Einstein and "spooky actions"

[Mr Spinkles]

Hi Legion,

Thank you for your replies. I read your first reply, and I am really just interested in one question: is it wrong to say 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?

It seems evident to me, even after reading your quotations, that this statement is accurate.

For example, your first reference Theoretical Foundations of Quantum Information Processing and Communication is talking about the use of a density operator. I believe you are confused about the ontological status of the density operator. A density operator is a statistical approximation useful when the actual state of the system is unknown, due to imperfect control of the system's preparation. Imperfect control of course would be expected to happen during "information processing and communication" and therefore density operators become extremely useful. It is the QM analogue of the "macrostate" in classical statistical mechanics; it does not contradict the existence, in principle, of some (unknown) "microstate" (the aforementioned ket). In other words, your reference is talking about quantum statistics, not quantum mechanics. The former is useful when imperfect preparation leads to an ensemble of systems with different (unknown) states. It is not the actual state of each microscopic system, which is some pure state. The actual state of an individual system that is prepared would be some ket, unknown to the experimentalists, and itself some superposition of eigenkets. That is why the authors say: "We stress the fact that statistical operators are actually to be associated to the considered statistical preparation procedure, rather than to the system itself." Why do the authors say that? Because the "system itself" has a ket which would, if known, completely represent it, just as a volume of gas in classical mechanics has a microstate, even though it's convenient to describe it using a statistical ensemble of microstates. (I can't arrange every molecule in a gas in some exact microstate very easily, can you? It doesn't get any easier with quantum systems.)

It's important to realize that we are not obliged to use density operators, i.e. quantum statistics, in cases where we really do have precise measurement and filtering of individual particles, protected from random environmental perturbations. Then we can really know the actual state (ket) and do quantum mechanics. This should drive home the point that quantum statistics is not fundamental.

Your reference Beyond the Quantum confirms what I said. Namely "Presently, this view of quantum mechanics is the most widely accepted among physicists". They are correct that this most-common interpretation leaves open the puzzling issue of the nature of measurement. This unresolved issue is reiterated in another one of your quotations, Classical and Quantum Information. But I am not interested in resolving the measurement problem or pursuing a formulation of QM that goes beyond what those authors call "standard textbook interpretations". That is beyond my depth and it is distinct from the point I have been trying to make. My point is that, according to what your quote calls "standard textbook interpretations", 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". Yes they do, by assumption, according to what your source calls "the Copenhagen interpretation ... the most widely accepted [view] among physicists".

Your sources Beyond the Quantum and Decoherence and the quantum-to-classical transition note that it is possible QM (or its most-common interpretation) is incomplete, or incorrect. In other words, we may need hidden-variable theories. You also quoted Einstein and cited Smolin, who argue for this. But I'm confused because you also say you are "definitely not" talking about hidden-variables. To reiterate a point I've already made: obviously in that case the "state" does not actually describe the state of the system. That's trivial, and it is not the fundamental difference between QM and CM that (I thought) you made it out to be. CM is an incomplete theory, so we could argue the classical microstate never "really" describes the state of the system (it's just a good approximation sometimes). There's nothing particularly "quantum" about the fact that any wrong theory cannot "really" describe a system's state.

Nevertheless, it seems Einstein was wrong about QM and, as you say, "The ket represents the complete state, yes." ... E.g. if the ket only represented our knowledge of the system, then it seems to me almost by definition it would not represent the complete state of the system.

Sure. Unless you regard the relationship between the system and a physical system as meaningless, which has been continually repeated since Bohr and is advocated to day by e.g., The Frontiers Collection editor Mark P. Silverman.

What is the difference between "the system" and "a physical system"? I'm confused by this distinction, I was using them interchangeably.

The alternative, however, seems to me quite straightforward: the ket represents the complete state of a physical system, not just our knowledge of it.

Ok. Then the naturally question is how did we obtain this knowledge? By running repeat experiments on an ensemble of systems, each time altering them in non-trivial ways, yet somehow we wind up with a complete knowledge of a physical system? How?

