Fun with Deltas and ##L^2##: Physicists vs Mathematicians

[rubi]

It depends what functions you allow.

The space of square-integrable functions is a well-defined object and it is definitely not a Hilbert space. There's no denying, this is a lost cause.

What you are trying to do is to define another space with fewer functions. However, you will only get a Hilbert space, if and only if you are able to pick exactly one representant from each equivalence class, which certainly requires the axiom of choice, so you will never be able to tell me what functions are actually contained in your space. Your proposal to pick continuous representants must fail, because most $L^2$ equivalence classes don't even have one. There is a reason for why we use equivalence classes. It's not to scare physicists, but rather because it is the simplest and most practical way to arrive at a Hilbert space. If there was a reasonable way to define a choice function on $L^2$, then mathematicians would already have done so.

Maybe I'm wrong? Please give me an example of a physically meaningful function with zero (square-integrable) norm, which isn't 0.

It's not enough to choose a function from the $0$ equivalence class. You must provide a choice for every equivalence class. I'm waiting for your proposal for a choice function. Until then, please don't make such unfounded claims anymore.

 

[stevendaryl]

It's not enough to choose a function from the $0$ equivalence class. You must provide a choice for every equivalence class. I'm waiting for your proposal for a choice function. Until then, please don't make such unfounded claims anymore.

I'm not sure who this is addressed to. But there actually is a sort-of canonical choice.

If $f(x)$ is square-integrable, then let $\tilde{f}(k)$ be the Fourier transform: $\tilde{f}(k) = \frac{1}{2\pi} \int dx e^{-ikx} f(x)$. Then we let $\bar{f}(x)$ be the Fourier transform of that: $\bar{f}(x) = \int dk e^{ikx} \tilde{f}(k)$. I think it'll be true that if:

If $\int dx |f(x) - g(x)|^2 = 0$, then $\bar{f}(x) = \bar{g}(x)$

I'm not absolutely sure that $f(x)$ being square-integrable implies that $\bar{f}(x)$ is well-defined, though.

 

[rubi]

I'm not sure who this is addressed to.

It was adressed to secur

But there actually is a sort-of canonical choice.

If $f(x)$ is square-integrable, then let $\tilde{f}(k)$ be the Fourier transform: $\tilde{f}(k) = \frac{1}{2\pi} \int dx e^{-ikx} f(x)$. Then we let $\bar{f}(x)$ be the Fourier transform of that: $\bar{f}(x) = \int dk e^{ikx} \tilde{f}(k)$. I think it'll be true that if:

If $\int dx |f(x) - g(x)|^2 = 0$, then $\bar{f}(x) = \bar{g}(x)$

This equality only holds almost everywhere. The definition of the Fourier transform on $L^2$ cannot easily be defined (see Reed&Simon II for details). One needs to perform a limit in the $L^2$ norm and it converges only to an equivalence class again, so you are still faced with the choice of having to pick a representant of the equivalence class.

I'm not absolutely sure that $f(x)$ being square-integrable implies that $\bar{f}(x)$ is well-defined, though.

It does, but as I said, the definition is a bit more complicated than one might think.

 

[secur]

Let's call the example function g: everywhere 0 except at origin where it's 1.

To make it (more or less) rigorous we integrate a function f against this parametrized family and take the limit as epsilon goes to 0.

To emphasize a point: first evaluate the integral, using the parametrized "not-yet-delta", then take the limit. I remember doing some exercises which carried out this procedure explicitly, with actual example functions, when I first learned about the Dirac delta many years ago.

That's right, thanks for the emphasis. Note, this is the answer to Kevin McHugh's OP.

The space of square-integrable functions is a well-defined object and it is definitely not a Hilbert space.

Please provide a reference to a physics (not math) textbook that says square-integrable functions don't form a Hilbert Space. Note, if we allow weird functions like g, they're not even a vector space (additive identity must be unique).

In fact "square-integrable" implies a scalar product on these functions, the normal Dirac bracket. But you can't even call it a scalar product unless the pre-image of zero is the unique function 0.

... which certainly requires the axiom of choice

Since we're doing applied math here let's assume AC.

By the way, are you a physicist or a mathematician?

