hand. In QM poker, I "collapse" my hand to be the winning one by observing the cards everyone else has.

 

[Mr Spinkles]

Hi Legion,

Sorry for not responding sooner. I've just been busy, and traveling and having a good time. Also, after thoroughly reading and considering all of your quotes and comments up to and including post #603, I have been unsure about how to proceed because I feel like I am saying something very modest, and although I respect your opinion, I don't agree with any of your objections (including your quotes, points about statistics, delta function limits of integration, etc.). In my opinion, none of your quotes contradict my claim that "QM postulates the existence of a ket which represents the complete state of a physical system, not just our knowledge of it, and this is the standard, most common way of formulating QM". The Griffiths quotes support that statement, IMO.

Instead of getting bogged down in lots of things and expanding the field of disagreement, why don't we simplify things and just try to answer one smaller, more manageable question: "Does Griffiths agree with Spinkles' statement?" If we can't agree on that question, then we may as well give up on trying to see eye-to-eye on any broader questions. Griffiths should essentially speak for himself, anyway (and you've done a good job of quoting the most relevant parts in context).

So here's you quoting Griffiths:

"The orthodox position: The particle wasn't really anywhere. It was the act of measurement that forced the particle to "take a stand" (though how and why it decided on the point C we dare not ask). Jordan said it most starkly: "Observations not only disturb what is to be measured, they produce it...We compel (the particle) to assume a definite position." This view (the so-called Copenhagen interpretation) is associated with Bohr and his followers. Among physicists it has always been the most widely accepted position. Note, however, that if it is correct there is something very peculiar about the act of measurement- something that over half a century of debate has done precious little to illuminate" pp. 3-4

I submit to you that this is fully supportive of what I said: "QM postulates the existence of a ket which represents the complete state of a physical system, not just our knowledge of it, and this is the standard, most common way of formulating QM." (Emphasis added.) When Griffiths says "the particle wasn't really anywhere" he is saying that the particle did not have a well-defined position. He is not saying the particle didn't have a ket. As you know, a perfectly well-defined, valid state (ket) in QM is not generally an eigenstate (eigenket). Upon measurement, the initial state becomes an eigenstate with a well-defined property (such as position), which is what Griffiths means by "taking a stand". We "compel" the system through measurement to have a well-defined position (for example) and to adopt a position eigenstate. But both before and after measurement, it always had a state (eigen or general) which was indeed supposed to describe "a physical system, not just our knowledge of it". Think about it: if we didn't "compel" the particle to "take a stand", then the general ket beforehand would describe only our limited knowledge; in that case, when we measure we would simply expand our knowledge about a position which the particle always possessed. But according to the orthodox interpretation, we are actually compelling it to change its state and adopt a well-defined position. Hence, the ket which describes the particle beforehand is not merely a description of our knowledge but the actual state of the physical system--a physical system which actually did not have a position until we compelled it to do so (and so of course its representation, the ket, was also compelled to change--to an eigenket).

Griffiths stresses this point in the introduction to spin when he's talking about the singlet state. He says something like, IIRC, "If the electron is not in an up or down spin state, does it have no state? No, that's a mistake, it does have a state: it's in the singlet state. Upon measurement it will be forced into one of the spin eigenstates." I don't have the book with me (traveling) so I can't verify my memory or quote it at the moment.

The second Griffiths quote:

Second, as your textbook points out, the "standard interpretation" holds that the measurement produces (literally) the outcome. If you look on p. 420, Griffiths returns to the original question:
"Did the physical system "actually have" the attribute in question prior to the measurement (the so-called realist viewpoint), or did the act of measurement itself "create" the property, limited only by the statistical constraint imposed by the wave function (the orthodox position)"

Right. I believe I can see how you would think this contradicts what I'm saying, but read carefully. The orthodox position is that the act of measurement itself created the "property" or "attribute"--the attribute of having a precise position or momentum or spin, for example. Before the measurement, the physical system did not "actually have" these things; hence, the ket which represents the physical system did not have those things, either. Makes sense. Since the measurement creates a new property or attribute in the physical system, the ket must change, too: namely to an eigenket which has the new property or attribute. So both before measurement, and after measurement, QM postulates "there exists a ket which represents the complete state of a physical system, not just our knowledge of it". In context this is what Griffiths is saying and he confirms this is the "orthodox" way of formulating QM, as I said.

it is not accurate to say, as you said earlier, that a fundamental difference between QM and CM is that quantum states don't describe "the physical system".

The interpretation which differs most fundamentally is the standard one. Per Griffiths: "the quantum state...does not uniquely determine the outcome of a measurement; all it provides is the statistical distribution of possible results." p. 420 (italics original, emphasis added).

I said QM states describe the physical system. Here you've quoted Griffiths saying QM states do not describe the outcome of a measurement. These two statements do not contradict each other. Suppose that the physical system really, truly, intrinsically does not possess beforehand a unique outcome of a measurement. Then it would be true to say that the QM state (1) represents the physical system, and (2) does not uniquely determine the outcome of a measurement. No contradiction there and that is the orthodox interpretation.

 

[LegionOnomaMoi]

Hi Legion,

Sorry for not responding sooner. I've just been busy, and traveling and having a good time.

No problem (pay no heed to my neurotic insecurities), and I hope you continue to have a good time.

Instead of getting bogged down in lots of things and expanding the field of disagreement, why don't we simplify things and just try to answer one smaller, more manageable question: "Does Griffiths agree with Spinkles' statement?" If we can't agree on that question, then we may as well give up on trying to see eye-to-eye on any broader questions.

That was my hope (or rather, my hope was that by using one source as a foundation we could come up with a manageable and single question that tells us whether there is any hope of an outcome we both agree upon). Your question is perfect and perfectly put.

However, because we're going to be referring to kets and physical systems, I'll start off with this: QM, in the most common formulation a.k.a. the standard interpretation, posits the existence of a ket which represents the complete state of what is often called a physical system.

In other words, my position (and again, this is only to define some terms that will frame the answer to the question) is that insofar as the standard/Copenhagen/orthodox position is correct, if the ket describes a physical system we have complete knowledge of, it is because that "physical system" literally is the ket. This holds true for other all equivalent mathematical notations: we do not know (though there are many theories) how any of them correspond to anything other than the mathematical representation (I would also say the standard interpretation is fundamentally and irreducibly statistical, but to simplify it suffices to note that the only possible way the ket describes a physical system we have complete knowledge of is when we refer to the notation itself as the physical system).

