 So I'm just going to start by recalling some things that we said from last time. So we defined this notion last time of this like blocking coding where we said that we have a unitary U which was a blocking coding of matrix A. If we can write U in this form where U has like an A and it's like top left block or top left as a sub matrix in the top left and then the one way I wrote this was I defined this like I sort of had this like selection operator that like went into this like top left corner and I said that this was equal to A or alternatively like if the dimensions match up you can write it in terms of like selecting the Q bits. And here the Q is like the gate complexity of U. And what we wanted to where we left off was I had this statement that I sort of asserted and I didn't prove which is that if you have a blocking coding sorry if you have a okay so we're looking at like polynomials applying polynomials to blocking coding. And we had some polynomial that's even or odd and it's bounded by one for all x in minus one one. And in this setting I stated that we can take a Q blocking coding of A and then we map it to something like I'm going to write something here it's like maybe D log D plus Q blocking coding of the polynomial applied to A. So recall that here this polynomial just this PSV just notes if you have x cubed it gets converted to like A, A dagger A and like if that's x squared it gets converted to like A dagger A. So what I'm going to do in this lecture is try to just prove this theorem. So yeah that's a great point. So here I might make this mistake more but so just keep that in mind if there's any more issues with this. But P is degree so I'm going to I made this mistake here so it should be like okay so I'm using I think I'm going to use D for the dimension and N for the degree yeah okay so I'm going to prove this. So it might be a little bit tedious at times so hopefully it'll be illuminating because I think it's a it's a what we're going to do is we're going to sort of analyze the expressivity of some types of quantum circuits and I think that's a useful exercise okay any questions before I before I begin B is a blocking coding yeah any other questions so how this is going to work is that first of all I'm going to consider this notion of quantum signal processing which is basically if you're looking at a unitary that's like two by two and my and it's blocking coding something that is just like a real number and I'm going to show how you can basically so here it doesn't make sense to call it a anymore I'll call it like X and then I'll show how you can get something that has X in the top left to something that has like pfx in the top left or something like this and then there will be some like lifting to get this to work for arbitrary matrices and this lifting will basically it'll show that the same protocol will work but if you just define it in the right way okay so first off I'm going to discuss quantum signal processing and the idea here is that we have some matrix that we can apply as a black box but sort of we think about it as like a blocking coding so like we know it's a blocking coding this X here but we don't actually know what this X is okay so I'm calling this R and what we can try to do to interact with it is sort of interspersed other rotations other single qubit rotations that are going to look like the form here and the idea is that like okay this comes from like signal processing intuition but so I don't know if this is like exactly right but like maybe R of X is their signal and then we're somehow like controlling our signal and then getting something that's like able to transform without necessarily knowing what the signal precisely is okay and so we're going to define a QSP circuit which is it has a bunch of has a bunch of individual real numbers which are going to be our phase factors I guess I should say we're gonna denote these 5j okay and how we define our QSP circuit is it's going to be equal to basically I'm going to interleave these E these rotations with these R of X which happens to be reflections you could also think about it as rotations in the X poly X basis in the rotations in the Z and then you interleave these here this is 5j and this is 5 0 and the sigma Z is okay here we're taking a rotation with respect to the Pali Z basis okay and okay where this is like a product we're imagining the product is being like going from and on the left-hand side to one on the right-hand side okay and so the question is what you can achieve with these sorts of circuits and so based on this we can say that a polynomial is QSP achievable if there exists some phase factors such that QS the corresponding circuit is going to give me something with like a P of X in the top of corner okay and we want to figure out which kinds of polynomials are QSP achievable and and the way that you can do this is if you sort of just look at the first few terms and then sort of keep working your way up you can do it as said there's some recurrence that my entries of the QSP circuit satisfy and what this is in particular is that if we consider if we consider QSP of some of let's say k plus k phases or something like this okay so if we only consider performing our QSP up to the case phase factor like this I guess the application of R then we can write this in the form so p of X so I'm gonna so it's gonna take this form so basically I have and here so I'm just saying it has some form where these p pks and qks are all polynomials qk this bar means conjugate so it means I'm taking the coefficients of the polynomial and conjugating all the coefficients and so I'm going to say that it looks like this and then these pk satisfy the current recurrence that I'm going to write down here and we have some sort of base case that comes from right with QSP of we only look at one phase factor this is going to look like e to the i phi e to the minus i phi because we're just we're not applying any R it's just one of the it's just this this final term up here and so we can get stats we get that this is the base cases like what is it p0 is equal to e to the i phi and then q0 is 0 yeah okay and then you can see that it does indeed