First, what I said is a postulate of QM, whose predictions may or may not be falsified by experiment. The "knowledge" that this postulate forms the basis of an accurate theory, has to be distinguished from "knowledge" of the individual "pure" state (or ket) of a system. I am guessing you are wondering how we achieve the latter, without resorting to the approximation of quantum statistics (using density operators)? There are many ways, but one example: perform an ideal measurement on the position 'x' of an electron. Then the wavefunction Psi(x) of that individual electron immediately after measurement is a delta function centered at 'x'. The only remaining unknown component of the state is the spin. This can be known e.g. by using electrons coming from pion decays, all of which are in the singlet spin state. We now know, with 100% certainty, the complete wavefunction describing the individual electron we measured, at the moment we measured it. The next electron we measure may collapse at a different position 'y', and it will therefore have a different state. But we will know that electron's state, too. There is no need to refer to statistical ensembles or probabilities in terms of the state (or ket or wavefunction) of each individual system immediately after measurement, in cases like this.

I'm saying that it is a postulate that all the information is contained in the ket. The problem is that that could mean that system can be in infinitely many states, or that this is just a probability function.

Right, I acknowledge it is a fundamental postulate of QM. In fact it's my main point. I don't want to quibble over semantics, but I prefer to say the system cannot be in infinitely many states, i.e., infinitely many kets. It can only be in one state at a time, although that one state is some linear combination (superposition) of eigenkets. To me, it's really much simpler to consider "the state" of the system as being its ket, and to consider as a special case the particular kets where only one eigenket has non-zero coefficients (i.e. "eigenstates" are just special cases of "states"). What the state will collapse into after measurement--there are many possibilities. But it will always be in one state at a time, even if experimentalists choose to do quantum statistics (as a convenience). I don't see it as a "problem" that a particle's state, and I mean it's REAL state, can be some superposition of eigenstates. To me, the s-orbital of an electron around a nucleus (say) is a "state", the p-orbital is a different "state", and they are just as good as a position eigenstate or a classical position state. They're just weirder to picture in one's mind.

So I guess I should say, at any given moment, there is "a mathematical representation [which] has what you call a "one-to-one correspondence" with the system".

And I would say this is generally regarded as clearly not true.

But your own source disagrees: "it is important to emphasize that (according to our current knowledge) quantum states represent a complete description of a quantum system, i.e., the quantum state encapsulates all there is to say about the physical state of the system." How could an actual thing and a mathematical description possibly correspond more than that?

This is key. There is a ket which contains all the information of the system's state. But how? There isn't even a system for this to be true. The reason it is true is because the ket is treated as the system and the information part is again a postulate. There's no way to check for such a correspondence but we know that there cannot be one.

I'm confused by this. Are you asking about how in practice we could know the ket? See above.

 

[PolyHedral]

I clearly picked the wrong notation. c_i is supposed to stand for the basis vector representing the ith card of the deck.

 

[LegionOnomaMoi]

I am really just interested in one question:

That's an unanswerable question as it is phrased. To illustrate:

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.

I can't say anything about Shankar, but I can about Griffiths (emphasis & italics in original):

"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

your first reference Theoretical Foundations of Quantum Information Processing and Communication

Mea culpa. I forgot to cite until I had copied the whole thing, and then like an idiot I cited it as if the editors were authors. I did quote from that volume, but the paper in it was Vacchini's: "Covariant Mappings for the Description of Measurement, Dissipation and Decoherence in Quantum Mechanics".

A density operator is a statistical approximation

If so, how do these operators relate to Hermitian operators and observables?
This:

A density operator is a statistical approximation useful when the actual state of the system is unknown, due to imperfect control of the system's preparation.

runs into 2 problems.

First, "the actual state" is always unknown: "We stress the fact that statistical operators are actually to be associated to the considered statistical preparation procedure, rather than to the system itself. More precisely they describe a whole equivalence class of preparation procedures which all prepare the system in the same state, even though by means of quite different macroscopic apparatuses." (Vacchini)

The "ensemble interpretation" exists because the state vector is obtained through preparation of a collection of systems that is then called the system. There is never a time in which we know that "actual state" without statistics.

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)"

When Griffiths follows this with the realist position that "quantum mechanics is an incomplete theory" the part about knowing "everything quantum mechanics has to tell you about the system (to wit: its wave function), still you cannot determine all of its feature" and thus some "other information" is required, its important to note what is meant here by "knowing...about the system". The realist position isn't that we know everything, but that if we did (as we say we do in the orthodox view), then we shouldn't need "some other information, external to quantum mechanics".

But we do in the orthodox version, as noted on pp. 420-421, because we "create an attribute that was not there previously". So if we assume we know everything there is to know, then we must accept the fact that in some way this is only because we can relate the state vector to the measurement using an operator which is entirely secondary yet necessary to actually confirm anything at all about the system through measurement.

your reference is talking about quantum statistics, not quantum mechanics.

