Younsay formal variable..az opposed to an INFORMAL bariabpe..wjat would even constitute an informal variable? Not a clear or particularly.techniczlly.defined variable?
15:45 Michael kind of breezed over it, but it's interesting to note that if you use substitution 2) you instead get the slightly different g(xq)Dᵩ[f(x)] + f(x)Dᵩ[g(x)] , basically the same formula but the extra "q" is under g instead of f. Which means that the two formulas are equal, you can effectively move that q back and forth between the corresponding f and g in the product and get the same result.
@@danielfarbowitz671 I think this remark implies that as long a D_q is well defined (and that the adding 0 and factoring steps make sense), we have this identity. But yes, it raises the question of what are the minimal conditions for D_q to be well-defined, for this identity to be true. i.e. if you want the two expanded product rule to be equal for your favorites f, g and q, you could start by checking if D_q is well-defined. If yes, bingo, otherwise, well, good luck 😉
Yeah Micheal, we definitely want to know about Quantum Groups! That’s so thrilling! But please Micheal, how about a whole Lie Theory/Algebraic Structures for Mathematical Physics playlist? You could cover Lie Algebras, Hopf Algebras, Supergroups and Superalgebras, Clifford Algebras, Jordan Algebras… I think they can be cool enough, less technical than Quantum Groups, and also, they provide a solid background before more technical topics, such as, say Quantum Groups.
"Recall: [n!] q"-in the summation above, the factorial is outside the square brackets! Also, where has this definition of a quantum factorial been defined or derived?
YES YES YES!!! Quantum Groups Please! That and noncomm algebra, algebraic geometry and quantum topology, plus HoTT: list of topics I would love to understand better. Cheers and love your channel! YES!
@@oliviermiakinen197 Doesn't it feel perfectly natural and intuitive for the *multiplicative* derivative to be undefined at 0? It's like complaining that the field axioms demand a multiplicative inverse for every element except the 0 element.
@@goclbert however, these are always removable singularities due to the correspondence to the usual derivative, because as 'formal' variables we can cancel the terms as polynomials, which include removing zeros from numerators and denominators, and failing that we can take the limit of x to 0 over the expression (and failing that, it's because the derivative really doesn't exist at 0).
I would have liked to see some description of the applications of the quantum derivative. Why is it useful if it's equivalent to the ordinary derivative? Does it make some calculations easier or provide some deeper insight?
these 2 derivatives are ONLY the same for q -> 1. Otherwise, they're nothing alike. And in the case q=1 the equivalence is trivial with the change of coordinates he made.
Thanks for the great video, and this series in general. I was wondering if there were any known solutions to q differential equations? Specifically, I'm studying D_q f = f^2. There seems to be some theory for linear q diff equations, though sadly not for nonlinear ones (even such as this which under normal circumstances would be separable!). Are there any results about equations of this form? Thanks again for your work on this channel
This probably doesn't work for x = 0, because the denominator is already zero for any q. The usual derivative works for every point (as long as the function itself is differentiable).
however, these are always removable singularities due to the correspondence to the usual derivative, because as 'formal' variables we can cancel the terms as polynomials, which include removing zeros from numerators and denominators, and failing that we can take the limit of x to 0 over the expression (and failing that, it's because the derivative really doesn't exist at 0).
@@MagicGonads I don't think this strategy will always work because the (usual) derivative is not necessarily continuous. For example, if f(x)=x²sin(1/x) for x≠0 and f(0)=0, then f'(x)=0, but the limit of f'(x) at x=0 doesn't exist.
im in the same boat as you but i assume that the term "quantum" is being applied differently here since it seems to be a purely mathematical context rather than anything to do with actual quantum mechanics
@@v_munu It isn’t entirely unrelated to quantum mechanics. I’m not really familiar with quantum groups, but, there are a number of “q-deformations” of things which show up in some quantum mechanical things. If you’ve ever heard of non-commutative geometry, I think that’s connected to both (and is like, a thing that some people think might be part of how to combine GR and quantum mechanics. I think the rough idea of that, is that like, instead of having like, functions of x and y described by power series of x and y where xy = yx, instead you might say, have things that are “kinda like functions of x and y”, which are power series in x and y, but where we have xy = qyx, So then, like, xyy = qyxy = qqyyx But, I haven’t really studied non-commutative geometry, so I could be wrong about much of what I said about it. But, also, there are plenty of q-analogs which *aren’t* a quantum mechanics thing. Or at least, quantum mechanics isn’t my main reason for being exposed to them. The combinatorics of counting subspaces of a given dimension, satisfying some properties, of a vector space over the field with q elements, is described using the q-factorial and q-binomial coefficients and such! For these reasons, people have speculated about a (non-existent) “field with one element”, where a “vector space” over this “field” would be basically just a set (but with a designated zero element) and generally trying to figure out, “seeing as there can’t actually be a field with one element, what thing could play the role of it, in a way that explains why the formula for “number of m-dimensional subspaces of an n-dimensional vector space over the field with q elements” , namely, (n choose m)_q , becomes the formula for the number of m element subsets of an n element set (I.e. (n choose m) ), when you take q->1 ? Fun stuff
Is the quantum factorial [n]_q! the same as the q-analogue of the natural component of n!, namely [n!]_q? If not, then the e_q(x) definition isn't in line with the general f_q(x) definition.