like take this form if I plug in these values this is precisely this okay so to prove this is just some annoying calculation but I'll go through a little bit of it and the idea is that just that like if you wanted QSP like you just want to compute what happens when you add another phase factor another piece of your circuit so I'm considering up to k plus one and then this is going to be I'm gonna peel off the the first the first element of the product and then I'm going to have some the rest of it okay and then we know that it takes this form from sorry we know that it takes this form from here I'm just going to copy it so it's this guy and then there's also this e to the i phi k plus one times R and what this guy is is it's yes I'm just gonna do the multiplication okay and here you can see that what happens when I multiply things together is that okay so I have like a term here this e to the i phi k plus one times px this gives me this like first term of my pk and then this second term the square one minus x squared they they multiply together and then so you don't get any root terms in my polynomial here I and if you want us to check you could see that like if I look at this like bottom corner term I have the same thing but instead I have this like one minus x squared e to the minus i phi k and then q bar of k of minus x so that's what you get if you take this px pk and then take the conjugate and then a sub in minus x so you're so in this bottom right corner you're getting like in this bottom right corner you are indeed getting something that looks like pk plus one bar okay so hopefully they should just confirm to you that this is like indeed what I'm what I'm saying it is and so now that we have this recursion we can sort of see what is happening to our polynomials and so one thing to there's a couple things to note here first of all it how it starts off is that at p zero at at k equals zero my p is even and q zero is okay it's a zero so it's even or even and odd but I'm gonna say it's odd and whenever you compute this pk plus one and qk plus one if pk is even and qk is odd then pk will flip the parity then you know so if this is if pk is even then x times pk is odd if qk is odd then one minus x squared times qk is odd so what's happening is that if p zero q zero are even odd then p one q one will be odd even and you can keep going and then see that for example you know like you know pk here if it's even and qk is odd then x qk is even and so the whole thing is even and so on okay so you can see that basically what's happening is that we're building up these polynomials so these polynomials are increasing in degree by one each time is the other thing to notice because we start off with p zero being degree zero q zero we're saying is degree minus one and here we add one degree to pk and then we add two degrees to qk minus one but because qk is one degree lower then they'll be degree they'll be degree k or pk plus one will be degree k plus one or at most k plus one so there's like some things that you can notice about this recursion and okay so what you can actually show is that the sort of set of requirements that you need is essentially like what you can just visibly see from these recurrences so if you have some degree polynomial it's qsp achievable so we can find phase factors for it if and only if the following holds if and only if there exists some q so here this is like the q from above such that first of all q has degree at most n minus one secondly p and q are like either even or odd or odd or even more precisely right because we said that p is degree n p will be whatever parity that n is and q will be the opposite parity from that and the final thing that we need is that this equation holds okay well where does this come from if we look at our lemma we can see that what this is is we're saying that this column has norm one that's what this equation means and so this is immediate so like if you have a qsp phase sequence then the polynomials the implements must satisfy this because this qsp thing is a product of unitary so it's unitary okay so so this has a norm of one for these qsp achievable polynomials and this is the constraint that we see that we see showing up okay so hopefully I've argued that this direction at least most of this direction so if you have some qsp achievable polynomial then you can look at the source corresponding p and q that you get from this above recurrence and then see that it satisfies okay that you have your q that's degree one lower at least one lower and if they're opposite parities p and q are opposite parities and that you have this property that this has a unit norm so this should just follow from the lemma and like unitary of qsp okay now in the reverse direction where we have p and q that satisfy these properties and we want to show we want to show what we want to give like an explicit construction for what the phases are or just show that some phases exist okay so how we can do this is we can do this by induction by the way are there any questions so far about what I've talked about it's like a little technical yeah so here I even odd this basically means p is odd or and q is even or p is even q odd and here by even I mean that p of x is equal to p of minus x and by odd I mean p of x is equal to minus p of minus x does that make sense yeah yeah I guess like the recursion is what's occurring like every time I do my reflection operator and then my rotation operator so like I'm imagining I'm doing these in in in sound however however many steps well your quantum walk operators well your I think I think typically the quantum walk operator is defined for two of these time steps I'm not I'm not sure though like the typical one that you might see for like rover or something that's what you mean right I think that's typically like two of these time steps but it's like the same principle I guess yeah so so what I did was I I defined this notion of achievable which means if polynomials achievable and I can get a blocking coding get from a blocking coding of a to a blocking coding of p of a so what I did as I constrained this to look at scalar guys so I now I'm looking at there are two