The question we're wasting neurons on is philosophical: should applied mathematicians assume the functions and the physical conditions they're analyzing are physical? My answer is yes. Dirac, the greatest Quantum Mechanic ever, agrees with me. Also Einstein, Everett, and anyone else up to 1980 (when I was last involved professionally). You disagree. That's fine. I think I'll stick with Dirac. However if you reference a modern physics textbook that demonstrates I, and Dirac, are now obsolete, I'll change this (merely) philosophical opinion.

rubi, and everyone, I thought someone made an obvious mistake (square-integrable not Hilbert), so I corrected him; but it turns out I was wrong. Clearly it's not obvious, because everybody else is making the same mistake. If I'd known that I would have let the issue alone. But if I hadn't mentioned it I would have gotten no response at all, so perhaps it worked out for the best.

@stevendaryl:

You're pointing out that a Fourier Transform "sanitizes" a weird function like g. You're right. If a function differs from 0 only on a set of measure 0, the Fourier integral for any value of k will be 0.

More generally, that's true for any integration at all. That set of measure 0 - in the case of our g, the origin - is being multiplied by the infinitesimal "dx". So any finite value vanishes, in any integration. In a word it's unphysical. That's why Dirac and every physicist prior to 1980 or so assumed they weren't dealing with such functions.

The issue is trivial, let's forget it.

Instead let's talk about dirac function. You made the mistake of saying the distribution was defined on L2, but actually it's on the space of Test functions, as I explained in my too-long post. rubi and micromass also told you about it. But in fact your mistake shows the right attitude. The whole point of defining dirac measure, and associated distribution, via test functions, is to apply dirac(f) to the L2 space of physical interest. With that mechanism understood, a physicist can go right ahead and use dirac(f) the way Dirac did - but now with a rigorous foundation to substitute for Dirac's flawless physical intuition.

Another mistake comes from Wikipedia. They say (on dirac function page) dirac(x) can be considered a measure or a distribution. As I showed in my post (hope you read it) that's not right - both measure and distribution are involved. First you consider dirac(x) a measure: now it's rigorously Ok to use it in an integral. There you treat it as a kernel, integrate against any f, to produce the distribution dirac(f). This is associated with dirac function - it's not, itself, dirac function. But it's a common abuse of notation to pretend it is.

My point was that this whole treatment was defined - dictated - by Dirac himself. He emphasized that the dirac function must be used as a distribution (in modern language), and showed how, creating the notion of "Test function". Left it for mere mathematicians to fill in the details.

Dirac did this often. For instance he laid out the essential ideas of QED ("Principles of QM", section 32: "Action Principle") and left it for others to work out. That's why I took the trouble to write that post, I never miss an opportunity to praise him. Instead of dabbling in pure math (remember a little knowledge is a dangerous thing) you guys should proudly follow in Dirac's giant footsteps.

Please give me an example of a physically meaningful function with zero (square-integrable) norm, which isn't 0.

Please tell me the physical situation where this function has meaning.

- Obviously we all agree there is no such physical situation.

 

[bhobba]

The definition of the Fourier transform on $L^2$ cannot easily be defined (see Reed&Simon II for details).

That's one advantage of distribution theory. Since L^2 a subset of tempered distributions using that its a snap.

To anyone reading this who hasn't studied this I highly recommend the following:
https://www.amazon.com/dp/0521558905/?tag=pfamazon01-20

It really should be part of the armoury of anyone into applied math.

Thanks
Bill

 

[rubi]

Please provide a reference to a physics (not math) textbook that says square-integrable functions don't form a Hilbert Space.

I don't see why you keep pursuing this idea. There is no point in telling people that $\mathcal L^2$ is a Hilbert space. It isn't. There is no need for QM books to mention $\mathcal L^2$, since they don't use it. You won't find a QM book that says it's using $mathcal L^2$. All books use the quotient space $L^2$.

Note, if we allow weird functions like g, they're not even a vector space (additive identity must be unique).

That's also false. $\mathcal L^2$ is a a vector space with identity element $0$. Any vector that differs on a null set is not an additive identity.

In fact "square-integrable" implies a scalar product on these functions, the normal Dirac bracket. But you can't even call it a scalar product unless the pre-image of zero is the unique function 0.

That's false. "Square-integrable" doesn't imply the existence of a scalar product.

