Ch 3: Why do we need a Hilbert Space? | Maths of Quantum Mechanics 

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BRA. I ALMOST FELL ASLEEP AND NOW I CANT CLOSE MY EYES. Thank you. But damn you. No, for real. Cool, never realized that. Have a nice day

It's also the entirely wrong explanation. The necessity for Hilbert spaces in physics does not come from the approximation theory of L_2 functions. It comes from Kolmogorov's axioms for ensembles. The creator of this video keeps barking up an entirely wrong tree here.

Except that there is nothing wrong with that limit. It is not physics is all, you'd impose the correct asymptotics to a computation of the amplitudes, so would never get an exponential in QM, except as a mistake or a bad approximation. Imposing correct conservation law asymptotics might still yield a transcendental wave function, it just won't be one blowing up at spatial infinity.

@@Achrononmaster I think you aren't getting the Mathematical concepts being explained / described. The reality of the limit has nothing to do with Physics at all, it stands in its own Mathematical right. As for your assertion that there are no exponentials in QM then that is quite interesting and merits a discussion as to what constitutes a necessary feature of any formulation of Quantum Theory. If GR and QM are ever to get closer then such dogma will have to be addressed. Or are there no exponentials in GR either?

I think the addition of the inner product condition makes the Hilbert space Cauchy complete. We only added one thing (as I understand).

- Functions that can be expressed as a power series (we call these analytic functions) do indeed form a vector space. However, it is a different vector space than that of polynomials. Also note that if we take an infinite sum of exponentials (which are vectors in a vector space of analytic functions), we might end up with a step function, which is not analytic, and in fact not even continuous! Then by taking a potentially infinite sum of step functions, we might end up with a different vector space still. Eventually the space we're working with looses all the nice properties that the original vector space started with. Not to mention that if we allow infinite sums, it's not even obvious what space we're actually working with. - Note that Caushy completeness uses a language of "converging". But what does converging even mean? We need some notion of distance on our vector space in order to even say that two vectors are close together, or that a sequence of vectors is approaching some other vector. Inner product gives us that notion of distance. Note that for R^n we define a length of a vector with a dot product, which is just a special case of inner product.

I just got into college and while watching this I was like “who’s Cauchy”

*completely

why, my Dear?..🍷😊

The problem is he never defined the metric induced by the inner product, a metric space is complete iff every Cauchy sequence converges( look up these terms).