It isn't. Hence the title "Covariant Mappings for the Description of Measurement, Dissipation and Decoherence in Quantum Mechanics".

we are not obliged to use density operators

It's important to understand what Vaccini is saying (emphases added):
"The basic idea that we would like to convey or at least draw to the reader’s attention is that quantum mechanics is indeed naturally to be seen as a probability theory, significantly different from the classical one, rather than an extension of classical mechanics. Experiments at the microscopic level are of statistical nature in an essential way and their quantitative description asks for a probabilistic model which is the quantum one, emerged in the twenties and first thoroughly analyzed by von Neumann, actually before the foundations of classical probability theory were laid down by Kolmogorov in the thirties."

The only way you can say we know everything there is to know about the system is if you treat it as an abstract mathematical entity which tells you everything you know about probabilistic measurement outcomes.

in cases where we really do have precise measurement and filtering of individual particles

Which, in the orthodox position, didn't exist before measurement.

Your reference Beyond the Quantum confirms what I said

Part of what you said, as either
1) the wave function contains all there is to know, because there is no way in which it corresponds to any physical system except through an additional mathematical operator that relates the preparation to measurement
2) The wave function describes a physical system, but our knowledge is incomplete

You seem to want to conflate the two.

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).

What is the difference between "the system" and "a physical system"? I'm confused by this distinction, I was using them interchangeably.

They typically are, which is really annoying. However, if you still have Griffiths' textbook, note that the standard interpretation is fundamentally, completely, and in all other ways an irreducibly statistical interpretation. That's it. Mark. P Silverman's way of dismissing the various the realist interpretations is to say that "quantum mechanics...is an irreducibly statistical theory, albeit unlike any necessitated simply by ‘incomplete knowledge’, with nonlocal features inexplicable from the perspective of classical physics." (emphases added).

I am guessing you are wondering how we achieve the latter

More the "we", as I believe that your field uses particular methods (STM, quantum dots, nanotubes, etc.) as a physicist. I also know that if I haven't totally mistaken what kind of physics you do, then you don't use the same methods used in experimental particle physics (although there is some overlap) and you are closer to my position (I'm not using a LHC, and my use of imaging technology is far less related to QM than yours)

perform an ideal measurement on the position 'x' of an electron.

How? Given the trade off of the uncertainty principle, what constitutes and ideal measurement?

Because it seems you start with a measurement on a position that can only be known very well by ensuring you have no idea where it goes, and then state you use a dirac/ δ function (we use the same for neural spikes) such that the position

is a delta function centered at 'x'.

This means "the electron is somewhere." I'm sure you're simplifying here and while I appreciate it, it makes your explanation amount to an unknown position in some arbitrary region. In QM terms, "the scalar product between position eigenstates is given by the Dirac delta function. Since δ(x-x') vanishes for x /= x' and equals ∞ if x = x', the limits of integration can be arbitrary, provided that the point x is enclosed within the interval of integration"
Quantum mechanics 3rd. ed. (Graduate Textbooks in Physics) by Bes.


Basically, what you've said is you've used a delta function to determine a region within which the electron is, although as formulated that could be anywhere in the solar system.

There is no need to refer to statistical ensembles or probabilities ...in cases like this.

You already have. That's what the delta function is for. It's why Mathematical Concepts of Quantum Mechanics by Gustafson & Sigal says it's "not really a function, but a distribution." (p 24 italics original). If it were simply a function, then it would be infinitely high and infinitely thin. It's a probability function in that it is the limit of a given sequence of functions which allows one to specify an bounded region that gives a spike or curve if the argument falls in that interval and 0 everywhere else.

 

[LegionOnomaMoi]

I clearly picked the wrong notation. c_i is supposed to stand for the basis vector representing the ith card of the deck.

Which (as we learned from the Pixar movie Despicable Me) has both magnitude and direction. More importantly, a vector is a set of coordinates in some dimensional space. But you've given me nothing that allows me to know what space these vectors span, only that they are linearly independent. So they span some dimensional space such that they are closed under the standard operations that define bases, yet I still have no clue what that space is, and no clue what the value of any element in these vectors are. I still have nothing.

 

[PolyHedral]

Which (as we learned from the Pixar movie Despicable Me) has both magnitude and direction. More importantly, a vector is a set of coordinates in some dimensional space. But you've given me nothing that allows me to know what space these vectors span, only that they are linearly independent. So they span some dimensional space such that they are closed under the standard operations that define bases, yet I still have no clue what that space is, and no clue what the value of any element in these vectors are. I still have nothing.