They are not. To see this: [6]_q = (1-q^6)/(1-q) = (1 + q + q^2 + q^3 + q^4 + q^5) while [1]_q [2]_q [3]_q = 1 (1+q) (1+q+q^2) = 1 + 2q + 2 q^2 + q^3 These are not the same. But, if instead of taking like, the q-analog of whatever a_n happens to be, separately for each n, you express a_n as a function of n in a normal way, and take appropriate q-analogs of those functions from natural numbers to natural numbers, then, that may do it? ... Hm, I wonder whether f(n) = sum_{k=0}^n k would have a nice q-analog. Well, sum_{k=0}^n [k]_q = sum_{k=0}^n sum_{m=0}^{k-1} q^m = sum_{m=0}^{n-1} sum_{k=m+1}^n q^m = sum_{m=0}^{n-1} (n - m) q^m Now, this of course will have limit equal to n(n+1)/2 as q goes to 1, but, will it be analogous to n(n+1)/2 in any reasonable way? Like, [n]_q [n+1]_q / [2]_q ? Hm, no, they don’t have the same degree. [n]_q has degree n-1, and [n+1]_q has degree n, so their product has degree 2n-1 but the sum above only has degree n-1. Disappointing. Perhaps there are some principles describing when q-analogs are reasonable.
Hm, well, (n choose 2)_q is a useful q-analog, so maybe the thing to explain is why (1+2+3+...+n) is not well-suited for taking a q-analog, even though it has the same value as something which is. The (n choose 2)_q , when we interpret it with q a prime power, counts the number of 2-d subspaces of an n-d space over the field with q elements. [n]_q = (n choose 1)_q counts the number of 1-d subspaces, which is basically the number of directions. Adding up [k]_q from k=0 to k=n , would then, I guess, be counting the number of ways to pick a dimension for the space, where it has to be at most n, and then pick a direction in that space. That’s, a very different question from counting 2-d subspaces. It also doesn’t seem like an especially natural question. I think maybe q-analogs maybe don’t really like being added to each-other, and prefer being multiplied? (I’m anthropomorphising, yes) (I’m not confident of this. It is just a guess.) If you have a function of the form “if you have an n-dimensional vector space over the field with q elements, how many ways are there to [do some thing that is natural to do with an n-d vector space]?” I think that will tend to give you a nice q-analog of something? (The q-factorial of n for example, iirc counts the number of ways to pick a complete flag in F_q^n . I.e. pick a 1-d subspace, then a 2-d subspace containing that 1-d subspace, then a 3-d subspace containing the 2-d subspace, and so on.) But, there are other things it counts, I think. The determinant of an n by n matrix can be computed as a sum indexed by permutations s of {1,2,...,n}, of (-1)^{|s|} prod_{j=1}^n A_{j,s(j)) this certainly seems like a linear-algebra-ish thing , and it involves a set of size n! is there maybe a way to get q-binomial coefficients to pop out of this? One thing that I am reminded of, is some measure of how out-of-order a given permutation is, Where like, (1,2,3,4) gets a 0, (2,1,3,4) gets a 1, (2,3,1,4) gets a 2 I think, And (4,3,2,1) I think gets a 6=3+2+1? I think this is called the inversion count of a permutation. And I think I remember it being used as an exponent for q. Well, what if we summed up like, q^{inversion count of s} over all permutations s of n elements? For n=2 that would be 1+q Which is [2]_q For n=3 that would be (1,2,3):0 (1,3,2):1 (2,1,3):1 (2,3,1):2 (3,1,2):2 (3,2,1):3 so, 1+2q+2q^2+q^3 which... is (1+q)(1+q+q^2) = the q-factorial of 3 (whoo!) And I expect the pattern continues. Though, I guess that maybe relates more closely to the permanant than the determinant? The inversion count is odd exactly when the permutation is odd, I think, so maybe we should see it as, (-q)^{inversion count of s} and so the sum would actually be the (-q)-factorial of n. How can we fit this in with the products of entries in the matrix though? If all the entries were 1, then that part would be 1, but the determinant of that is 0, and so not very interesting? What if the entries of the matrix were from some non-commutative ring, so that it could be the cause of those powers of q? Hm, say we take R[x_1,x_2,...,x_n | x_i x_j = -q x_j x_i for j
OK, given your comment at 10:10 I absolutely need to hear about quantum groups. The similarity between $xq$ and $x+dx$ is indeed striking. Where does this end up going?
Neat seeing how multiples of 1 get exploded when you leave the derivative parameter as it is rather than just taking the limit. On the other hand, while this is an interesting generalization of the derivative, at no point did it feel like there was anything "quantum" about it. Maybe in quantum mechanics, q instead approaches some quantum operator rather than 1, but it could just as easily approach the identity matrix, or just some number that isn't 1 like 0, 2, maybe even i. Actually, I'm pretty sure the discrete derivative is what you get when you take the normal derivative and have h approach 1 instead of 0, which is another instance of changing how the the parameter is interpreted. Having q approach 2 (or maybe e) is probably the closest thing for the relative derivative to what the discrete derivative is for the absolute derivative. (Because they come down to relative change in x vs absolute change in x, and because it makes more sense than slapping "quantum" onto it.)
The q-deformed derivative (the one shown here) is used in what is known as “quantum algebra” which is used in some quantum mechanical stuff (for topics relating to anyons and such). A special case in “quantum groups” (at least in the case of su_q(2) ) is when q is a primitive root of unity. The Fibonacci category (related to Fibonacci anyons) can be constructed by starting with getting rep_q(su(2)) and setting q=exp(2pi i / 5) , and then doing some other stuff with quotienting.
Dear Michael. Regarding fractional derivatives, is there a 1/pi -th derivative or generally irrational fractional derivatives? I have the naive idea that if there's a half derivative, then there's also quarter, eighth etc. derivatives. But irrational numbers might change that.