by twos with like the top top left being a scalar and so you can still define this notion of achievability here but only for like scalars I'm looking at a subclass of this which is which polynomials are achievable with these particular kinds of circuits so we're going to prove which ones are QSP which polynomials are QSP achievable this will all of the QSP achievable polynomials will turn out to be achievable and furthermore yeah I guess this is like not this is this will follow by some like lifting argument and there's some additional extra step what we what we want is actually some things that are like slightly outside of this achievable QSP achievable class and I'll show how you get those okay okay any other questions okay so arguably I have one more annoying sort of matrix computation to do but the basic idea is that like okay I have my p and q and I want to figure out how do I construct phase factors from this and so how I start is I suppose I suppose p and q satisfy he's like one two and three then what I can do is I can sit I can try to figure out I can try to work backwards and try to see like if I have some degree n polynomial can I pull off one of my steps of my quantum walk and then be left with an n might degree n minus one polynomial and so if I can do that then I can sort of apply induction to say that my whole set of my whole polynomial can be described via phase factors and the thing to note is that like there's like a base case is like a base case of n equals zero and as I mentioned before what this what this what this corresponds to is like p being like e to the i phi some like root of unity and q being zero so this is so so this is this is my base case and so then this corresponds to sort of as I might have alluded to you can just look at the look at these p and q and then you see that this is a qsp with a phase factor of like phi where you where you're just not applying x at all okay so what did I say I wanted to do I said I wanted to take like sort of pretend that I could implement my function and then I would get some some matrix of this form and I want to peel off one layer of like my supposed qsp circuit so that's one layer and I'm taking the inverse so I'm peeling off one layer and then what I want to say is that this with this for particular choice of of var phi that I can get this to be equal to some I'm calling it p p down arrow and q down arrow here but these guys hopefully will be one degree lower and so the thing to notice is that if these p down arrow and q down arrow if these are one degree lower and however I define them if they happen to be like the opposite parodies you know even an odd or or respectively or like vice versa then then I can imply my inductive hypothesis to get that this is equal to qsp of some phi so here I'm okay so here this case is like so inductive hypothesis I'm taking p to be degree n plus one here okay and then I'm looking I my inductive hypothesis as I can find face factors for this degree n thing okay and the reason I can do this is because it'll satisfy all the criteria so the main thing to check here is that your your equation here satisfied for the p and q down arrows and this will follow because by my assumption here that p q satisfy three this matrix is unitary and so I'm multiplying unitaries by unitaries and so this matrix is unitary and so you'll have that this column satisfies your equation and so you can apply the inductive hypothesis and then you'll get like then that then this will imply that that PSP is a qsp achievable and the face factors will be well it'll be phi zero through phi n and then there will be this additional var phi if I can find what it is okay because I like I take this qsp and then I I move this this piece to the other side and then I'll get an additional factor of face factor of phi or phi okay right so what is this expression well I'm going to what this is is I'm going to what you end up doing is to do this is you just do the computation you like take this e phi of r phi I see you take this one step of my qsp circuit I use that ours a reflection so it's its own inverse and so this will look like the following okay and if you multiply these guys together you'll see that what you need your what these expressions are or what you what these p down arrow and q down arrow if you like pattern match between the two you'll see that they need to be the following okay so this is like I guess pretty similar to the recurrence relation from before if you remember that but the question is like can we choose some phi so that these two have degree lower than the degree or this p down arrow has a degree lower than the degree of p and this q down arrow has degree lower than the degree of q or at least like lower than degree and now the way that you can see this is so by my assumption I know that p is some degree and plus one thing and it is has parody the same as parody the parody of n plus one so namely this means that I can write this as a n plus one x to the n plus one plus I just basically I just want to look at like the highest monomial like sort of the leading coefficient and seeing if it can become zero and so on and q of x is going to be equal to some bn x to the n and so on and okay what's going on here is that if you look at the leading coefficients of p down arrow and q down arrow the leading coefficient so it's like this is like degree this would be the degree n plus one thing for for the p and then this would be the degree n coefficient for q these turn out to be the same and this is e to the minus i phi of n plus one plus e to the i phi of bn there's a minus sign okay and so the question is can I find some var phi to make the zero and the answer is yes the thing to use is that what you can just say is that a n and a n plus one and bn have the same magnitude and the reason to see this is that in this equation okay if you look at the leading coefficient of here the leading coefficient is a n plus one squared and the leading coefficient here is minus bn squared and we know that they cancel out right so there's a term here that looks like x to the 2n plus 2 and this will have a term that's x to the 2n plus 2 and because this is equal to