Yes you do - you know that the space they're in is a Hilbert one (by context) and that it is 52-dimensional (because otherwise there couldn't be 52 distinct basis vectors.)

 

[marvek32]

It was Einstein who created the conversation in those conditions. It doesn't mean it is appropriate. Gong created two assumptions: 1) authenticity or reasoning, and 2) area. If his inequality is breached, which was shown experimentally consequently by Element, it indicates one of those two is incorrect.

 

[PolyHedral]

Also,

Which, in the orthodox position, didn't exist before measurement.

How on earth did anyone square this with the conservation laws?

 

[idav]

Also,
How on earth did anyone square this with the conservation laws?

Wouldn't MWI have issue with conservation when there can only be one correct answer?

 

[PolyHedral]

Wouldn't MWI have issue with conservation when there can only be one correct answer?

I don't think so, since there's only one set of particles/energies involved, they're just in a really complicated superposition.

 

[LegionOnomaMoi]

Yes you do - you know that the space they're in is a Hilbert one (by context) and that it is 52-dimensional (because otherwise there couldn't be 52 distinct basis vectors.)

1) Within 2D Euclidean space, there are infinitely many pairs of vectors that that span that space. Think about the properties of a vector space in general: any linear combination spans the space. In the Cartesian coordinate system, pick any two points such that you could draw 2 lines that each pass through the origin (point ( 0, 0 ) in the 2D case) and that pass through the points you picked. Now use these points to create 2 vectors drawn from the origin to the points. You have a vector space (or rather, you have a basis such that any linear combination spans the entire plane and Ax= 0 iff the only solution is the trivial one). There are infinitely many more. In fact, if you start with the standard basis for a the Cartesian plane, [0,1] & [1,0], then you can form pairs of vectors which span this space simply by multiplying either or both by an arbitrary scalar, and you'll always have a basis.

2) For 52D, create a matrix with 52 columns. Let all the columns be independent. You now have a basis. There are infinitely many more.

3) Hilbert space need only satisfy certain properties (just as it extends a real vector space to include these spaces, every finite Hilbert space is a Banach space but a Hilbert space requires a particular norm that a Banach space does not). We can satisfy these within Euclidean space such that the we've already shown an infinite number of bases for a 52D Hilbert space. Basically, we need only ensure our norm is Cauchy convergent (and the reals are) and that we have a vector space. However, we can do more, because we can add a complex conjugate such that once again we have a complete metric space.

4) There is no reason to assume that we are using a 52-dimensional space just because there are 52 cards. Off of the top of my head I'm not even sure how you'd do this. How would you define the inner product using a deck of cards? Even a quantum shuffled deck? What operations are defined and how? To have a basis I need to show that I have a set of vectors closed under scalar multiplication and vector addition. The fact that the scalars can all be 0s means I must ensure the null space is included. So how do I multiply an ace by any scalar c such that I can even begin to show I have a basis?

 

[PolyHedral]

1) If any set of basis vectors has 52 elements in it, as this one does, that implies the space is 52-dimensional.
2) So? We already know a basis exists.
3) I know that. This is the same Hilbert space you do QM in - complex projective function space.

4) There is no reason to assume that we are using a 52-dimensional space just because there are 52 cards. Off of the top of my head I'm not even sure how you'd do this. How would you define the inner product using a deck of cards?

The inner product is the same as it always is, a.b = <a|b>

What operations are defined and how?

The operation we care about at the moment is equivalent to flipping over the top card of the deck.

To have a basis I need to show that I have a set of vectors closed under scalar multiplication and vector addition. The fact that the scalars can all be 0s means I must ensure the null space is included. So how do I multiply an ace by any scalar c such that I can even begin to show I have a basis?

The space is projective, so all vectors pointing in the same direction are equivalent and the final answer by a scalar doesn't make a difference. Multiplying some component of a sum by a scalar just adjusts that component's contribution to the sum.

 

[LegionOnomaMoi]

1) If any set of basis vectors has 52 elements in it, as this one does, that implies the space is 52-dimensional.

No, it implies that you are redefining linear algebra and all extensions of it. You have a set of basis vectors with 52 elements. You state this requires the space be 52-dimensional. Each vector in this set, therefore, must have 52 entries. The only reason to have that number is because of the number of cards (52). Which means that each card is a vector in your set, as well as a linear combination of vectors that spans some (sub)space in H52. As that is impossible, because a vector in H52 can't be a vector space, you have to now develop an entirely new algebra and somehow work Hilbert space into it.

Spoiler: Why 52 cards doesn't entail 52 dimensional space

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.