Since we're doing applied math here let's assume AC.

That is not practical, since it would mean that you can't even even tell which functions are contained in your Hilbert space. Also, how do you define addition and scalar multiplication on your space? Pointwise addition won't work, since AC doesn't guarantee that the choice function chooses the correct pointwise sum from the equivalence class of the pointwise sum. This is really beyond impractical and every physicist would call you crazy if you forced them to deal with such issues. There is really only one practical way to obtain a Hilbert space and it's using equivalence classes. Your proposal is so much more complicated and you don't seem no notice this.

By the way, are you a physicist or a mathematician?

I'm a working physicist. Mathematically rigorour is indispensable in modern research in physics.

The question we're wasting neurons on is philosophical: should applied mathematicians assume the functions and the physical conditions they're analyzing are physical? My answer is yes. Dirac, the greatest Quantum Mechanic ever, agrees with me. Also Einstein, Everett, and anyone else up to 1980 (when I was last involved professionally). You disagree. That's fine. I think I'll stick with Dirac. However if you reference a modern physics textbook that demonstrates I, and Dirac, are now obsolete, I'll change this (merely) philosophical opinion.

Almost none of the functions in $L^2$ or even your highly pathological AC Hilbert space are physical. Yet, QM needs a Hilbert space to work. All the geniuses you mentioned would acknowledge this. I also note that you still haven't given us a definition of your pathological Hilbert space.

Clearly it's not obvious, because everybody else is making the same mistake.

On the contrary, you are the only person I know who makes this mistake. $L^2$ spaces are part of every undergraduate education in physics.

That's one advantage of distribution theory. Since L^2 a subset of tempered distributions using that its a snap.

Right, introducing it using distributions is very clean and Reed&Simon also do it this way. What I'm saying is that on $L^2$, the Fourier transform isn't given by an integral, like it is on $L^1$ or $\mathcal S$. Instead, we need to define it on a dense, absolutely integrable subset of $L^2$ and extend it to $L^2$ by taking $L^2$ limits. That means that the Fourier transform on $L^2$ produces only an equivalence class, rather than a function, so it can't be used to single out a representant.

 

[bhobba]

Right, introducing it using distributions is very clean and Reed&Simon also do it this way. What I'm saying is that on $L^2$, the Fourier transform isn't given by an integral, like it is on $L^1$ or $\mathcal S$. Instead, we need to define it on a dense, absolutely integrable subset of $L^2$ and extend it to $L^2$ by taking $L^2$ limits. That means that the Fourier transform on $L^2$ produces only an equivalence class, rather than a function, so it can't be used to single out a representant.

Thanks
Bill

 

[PeroK]

Please provide a reference to a physics (not math) textbook that says square-integrable functions don't form a Hilbert Space. Note, if we allow weird functions like g, they're not even a vector space (additive identity must be unique).

This is not the sort of thing a physics book would be interested in. You would have the same issue with the common integral. The anti-derivative of a function is an equivalence class of functions, but that is something you would only expect to see in a maths book. A physics book would use the constant of integration and move on, without worrying about equivalence classes.

In some of your posts, you are trying to make up maths on the fly.

 

[bhobba]

Please provide a reference to a physics (not math) textbook that says square-integrable functions don't form a Hilbert Space. Note, if we allow weird functions like g, they're not even a vector space (additive identity must be unique)..

Don't think the zero element has to be unique - here are the axioms:
https://math.kennesaw.edu/~plaval/math3260/vectspaces.pdf

I suspect all square integrable functions form a vector space but not a Hilbert space because its not isomorphic to sequences ai such that ∑ |ai|^2 exists.

Still it depends on the interpretation of 'a zero element' in those axioms - is that meaning just one element?

Its nearly 40 years since I studied linear algebra an its the first time I even noticed that.

Thanks
Bill

 

[PeroK]

Don't think the zero element has to be unique - here are the axioms:
https://math.kennesaw.edu/~plaval/math3260/vectspaces.pdf

I suspect all square integrable functions form a vector space but not a Hilbert space because its not isomorphic to sequences ai such that ∑ |ai|^2 exists.

Still it depends on the interpretation of 'a zero element' in those axioms - is that meaning just one element?

Its nearly 40 years since I studied linear algebra an its the first time I even noticed that.