Here is my (verbose) motivation for completeness in general, as well as how it manifests in terms of the relevant spaces to these videos. The starting point is to consider the rational numbers, which have many "gaps" - for example, sqrt(2) is irrational, but it is a real number. The real numbers fill the gaps that the rationals possess, lending to their pictorial manifestation as an infinite, unbroken line. How do we formalize the fact that the rationals have gaps, but the reals don't? Let's consider the sequence (1, 1.4, 1.41, 1.414,...) of rational numbers which approaches sqrt(2) in decimal expansion. Clearly this can't converge within the rationals - we know sqrt(2) is irrational! Yet its terms are just (some rational)/10^k, and the terms get arbitrarily close to one another: they dance ever-closer to sqrt(2) without ever converging to it as the limit. This phenomenon does not happen in the real numbers; sequences whose terms get arbitrarily close to one another will always have each of its terms converge to a limit, because there are no holes to obstruct this from happening. Let us call a sequence whose terms get arbitrarily close to one another Cauchy, and think of convergent sequences as those whose terms eventually get arbitrarily close to a limit. Note that every convergent sequence is Cauchy, since terms that get close to a certain point will necessarily grow closer to one another. In the rationals, the converse doesn't hold, but in the reals, it does: a space is defined to be complete when every Cauchy sequence within it converges. (In fact, one can construct R from Q this way: you force the Cauchy sequences to converge, and impose an equivalence relation on the sequences that converge to the same value; the collection of these limits modulo that identification comprises the real numbers) Implicit in our discussion is the very notion of distance. The general notion of distance comes down to defining something called a metric (or distance function) on what is a priori a set, which takes as its entries pairs of points in the space, returning a nonnegative real number that we can think of as the distance between said points. It must satisfy three axioms that are more intuitively familiar to us than we may at first realize. First of all, the distance between two points is zero if and only if the two points are actually the same (distance is only zero from a point to itself), positive otherwise. Furthermore, it is symmetric in its arguments: the distance from point x to point y should be the same as that measured from point y to point x. Finally, it satisfies a triangle inequality. Spaces equipped with a metric are referred to as metric spaces: for example, R with its absolute value d(x, y) = |x - y|, which Q inherits (and with respect to which we conducted our first paragraph's discussion). It is with respect to a metric that notions like limits and convergence take root. Now, let's shift our attention to vector spaces, which are the natural domains of definition for quantum-mechanical processes as explained in the first video (and are fundamental in mathematics much more broadly). To start performing analysis on algebraic structures such as these, we would like to have some notion of "size" or "distance" as before. The answer comes in defining a norm ||•|| on a vector space: a function which takes in vectors and returns real numbers as "lengths", satisfying positive-definiteness, homogeneity (pulling out scalar multiples), and an analogous triangle inequality which states that the sum of norms bounds the norm of sums. This natural identification above (and the heuristic that size gives rise to distance) is not in vain: a norm on a vector space yields a metric, defined by d(v, w) = ||v - w||. Now we can start considering analytic ideas like convergence and completeness on normed vector spaces. Not all normed vector spaces are complete, and we're often interested in the ones that are. These are termed Banach spaces, and their study furnishes many interesting and beautiful results. Notable Banach spaces that are not Euclidean (e.g. R^n or C^n) include l^p and L^p of p-summable/integrable functions on a measure space respectively. We want to specialize further. As the next video explains, we are also looking for notions of angle and orthogonality to abstract from R^n to the spaces of functions we'll want to consider. The answer here comes from inner products which in fact always produces a norm (take ||v||^2 = ) and the special kinds of Banach spaces from which the norm comes from an inner product are called Hilbert spaces. It is a result from this theory that every finite-dimensional inner product space actually turns out to be Hilbert (we get completeness "for free"), but we aren't interested in finite-dimensional vector spaces here due to the physical constraints that this video elaborates on. There are incredible theorems that display equivalences between specific Hilbert spaces (actually, the "only" infinite-dimensional Hilbert space is l^2 of square-summable sequences, up to isometric isomorphism) but quantum mechanics is largely concerned with the infinite-dimensional space L^2 of square-integrable functions, which is the only L^p space that carries Hilbert space structure (reading more on these in-depth might involve a bit of background in measure theory). It also lays down the groundwork for Fourier analysis, has interpretations in probability theory, applications in partial differential equations, and much more.

Hello! Thanks for watching the video. The version that I previously uploaded was a teaser for the series. I’ve since finished up the series, so I decided to start fresh with episode one, while fixing a few typos with my previous upload. Thanks for sticking around since my first few uploads! -QuantumSense

Hello, thanks for watching, and being an early supporter! And yes, I would agree in saying that a Hilbert space is more restrictive than a vector space, since we only allow vector spaces that already include their limit points. You can retroactively add the limit points to a vector space, however, in what is known as the "closure" of the set. This changes your vector space though, and is more suited to be understood in the context of topology. And yes, I got your email! I haven't had the time to sit down and think through the answer, but once I do, I will let you know. -QuantumSense

@@quantumsensechannel Thank you very much for your reply!

@@narfwhals7843 A Hilbert space is a special type of inner product space. All Hilbert spaces are inner product spaces. But only some inner product spaces are Hilbert spaces.

you are correct. e^x is polynomial.