Alternatively, I can use a 13 by 4 matrix, with entry Aij giving me the value of the card i, and the suit (1 for diamonds, 2 for spades, 3 for hearts, 4 for aces, or any permutation of this) given by j.

The point is that a set with 52 elements doesn't suddenly mean you are working in 52-dimensional space. And if you are saying we are working in this space, such that each card has a basis in H52, then you need to tell me what exactly I'm supposed to do with the 52 entries of the vector corresponding to the 2 of diamonds card.

2) So? We already know a basis exists.

No, we don't. We only know what you've insisted is required. Let's make this explicit. I pick up a queen of hearts. This card is represented by a vector [a1, a2, a3,...a52]' or (simply put) is a 52-tuple ordered sequence. Great. Now we can look here:

The inner product is the same as it always is, a.b = <a|b>

The inner product you've written above is a generalization, just like the a1 to a52 I represented above. Neither one tells me anything useful, as in my example every component or entry in the vector is a variable not a value. I can't say it is orthogonal to anything, because I have no idea what point in H52 space this vector corresponds to.

The operation we care about at the moment is equivalent to flipping over the top card of the deck.

I can't have a basis for flipping a card in R52 or H52. In order to have a basis B I need a linear combination of vectors that span B and these must be linearly independent vectors.

To illustrate: while a vector space has infinitely many bases, it has only one standard basis. For R2 that's e1 and e2 or [1,0]' and [0,1]'. For R3, that's [1,0,0]', [0,1,0]', and [0,0,1]'. Given any dimension, I can always form a vector space using the standard bases and scalar multiplication. Moreover, if I want 2 vectors to span a vector space in R3, then any linear combination of these vectors must stay within the same plane in R3. This generalizes to higher dimensions.

You are telling me that we're in H52 and that we have 52 distinct vector spaces. I draw a card. In order to have a basis, I need a subspace of H52 (which can but need not span H52, yet it cannot span any higher dimension), and I need vectors such that the inner/dot product (as we're in Hilbert space) of these vectors is 0. I also need to have the dot product corresponding to the drawn card be 0. There's just one tiny problem. I'd need another vector to have a dot product. I don't have one by flipping over the first card.

The space is projective, so all vectors pointing in the same direction are equivalent and the final answer by a scalar doesn't make a difference. Multiplying some component of a sum by a scalar just adjusts that component's contribution to the sum.

1) Spaces aren't "projective". A projection is a mapping from one space to another, typically represented by Ax=b where the matrix A is a linear transformation mapping x to b.
2) If you multiply by a scalar, you adjust all the 52 components or entries of the vector, which in this case would mean that we somehow take e.g., the ace of spades vector and multiply each and every one of its 52 entries by a scalar that "adjusts" these entries (via this multiplication) from whatever values we had to some others. As we have no idea what the values are, and (see the hidden portion above) this method of representing cards by vectors is needlessly complicated and involves a lot of arbitrary entries, this is a rather difficult procedure.
3) You didn't answer my question. This is your claim:

you know that the space they're in is a Hilbert one (by context) and that it is 52-dimensional (because otherwise there couldn't be 52 distinct basis vectors.)

How exactly do I get 52 distinct basis vectors if I have cards (each a vector space in H52), when a basis is defined in terms of a linear combination of vectors?
4) You want to think of the bases in these terms:

if we then draw a card from the deck, what's so impossible about the player superpositioning into 52 orthogonal states?

Quite apart from the fact that you cannot have any orthogonal state unless we have at least 2 vectors, you state:

so all vectors pointing in the same direction

ensuring we don't get orthogonal states.

 

[PolyHedral]

You have a set of basis vectors with 52 elements. You state this requires the space be 52-dimensional. Each vector in this set, therefore, must have 52 entries. The only reason to have that number is because of the number of cards (52). Which means that each card is a vector in your set, as well as a linear combination of vectors that spans some (sub)space in H52. As that is impossible, because a vector in H52 can't be a vector space, you have to now develop an entirely new algebra and somehow work Hilbert space into it.

That's a non-sequitor. A linear combination of vectors is a vector - I never claimed the elements were vector spaces in themselves.

And we have 52 basis vectors because we're representing a state that returns 52 different values when measured. I think that additionally implies that the vectors are orthonormal, but I'm not sure.

The point is that a set with 52 elements doesn't suddenly mean you are working in 52-dimensional space.

It does when you specifically define the set as a set of (orthonormal) basis vectors.