But let us allow Griffiths to speak for himself:

Here I believe is the crux of the matter. When Griffiths says "the answer is provided by Born' statistical interpretation", what does he mean by this?

He is not saying the particle didn't have a ket.

I will address kets in shortly, but I can't yet because:
1) I am going to quote what I have already, only now with more context
2) Griffiths doesn't use Dirac's notation much, so I had to make some arrangement choices and I chose to continue to try to answer what Griffiths means by "the answer" and then cover what he says about kets.



From the page following that depicted in the image above of sect. 1.2:
"I'll return to this story at the end of the book, when you will be in a better position to appreciate Bell's argument; for now, suffice it to say that the experiments have decisively confirmed the orthodox interpretation: A particle simply does not have a precise position prior to measurement, any more than ripples on a pond do; it is the measurement process that insists on one particular number, and thereby in a sense creates the specific result, limited only by the statistical weighting imposed by the wave function"

As you state, nothing about localizing a wavefunction via measurement is inconsistent with complete knowledge. However, when Griffith says that the wave function describes a probability, then this sort of precludes an interpretation that the wave function describes the actual system.
But your question was framed in terms of kets, which Griffiths does at least cover. Because you are travelling I have scanned the pages so that no misquote or selective quote or any other deliberate or accidental misrepresentation of Griffiths is possible (at least until I try to interpret it):

So the question becomes what "hitting" a vector means? Are we really talking about physical systems as abstract mathematical operations and notations that, even if we call refer to a ket in terms of some "physical system", there is no physical system to which we can related that ket (vector) to? More importantly, that ket has a complete set of specific entries (hence our complete knowledge), so what do we gain from a bra or an operator "hitting" a ket? Also, how does one "hit" a physical system with mathematical notations to obtain measurement outcome values?

Granted we have

a perfectly well-defined, valid state (ket) in QM

how does a state "span" anything such than an operator is needed and what does the operator do?

Hence, the ket which describes the particle beforehand is not merely a description of our knowledge but the actual state of the physical system--a physical system which actually did not have a position until we compelled it to do so (and so of course its representation, the ket, was also compelled to change--to an eigenket).

What is the function f that, as Griffiths puts it, "the ellipses [...] waiting to be filled" then become filled once "the bra encounters [this function] in the ket to its right"? Why does problem 3.21 ask us about eigenvalues and eigenvectors of an operator? The next problem gives us kets that have values, but if these are supposed to be states of some system, how might we interpret them in terms of this system?

Think about it: if we didn't "compel" the particle to "take a stand", then the general ket beforehand would describe only our limited knowledge; in that case, when we measure we would simply expand our knowledge about a position which the particle always possessed.

Which is exactly what we do. That's why the ellipses can't be filled until we apply a function. If the ket is supposed to be the state of a physical system, why do we require a projection operator and why is it this, not measurement, which "picks out the portion of any other vector that 'lies along'" the ket? And why, if the ket describes a physical system, does the ket span a 1-dimensional subspace that the operator projects the "bra" onto?

But according to the orthodox interpretation, we are actually compelling it to change its state and adopt a well-defined position.

Using an operator as well as physical measurement. What is the operator for?

Griffiths stresses this point ...I don't have the book with me (traveling) so I can't verify my memory or quote it at the moment.

I couldn't find what you refer to, but on page 172 note 29 he does say "People often say that

is "the probability that the particle is in the spin-up state," but this is sloppy language; what they mean is that if you measured

is the probability you'd get &#295;/2." (italics in original)

How do you measure a hermitian operator? That Sz is the Pauli spin matrix &#963; for the z axis multiplied by &#295;/2. So what instrument other than a calculator can "measure" a matrix multiplied by a constant?

Before the measurement, the physical system did not "actually have" these things; hence, the ket which represents the physical system did not have those things, either. Makes sense. Since the measurement creates a new property or attribute in the physical system, the ket must change, too: namely to an eigenket which has the new property or attribute.

1) If the measurement alone created these attributes, why do we get them using an mathematical function which interacts with the ket? Griffiths makes the metaphorical interpretation of the ket as a physical system clear by describing it literally as being "hit" by a mathematical operator. You can't hit a system with a function unless we are being metaphorical and both the system and the function are "hitting" metaphorically.
2) Why does Griffiths stress that the mod square of a is not a probability of the particle spin itself? And what does measuring an operator entail?

 

[PolyHedral]

That's it. The functions themselves are not a field, because they are functions. A field is taking some set or collection of sets (ad infinitum) and determining that it satisfies these criteria. The reason we don't say these functions are a field is because they need not be. I can impose other structures. Just because you can have a field over the reals doesn't make the reals (strictly speaking) a field.

I don't see how you can possibly define addition, subtraction, multiplication and division differently for real/complex-valued functions in any useful way.

That's why we refer not to an arbitrary field (there's not really any such thing) but some arbitrary set and either let or determine whether that set meets the conditions required of a field.

I think you're just making distinctions without a difference if you say that a structure that is defined solely by its field properties isn't "an arbitary field."

Every Boolean algebra is a field of sets.

Oh, I see, you're right, but: (Many authors refer to

itself as a field of sets. The word "field" in "field of sets" is not used with the meaning of field from field theory.)

The only reason (I think) that you come across examples in which the rational numbers are said to be a field is because in that case the set is practically defined by the algebraic structure created by a field. Fields came after, and as generalizations of algebra to algebras (like rings, groups, etc.).

The concepts of addition, subtraction, multiplication, division and the two identities certainly pre-date any notion of formal algebra, rational, real or otherwise.

I actually like the other one better, but let's go with the first, and lets say I'm numbering the cards as such that aces are 1's and face cards go from 11to 13.

There are a couple ways to do this. One would be just to partition the vector into four subvectors 1-13, 14-27, etc., and then designate which partition corresponds to which suit.
[huge image]
This is actually not what partitioning would look like. It should just be one vector with horizontal bar separating the entries into 4 partitions, but I don't know how to do that, so we'll pretend that instead of what looks like a vector in a vector is really just one set of () partitioned with something like this (--) only with 4 bars.'

That's vector syntax, but the thing you get doesn't feel like a vector. For instance, what does -(1,2,3,4...52)^T mean? Is it a valid order of the deck? It's a valid vector in this syntax. (Very vector has a negation!)