one it's equal to one as like polynomials so you know that these two cancel in so that you know they have the same magnitude and so you can just choose just choose e to the i phi to be like like square roots of bn over a n plus one something like this and this is going to this is going to be some root of unity and so or not root of uni but like some something on the unit circle and so you can actually cancel okay um so that's like the proof um yeah where yeah yes um yeah I think that's a typo in my notes okay yeah yeah thanks any other questions yeah so what I want to do is I want to take some my degree and polynomial I want to find some degree n minus one polynomial such that they're the same properties such that if I apply one more step of my qsp thing I get to p and q and so I just want to find some value of phi that will achieve that'll achieve this and I sort of like worked in reverse a little bit but this will show you how you get this value of phi and you know about that if you wanted to at the top of the statement you could say we choose this phi to be this this particular value based on the monomials and this will give you like then you could like you know if you wanted to actually do this which you shouldn't because it's not numerically stable but you could then like you could take your p and q and then you could work backwards work all the way down you get to down arrow and then further down any other questions okay so there's one more characterization that I wanted to say which I'm not going to prove because of time and what this says is that so I know what my p and what p and q are that are like what my p and q can be such that these are p is qsp achievable achievable but this is actually like pretty constraining it's not clear that if you give me a polynomial p that I can find such a q that satisfies this but it turns out if I relax if I allow myself some more room if I like give up control over the imaginary part of my complex polynomial then I can actually get any real polynomial to be essentially achievable so I'm going to consider two polynomials to call p real and q real and these are going to be real valued polynomials so so I'm going to say given this there exists some p and q that are qsp achievable such that p the p real is like the real part of p so so here these two can be complex valued or have complex coefficients and so basically I can find a comp like complex I can I can sort of take these real polynomials I can add some imaginary part to them and when I add my imaginary part they'll be qsp achievable that's why I'm saying I can do and I can do this if and only if well first of all like there'll be two properties but the two properties will be the same as these two properties up here so the properties will be q is degree and most n minus one secondly you'll have pq are you know even or odd or odd or even and the final thing that you need is that you should all be the real parts it should be the exact same thing as before exact same equality but instead of being an equality it's an inequality so I'm just saying like I have my p real and my q real they satisfy everything that I want except they're maybe not something to one and then what I can do is I can add imaginary parts to make it some to one and if I don't really care about what my value of q is I can just set this q to be like zero or yeah and so this will give me some p that's qsp achievable and the real part of p is this p real of my choice provided that it is a most one okay so this is like how you get so that you think from this you can get like sort of how you get that p real is a cheat is a achievable in this like two by two setting is that you get some p that's qsp achievable such that the real part of p is this p real and remember that this the only constraint that we really needed is that this is even or odd and also that this is bounded by one so this is okay this is for all x and minus one one by the way okay and then what we could do is we could just say that like so p is going to be equal to some p real plus like some imaginary part help p imaginary and so I can achieve this with some phase factors phi I can also achieve the conjugate with the phase factors minus phi so I don't know if that's obvious but and then once you have that both p and p conjugate are qsp achievable you do like the linear combinations of unitaries to get that one half of p plus p bar which is your p real is okay well you're doing a linear combination of unitaries of a qsp circuit so it's not like strictly qsp achievable but it is achievable and so this is how you get that any real polynomial you can achieve at least in this two by two case where you're like doing a blocking coding of a scalar guy are there any questions I'll show how to lift this okay right so I don't have much time but I'll try to explain so recall that what we had is we had some in the setup for the beginning we had that you is a blocking coding of a which means that we can write you is some block matrix where a is in the top left corner and we have some other matrices in the other parts and how can we and what we have is okay I'm going to define a couple operators so I'm going to define pi r as being the identity so here this is r by c and so pi r is just gonna be the projector onto the part the a part of the right-hand side and I'm going to define pi l to be the projector onto the left-hand side and the way that I define this is so that pi l u pi r is equal to a zero zero zero so I'm going to define these operators and so what I was doing before is I was sort of taking my original matrix or start taking my scalar x and I was doing some rotation sort of between x and these like other parts of my matrix and so you can sort of pattern match and try to figure out what the right thing is in the more general setting I'll just tell you so if we have some phase factors we can define this quantum singular value transformation circuit as being basically the product of I'm going to write it as the following there's some now some issues of like parity so I'm I'm just gonna try to write it for the even case here so again I'm I'm like interspersing I'm interleaving my applications of the u with