And if you are saying we are working in this space, such that each card has a basis in H52, then you need to tell me what exactly I'm supposed to do with the 52 entries of the vector corresponding to the 2 of diamonds card.

Why do the exact details of which basis vector encodes which card matter? The fact that the vectors span the space, are linearly independent, and are orthogonal is all we need. If it helps, assume that the ith card is represented by [a0, a1, ..., a52], where ai = 1 and all other components are 0.

The inner product you've written above is a generalization, just like the a1 to a52 I represented above. Neither one tells me anything useful, as in my example every component or entry in the vector is a variable not a value. I can't say it is orthogonal to anything, because I have no idea what point in H52 space this vector corresponds to.

Of course the inner product is a generalization - it's a function! You can't specify it by example!

1) Spaces aren't "projective".

Aren't they?

2) If you multiply by a scalar, you adjust all the 52 components or entries of the vector, which in this case would mean that we somehow take e.g., the ace of spades vector and multiply each and every one of its 52 entries by a scalar that "adjusts" these entries (via this multiplication) from whatever values we had to some others. As we have no idea what the values are, and (see the hidden portion above) this method of representing cards by vectors is needlessly complicated and involves a lot of arbitrary entries, this is a rather difficult procedure.

And you never need to do it, because scalar multiplication doesn't mean anything

Quite apart from the fact that you cannot have any orthogonal state unless we have at least 2 vectors, you state:

ensuring we don't get orthogonal states.

You are quote mining.

 

[LegionOnomaMoi]

That's a non-sequitor. A linear combination of vectors is a vector - I never claimed the elements were vector spaces in themselves.

Bases for Vector Spaces​

"A set is independent if, roughly speaking, there is no redundancy in the set: You can't "build" any vector in the set as a linear combination of the others. A set spans if you can "build everything" in the vector space as linear combinations of vectors in the set. Putting these two ideas together, a basis is an independent spanning set: A set with no redundancy out of which you can "build everything"." (source; emphasis added)

Linear Combinations, Basis, Span, and Independence

"We're interested is pinning down what it means for a vector space to have a basis, and that's described in terms of the concept of linear combination. Span and independence are two more related concepts." (source; emphasis added)

I was going to go with one more quote until I found this: Linear Algebra Review

You really might want to review your linear algebra.

And we have 52 basis vectors because we're representing a state that returns 52 different values when measured

You don't need a vector space for that. Just a vector.

I think that additionally implies that the vectors are orthonormal, but I'm not sure.

It doesn't. But we can get back to that when we've moved beyond what a vector space is.

It does when you specifically define the set as a set of (orthonormal) basis vectors.

If you are going to define a basis of some vector space, you have to know what that is. Orthonormal bases are collections of unit vectors that are orthogonal.

Why do the exact details of which basis vector encodes which card matter? The fact that the vectors span the space, are linearly independent, and are orthogonal is all we need.

It matters because of what you said earlier:

you know that the space they're in is a Hilbert one (by context) and that it is 52-dimensional (because otherwise there couldn't be 52 distinct basis vectors.)

The operation we care about at the moment is equivalent to flipping over the top card of the deck.

This is why the Dirac notation is such a pain for those of us who were used to much more precise notational schemata. In fact, even without the bra-ket notation, the textbooks are imprecise. It's the exception, rather than the rule, to find a text (graduate level, undergrad, reference materials for grads and post-docs) that goes beyond noting how some vector &#936; can represent a linear combination of vectors and then use this "vector" as a basis vector because it is a sum, when in reality this is a simplified and imprecise (or inaccurate) statement. A "basis vector" in that sense isn't actually a vector. It's shorthand: "The same vector &#936; can be expressed as the sum of two different systems of basis vectors". Dirac notation makes it worse: "Suppose we have basis vector |i>, analogous to the &#710;ei, which form a complete orthonormal set" (source). A basis vector forms a set? Of what? Well, of basis vectors. If you look at the source and see that the "orthonormality" and "completeness" criteria require
1) Summation of numerous vectors
2) Getting a matrix

Only it isn't expressed as a matrix usually, and Dirac notation throws mathematical precision out the window. Note how write after 11, the author writes what I quoted above, and then states "Then any vector |V> may be expanded in this basis..." followed by a similar summation of vectors to create a basis. Finally, as an extra treat (emphasis in orginal) "we can use the eigenvectors of a hermitian operator for our basis vectors. Matrices become operators in this language"

The reason that "Functions can be considered to be vectors in an infinite dimensional space" when vectors aren't functions is because matrices are functions (that's why, like functions, adding two of them is easy but multiplying to is counter-intuitive and takes practice). And the reason "wave functions can be thought of as vectors in this space" is because once again we aren't dealing with vectors but matrices (which are made up of vectors, but the reason we call vectors that act as function matrices is because that was the whole point of matrix notation in the first place: making systems of equations compact without any loss of information).