I'm now idly wondering what happens if q, instead of 1, approaches another root of unity. I see that q approaching i gives a nice periodicity to the q-analogs, for starters.
the unit circle is dense with roots of unity, so if the behaviour depends on being a root of unity specifically, it would be impossible to take the limit about any point along the circle since it would also be dense in non-roots-of-unity (e.g. if it depends on what order of root of unity it is, then it would definitely not be possible to pass through the circle and get a continuously morphing picture). For a given q-analogue of a complex function (bivariate as f(z,q)), if that happens, we can't analytically continue over q through the unit circle, but if that doesn't happen we might be able to.
The explanation of the quantum derivative was pretty much perfect. Nothing to wine about. But, what is the quantum version of a function? I am sorry, but that's where you lost me. I suppose that the limit of fq(x) for q to 1 should be equal to f(x). But, if this is the definition, then you can all kinds of quantum versions... Sorry, it's probably very simple to explain, but I lost it somewhere!?
This is a wild guess: The standard derivative of the quantum version of your function is equal to the quantum derivative of the non-quantum function equivalent? Is this correct?
I don't know either, but it seems a bit arbitrary. For e^x as a power series, why take ([n]_q)! instead of [n!]_q? The answer is because it works. It solves f'=f in this q-analog world. Not sure if there's an underlying structure that determines this, but the fact that this works is motivation enough. - unless it's not supposed to work, but we should try to make it work at least until we know better
Wait, there is a problem with the second proof, you assumed that x is not zero, otherwise the change of variables is not defined. Looking at it, seems that the quantum derivative is not defined when x is zero
Mm, I wouldn’t say “nothing”, but, I do think it would be better to call it “q-deformed” or “q-analog” rather than “quantum”? The q-deformed derivative *is* used in a specific sub-field of quantum physics, but I don’t know that it is exactly because of it being quantum mechanical? It might be possible to use what is known as “quantum algebra” (the q-deformed derivative is used in “quantum algebra”) in a physics-but-not-quantum-mechanics context. Or, even if that doesn’t work so well, the q-deformed derivative isn’t particularly closely tied to the core ideas of quantum mechanics. It’s a simple enough, and widely-applicable enough, idea, that naming it “the quantum derivative” is imo not the best name, But, it *is* used in some quantum mechanics based stuff.
Hmm, x^a = x x^(a-1) , x^a - x = x (x^(a-1) - 1) f(x x^(a-1) ) - f(x) If we set q_x=x^(a-1) , your expression becomes (f(q_x x) - f(x))/(x (q_x - 1)) Which is (1/x) times D_{q_x} f(x) ? I think probably it would be better to have, instead of x^a - x as the denominator, to instead have x^(a-1) - 1 , as the denominator. That way you do get something derivative-like. Specifically, you get the q-derivative except that q depends on x and on a, specifically, q = x^(a-1) . I don’t mean that to say “and therefore it is just the q-derivative, and not a usefully separate thing”. The q-analog can also be thought of as the (f(x+h)-f(x))/((x+h)-x) thing except with h depending on x, but they are both useful for different use-cases. (Similarly the h version can be expressed as the q version with the q depending on x (except for at x=0 but shhh).) So, the version with x^(a-1) in place of q might also be useful for something, idk. Uhh… (ln(x^a) - ln(x))/(x^(a-1) - 1) = (a ln(x) - ln(x))/(x^(a-1)-1) =(a-1) ln(x) / (x^(a-1) - 1) Hm, weird, I’m not getting the right derivative for ln(x) when I take the limit as a-> 1; I must be making some mistake.
There are iirc two main conventions. One of them is [n]_q = (q^n - 1)/(q-1) for q≠1 and n for q=1 (Or, equivalently in both cases, \sum_{k=0}^{n-1} q^k ) But there’s also a more symmetric version, which is iirc, [n]_q = (q^n - q^{-n})/(q^{1} - q^{-1}) And also I think a slight variation on that one. The first one, (q^n - 1)/(q-1) , is probably the one being referred to here.
My feeling is that when he wrote [n]_q! he meant [n!]_q, as he wrote on the following line, so [n]_q! was a typo. EDIT: I have since changed my mind about this, as per my next reply below.
@@jounik actually, since I wrote this, I came to the opposite conclusion - he is defining [n]_q!. The reason is that [n]!_q is already defined, but doesn't give what we want, which is given on the next line. Therefore, I now believe that, in fact, the [n]!_q on the next line is a typo, and should be [n]_q! :)
Here is an issue: The quantum derivative showed in the video is undefined at x = 0. Since it is `q*x - x` aka `0q - 0` = `0`, we just get a literal `0/0` and it just breaks. So when q->1, D_q(f(x)) = f'(x) except at x = 0. What will other quantum functions we derived behave at x = 0?
I think df(x)/dx = lim y→x Dq f(y) for all x; I wonder if that limit is actually part of either the definition of Dq and Michael just skipped over it to slightly simplify things.
That's a sound objection. The definition doesn't work for x=0. But you may be capable of fixing that if you set x=y-e, e being some arbitrary number greater than zero. f(x) = f(y-e). Then you calculate the quantum derivative for y=e and define this to be the qD at x=0. In the denominator you've got e(q-1), which is not equal to zero.
Why do we need quantum derivative in quantum physics can someone wants to explained to me? I assume the q introduces some kind of probabilistic efffect.