Thanks
Bill

Yes, the zero element is unique. You can, if you wish, specify that as part of the axiom: "There exists a unique element $0$ ...". But, if you don't, you can simply prove it is unique by considering two elements with the $0$ property and showing they are equal.

 

[bhobba]

But, if you don't, you can simply prove it is unique by considering two elements with the $0$ property and showing they are equal.

Then what axiom do L^2 functions fail on?

Thanks
Bill

 

[PeroK]

Then what axiom do L^2 functions fail on?

Thanks
Bill

None. $L^2$ is a vector space.

But, a Hilbert space has analytical structure in the shape of an inner product and hence a vector norm. And the norm of every non-zero vector must be non-zero.

So, to make $L^2$ into a normed vector space, you would have to find a way to define a non-zero norm for a function that is zero everywhere except at one point. This creates a problem (specifically if you are trying to use an integral to define the vector norm).

Therefore, all functions whose square integral is 0 must be associated with the zero function through an equivalence relation. This creates the equivalence classes on $L^2$ that allow an inner product and norm to be defined on the quotient space. And in that quotient space, only the equivalence class of the zero function has zero norm. That solves the problem.

 

[rubi]

Then what axiom do L^2 functions fail on?

Thanks
Bill

None. The addition in $L^2$ is defined as $[f] + [g] := [f+g]$ with additive identity $[0]$. secure was concerned with $\mathcal L^2$, where the addition is just pointwise: $(f+g)(x) := f(x)+g(x)$. The additive identity is just the $0$ function $0(x) = 0$. The other functions in the equivalence class of $0$ aren't additive identities (contrary to what secur was suggesting).

 

[bhobba]

Thanks guys. Got it.

Like I said its 40 years since I learned this stuff and something basic can still surprise.

Thanks
Bill

 

[rude man]

I don't understand the question. $\delta(x)$ is not a function. It doesn't have a derivative.

It does have a derivative:
In the theory of electromagnetism, the first derivative of the delta function represents a point magnetic dipole situated at the origin. Accordingly, it is referred to as a dipole or the doublet function.[41]

The derivative of the delta function satisfies a number of basic properties, including:

• 
• 
• [PLAIN]https://upload.wikimedia.org/math/f/6/8/f6892fd7a5ad81141647eb1307c97ab1.png[URL='https://en.wikipedia.org/wiki/Dirac_delta_function#cite_note-42'][42][/URL]

cf. https://en.wikipedia.org/wiki/Dirac_delta_function

 

[George Jones]

It does have a derivative

This is one of the nice properties of distributions. From "Fourier Analysis and its Applications" by Folland

"It is possible to extend the operation of differentiation form functions to distributions in such a way that every distribution possesses derivatives of all orders that are also distributions."

 

[secur]

In some of your posts, you are trying to make up maths on the fly.

It may seem that way at times. In fact, I'm trying to remember math on the fly. I need to find my old textbooks; pretty sure they're in the attic somewhere. However no doubt my math is usually right. Much bigger problem, I have to adjust to your language, which is often different from the way we used to do it.

But the biggest problem I'm having is adjusting to the fact that, sometime in the last 40 years, all the physicists morphed into pure mathematicians! And, predictably, you take these details more seriously than real mathematicians do.

Here's an analogy. Real mathematicians are like professional soldiers who might casually chuck a hand grenade to each other, since they know precisely when it's safe to do so. Whereas you're like reserves who've trained only on duds. When you have to deal with a real one you put on bomb-handling equipment before even looking at it.

Thus when I'm in a situation where a function like our example "g" above can't possibly occur (like, physics), I might casually assume the additive identity is unique, knowing that if the square of a physical function integrates to 0 it has to be 0 everywhere. But a week-end warrior has to first take equivalence classes (under the SI norm) - then invoke the Axiom of Choice to select a representative member - then carefully use the result, praying it won't blow up in his face.

My opinion is: when a physicist starts worrying about AC he's seriously off track! But if I'm wrong I want to be set straight.

Where in all of physics do you meet with a non-physical function like our example g, because you need infinite precision? And, tell me a real physical situation where you'll get the wrong answer if you assume - or, don't assume - AC. (By the way in most of math we simply assume it.)