You can't specify it by example!

Of course you can. Not only that, it's part of the problem sets in textbooks. You have a system of cards and you know the types of cards, the suits, everything you'd need so that you could be specific.

Aren't they?

No. Projective geometry, in which you do have a projective space, is not what you are referring to, and even here a projective space P(V) is over a left vector space.

When you can explain to me how you want to work Desarguesian projective spaces into quantum shuffling, then we'll deal with projective spaces, algebraic geometry, and the worst reason I ever had for studying a subject: I get bored in art museums, as I lack the ability to appreciate most art. However, looking at it from the point of view of projective geometry makes it more interesting to me, and projective geometry began as an applied "art" math:

You are quote mining.

Recall this?

Well, if Matlab is what you're thinking of, no wonder you're underestimating the power of AI.

MATLAB stands for Matrix Lab. It's designed to make using vectors and matrices, which are the backbone of not only neural networks and other machine learning but multivariate statistics. The people who use it tend to want something more powerful than SAS when it comes to models and more user-friendly than R (which I prefer).

This is what I do. Factor analysis, PCA, feature extraction, and basically all statistical or mathematical procedures designed to reduce dimensionality or extract from large data sets particular relevant features is all done using vectors and matrices. If you think I'm quote-mining, aside from when I actually use quotations, then show me whence my quote came. Otherwise, we're going to end up talking about particular types of abstract algebra which partially become important in relativistic physics (e.g., affine spaces) just because you need a refresher in basic linear algebra.

 

[Reptillian]

Lol, I leave RF for a month and come back to find y'all discussing playing cards as a potential vector space. Apparently I must have missed something. Anyone care to summarize what's being discussed for me so I don't have to read through tens of pages of posts?

By the way the dimension of a vector space is equal to the number of basis vectors for the space and basis vectors don't have to be orthonormal (that's why we have the Gram-Schmidt orthonormalization procedure)...they just have to have components which are orthogonal.

 

[Mr Spinkles]

Hi Reptillian,

They were debating the multiple worlds interpretation of QM. I objected to Legion's claim that, unlike in classical mechanics, the QM state (namely the ket or wavefunction) does not represent the state of a physical system.

 

[PolyHedral]

Lol, I leave RF for a month and come back to find y'all discussing playing cards as a potential vector space. Apparently I must have missed something. Anyone care to summarize what's being discussed for me so I don't have to read through tens of pages of posts?

By the way the dimension of a vector space is equal to the number of basis vectors for the space and basis vectors don't have to be orthonormal (that's why we have the Gram-Schmidt orthonormalization procedure)...they just have to have components which are orthogonal.

The reason we're talking about cards as vector spaces is because I was trying to model picking a card off of a deck as a quantum mechanical operation.

 

[LegionOnomaMoi]

Hi Reptillian,

They were debating the multiple worlds interpretation of QM. I objected to Legion's claim that, unlike in classical mechanics, the QM state (namely the ket or wavefunction) does not represent the state of a physical system.

Actually, centuries ago when PolyHedral and I first started to disagree about QM, our debate was pretty much the same as what you objected (and what I think you both do still). It was a long time before I realized that he supported the MWI. Usually, we never quite got past the point you and I did and (as is clear from our quantum shuffling discussion in which we can't even agree on how to pick up a card to start quantum poker) that's still the central issue. Which was why I was glad you stepped in before.

I was sort of hoping you could point out what I got wrong or misunderstood in my latest reply to you, as neither I nor PolyHedral are physicists. Also, you mentioned your textbooks, one of which I have too. I thought that gave us an unprecedented starting point, or basis (or basis vector), to build a discussion on. Finally, I was able to use specific lines that you've read too, and supplied my interpretation of them. Rather than explaining, or quoting from something you haven't read and thus are forced to speculate what the context is, we finally had a text we've both read and you have the context. Thus any quote I've interpreted incorrectly is one you can correct otherwise explain.

PolyHedral and I were keeping the thread alive with a card game until you could jump in again and bring it back to how kets or vector states represent the physical system if we know everything there is to know already yet require an operator to connect the prepared state to the measured outcome. In particular, that the Copenhagen interpretation or orthodox interpretation is fundamentally statistical (as I understand it, anyway), and thus the "physical system" description doesn't actually correspond in a way we know to any physical system but does (in the orthodox view) correspond to predictable outcomes.