That would be one way. Another way would simply to have a entries 1 through 52 represent variables and one of 4 scalars in front of each variable (card, whether represented numerically or by letters or whatever), like d3 for the 3 of diamonds or sk for the king of spades. Or I could have a vector of ordered pairs:


such that each entry gives the suit and then a number from 1-13 representing ace to king.

Whereas in this notation, you'd need to define the field operators on your set of (suit, value) tuples. If I add the four of diamonds to the king of clubs and then take away an ace of spades, what does that give me?

You didn't do that. At all. You gave some series with no actual polynomial to use, and even then the terms were marked differently and clearly according to a notation used for at least a century. In other words, you supplied not only the wrong context by saying "a polynomial" when that clearly isn't one, yet I was still able to read the notation well enough to realize that
1) it was a series
2) you were either providing indices weirdly or an argument of 0
3) that it was for an approximation

All without actually even having a polynomial and being told I did.

I screwed up, because what I meant to say was,

But that is not an approximation at all, unlike if we were working with an arbitary function. In the case of an n-degree polynomial, the equality is exact, and I can prove that to you if I have to. The above equation is (the definition of!) a polynomial, not an approximation or converging series.

Wrong? No. Imprecise and misleading? Yes. Look what you compared it too. I had less info in your example which was a written out as only part of a standard way of describing a series and I knew anyway. Why? Notation.

And you misunderstood, because I was not actually using that notation to mean that thing - as the English explanation told you. Just as you misunderstood the vector example for the same reason.

??? Are you confusing the identity matrix with something else?

No, I'm talking about matrix identity, as in the process we use to compare whether two matrices are "the same."

Here's an example. Let's represent a system in two equations like so:

Now, swap the rows:

This is clearly the same system of equations - after all, the answers we get don't depend on the other we write the equations!

However,

Therefore, it's bad practice to say a matrix "is" a system of equations, because matrices don't map uniquely to systems. We can perform operations on them that clearly give us the same system, but do not give us a matrix with the same arithmetical value.

You can built infinitely many linear combinations with two vectors (actually you can do it with one by some definition). Nothing about that definition is accurate or close to accurate. A space is spanned by linear combinations and these concepts combine in ways that have to meet particular conditions.

I forget that |R^n| = |R|.

It's not that I can span a space, but that I can and I cannot get out of that space by the 2 operations.

"Each of those vectors exists in R3. To span R2, when we'd need closure under the 2 operations required. However, for any plane spanned by these vectors we could use either operation and not be on that plane. We do not have the required closure..."
:help:

Any set of vectors of any dimension which contains precisely 2 linearly independent members will span R^2.

As I said in my other latest reply to you here, I'm having a problem understanding why you are saying the things you are about vector spaces, fields, etc.

Because I'm being somewhat loose about identity. The real numbers have unique, natural addition and multiplication operators, so for all practical purposes, they are a field. Square-integrable functions ditto, so they are a field, not merely a generic set.

 

[PolyHedral]

This is the second time you have implied I am quote mining or otherwise unable to understand what you have said.

I mean no offense, nor do I mean to accuse you of intellectual dishonesty.

However, your responses to my posts seem logically disconnected from what you're responding to. Perhaps you are so much more intelligent than me that you can head my argument off at a pass I didn't realize I was heading towards, or I am incredibly bad at explaining myself [but for external reasons, I don't think its that], or something else*, but I don't think you're responding to what I'm saying, which means that carrying on a coherent discussion is incredibly difficult. Of course, the problem might be on my end, in which case you need to explain why you're taking away what you are from what I said.

My best guess as to the reason for this: the forum software chops out nested quote blocks, and so makes completely distinct points look as though they're two paragraphs talking about the same thing. I don't know about you, but I have the same thread open in two tabs - one with the reply box, the other with the post I'm replying to open, so I can see what any given part of your post is actually talking about. If you have overlooked this effect when composing your previous replies, please take more care replying to this one.

When you first posed this question, you asked that I explain using my words (presumably rather than scanned pages or lines copied out of my library or from some internet source). I have tried to limit my use of others' words to links and a quote or 2 to support something I have usually already said, and to find reputable material for you rather than just grabbing a few of my books (which would mean quoting sources that you can't access without owning the book). I have gone out of my way to use quotes never just to support what I have said, but always to give you a source to read that has more than what I quoted. I have done that so you can perhaps review something you haven't studied in some time or perhaps which was explained badly or too informally at some time.

I appreciate that you go to the effort, but I asked you specifically to answer in your own words because when you answer with others', IMO, you don't support the point you're trying to make. You support closely related but uncontroversial points, and not the one you're actually asserting.

For instance: I say that a vector space, V<F> over some field F is not itself a field. You respond with this section. This is very informative, but it does not actually answer my point, AFAIK - it does not tell me if V is a field or not. If you believe it does actually assert that V, and not F, is a field, can you quote the exact sentence?

Rather than insulting my intelligence (which you did before you rightfully pointed out I was being unnecessarily offensive), if you think some statement I make indicates I don't know what I'm talking about, you can ask me to support it and I will.

"So which linear combination of [1,1,2] and [1,2,1] does not lie on the plane -3x+y+z=0?"
You then respond with an expression that will not be a such a linear combination except for quite specific values of y and z.

 

[LegionOnomaMoi]

Any set of vectors of any dimension which contains precisely 2 linearly independent members will span R^2.

A vector's dimension is determined by the number of "rows" it has (that is, when written correctly, it will be vertical and the dimension corresponds to number of entries). A vector with three entries cannot span R2 because it doesn't exist in R2. It can (via linear combinations with other vectors) span a plane in R3.

I don't see how you can possibly define addition, subtraction, multiplication and division differently for real/complex-valued functions in any useful way.

"Definition (sum and product): Let f and g be functions from a set A to the set of real numbers R.
Then the sum and the product of f and g are defined as follows:
For all x, ( f + g ) (x) = f(x) + g(x) , and
for all x, ( f*g (x) = f(x)*g(x) ,
where f(x)*g(x) is the product of two real numbers f(x) and g(x)." (source)


Is it true that for any real number x, (cos +sin) (x) = cosx +sinx?

Alternatively:

Each trig function above is defined using all the operations you named and using them on exponential functions.