My suspicion is that "quantum" here refers to discrete jumps in the variable, not about probabilities. If you take q > 1 you get a finite difference (a quantum jump) instead of the instantaneous derivative (when q goes to 1)
@@radadadadeeI suspect it is called this because of its relevance in quantum groups, which are deformations of the enveloping algebras for certain groups. Deformations in terms of a parameter q which, as q goes to 1, recovers the enveloping algebra for the original group. These are studied in quantum mechanics for… reasons, but generally not at the start of studying quantum mechanics? Sometimes the q is thought of as being like e^hbar , except not literally hbar because hbar has units, but when one forgets about units *and* thinks of classical things as being the limit as hbar goes to zero, then that has e^hbar going to 1. … but, also, really q in quantum algebra is often complex valued, which doesn’t really make sense for hbar either, as hbar should be real valued and non-negative. One place that it is relevant is in stuff about anyons (which are an alternative to Fermions and Bosons, but which only make sense in 2 or fewer dimensions of space, but quasiparticles in a flat surface medium can be anyons). One can take the category of representations of the q-deformed version of (the universal enveloping algebra of) SU(2) , (the representations of SU(2) are/correspond-to the different possible values of “spin” for a particle), and if one picks q=e^(2 pi i/ 5) , and then quotients out by some stuff, one ends up with “the Fibonacci category”, which describes the Fibonacci anyons. Iirc, these theoretically could be found if you tune things just right, in fractional quantum Hall effect demonstrations/setups (but it is very difficult, as the region to get specifically Fibonacci anyons is really small I think?) Despite all that, I think calling it “the q-deformed derivative”, or “the q-analog of the derivative” would probably be a better name? It doesn’t sound as cool, but it is less confusing, easier to explain why it is called that, and, also, it doesn’t sound cooler or more mysterious than it should.
Michael, is the quantum derivative isomorphic to the partial derivative in any cases, maybe with normalization? I’ve never heard of this before but it seems like it could even be applied to polynomial basis functions for fitting a regression to data when you have q as a constraint. Any thoughts?
@@drdca8263 I was thinking there could be a correlation to the gamma function half-derivative (or partial derivative) for some class of functions. I haven’t thought any more deeply about it than that. Since we have a polynomial with rational coefficients maybe we could make a basis like the Bernstein basis for fitting curves over data with a constraint, or the Grobner basis.
@@michaelzumpano7318 Oh! *Fractional derivatives* Yeah, calling those “partial derivatives” confused me, because that’s the name of a different thing. Hm... Well, the fractional derivatives iirc aren’t local, in that their values near a given input depend on the whole function, while this, the quantum derivative.. well, it kind of only depends on the value of the function at 2 points, but one of those two points depends on q, so I guess that might not really be too different? Though, if you suppose that q is in a neighborhood of 1, then it only depends on a neighborhood of the point, so it is kinda local? Hm, idk, maybe they could be connected.
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Been loving your videos for a while. I'm so happy you're getting sponsors!! :) Keep it up!
Younsay formal variable..az opposed to an INFORMAL bariabpe..wjat would even constitute an informal variable? Not a clear or particularly.techniczlly.defined variable?
Heck yeah, we wanna learn about QUANTUM GROUPS!
So much!
Heck yeah
For sure
Indeed!
requiredmandatory
I'd like to see a video about the quantum groups, although they are quite technical
But that would be a good place to stop.
Love it!! I would especially love the quantum group video, all the gritty details included!
15:45 Michael kind of breezed over it, but it's interesting to note that if you use substitution 2) you instead get the slightly different g(xq)Dᵩ[f(x)] + f(x)Dᵩ[g(x)] , basically the same formula but the extra "q" is under g instead of f. Which means that the two formulas are equal, you can effectively move that q back and forth between the corresponding f and g in the product and get the same result.
I was thinking of that as well -- I wonder if that puts any particular constraints on the quantum derivative.
Before reading this I was about to denounce that the product rule turned out non-commutative. 😂
@@LucasCAPS it has to be commutative for D_q to be a well-defined operator (since f•g is equal to g•f as function from a field to a field)
@@danielfarbowitz671 I think this remark implies that as long a D_q is well defined (and that the adding 0 and factoring steps make sense), we have this identity. But yes, it raises the question of what are the minimal conditions for D_q to be well-defined, for this identity to be true.
i.e. if you want the two expanded product rule to be equal for your favorites f, g and q, you could start by checking if D_q is well-defined. If yes, bingo, otherwise, well, good luck 😉
Micheal , please make a video about quantum groups and Hopf Algebra.
Yeah Micheal, we definitely want to know about Quantum Groups! That’s so thrilling!
But please Micheal, how about a whole Lie Theory/Algebraic Structures for Mathematical Physics playlist?
You could cover Lie Algebras, Hopf Algebras, Supergroups and Superalgebras, Clifford Algebras, Jordan Algebras…
I think they can be cool enough, less technical than Quantum Groups, and also, they provide a solid background before more technical topics, such as, say Quantum Groups.
Yes, i dream of a video series on mathematical physics taught by someone as talented as this!
Hey!!
Make a video on Quantum groups .
"Recall: [n!] q"-in the summation above, the factorial is outside the square brackets! Also, where has this definition of a quantum factorial been defined or derived?
YES YES YES!!! Quantum Groups Please! That and noncomm algebra, algebraic geometry and quantum topology, plus HoTT: list of topics I would love to understand better. Cheers and love your channel! YES!
I love math man. Someone goes "hey what if the derivative was multiplicative?" and it just works so well.
However it does not seem to work so well for the quantum derivative at x=0, does it?
@@oliviermiakinen197 Doesn't it feel perfectly natural and intuitive for the *multiplicative* derivative to be undefined at 0? It's like complaining that the field axioms demand a multiplicative inverse for every element except the 0 element.
That is a good point.
@@goclbert however, these are always removable singularities due to the correspondence to the usual derivative, because as 'formal' variables we can cancel the terms as polynomials, which include removing zeros from numerators and denominators, and failing that we can take the limit of x to 0 over the expression (and failing that, it's because the derivative really doesn't exist at 0).