Hello! This is a really good question! You say that it seems like a matter of language as opposed to math, and that’s sort of the key. Remember that math is partially the art of precise language; what distinguishes a polynomial from an arbitrary function is how we define a polynomial. From the definition, we can derive several properties that *must* hold true of any polynomial. For example, it’s easy to show that for every polynomial, there exists some integer N such that after taking the Nth derivative, you get zero. e^x satisfies no such property. Likewise, you can prove that any non constant polynomial approaches either + or - infinity as x goes to +- infinity. e^x approaches zero as we go to -infinity, so this property is broken. So yes, in a way it is a matter of language, but we need to be precise with our language in order to be consistent (otherwise, we could say that any function with a Taylor series is a polynomial…which isn’t a very good definition for a polynomial). Hopefully this partially addressed your question! -QuantumSense

@@quantumsensechannel The Nth derivative rule pretty cleanly rules out any polynomial of infinite degree. The divergence rule already has an exception, so it's not really as convincing as a rule rather than just a trend. As I mentioned in my initial comment, allowing infinite polynomials still excludes many functions, namely those with finite radii of convergence, like the aforementioned logarithm. The inverse function also doesn't have a taylor series. _Most_ inverses of polynomials, infinite or otherwise, probably aren't polynomials. Luckily the definition of a vector space doesn't say anything about needing an inverse.

@@angeldude101 I can't post a link in the comments, but search for "Why is the exponential function not in the subspace of all polynomials?" on Math Stackexchange. Your doubt seems to be addressed there. In the video, the analogy of vectors inside the box and their limit being on the edge is relevant. We need to invoke topology at some point to make complete sense of it, as far as I understand.

@@angeldude101 The major problem with allowing "polynomials with infinitely many non-zero coefficients" is that, as you yourself stated, this requires a lot of additional structure: It only really makes sense, when these polynomial approximations converge, which requires a notion of convergence, which in itself requires a metric, which induces a topology and so on. This space can be defined (it is usually called a closure of the underlying set), but it is usually not thought of as the set itself, since polynomials can exist in a space without all this extram machinery. Hope this helps

We are not barring out functions which are not polynomials. We are getting rid of the vector space of polynomials altogether, and replacing it with a better vector space.

A Hilbert space isn't always (almost never) a differentiable manifold since the homemorphism to R^n is ill-defined for a lot of cases. For example, the Hilbert space L^2(R), which is the usual function space used with position and momentum operators, isn't a differentiable manifold.

@@williammendez5209 this Hilbert space, when you get just a piece, but of course, if you go to infinite, no good. It's the application for quantum. We get those fields, not the whole space, and there are boundaries of course, the most important one is H=T+U (or V) =1, invariant, 100% energy. Pseudoriemmanian manifaltigkeit on each point of the fields, but with QFT we have those deltas, and no way they get to infinity. They have a limit. Ok the rest would be non-linearities... Anyway, we get the trajectory in QFT in configuration space, so a distribution trying to take the whole fields ... 🤔 😬 🖖🤓 So there's basically a little sample of adjacent manifolds, Pseudoriemmanian, which are in this quantum scenario our pieces of fields, all converges inside the same "subspace" any linear combo.

Also, I would like to know why these spaces must be projective. Thank you in advance

Perhaps what you are trying to say is that countable infinity is a cardinality. It isn't a real number, or integer, or anything similar, because all of these are more strict than simply having a total order, or even a well order (as you say, they are all comparable, which I would interpret as a total order or a well order). Rather, all of these "numbers" have useful binary operations like addition and multiplication. These aren't just arbitrary either; addition is an abelian group operation, and multiplication distributes over addition. Infinity does not naturally fit into these operations, unless you conceed to essentially turn the ring/field into a polynomial ring over infinity (in which case, infinity may aswell be "X"), and the meaning of "infinity" is essentially lost to just being an extra element.

@@stanleydodds9 “Infinity is all numbers simultaneously.” Proof: sin(x) has unlimited domain, yet range is [-1,1]. Therefore, x is any number then sin(x) can also be found using only one period, and therefore infinity is always possible in sin(x) period also. Mapping any finite number onto the circumference is thought possible, though further proof is needed (I can attest). Therefore, given some finite number then infinity is indistinct and possible, using sin(x). Follows that infinity is indistinct with all finite numbers (et blau, above) and is uniquely itself. It is therefore impossible to measure infinity as better one finite number, or another, yet infinity is approximated, understood, a value like other numbers, real though now imagined in specific value. I love infinity, and use it all the time in math and science. “Prove all uniqueness of values: create all numbers. Infinity is all numbers, so the full-set of numbers is exactly infinity long. From 0, infinity is therefore all numbers, each only unique. QEDunum.” This also shows all numbers are ordered by infinity and only. Without infinite elements then some are missing! Nice discussion! Edit: I’m leaving out a discussion of how to multiply and divide using infinity, but the math is clearly seen in a famous physics equation: E=Mc^2. Using the universally constant nature of ‘c’ as a sign it implies infinity (since only infinity means everything, without changing value): multiplication must be “finite = finite * infinity” (leaving off the square for discussion), division must be “finite / finite = infinity”. Take care!