Then there is basic calculus:

If we limit ourselves to merely to basic algebraic, trig, and calculus functions than we can see, quite clearly, that there are different rules when it comes to the operations you describe on real-valued functions.

So math most students learn in high school would entail operations with functions that you "don't see how" we "can possibly define" differently yet we do.

I think you're just making distinctions without a difference if you say that a structure that is defined solely by its field properties isn't "an arbitary field."

I think you are misunderstanding fields:

from Advanced Linear Algebra(p. 36 is given in the preview).

Oh, I see, you're right, but:

From Wikipedia (emphasis added)

"The property x &#8853; x = 0 shows that any Boolean ring is an associative algebra over the field F2 with two elements, in just one way. In particular, any finite Boolean ring has as cardinality a power of two. Not every associative algebra with one over F2 is a Boolean ring: consider for instance the polynomial ring F2[X]."

The word "field" on the wiki page links to Field (mathematics)

That's vector syntax, but the thing you get doesn't feel like a vector.

When you think a vector with 3 entries can do anything in R2, then what "feels" like a vector to you is perhaps an inadequate litmus test.

Here's an example. Let's represent a system in two equations like so:

Now, swap the rows:



This is clearly the same system of equations - after all, the answers we get don't depend on the other we write the equations!

However,

This is invalid. The two matrices are equal, for exactly the reason we have this notation in the first place. They are the representations of coefficients with the variables removed so that only the coefficients remain. This is one of the first thing one learns in linear algebra, as it's how we understand echelon form: "By means of a finite sequence of elementary row operations, called Gaussian elimination, any matrix can be transformed to row echelon form. Since elementary row operations preserve the row space of the matrix, the row space of the row echelon form is the same as that of the original matrix." See also Row Operations.

Therefore, it's bad practice to say a matrix "is" a system of equations, because matrices don't map uniquely to systems.

Hence the name: coefficient matrix.
You said:

But matrices represent something apart from the systems of equations.

I never disagreed with that (I called a matrix a transform right after I asked you whether you were confusing matrix identity with something else).

For instance, what does -(1,2,3,4...52)^T mean?

There are 52 cards in a deck. Does the number 52 necessarily refer to cards in a deck? No. It could refer to number of cars in a parking lot. Or infinitely many other things. Matrices and vectors are no different. An equation such as 3q + 4o= 18b has no inherent meaning. Systems of equations have no inherent meaning, nor do they suddenly obtain one by removing variables. In my analogy, I didn't just give the number 52. I talked about a deck of cards.

you'd need to define the field operators on your set of (suit, value) tuples

I'm not the one who wanted to talk about the eigenvectors of a deck of cards. You'd need far more for your quantum shuffle than I'd need to program a card game with actual vectors and matrices and actual operations in MATLAB

I screwed up, because what I meant to say was,

But that is not an approximation at all, unlike if we were working with an arbitary function. In the case of an n-degree polynomial, the equality is exact, and I can prove that to you if I have to. The above equation is (the definition of!) a polynomial, not an approximation or converging series.

The above isn't an equation. There is nothing that anything is equal to. Those...are used in mathematics to indicate a series and they mean n can be any natural number. Nor is it a polynomial. A polynomial has a specific definition, and so does a polynomial function (the two are not the same thing). You have n functions given by the n derivatives in your formula, the function p(x) is defined by a list of undetermined functions using the same letter all with arguments of 0's, which means that the function itself is a composite of functions (how many we don't know) used to evaluate some expression, value, or anything we could plug in such that we'd have a value.

I forget that |R^n| = |R|.

We are not talking about set cardinalities, but about spaces and topology. There is nothing whatsoever that makes the space spanned by R1 equal to R2.


The dimensionality of a space is determined by the number of coordinates you need to describe a point in that space. In R1, you need only one point (the number). In R2, you need a pair. In Rn, you need n coordinates to describe any point. Imagine a regular 3D coordinate system. Let a plane extend infinitely in all directions y and z but let ever x at every point on the plane be 2.

Let v1= [1, 1, 1], v2= [2, 2, 2] and v3 = [x, y, z]

The linear combination of the vectors can span the requisite plane, because I make the first scalar 2, the second 1, and the third whatever I wish to ensure that x=2 always. Now I make linear combinations of these vectors into all those needed to correspond to every point on the plane as described above (where x=2 always, but y and z extend infinitely). However, I can also do the exact thing for a plane where x always equals 1. In fact, I can span tons of planes in R3 by using linear combinations of these vectors, but I need 3 linearly independent vectors to span R3 and get a vector space (I need basis vectors).

 

[Mr Spinkles]

Legion,

I think all your questions in your last post can essentially be answered by noting the following: an operator which "hits" a ket is the mathematical representation of a measurement being performed on a physical system. To summarize:

Physical reality: some physical system (e.g. an electron)
Mathematical representation: ket

Physical reality: physical measurement of a system
Mathematical representation: operator "hitting" a ket

A physical system in a state with no well-defined position (say) changes, in a statistical way, to a state with a well-defined position when it is measured (according to the orthodox interpretation). Hence, an accurate mathematical representation of the physical system must reflect that. Therefore, the mathematical representation involves an initial ket with no well-defined position (say), which is "hit" by an operator (measurement), which changes the initial ket in a statistical way into an eigenket with a well-defined position.

I'm just failing to see at what point, according to standard QM, the system lacks a ket which completely describes it.

As you state, nothing about localizing a wavefunction via measurement is inconsistent with complete knowledge. However, when Griffith says that the wave function describes a probability, then this sort of precludes an interpretation that the wave function describes the actual system.

No, it does not preclude that. Suppose the actual system is really, truly not deterministic. Suppose the actual system's future condition/behavior (upon measurement, for example) is only constrained by probabilistic tendencies which the system possesses, and each tendency is stronger or weaker based on the system's state or condition at the moment it was measured. Suppose that is the reality. Then, naturally, a mathematical representation (wavefunction) which describes that reality should also describe probabilities representing the system's tendencies. Ergo, the fact that a wavefunction describes probabilities does not preclude the wavefunction from describing an actual system. In fact, under these assumptions, the latter cannot be accomplished without the former. And these are (essentially) the assumptions of the orthodox interpretation.

Here I believe is the crux of the matter. When Griffiths says "the answer is provided by Born' statistical interpretation", what does he mean by this?