Yes, we NEED a vidoe on quantum groups
As a physicist, I was hooked by the title! Brilliant video!
I would have liked to see some description of the applications of the quantum derivative. Why is it useful if it's equivalent to the ordinary derivative? Does it make some calculations easier or provide some deeper insight?
There is a research paper about using it in Neural Networks by Frank Nielsen and Ke Sun.
Exactly! I wanted to know what motivated its existence and where and how it's used
I'd like to see a video about quantum groups although they are quite technical
Very cool! I'm curious if there are some functions that are easier to differentiate like this rather than with the standard definition?
these 2 derivatives are ONLY the same for q -> 1. Otherwise, they're nothing alike. And in the case q=1 the equivalence is trivial with the change of coordinates he made.
Thanks for the great video, and this series in general. I was wondering if there were any known solutions to q differential equations? Specifically, I'm studying D_q f = f^2. There seems to be some theory for linear q diff equations, though sadly not for nonlinear ones (even such as this which under normal circumstances would be separable!). Are there any results about equations of this form? Thanks again for your work on this channel
this is an interesting generalization of the derivative, would love more on this
This probably doesn't work for x = 0, because the denominator is already zero for any q. The usual derivative works for every point (as long as the function itself is differentiable).
The caveat x≠0 needs to be added to all the formulae in the video.
however, these are always removable singularities due to the correspondence to the usual derivative, because as 'formal' variables we can cancel the terms as polynomials, which include removing zeros from numerators and denominators, and failing that we can take the limit of x to 0 over the expression (and failing that, it's because the derivative really doesn't exist at 0).
@@MagicGonads I don't think this strategy will always work because the (usual) derivative is not necessarily continuous. For example, if f(x)=x²sin(1/x) for x≠0 and f(0)=0, then f'(x)=0, but the limit of f'(x) at x=0 doesn't exist.
@@MichaelRothwell1 ok, then we can just shift the function then take this derivative then shift it back
QGs, yes please! ^.^
Love your videous
All math fun without exam
I don't know why, but I kinda love this. Why haven't I heard about it before Idk.
Great video. Thank you
3.35
He said A^3 minus B^3, but he wrote A^3-B^2
Would love to learn about quantum groups and Hopf Algebras I've always wanted a clearer motivation for their use
10:55 That only works if x is nonzero.
If x=0, then the quotient is 0/0 undefined before any limit takes place.
yes please quantum groups!
Ok, so I get all the algebra, but I'm not really sure what makes this "quantum" I guess it's a tool used over there, and would love to hear more.
im in the same boat as you but i assume that the term "quantum" is being applied differently here since it seems to be a purely mathematical context rather than anything to do with actual quantum mechanics
@@v_munu It isn’t entirely unrelated to quantum mechanics.
I’m not really familiar with quantum groups, but, there are a number of “q-deformations” of things which show up in some quantum mechanical things.
If you’ve ever heard of non-commutative geometry, I think that’s connected to both (and is like, a thing that some people think might be part of how to combine GR and quantum mechanics.
I think the rough idea of that, is that like,
instead of having like, functions of x and y described by power series of x and y where xy = yx, instead you might say,
have things that are “kinda like functions of x and y”, which are power series in x and y, but where we have xy = qyx,
So then, like,
xyy = qyxy = qqyyx
But, I haven’t really studied non-commutative geometry, so I could be wrong about much of what I said about it.
But, also, there are plenty of q-analogs which *aren’t* a quantum mechanics thing. Or at least, quantum mechanics isn’t my main reason for being exposed to them.
The combinatorics of counting subspaces of a given dimension, satisfying some properties, of a vector space over the field with q elements, is described using the q-factorial and q-binomial coefficients and such!
For these reasons, people have speculated about a (non-existent) “field with one element”, where a “vector space” over this “field” would be basically just a set (but with a designated zero element)
and generally trying to figure out, “seeing as there can’t actually be a field with one element, what thing could play the role of it, in a way that explains why the formula for “number of m-dimensional subspaces of an n-dimensional vector space over the field with q elements” , namely, (n choose m)_q , becomes the formula for the number of m element subsets of an n element set (I.e. (n choose m) ), when you take q->1 ?
Fun stuff
Is the quantum factorial [n]_q! the same as the q-analogue of the natural component of n!, namely [n!]_q? If not, then the e_q(x) definition isn't in line with the general f_q(x) definition.
They are not. To see this:
[6]_q = (1-q^6)/(1-q) = (1 + q + q^2 + q^3 + q^4 + q^5)
while [1]_q [2]_q [3]_q = 1 (1+q) (1+q+q^2) = 1 + 2q + 2 q^2 + q^3
These are not the same.
But, if instead of taking like, the q-analog of whatever a_n happens to be, separately for each n, you express a_n as a function of n in a normal way, and take appropriate q-analogs of those functions from natural numbers to natural numbers, then, that may do it?
...
Hm, I wonder whether f(n) = sum_{k=0}^n k
would have a nice q-analog.
Well,
sum_{k=0}^n [k]_q
= sum_{k=0}^n sum_{m=0}^{k-1} q^m
= sum_{m=0}^{n-1} sum_{k=m+1}^n q^m
= sum_{m=0}^{n-1} (n - m) q^m
Now, this of course will have limit equal to n(n+1)/2 as q goes to 1,
but, will it be analogous to n(n+1)/2 in any reasonable way?
Like, [n]_q [n+1]_q / [2]_q ?
Hm, no, they don’t have the same degree.
[n]_q has degree n-1, and [n+1]_q has degree n, so their product has degree 2n-1
but the sum above only has degree n-1.
Disappointing.
Perhaps there are some principles describing when q-analogs are reasonable.