I'm worried that you've decided my view is so off I haven't so much as a vector basis to stand on, and so you've given up trying to explain what I've misunderstood.

 

[LegionOnomaMoi]

For once I think I may have done it: "quote-mined" something that can settle 1 issue:

"For the purposes of solving the electronic Schrödinger equation on a computer, it is very convenient to turn everything into linear algebra. We can represent the wavefunctions

as vectors:

where

is called a "state vector,"

are the expansion coefficients (which may be complex), and

are fixed "basis" vectors." (source)

 

[PolyHedral]

You really might want to review your linear algebra.

Possibly, but nothing you quoted appear to contradict me.

You don't need a vector space for that. Just a vector.

If we just wanted to represent each card as "a vector," sure, we could just have a 1D vector in the space [0,52], but that doesn't have the properties we want. The framework of quantum mechanics says that the vectors representing our eigenstates (of which there are 52, because there are 52 eigenvalues, i.e. measurable outcomes) should be basis vectors, and thus we need the vectors to be 52D.

If you are going to define a basis of some vector space, you have to know what that is. Orthonormal bases are collections of unit vectors that are orthogonal.

Do you mean what basis we choose, or what the space is? The space is C^53 with the equivalence class that, for some entry A, A=xA for any complex value x.

This is why the Dirac notation is such a pain for those of us who were used to much more precise notational schemata.

Do the CS thing and invent your own notation.

Actually, can we do that anyway and strongly-type what we're saying?

In fact, even without the bra-ket notation, the textbooks are imprecise. It's the exception, rather than the rule, to find a text (graduate level, undergrad, reference materials for grads and post-docs) that goes beyond noting how some vector &#936; can represent a linear combination of vectors and then use this "vector" as a basis vector because it is a sum, when in reality this is a simplified and imprecise (or inaccurate) statement.

I don't follow. A linear combination of vectors is, by the definition of what a vector is, also a vector.

A "basis vector" in that sense isn't actually a vector. It's shorthand: "The same vector &#936; can be expressed as the sum of two different systems of basis vectors".

I read that as there existing some set of coefficients to represent &#936; in one basis, and there also exists another (implicitly distinct) set to represent it in another basis. Of course that's true.

The reason that "Functions can be considered to be vectors in an infinite dimensional space" when vectors aren't functions is because matrices are functions (that's why, like functions, adding two of them is easy but multiplying to is counter-intuitive and takes practice).

For deriding a lack of rigiour, that's rather unrigorous.

The reason (IMO) that, e.g. the set of polynomials is an infinite dimensional function space is that this works:

And the reason "wave functions can be thought of as vectors in this space" is because once again we aren't dealing with vectors but matrices (which are made up of vectors, but the reason we call vectors that act as function matrices is because that was the whole point of matrix notation in the first place: making systems of equations compact without any loss of information).

Wave functions can be thought of as vectors because the field we are building the vectors out of is the square integrable functions.

Of course you can. Not only that, it's part of the problem sets in textbooks. You have a system of cards and you know the types of cards, the suits, everything you'd need so that you could be specific.

So, how do you specify a function over an infinite space by example?

When you can explain to me how you want to work Desarguesian projective spaces into quantum shuffling, then we'll deal with projective spaces, algebraic geometry, and the worst reason I ever had for studying a subject: I get bored in art museums, as I lack the ability to appreciate most art. However, looking at it from the point of view of projective geometry makes it more interesting to me, and projective geometry began as an applied "art" math:

We are dealing with a projective space because the absolute length of the state vector is not physically meaningful, therefore all scalar multiples of a given vector are equivalent. That's what "projective space" means.

If you think I'm quote-mining, aside from when I actually use quotations, then show me whence my quote came.

I think you're quote mining because you're using quotes in a completely out of context fashion.
e.g.

4) You want to think of the bases in these terms:
if we then draw a card from the deck, what's so impossible about the player superpositioning into 52 orthogonal states?

Quite apart from the fact that you cannot have any orthogonal state unless we have at least 2 vectors, you state:
Quote:
so all vectors pointing in the same direction
ensuring we don't get orthogonal states.

The full sentence reads,
The space is projective, so all vectors pointing in the same direction are equivalent and the final answer by a scalar doesn't make a difference.
It has nothing to do with basis vectors, or superpositions as you seem to imply.