I don't think the meaning of the statistical interpretation is the crux of the matter. I think the crux of the matter is that the statistical interpretation, whatever it means, is (according to Griffiths) the answer to the question, "How can [the wavefunction] represent the state of a particle?" Clearly, then, it is implicit that Griffiths does indeed claim the wavefunction represents the state of a particle. How/why he arrives at that answer to his own question is another matter. But for now, let's just establish that Griffiths believes there exists some answer to his question and, therefore, the wavefunction does somehow "represent the state of a particle" (according to Griffiths).

In other words, my position (and again, this is only to define some terms that will frame the answer to the question) is that insofar as the standard/Copenhagen/orthodox position is correct, if the ket describes a physical system we have complete knowledge of, it is because that "physical system" literally is the ket.

It literally is completely represented by the ket. Obviously a thing I write down on paper is not literally, 100% equivalent in every way to a thing that I measure in an experiment. But the former is completely representative of the latter, yes. That is the orthodox interpretation. The main alternative to this interpretation is that there is more going on in our experiments that we don't know about, and the ket merely represents what we know. Then the physical system would not be literally completely represented by the ket, there would be (in principle) some other (undiscovered) mathematical thing which would more completely represent the physical system.

 

[LegionOnomaMoi]

an operator which "hits" a ket is the mathematical representation of a measurement being performed on a physical system.

Measurement is described as "in the sense of the statistical interpretation".

This:

Physical reality: physical measurement of a system
Mathematical representation: operator "hitting" a ket

Does not describe a statistical interpretation of measurement. It separates the mathematical from the physical.

No, it does not preclude that. Suppose the actual system is really, truly not deterministic.

If a system is really, truly not deterministic, then how can we have complete knowledge of it? As you know, complex systems involve merely epistemic indeterminacy. Our mathematical representations are incomplete, but we could at least in principle describe the system precisely. If the system is ontologically indeterministic, then by definition we cannot do this: either our models are probabilistic, or we simply conflate the math with the physical system, calling equations and functions "physical".

You mentioned before:

He says something like, IIRC, "If the electron is not in an up or down spin state, does it have no state? No, that's a mistake".

What is it Griffiths asserts that the system "does not have" below:

Recall your disagreement with my stance on correspondence in classical vs. quantum physics:

The correspondence I refer to is this:
"In classical physics, the notion of the &#8220;state&#8221; of a physical system is quite intuitive...there exists a one-to-one correspondence between the physical properties of the object (and thus the entities of the physical world) and their formal and mathematical representation in the theory...With the advent of quantum theory in the early twentieth century, this straightforward bijectivism between the physical world and its mathematical representation in the theory came to a sudden end. Instead of describing the state of a physical system by means of intuitive symbols that corresponded directly to the &#8220;objectively existing&#8221; physical properties of our experience, in quantum mechanics we have at our disposal only an abstract quantum state that is defined as a vector (or, more generally, as a ray) in a similarly abstract Hilbert vector space."

Your response:

It's still acceptable to say (or at least I thought it was) that if you prepare a system even once, and never prepare any other system identically, there still exists a ket which does indeed describe the complete quantum state of that system, and that mathematical representation has what you call a "one-to-one correspondence" with the system.

Griffiths, however, closes his book with the following: "In this book I have tried to tell a consistent and coherent story: The wave function (&#936 represents the state of a particle (or system); particles do not in general possess specific dynamical properties (position, momentum, energy, angular momentum, etc.) until an act of measurement intervenes; the probability of getting a particular value in any experiment is determined by the statistical interpretation of &#936;; upon measurement the wave function collapses, so that an immediately repeated measurement is certain to yield the same result." p. 434

It seems to me that even though there are differences between QM and classical physics, this is not one of them. Both say in principle that any given physical system has a one-to-one correspondence with its mathematical representation. In classical mechanics the mathematical representation is a list of all the positions and momenta of all the particles, i.e. a point in a 6N-dimensional phase space; in QM it's a vector in Hilbert space.

In classical physics, I can refer to the mathematical description of "position, momentum, energy, angular momentum, etc." of some physical system, and I can also refer to the actual physical system that has these properties. As Griffiths states, such properties don't exist until I determine them. Until measurement, there are none of the properties of physical systems we use in classical physics. There only the mathematical representation of the system, and to get values I don't just need to apply a mathematical operator to get observables, these observables are mathematical functions (see below). The difference is your dichotomy between representation and physical reality. In QM, there's nothing, according to Griffiths, for the math to correspond to.

The closest Griffiths gets to ontological commitments (p. 422) is when he states "Evidently the collapse of the wave function- whatever its ontological status- is instantaneous."


He is clear that "observables are represented by Hermitian operators" (eq. 3.18). On the surface, this is what you've said: observable properties are represented mathematically. But this interpretation quickly becomes problematic:

If observables are represented Hermitian operators, and observables are therefore the measurement results (in your schema, the operator is the math, the observable the physical part), why do measure observables? And on an ensembles of systems no less? In fact, in sect 3.2.1 Hermitian Operators we find:

"The expectation values of an observable Q( x, p ) can be expressed very neatly in the inner-product notation"

Why do observables have expectation values? The observables are functions. If this is accurate:

Physical reality: some physical system (e.g. an electron)
Mathematical representation: ket

Physical reality: physical measurement of a system
Mathematical representation: operator "hitting" a ket

then where is the physical measurement of a system that the mathematical representation corresponds to? Why are observables measured using what tools such that they have expectation values rather than, well, what we actually observe?

And an ensemble again plays a role:

I'm just reiterating what Shankar and Griffiths say in their QM books. Griffiths acknowledges alternative interpretations and describes what I've said as the "orthodox" one. They say the ket or wavefunction represents the state of the system, period. They do not adorn this statement with the objections you've raised, namely that it doesn't represent "a physical system", or it's just "our knowledge",or it's really "an ensemble".

The expectation value, as Griffiths emphasizes above, is the "average of repeated measurements on an ensemble of identically prepared systems, not the average of repeated measurements on one and the same system". We are also told that observables have expectation values, and that these are repeated measurements on an ensemble of systems.

How, then, can an operator be what you say, the mathematical representation of measurement, when Griffiths says they represent observables and observables are not something actually observed at all but are still mathematical formulations?

the fact that a wavefunction describes probabilities does not preclude the wavefunction from describing an actual system.