Suppose I boil water, stick in a thermometer, and see the temperature is 100. Now, suppose I realize - with a shock - that AC is not implied by Zermelo–Fraenkel axioms! Oh no - better check that temperature again. So I re-boil the water, and look at the thermometer. Will it still read 100? What if I hold the thermometer at precisely pi inches (with infinite precision) below the surface - will it make a difference?

Actually I know the answers to those questions: "no".

But there must be some physical situation where infinite precision and Axiom of Choice play a part, or else you wouldn't care about such minutiae. So, please tell me where that situation arises.

A physics book would use the constant of integration and move on, without worrying about equivalence classes.

Now you're making sense

L2L2L^2 spaces are part of every undergraduate education in physics.

I hope, in the midst of all this abstract math, you find time to teach those poor undergraduates a little physics as well. But I can't help wondering, where in all of physics do you use, as a norm, the 5th root of the 5th powers? Or L3, or anything other than L1, L2, and L-infinity?

By the way, I didn't notice the difference between regular L and script L, so I probably said something incomprehensible. Sorry about that.

Mathematically rigorour is indispensable in modern research in physics.

Sure it is - within reason. Dirac was a great mathematician, and he was just as rigorous as he had to be - no more, no less. Please tell me what physical situation requires infinite precision, or Axiom of Choice, to analyze. No details needed, just a general description of where, and why, you'll get the wrong physical answer if you don't handle these details right.

Here's the type of answer I'd like to see:

"Well, the number of multiple universes, considering Guth's inflation and Everett's work, is clearly Aleph-2. And, in order to show that QM probabilities really come out to the (complex) square of the wave function, we have to integrate over all the infinite universes in which the different experimental outcomes occur. To select those universes from this (very large) set, we need the Axiom of Choice. See how obvious it is?"

Or,

"we know exactly what conditions occur 10-33 meters from a Kerr BH ring singularity. But we need to see what happens at the singularity itself, whose coordinates are known with infinite precision (of course). If we don't exclude functions like g - we might get any answer at all, due to spurious functions in our solution space. We might calculate that the wormhole takes us back to August 12, 1066, at 10:30 in the morning. But when we actually go through the wormhole, it turns out to be 10:31! That's what can happen without rigorous mathematics."

Please provide a common-sense, real-world, 3-line example like these samples, with as much hand-waving as necessary. Please don't tell me I'll understand these things when I'm older, now eat those veggies.

 

[micromass]

OK secur, you clearly believe physicists should be able to say incorrect things and get away with it. Got it.

Why is it important to say that $L^2$ is a Hilbert space and $\mathcal{L}^2$ is not, etc? It's because being precise means easier communication between experts. Saying that $L^2$ is defined by equivalence classes isn't even heavy math! It's something every physicist can easily understand and grasp.
I agree we shouldn't get lost in mathematical details. But sometimes those details really do shine a nice light on certain stuff.

 

[samalkhaiat]

Mathematically rigorour is indispensable in modern research in physics.

Really? So, how come our most accurate and successful theory (QED) is built on mathematical “gibberish”? Why are there no room for improvement left concerning the purely electromagnetic phenomena?
The “wish” to understand the QED-miracle in terms of a rigorous mathematical structure was shared by many physicists in the early 1950’s, (so your view is rather old not modern). That was the motivation for a development subsumed under the name “axiomatic QFT”, merging into “constructive field theory” in the late 1960’s. In the last 70 years, an army of great mathematicians tried but failed to provide a rigorous reformulation of QED that can account for its miracles. That, I believe, was a waste of effort because the great and essential progress came ultimately from elsewhere, namely from applying the same mathematical “gibberish” of QED to non-abelian groups, and thus covering a wider range of high energy physics.
Physics, my friend, is an ill-defined mathematical structure, for one cannot associate a unique and well-posed mathematical problem for every experimental question. Rigorous mathematic is good, very good indeed. However, physicists should not expect too much of it all the times.

 

[secur]

OK secur, you clearly believe physicists should be able to say incorrect things and get away with it. Got it.

Great! All I, or any reasonable person, has to insist on is that his point be understood - not agreed to.

Why should physicists - and mathematicians, and any group of experts - be allowed to say incorrect things? Two conditions must be satisfied. One, it's simpler that way. Why be incorrect if it's more trouble? Two, we all know what we mean.