Hello! Thank you for watching, and the kind words. And to answer your question, polynomials are indeed closed under addition. Adding two polynomials together will always give you another polynomial. Is there some example that seems to be contradictory? Let me know, and I can try and clear it up! -QuantumSense

​@@quantumsensechannel thanku for ur response my confusion generated while i was digging in the propreties of vector spaces and i found those affirmations so i decided to ask chat gpt he firt was affirmative about polynomials being vector spaces but then he said the opposit i am sorry is there a way i could send u the screenshots of what he said The main concern is that adding polynomials of the same degree could give u an other polynomyal of a different degree and m not sure if that violates the closure axiom

Hello, Ah, I see. AI may be quite intelligent, but we should not rely on it as the sole source of mathematical understanding! The set we are considering is the set of polynomials with real number coefficients, we make no mention of the degree. So adding two polynomials with real coefficients, will always give you another polynomial with real coefficients, hence closure under addition. ChatGPT may be considering the set of all polynomials of degree n. This, indeed, would NOT be a vector space, since adding two polynomials of degree n could give you a polynomial of different degree (e.g. (1+x^5) + (1-x^5) = 2). So ChatGPT seems to be considering a different set than the one I mention in the video. Let me know if this doesn’t clear it up! -QuantumSense

@@quantumsensechannel i am really greatfull for ur response it cleared al my confusion your videos really helped me and you helped understad this concepts further i honestly hope you succeed i actully started studying qm for curiosity from a young age i had a solid grasp of the abstract concepts but i needed to construct mathematical intuition and your videos are perfect thank you so much 😊

Frankly, I've already spent waaay too much time trying to formulate a precise and accurate answer to your question. I'll cheat a bit saying this: Boundaries are not an easy thing to define, since this requires the notion of a metric on the given space (which I believe is why Hilbert spaces necessarily require an inner product, inducing the metric). Given a metric on a space: If that space is compact, meaning that it is bounded (there is a maximum to its norm) and it is closed (meaning that it contains its boundary) then it is Cauchy closed. But the other way is not necessarily true: A space can be Cauchy complete without being compact. The set of polynomials is not considered Cauchy complete, because there exist sequences of elements of the set (e.g. Taylor Polynomials) which converge to things that are not members of the set themselves (e.g. the exponential function). The functions that can be expressed as Taylor series can therefore be thought of as some sort of "boundary" of the set of polynomials, though I would not be able to show that this forms the complete boundary.

What may help here is the analogy with the rationals. They form a Q-vector space, but they are not Cauchy complete. The Cauchy-completion of the rationals is the Reals, i.e. these fill up the gaps between the rationals. So you could view the irrationals as the boundary of the rationals, but I think it's better to think of these as filling the "holes" in the rationals. In terms of polynomials, the Stone-Weierstrass theorem tells you that every continous function on a closed interval can be uniformly approximated by polynomials. Stated diifferently, the continous functions on a closed interval are the Cauchy-completion of the polynomials, for the sup norm (uniform norm, i.e. uniform convergence). (The sup norm doesn't come from an inner product so we don't have a Hilbert space, but the idea is analogous)

That isn't something that can be explained with a single video.

I also finished with that question in mind but couldn't find the answer on Google

Hello, thank you for the kind words! Any music I use will always be listed in the description along with a link to the artist’s page. -QuantumSense

Hello! Thank you so much for watching the video. All animations were created in python using Manim. In the description, there is a link to the community page for Manim. -QuantumSense