Every identically-prepared system would have this same ket, and hence there is no need to refer to any 'statistical distribution' if we are simply talking about the complete quantum state. It is only the outcomes of measurements, and not the state itself, that must refer to a 'statistical ensemble', since outcomes would vary even in hypothetical identically-prepared systems.

So now a physical system is a bunch of probabilities? How is that a physical system? And in whatever way you wish to assert that it is, how can you say that it is a physical system as in classical physics?

I don't think the meaning of the statistical interpretation is the crux of the matter.

Griffiths interprets measurement "in the sense of the statistical interpretation". As measurement is the only way we get results, I'd say he thinks that's pretty important.

Clearly, then, it is implicit that Griffiths does indeed claim the wavefunction represents the state of a particle.

You talked about both a mathematical representation and physical reality. I've never denied that the mathematical representation exists. Of course it does. I've denied that we know how it corresponds to the physical system, unless we literally conflate the mathematical representation with the physical and assert that the system is a mathematical one. Griffiths clearly agrees we can represent the state mathematically. I've seen nothing in his books that says anything about what this means for a physical system other than that it has no properties until measurement, the "collapse" from measurement has no clear ontological status, and contrary to your interpretation of operators, they do not represent mathematically any measurement but are operators on observables, which themselves are mathematical functions yielding expectation values.

The main alternative to this interpretation

There are classes of main alternatives: hidden variables, relative state, and incompleteness of the physics itself.

 

[PolyHedral]

If a system is really, truly not deterministic, then how can we have complete knowledge of it? As you know, complex systems involve merely epistemic indeterminacy. Our mathematical representations are incomplete, but we could at least in principle describe the system precisely. If the system is ontologically indeterministic, then by definition we cannot do this: either our models are probabilistic, or we simply conflate the math with the physical system, calling equations and functions "physical".

Our "complete" knowledge of the system cannot give us a deterministic answer, therefore it is not complete?

 

[Mr Spinkles]

What is it Griffiths asserts that the system "does not have" below:

"a particular x-component of spin". What is it that Griffiths asserts the system does have, in that same quote? A state. And he further asserts we know the state of the system precisely: ""Are you telling me you don't know the true state of the particle?" On the contrary; I know precisely what the state of the particle is: X+" That absolutely agrees with my memory and that was my point.

Recall your disagreement with my stance on correspondence in classical vs. quantum physics:

The things you quoted are disconnected. The part you quoted from me wasn't responding to the part you quoted from you. Let's stick to the limited, manageable question we agreed on (or so I thought):

Does Griffiths agree that "QM postulates the existence of a ket which represents the complete state of a physical system, not just our knowledge of it, and this is the standard, most common way of formulating QM"?​

I think we have seen more than enough quotes from Griffiths to answer this question. Do you maintain that the answer is 'no'?

 

[LegionOnomaMoi]

"a particular x-component of spin". What is it that Griffiths asserts the system does have, in that same quote? A state.

I have always maintained that we describe the system as having at state. What I have denied is that we have complete knowledge of a physical system such that this:

Physical reality: some physical system (e.g. an electron)
Mathematical representation: ket

Physical reality: physical measurement of a system
Mathematical representation: operator "hitting" a ket

is meaningful. Not only is Griffiths quite clear that the operator in no way corresponds to a mathematical representation physical measurement or the ket to a physical reality, he specifically states that his answer to what the wave function represents is "Born's statistical interpretation of the wave function, which says that [it] gives the probability of finding" outcomes.

And his correspondence between the math and physical reality?
"All I am asserting is that there must be something- if only a list of possible outcomes of every possible experiment- associated with the system prior to measurement"

That's what the state is for Griffiths. Possible outcomes. He interprets measurement "in sense of the statistical interpretation"

Of the "state" he says "The wave function (&#936 represents the state of a particle (or system); particles do not in general possess specific dynamical properties (position, momentum, energy, angular momentum, etc.) until an act of measurement intervenes; the probability of getting a particular value in any experiment is determined by the statistical interpretation of &#936;"

At no point does he ever say anything about the a wave function or a ket representing complete knowledge of a physical system, but specifically states that it just somehow represents something.

The things you quoted are disconnected. The part you quoted from me wasn't responding to the part you quoted from you.

I was trying to clarify my position and why we started the discussion: I denied that there was the one-to-one correspondence between the mathematical representation of a physical system and the physical system in QM.

Let's stick to the limited, manageable question we agreed on (or so I thought):

Does Griffiths agree that "QM postulates the existence of a ket which represents the complete state of a physical system, not just our knowledge of it, and this is the standard, most common way of formulating QM"?
​

I think we have seen more than enough quotes from Griffiths to answer this question. Do you maintain that the answer is 'no'?

This is Griffiths answer:

If you want to interpret a physical system as being nothing more than this: "All I am asserting is that there must be something- if only a list of possible outcomes of every possible experiment- associated with the system prior to measurement", then Griffiths posits we have complete knowledge of a physical system. It's not physical in any sense that we are familiar with nor with any known correspondence with anything physical, and the only thing we have are probability functions to describe the physical states of this physical system. However, if you think that what Griffiths states above (the wave function is a probability function), means that the physical system is a mathematical function that yields outcomes, then yes we do have complete knowledge of a "physical system".

 

[LegionOnomaMoi]

Our "complete" knowledge of the system cannot give us a deterministic answer, therefore it is not complete?

There is a reason we distinguish between epistemic indeterminacy and ontological indeterminacy. One asserts that the dynamics of the system are in principle deterministic and thus we could in principle give a deterministic answer. Ontological indeterminacy asserts that there is no possible way we could ever determine outcomes.

So, if our knowledge is complete, and we can never determine outcomes, what do we have complete knowledge of? Possible outcomes.

Ergo, the fact that a wavefunction describes probabilities does not preclude the wavefunction from describing an actual system. In fact, under these assumptions, the latter cannot be accomplished without the former. And these are (essentially) the assumptions of the orthodox interpretation.

Depending on how one interprets "actual system", the above is, I believe, the orthodox interpretation: the wave function, state vector, ket, etc., describes probabilities. The problem then becomes what we refer to by "state of a physical system"

Suppose the actual system is really, truly not deterministic. Suppose the actual system's future condition/behavior (upon measurement, for example) is only constrained by probabilistic tendencies which the system possesses, and each tendency is stronger or weaker based on the system's state or condition at the moment it was measured Suppose that is the reality. Then, naturally, a mathematical representation (wavefunction) which describes that reality should also describe probabilities representing the system's tendencies.