applications of some rotation and what this is going to be is it's going to be with respect to the projectors so so here to pi l minus identity this is a reflection across the subspace defined by pi l same for pi r and we're sort of applying you and then we're doing this rotation and then we're applying you dagger and then applying another rotation and so on and this pi is actually a product here so I this like it should be like J and it's like should be like J from 1 to n over 2 or something okay and okay how should I be looking at this yeah question this is Phi 2 J minus 1 this is Phi 2 J is that what you are asking okay if I remember writing and it's confusing they're also the lecture notes which I'm copying off of any other questions what's the goal of which one yes I'm defining this u Phi to get a blocking coding of polynomials of a yeah so right the hope is like I can pick some value of Phi such that what I get is p of a and there's a result that you can show okay first of all you should notice that okay if indeed I'm looking at my you and my you is r of x blocking coding x then this u Phi is precisely the same thing as the qsp circuit you can stare at this and like see that this is the case and the main result here is that if p is qsp achievable with these phase factors Phi then u Phi is going to be a blocking coding of p of a so I'm saying that if p is qsp achievable then it is also achievable in the blocking coding sense right okay so how am I going to prove this I'm just going to show you what happens when I write everything out in terms of block matrices and you should hopefully see sort of what's going on okay so I'm gonna take specifically I'm gonna look at what's happening with this qsp t this u Phi that I just defined and something I'm going to notice I'm just first of all I'm going to write down what these these guys are okay and I'm going to not really I'm gonna sort of be a little bit loose with defining dimensions here because when I define these they're all gonna be different dimensions but the products are going to work out so it'll be fine in the end what this is is it's e to the Phi 2 j times I e to the minus I if you just look at what this guy is it's a block matrix where you have some it's basically if you think about it it's sort of like the same thing as my rotation from before except I do it on a large subspace all at once okay similarly I can say that e to the I with the sorry this is like this is in the exponent just to be clear 2 pi r minus I and what I'm getting is it's gonna be exactly the same picture but the dimensions are gonna be different so hi Phi 2 j plus 1 times identity okay and the final step here is what I'm going to say is like okay I'm gonna call okay for the final step I'm going to use this decomposition called the CSD composition and what this states is that if we consider a block matrix of the form if I have this like blocking coding that I mentioned before then we can actually write it in terms of a bunch of simultaneous singular validity compositions okay so what's happening is I am taking this so V what I'm doing is I'm getting a an SVD for every block and they sort of pair nicely together in the sense that my left singular that vectors of a are also my left singular vectors of d1 of like you want to like my I have the same left singular vectors for both of these matrices and I have the same right singular vectors for both of these matrices that's what's happening there and now what I can do is I can notice that if I look at what this guy is this V so what I have here is I have this matrix you okay my matrix you I'm sort of interleaving with these polynomials here so you and then on the left hand side is going to be one of these pi L's right so you have a U and then a pi L and then a U dagger and then a pi R the thing to notice is that to make it simpler what's happening is that this V can actually commute with this they're saying this pi L and if you look at what's happening and you compute this times this versus this times this they will commute again I'm being messy with dimensions but if you were trying to be formal this a here is this this these blocks are the same as these blocks in terms of sizes and then here V1 and V2 are square W1 and W2 are square and then these have the same dimensions as these so they're actually commuting genuinely okay and similarly W and my pi R commute and so if you look at what's happening right I had this expression that's sort of like U and then this e to the i phi pi R or pi L and then U dagger and then e to the i pi pi pi R okay and then I'm writing this as VDW dagger and then WD dagger V dagger this is the conjugate transpose I can see that these terms are in fact cancel out so U phi is going to be basically the same as like D phi up to some change of basis and then finally the thing to note is that what this guy D is if you look at what it is it's going to be a bunch of entries and the entries are going to be of the form sigma 1 and so on they're going to be of this form and so on so so I guess a way to write this is that it's going to be CSS minus C or S square root of 1 minus C squared so C and SR diagonal matrices and then if I look at every particular 2 by 2 block it'll be of this form of like a reflection of the corresponding sigma i where the sigma i's are the singular vectors of a singular values of a right so here my D11 which I'm calling C now are my singular values of B of a and then I'm doing this rotation operator with effective sigma i so if you work it out what's happening is that when you're doing this QSVT you're doing simultaneous QSP on all of the sigma i's all of your singular values at once so that's like the most important thing and if you want to actually prove this what you would do is you would go back up here so you would notice the CSS minus C you would go back up here look at your proofs and then saying tactically replace all of the X's with C's and all of the square root of 1 minus X squared with S's and then it would everything would go through immediately so that's like how you would actually prove such a thing more than the lecture notes that I don't have time to cover but that's going to be it thanks