Hm, well, (n choose 2)_q is a useful q-analog,
so maybe the thing to explain is why (1+2+3+...+n) is not well-suited for taking a q-analog, even though it has the same value as something which is.
The (n choose 2)_q , when we interpret it with q a prime power, counts the number of 2-d subspaces of an n-d space over the field with q elements.
[n]_q = (n choose 1)_q counts the number of 1-d subspaces, which is basically the number of directions.
Adding up [k]_q from k=0 to k=n , would then, I guess, be counting the number of ways to pick a dimension for the space, where it has to be at most n, and then pick a direction in that space.
That’s, a very different question from counting 2-d subspaces.
It also doesn’t seem like an especially natural question.
I think maybe q-analogs maybe don’t really like being added to each-other, and prefer being multiplied? (I’m anthropomorphising, yes)
(I’m not confident of this. It is just a guess.)
If you have a function of the form “if you have an n-dimensional vector space over the field with q elements, how many ways are there to [do some thing that is natural to do with an n-d vector space]?”
I think that will tend to give you a nice q-analog of something?
(The q-factorial of n for example, iirc counts the number of ways to pick a complete flag in F_q^n . I.e. pick a 1-d subspace, then a 2-d subspace containing that 1-d subspace, then a 3-d subspace containing the 2-d subspace, and so on.)
But, there are other things it counts, I think.
The determinant of an n by n matrix can be computed as a sum indexed by permutations s of {1,2,...,n}, of (-1)^{|s|} prod_{j=1}^n A_{j,s(j))
this certainly seems like a linear-algebra-ish thing , and it involves a set of size n!
is there maybe a way to get q-binomial coefficients to pop out of this?
One thing that I am reminded of, is some measure of how out-of-order a given permutation is,
Where like, (1,2,3,4) gets a 0, (2,1,3,4) gets a 1, (2,3,1,4) gets a 2 I think,
And (4,3,2,1) I think gets a 6=3+2+1?
I think this is called the inversion count of a permutation.
And I think I remember it being used as an exponent for q.
Well, what if we summed up like,
q^{inversion count of s} over all permutations s of n elements?
For n=2 that would be 1+q
Which is [2]_q
For n=3 that would be
(1,2,3):0
(1,3,2):1
(2,1,3):1
(2,3,1):2
(3,1,2):2
(3,2,1):3
so, 1+2q+2q^2+q^3
which...
is (1+q)(1+q+q^2) = the q-factorial of 3 (whoo!)
And I expect the pattern continues.
Though, I guess that maybe relates more closely to the permanant than the determinant?
The inversion count is odd exactly when the permutation is odd, I think, so maybe we should see it as, (-q)^{inversion count of s}
and so the sum would actually be the (-q)-factorial of n.
How can we fit this in with the products of entries in the matrix though?
If all the entries were 1, then that part would be 1, but the determinant of that is 0, and so not very interesting?
What if the entries of the matrix were from some non-commutative ring, so that it could be the cause of those powers of q?
Hm, say we take R[x_1,x_2,...,x_n | x_i x_j = -q x_j x_i for j
Hope to see more maths (green function, fourier transform etc) used in quantum physics
OK, given your comment at 10:10 I absolutely need to hear about quantum groups. The similarity between $xq$ and $x+dx$ is indeed striking. Where does this end up going?
I really want to know if this has analogues to differential forms
The right bracket in the recall should be switched with the factorial symbol, as in the line above.
yes for quantum group videos!
Quantum groups meets clofford algebras and representation theory. I want a series on that. Not just a video 😈☠️💀☠️😈
I need four series: Clifford algebras, representation theory, quantum groups, and the combination of them altogether.
Hey @MichaelPennMath, following the same logic, could we define an "Exponential Derivative"? something like the Lim (a->1) of (f(x^a)-f(x))/(x^a-x)
And how about quantum sum and chain rule?
21:47
3 minutes late
I wanna learn about Quantum Groups, please!
Head to squarespace.com/michaelpenn to save 10% off your first purchase of a website or domain using code michaelpenn
Group contraction and quantum groups
Quantum Groups sound very interesting and curious to me ( I heard that curiosity might have killed the cat or not - nobody is sure) 🙂
Why did the Chicken cross the road?
I looked at quantum groups. I understood about 10% of it. Please do a video on them.
[+] Add to... "Math References"
Nicely done!
Let's see some quantum groups
Neat seeing how multiples of 1 get exploded when you leave the derivative parameter as it is rather than just taking the limit. On the other hand, while this is an interesting generalization of the derivative, at no point did it feel like there was anything "quantum" about it. Maybe in quantum mechanics, q instead approaches some quantum operator rather than 1, but it could just as easily approach the identity matrix, or just some number that isn't 1 like 0, 2, maybe even i.
Actually, I'm pretty sure the discrete derivative is what you get when you take the normal derivative and have h approach 1 instead of 0, which is another instance of changing how the the parameter is interpreted. Having q approach 2 (or maybe e) is probably the closest thing for the relative derivative to what the discrete derivative is for the absolute derivative. (Because they come down to relative change in x vs absolute change in x, and because it makes more sense than slapping "quantum" onto it.)
The q-deformed derivative (the one shown here) is used in what is known as “quantum algebra” which is used in some quantum mechanical stuff (for topics relating to anyons and such).
A special case in “quantum groups” (at least in the case of su_q(2) ) is when q is a primitive root of unity.
The Fibonacci category (related to Fibonacci anyons) can be constructed by starting with getting rep_q(su(2)) and setting q=exp(2pi i / 5) , and then doing some other stuff with quotienting.