If the system "is really, truly not deterministic", then how can we create a dichotomy like that below between the mathematical representation of the system and measurement and their respective physical instantiations:

To summarize:

Physical reality: some physical system (e.g. an electron)
Mathematical representation: ket

Physical reality: physical measurement of a system
Mathematical representation: operator "hitting" a ket

Granted our complete knowledge of the system and granted that the system is truly indeterministic and represented mathematically by probabilities, what does the state of this "physical system" in "physical reality" mean?

If it is a physical system, then complete knowledge of the state entails the ability (in principle) to completely predict the systems evolution. We frequently say complex systems are indeterministic, but we mean they are deterministic in principle, but in practice we are unable to determine their dynamics.

If we can't, even in principle, describe the evolution of a physical system, then either our knowledge of the system's state is incomplete, QM is incomplete, or (the most widely accepted view) the state of the system doesn't correspond to a physical system at all, merely a mathematical device to tell us given preparation a and measurement b, we'll get outcome x with y probability.

 

[idav]

There is a reason we distinguish between epistemic indeterminacy and ontological indeterminacy. One asserts that the dynamics of the system are in principle deterministic and thus we could in principle give a deterministic answer. Ontological indeterminacy asserts that there is no possible way we could ever determine outcomes.

So, if our knowledge is complete, and we can never determine outcomes, what do we have complete knowledge of? Possible outcomes.


Depending on how one interprets "actual system", the above is, I believe, the orthodox interpretation: the wave function, state vector, ket, etc., describes probabilities. The problem then becomes what we refer to by "state of a physical system"



If the system "is really, truly not deterministic", then how can we create a dichotomy like that below between the mathematical representation of the system and measurement and their respective physical instantiations:

Granted our complete knowledge of the system and granted that the system is truly indeterministic and represented mathematically by probabilities, what does the state of this "physical system" in "physical reality" mean?

If it is a physical system, then complete knowledge of the state entails the ability (in principle) to completely predict the systems evolution. We frequently say complex systems are indeterministic, but we mean they are deterministic in principle, but in practice we are unable to determine their dynamics.

If we can't, even in principle, describe the evolution of a physical system, then either our knowledge of the system's state is incomplete, QM is incomplete, or (the most widely accepted view) the state of the system doesn't correspond to a physical system at all, merely a mathematical device to tell us given preparation a and measurement b, we'll get outcome x with y probability.

There has to be something missing when it can't even predict a 50 50 chance variable. Of course the outcome will only be one or the other but doesn't mean wasn't determined to be so.

Using weak measurements they've shown that the photon does take a path so there is evidence it goes through only one slit on top of the evidence that it goes through both slits.

New &apos;Double Slit&apos; Experiment Skirts Uncertainty Principle: Scientific American

 

[PolyHedral]

If it is a physical system, then complete knowledge of the state entails the ability (in principle) to completely predict the systems evolution.

Only if physics is ontologically deterministic.

 

[LegionOnomaMoi]

Only if physics is ontologically deterministic.

Physics is a systematic approach, like all sciences, and thus can't be ontologically anything. However, if by "physics" you mean the entities, processes, systems, properties, subsystems (and their dynamics), etc. (i.e., the things which the physical sciences, among others, study), are ontologically indeterministic, then the entire scientific enterprise is baseless, unfounded, and will never be more than fantasies built on assumptions.

The core of the sciences is the belief that if we can repeatedly get the same results from empirical study, then we can safely assume there is some underlying determinism even if it is not fully understand yet (Newtonian mechanics describes things pretty well, but more importantly it wasn't "wrong", just not complete; that's why it's still used). Saying that physicists (those whose work is in various fields of natural sciences called physics because their chief concern is physical reality), are using an empirical system to understand processes that are always and will always be indeterministic no matter what is to assert that physicists have no basis for their approach. If all physical systems, physical processes, constituent parts of physical systems or entities, etc., are ontologically indeterministic, which means there is no reason to assume that any order exists at all.

Planck's constant, like all other constants, has no basis because it assumes determinism of a certain kind. This constant is not only vital within quantum physics, but to understanding that fuzzy divide between the apparently classical world and the quantum world. Remove it, and quantum physics has no basis. Whether the ket describes a physical system or not is of no importance because even if all quantum systems are ontologically indeterministic, the formalisms we use to describe them involve constants, like Planck's, as well as the devices and the assumptions (the theoretical framework) is all baseless. Whether QM is irreducibly statistical, incomplete, or whatever is completely meaningless because QM's formalism is probabilistic for a very specific reason. It is assumed that someone can always perform experiments that can be repeated in some way such that the regularity (even if probabilistic) allows us to generalize from the collective set of such experiments.

The origins of QM (the study of light and the contradictory findings of Young and Einstein) assume determinism. If we fire electrons or photons at a detection instrument with two-slits on our "splitting" screen open, we won't suddenly fail to get interference patterns. For almost ~100, we have continually demonstrated through empirical study that photons, depending on our experimental design, instruments, etc., can behave like classical waves or classical particles or neither (e.g., localized detections which reveal a wave-like interference pattern). If physics is indeterministic, there is no reason to think any of that matters. We can't even trust that our perceptual faculties are able to accurately observe the read-out from some device or the pattern from some detection screen, let alone that the actual instruments can be trusted.

Quantum systems may be ontologically indeterministic, but the only reason for thinking this is because we assume that physics itself is not. If it is, then continually demonstrating for a century that the same experiments show the same things means nothing, as the systems, the instruments, and even our perception of any part of any aspect of the experiment, are ontologically indeterministic, and therefore we have no baseline to determine anything at all.

 

[LegionOnomaMoi]

There has to be something missing when it can't even predict a 50 50 chance variable. Of course the outcome will only be one or the other but doesn't mean wasn't determined to be so.

A "variable" that has a 50-50 chance is by definition predictable. I don't see how that's relevant here.

Using weak measurements they've shown that the photon does take a path so there is evidence it goes through only one slit on top of the evidence that it goes through both slits.