Why is it important to say that L2L2L^2 is a Hilbert space and L2L2\mathcal{L}^2 is not, etc? It's because being precise means easier communication between experts. Saying that L2L2L^2 is defined by equivalence classes isn't even heavy math!.

It's precisely among experts where precision is not necessary, because they (should) all be on the same page.

Anyway, we all see you can't allow something like the example g in Hilbert function space; it can't even be a linear vector space (under SI) because the kernel's non-trivial. Dirac and I simply assume those functions aren't there, because this is physics. Another approach is to assume they are there at first and get rid of them by factoring over the kernel. To me, the second approach seems pedantic. To you, the first seems sloppy. That's alright; we both wind up at the same place, with the same understanding, and can just move on.

Note there are cases where that's no good. If someone asks "why does the force of gravity cause clocks to run slow", you must point out that gravity's not a force (like EM) but curvature of spacetime. On the other hand, consider "If a ball falls for 3 seconds, and the force of graviy is 9.8 m/s/s, how fast will it be going?" It's ridiculous to insist on GR to answer this question.

It's something every physicist can easily understand and grasp..

I hope so!

I agree we shouldn't get lost in mathematical details. But sometimes those details really do shine a nice light on certain stuff.

I respect your attitude and agree with it maybe 60%. We should have little trouble communicating.

 

[micromass]

Anyway, we all see you can't allow something like the example g in Hilbert function space; it can't even be a linear vector space (under SI) because the kernel's non-trivial.

But the square integrable functions do form a linear vector space. No problem there. They just don't form a Hilbert space.

 

[strangerep]

Hmm. Since the OP seemed to disengage from this thread back at post #5, perhaps everything from post #6 onwards should be split off into a new thread. (The original thread was "I", but it has now become "A".)

 

[secur]

Since the OP seemed to disengage from this thread back at post #5, perhaps everything from post #6 onwards should be split off into a new thread.

That illustrates my main point perfectly. Post 5, Kevin McHugh asked stevendaryl how to get the limit of the test functions to show dirac(f) = f(0). stevendaryl had explained it in terms of distributions, which is more math than Dirac used, and not necessary for this elementary discussion; but OP seemed to be OK with it. Unfortunately stevendaryl had said it was defined on the square-integrable functions NOT the Test Function Space! micromass pointed out the mistake - stevendaryl made one last post to OP - rubi seconded micromass - and they were off and running about Test Function spaces and distributions for pages. That's when I tried to clear up the confusion, but (mistakenly, more-or-less) said square-integrable functions were a Hilbert Space, ignoring the equivalence classes. Everyone jumped on that essentially trivial "error". Etc.

Perfect illustration how physics, i.e. actually using dirac function, gets buried under a mass of nit-picky abstract math.

But the square integrable functions do form a linear vector space. No problem there. They just don't form a Hilbert space.

Ok, this is an illustration that math rigor is important (not as much as physics, but still important). Only when you actually know/remember the terminology are you entitled to casually blow by details. Not good enough to have known the terminology 40 years ago.

I meant square-integrable functions with SI norm aren't a normed vector space, because norm of functions like our example g is 0 - but g is not 0. Now, I'm hoping I remember correctly that normed vector space requires non-trivial kernel. ... If I can't find my old copy of Royden I'll buy one!

 

[vanhees71]

It's a little confusing. When people talk about $L^2$, they do mean functions modulo equivalence. But it seems to me that we can certainly talk about the set of square-integrable functions without taking equivalence relations.

Yes, and from this is clear that $f(0)$ is not well defined, because you can vary $f(0)$ arbitrarily without changing the Hilbert space vector in $L^2$. I can only stress it again, the $\delta$ distribution is defined on some dense subspace like the Schwartz space of $C^{\infty}$ functions that fall faster than any power. In QM the test-function space is the maximal domain of the position operator in the sense of an essentially self-adjoint operator, and $\delta$ is a distribution in the dual space of this "nuclear space". That becomes very clear with the modern formulation in terms of the rigged Hilbert space (see, e.g., Ballentine's textbook for a physicists' introduction or the one by Galindo an Pascual, which is mathematically more rigorous).