Dear Michael. Regarding fractional derivatives, is there a 1/pi -th derivative or generally irrational fractional derivatives? I have the naive idea that if there's a half derivative, then there's also quarter, eighth etc. derivatives. But irrational numbers might change that.
I mean, I'm not gonna understand a single sentence, but of course make a video on quantum groups. How could you not with a name like that.
I'm now idly wondering what happens if q, instead of 1, approaches another root of unity. I see that q approaching i gives a nice periodicity to the q-analogs, for starters.
the unit circle is dense with roots of unity, so if the behaviour depends on being a root of unity specifically, it would be impossible to take the limit about any point along the circle since it would also be dense in non-roots-of-unity (e.g. if it depends on what order of root of unity it is, then it would definitely not be possible to pass through the circle and get a continuously morphing picture). For a given q-analogue of a complex function (bivariate as f(z,q)), if that happens, we can't analytically continue over q through the unit circle, but if that doesn't happen we might be able to.
15:48 why is the product rule seemingly not-symmetric wrt f and g? I would have expected to get the same thing if I swapped f and g
The explanation of the quantum derivative was pretty much perfect. Nothing to wine about.
But, what is the quantum version of a function? I am sorry, but that's where you lost me. I suppose that the limit of fq(x) for q to 1 should be equal to f(x). But, if this is the definition, then you can all kinds of quantum versions... Sorry, it's probably very simple to explain, but I lost it somewhere!?
This is a wild guess: The standard derivative of the quantum version of your function is equal to the quantum derivative of the non-quantum function equivalent? Is this correct?
I don't know either, but it seems a bit arbitrary.
For e^x as a power series, why take ([n]_q)! instead of [n!]_q? The answer is because it works. It solves f'=f in this q-analog world. Not sure if there's an underlying structure that determines this, but the fact that this works is motivation enough. - unless it's not supposed to work, but we should try to make it work at least until we know better
How do you deal with things breaking when x is 0?
Can we find the q-derivative of a function at x=0?
Looks like to me that Dq(f(x)) != f'(x) when x=0 as then Dq(f(x)) is undefined
Intriguing :-)
Whats is the integration of lnx in q-calculus
I feel forced to ask about Quantum Groups. Please explain!
idk what this is
I can only tell that D_1 is prob just normal old D
D_-1=???
Wait, there is a problem with the second proof, you assumed that x is not zero, otherwise the change of variables is not defined. Looking at it, seems that the quantum derivative is not defined when x is zero
Take limit as x->0 I guess
QGroup = Group / Q
is this quantum physics x calculus???
How does the substitution work when x=0 since we’re dividing by x?
It doesn't, it's undefined. But that makes sense since 0 has no multiplicative inverse in the first place
@@eqwerewrqwerqre so functions have no quantum derivative at x=0 and f_q’(0) doesn’t exist?
@@eqwerewrqwerqre and what does multiplicative inverses have to do with it?
i'm guessing quantum in this case is just a name for these types of functions and have nothing to do with quantum physics. am i right?
Mm, I wouldn’t say “nothing”, but, I do think it would be better to call it “q-deformed” or “q-analog” rather than “quantum”?
The q-deformed derivative *is* used in a specific sub-field of quantum physics,
but I don’t know that it is exactly because of it being quantum mechanical? It might be possible to use what is known as “quantum algebra” (the q-deformed derivative is used in “quantum algebra”) in a physics-but-not-quantum-mechanics context. Or, even if that doesn’t work so well, the q-deformed derivative isn’t particularly closely tied to the core ideas of quantum mechanics.
It’s a simple enough, and widely-applicable enough, idea, that naming it “the quantum derivative” is imo not the best name,
But, it *is* used in some quantum mechanics based stuff.
Has anyone commented on (f(xª)-f(x))/(xª-x) as a->1? Just curious.
Hmm, x^a = x x^(a-1) ,
x^a - x = x (x^(a-1) - 1)
f(x x^(a-1) ) - f(x)
If we set q_x=x^(a-1) , your expression becomes
(f(q_x x) - f(x))/(x (q_x - 1))
Which is (1/x) times D_{q_x} f(x) ?
I think probably it would be better to have, instead of x^a - x as the denominator, to instead have x^(a-1) - 1 , as the denominator.
That way you do get something derivative-like.
Specifically, you get the q-derivative except that q depends on x and on a, specifically, q = x^(a-1) .
I don’t mean that to say “and therefore it is just the q-derivative, and not a usefully separate thing”. The q-analog can also be thought of as the (f(x+h)-f(x))/((x+h)-x) thing except with h depending on x,
but they are both useful for different use-cases. (Similarly the h version can be expressed as the q version with the q depending on x (except for at x=0 but shhh).)
So, the version with x^(a-1) in place of q might also be useful for something, idk.
Uhh…
(ln(x^a) - ln(x))/(x^(a-1) - 1) =
(a ln(x) - ln(x))/(x^(a-1)-1)
=(a-1) ln(x) / (x^(a-1) - 1)
Hm, weird,
I’m not getting the right derivative for ln(x) when I take the limit as a-> 1; I must be making some mistake.
How is the quantum analogue [n]_q defined?
What do you mean by [n]? I've come across multiple definitions of that syntax.
There are iirc two main conventions. One of them is [n]_q = (q^n - 1)/(q-1) for q≠1 and n for q=1
(Or, equivalently in both cases, \sum_{k=0}^{n-1} q^k )
But there’s also a more symmetric version, which is iirc,
[n]_q = (q^n - q^{-n})/(q^{1} - q^{-1})
And also I think a slight variation on that one.
The first one, (q^n - 1)/(q-1) , is probably the one being referred to here.
What the F is a "quantum function" in Maths? LOL
It is the first time i hear of this concept.
I guess I have to watch the earlier videos too...