That article was based on actual research and summarized for simplicity. One of the two studies it was based on was published in Science. This was what the researchers did:
"In our experiment, we sent an ensemble of single photons through a two-slit interferometer and performed a weak measurement on each photon to gain a small amount of information about its momentum, followed by a strong measurement that postselects the subensemble of photons arriving at a particular position. We used the polarization degree of freedom of the photons as a pointer that weakly couples to and measures the momentum of the photons. This weak momentum measurement does not appreciably disturb the system, and interference is still observed. The two measurements must be repeated on a large ensemble of particles in order to extract a useful amount of information about the system. From this set of measurements, we can determine the average momentum of the photons reaching any particular position in the image plane, and, by repeating this procedure in a series of planes, we can reconstruct trajectories over that range. In this sense, weak measurement finally allows us to speak about what happens to an ensemble of particles inside an interferometer"

They did not show anything about photons going through one or both slits. What they determined was that single particles don't behave according to any laws of physics nor any reality was are accustomed to. Rather, the study shows a method to illustrate how "particles" have no real trajectory in a more intuitive and empirically based method. That's what the researchers stated themselves.

 

[idav]

A "variable" that has a 50-50 chance is by definition predictable. I don't see how that's relevant here.



That article was based on actual research and summarized for simplicity. One of the two studies it was based on was published in Science. This was what the researchers did:
"In our experiment, we sent an ensemble of single photons through a two-slit interferometer and performed a weak measurement on each photon to gain a small amount of information about its momentum, followed by a strong measurement that postselects the subensemble of photons arriving at a particular position. We used the polarization degree of freedom of the photons as a pointer that weakly couples to and measures the momentum of the photons. This weak momentum measurement does not appreciably disturb the system, and interference is still observed. The two measurements must be repeated on a large ensemble of particles in order to extract a useful amount of information about the system. From this set of measurements, we can determine the average momentum of the photons reaching any particular position in the image plane, and, by repeating this procedure in a series of planes, we can reconstruct trajectories over that range. In this sense, weak measurement finally allows us to speak about what happens to an ensemble of particles inside an interferometer"

They did not show anything about photons going through one or both slits. What they determined was that single particles don't behave according to any laws of physics nor any reality was are accustomed to. Rather, the study shows a method to illustrate how "particles" have no real trajectory in a more intuitive and empirically based method. That's what the researchers stated themselves.

50 50 should be predictable but QM has it so you couldn't say whether it went through slot one or two. It didn't go through both slits just as if you tried to observe it going through one slit over the other. And you see the particle hit the screen passed the slits showing further that it only went through one slit. Further the article shows more evidence of the photon going through only one slit. Did we read the same article, lol, granted it was a repeated experiment but different in some ways to show what the title insinuates.

Edit: with what is being described, saying qm is indeterministic is like saying a 6 sided die is inderministic because we are unaware of how classical physics does that, must be truly random.

 

[LegionOnomaMoi]

50 50 should be predictable but QM has it so you couldn't say whether it went through slot one or two.

Predictable doesn't necessarily entail that we can predict one specific outcome accurately. If I flip a fair coin, my prediction need not be that I will get tails. I can say that my flip is predictable because I will either get heads or get tails and will not get anything else.

What I meant by "A "variable" that has a 50-50 chance is by definition predictable" is that you've given a probability function that I can use to predict whatever that variable will turn out to be. If you tell me I have a 50% chance of getting heads, and a 50% chance of getting tails, I can perfectly predict the outcome: it will be either heads or tails.

By contrast, if you give me a coin which isn't "fair" (both sides are the same, the coin is weighted to be more likely to never land on a side but to land on an edge, or you have a secret device that can determine how the coin lands), then I don't have my 50-50 chance and I don't know what chances I do have. That is unpredictability.

It didn't go through both slits

Why not?

just as if you tried to observe it going through one slit over the other.

If I observe it after photons as they hit a detection screen, I will get result A. If I observe them at the slits, I will get result B. In fact, if I observe them at the slits I am guaranteed a 50-50 chance. If I observe it at the screen, I am guaranteed to get results that cannot be explained by the photon going through one slit or the other. I can, however, guarantee that I will never ever see what is expected of classical particles.

And you see the particle hit the screen passed the slits showing further that it only went through one slit.

If you fired pellets at the splitting screen, some would bounce, some go through, and some go through but hit the sides of the slits sending the pellet off at weird angles. However, what will after you do this experiment over and over again (probably each time), is that you will see a cluster of spots where the pellet travelled through the slit and hit the detection screen. That is, you will find two spaces on the detection screen, both of which are where something that travelled through either slit would land, and corresponding to the spots that most of the pellets hit.

If you fire electrons through the same screen, the one place you won't ever see them land is where the pellets almost always do. They will arrive, one at a time, in a series of locations that indicate they didn't travel through either slit.

However, as they do show up on our detection screen, they must have travelled through the splitting screen. So they had to have travelled through the splitting screen, but our detection screen shows us the couldn't have travelled through either slit.

Further the article shows more evidence of the photon going through only one slit. Did we read the same article, lol, granted it was a repeated experiment but different in some ways to show what the title insinuates.

Your article was (like all such articles are) a description of an experiment that others did and which is described in full by the link to it at the bottom of your article (I believe you need access to the AAAS site to view it, as the AAAS publishes Science among other material). It is simplified, couched in non-technical language, and sensationalized to make it interesting because the actual study is the kind of thing that is not only largely unreadable to most people, but readable or not is boring.

As a result, articles that describe findings within research areas almost always make claims about the research that simply isn't true.

Edit: with what is being described, saying qm is indeterministic is like saying a 6 sided die is inderministic because we are unaware of how classical physics does that, must be truly random.

But we are aware of what is involved in getting a particular outcome of rolling die. We can construct a device in a room that we have customized (e.g., we've ensured that the "air" is made up of one or more gases in a specified way and that other than the movement of the "air" caused by rolling the die, the air doesn't move.

More importantly, we don't even need all of that to say with absolute certainty that we will get one of the 6 sides. We roll, we look, and we find out which one. In QM, looking at the die alters it. It's as if we have rolled the die, it has landed on one of the 6 sides, but merely by looking at the side it landed on makes the die shift (e.g., from the 5 side to the 2 side).

 

[PolyHedral]

Watch this, and be enlightened:
[youtube]dEaecUuEqfc[/youtube]
The Quantum Conspiracy: What Popularizers of QM Don't Want You to Know - YouTube

It explains quite neatly how measurement and entanglement are the same thing [ohai Legion], and also why the universe we see is consistent. Also, it is possible to partially measure things!