Does [n!] = [n]! ?
My feeling is that when he wrote [n]_q! he meant [n!]_q, as he wrote on the following line, so [n]_q! was a typo.
EDIT: I have since changed my mind about this, as per my next reply below.
@@MichaelRothwell1A factorial of a formal sum does seem a bit less defined for sure.
@@jounik actually, since I wrote this, I came to the opposite conclusion - he is defining [n]_q!.
The reason is that [n]!_q is already defined, but doesn't give what we want, which is given on the next line. Therefore, I now believe that, in fact, the [n]!_q on the next line is a typo, and should be [n]_q! :)
@@MichaelRothwell1 You do have a point. Taking the notation as a _definition_ of an accumulated product of sums at the limit is of course an option.
+1 quantum groups. And that’s a good place to stop😁😁😁
Here is an issue:
The quantum derivative showed in the video is undefined at x = 0.
Since it is `q*x - x` aka `0q - 0` = `0`,
we just get a literal `0/0` and it just breaks.
So when q->1, D_q(f(x)) = f'(x) except at x = 0.
What will other quantum functions we derived behave at x = 0?
It doesn't "just break"; there is still a limit as x approaches 0, so it's a "removable discontinuity"/"removable singularity".
The derivative isn't defined for all functions either but apparent "breaks" like this are sometimes resolved by simply applying l'Hôpital's rule.
I think df(x)/dx = lim y→x Dq f(y) for all x; I wonder if that limit is actually part of either the definition of Dq and Michael just skipped over it to slightly simplify things.
That's a sound objection. The definition doesn't work for x=0.
But you may be capable of fixing that if you set x=y-e, e being some arbitrary number greater than zero. f(x) = f(y-e). Then you calculate the quantum derivative for y=e and define this to be the qD at x=0. In the denominator you've got e(q-1), which is not equal to zero.
@@schweinmachtbree1013 It a limit of q->1 NOT x->0 so it is undefined in the context of the video.
Why do we need quantum derivative in quantum physics can someone wants to explained to me? I assume the q introduces some kind of probabilistic efffect.
Such as the most probable event is when its equals to the derivative.
My suspicion is that "quantum" here refers to discrete jumps in the variable, not about probabilities. If you take q > 1 you get a finite difference (a quantum jump) instead of the instantaneous derivative (when q goes to 1)
@@radadadadeeI suspect it is called this because of its relevance in quantum groups, which are deformations of the enveloping algebras for certain groups. Deformations in terms of a parameter q which, as q goes to 1, recovers the enveloping algebra for the original group.
These are studied in quantum mechanics for… reasons, but generally not at the start of studying quantum mechanics?
Sometimes the q is thought of as being like e^hbar , except not literally hbar because hbar has units, but when one forgets about units *and* thinks of classical things as being the limit as hbar goes to zero, then that has e^hbar going to 1.
… but, also, really q in quantum algebra is often complex valued, which doesn’t really make sense for hbar either, as hbar should be real valued and non-negative.
One place that it is relevant is in stuff about anyons (which are an alternative to Fermions and Bosons, but which only make sense in 2 or fewer dimensions of space, but quasiparticles in a flat surface medium can be anyons).
One can take the category of representations of the q-deformed version of (the universal enveloping algebra of) SU(2) , (the representations of SU(2) are/correspond-to the different possible values of “spin” for a particle),
and if one picks q=e^(2 pi i/ 5) , and then quotients out by some stuff, one ends up with “the Fibonacci category”, which describes the Fibonacci anyons.
Iirc, these theoretically could be found if you tune things just right, in fractional quantum Hall effect demonstrations/setups (but it is very difficult, as the region to get specifically Fibonacci anyons is really small I think?)
Despite all that, I think calling it “the q-deformed derivative”, or “the q-analog of the derivative” would probably be a better name?
It doesn’t sound as cool, but it is less confusing, easier to explain why it is called that, and, also, it doesn’t sound cooler or more mysterious than it should.
Ok. Seriously bro. Is there any math you aren’t well versed in? What is the hardest math there is?
Yes, I want to see videos about anything, thank you.
Michael, is the quantum derivative isomorphic to the partial derivative in any cases, maybe with normalization? I’ve never heard of this before but it seems like it could even be applied to polynomial basis functions for fitting a regression to data when you have q as a constraint. Any thoughts?
The partial derivative? As in, for functions of multiple variables? I don’t see a connection
@@drdca8263 I was thinking there could be a correlation to the gamma function half-derivative (or partial derivative) for some class of functions. I haven’t thought any more deeply about it than that. Since we have a polynomial with rational coefficients maybe we could make a basis like the Bernstein basis for fitting curves over data with a constraint, or the Grobner basis.
@@michaelzumpano7318 Oh! *Fractional derivatives*
Yeah, calling those “partial derivatives” confused me, because that’s the name of a different thing.
Hm...
Well, the fractional derivatives iirc aren’t local, in that their values near a given input depend on the whole function,
while this, the quantum derivative..
well, it kind of only depends on the value of the function at 2 points, but one of those two points depends on q, so I guess that might not really be too different?
Though, if you suppose that q is in a neighborhood of 1, then it only depends on a neighborhood of the point, so it is kinda local?
Hm, idk, maybe they could be connected.
Am ENTIRE 20min video and not ONCE is there an intuitive explanation of what quantum functions are or why quantum derivatives are useful.
-~~-
better write q as Wau
Hey can you do a video on manifolds, metric spaces and the like. I can’t find much content in that area online.
🚬
Wow, now i know the term "quantum" means
Pretty boring video. The mention of its uses applied to Combinatorics Groups might help prove this video useful.
I've never heard of this. What is this stuff used for?