Transcript for:
Quantum Computing Insights and Upcoming Topics

Good morning everyone. Yeah, good morning uh everyone. Uh hope you are all doing good. So today we are on week three day one of this course and today uh Mr. Joti Shwarenji is going to start the session for you all. So Joti Shwari we welcome you all and please uh start the session accordingly. Thank you. Yeah. So, am I audible? Yes. Yes, you're audible. Yeah. Okay. Yeah. Thank you. So, yes. Uh, welcome everyone to uh the third week of our uh one month uh short-term uh quantum course. So, okay. So I have one question right off the bat. I will just address a couple of questions. So this is from uh one Sophia who is asking will tomorrow's quiz cover topics of the previous week only or this week two? It will cover only the previous week. This week you'll give it next week and the final week you'll give it the week after that. At least that's how we have planned uh things to go. Okay. So yes uh so I shall be handling uh today's uh uh lectures. Uh before I begin though a couple of announcements uh one tomorrow as we mentioned last week professor Bishin Singh Puna from the electronics and communication uh engineering department of IIT Rudiki will be joining us for the uh uh sessions. So he shall be dealing with some uh uh first he shall be talking about some physical uh consequences of the quantum effects and uh you know maybe talk a little bit about a particular kind of physical cubit as well. So whatever questions you have on the physical devices front and the physics side of things you can uh ask those questions tomorrow. So with that I shall begin my part of the session today. Let me share my screen. So yeah. So are my slides visible? Yes, they are visible. Yeah. Thank you very much. Let me open the uh chat window. I think I can close this. I don't need this. Oops. What else? Yeah. So, and I've also noticed another uh question from uh one of you asking me to go through some things. So, let me see how. So what we are going to do in this week is essentially a consolidation like in today's session I'll try to cover almost everything that was covered in the last two weeks giving some more context and uh realigning uh the ideas and so as to gear us up for the next week where we can talk about uh uh a uh another quantum algorithm. So we shall be looking at only two quantum algorithms in uh this course. The first one you've already seen last week which was the Deutsch problem. It can be extended to what is known as the Deutsch Josa algorithm. So I will maybe do that today. Uh but other than that we are going to look at another algorithm which is called the uh HHL uh uh algorithm. So it is the uh what is the first H? I keep forgetting. I think it's arrow hassidim and Lloyd algorithm. So we shall be looking at that in the next week. So in this week, yes, let me now go back to week one where I had shown these slides to you and again I briefly ran through these in week two as well. So I kind of stopped at this point. So yeah, so what I shall do is go through these steps and now I I will talk about what is called the block sphere representation of the uh single cubit state and from there I will you know start moving gradually one by one to other related topics. So give me just one second before I begin. [Music] Okay. Yeah. All right. Yes. So, let me go to the previous slide or let me go right here. Yes. So we know that we have a quantum state that is represented as a column vector AB where a and b are complex and in addition to that we know that mod a² + mod b² is equal to 1. Now I can write a as r0 e ^ i 50 r1 e ^ i 51 and then what I can do is uh what I can do is remove the e ^ i50 common and we now know that this quantity is called the global phase and it can be ignored without uh losing any uh physical uh properties of the system like a system with this global face and without this global face or with some other global face they are all equivalent physically they are all equivalent so then I just get rid of it and write my effective cubid state like this r0 get 0 plus r1 e ^ 51 - iota * 51 - 5 0 get 1 r0 is purely real r1 is purely real the only complex number here is a unit complex number e ^ i5 five as mentioned here. Now we know that r0 square + r1 square now has to be 1 because the magnitude of this complex number is r1 by definition because we are writing them in the magnitude and phase form. So r² + r1² has to be 1. Now what we have is this. We know for a fact that R0 and R1 has to lie between 0 and one because if the sum of their squares is equal to one then neither one of them can be greater than one and neither of them can also be negative because this is the modulus of a complex number. This is the magnitude of a complex number. This is always positive semidefinite. This is at least uh zero. This cannot be less than zero. So from this I can say that I will you know choose to represent R0 as cos thetax 2 and R1 as sin theta by2. Why do I do sin theta 2 and cos theta by2? Okay I have a quick question where can we get these slides from? These slides should already be available on the git repository. So as and when we cover these slides you know we just upload them to the repository. Please uh look up the repository. Okay. And I have another question. Global phase and the north won't change the state. Should the given be represented just by C and C * C? Okay. Uh uh. Okay. Superar, I will come to your question in a few minutes. Maybe I will cover the block sphere part and then come back to your question. Okay. So yes, now I'm representing R0 by cos theta 2 and R1 by sin thetax 2 and I'm saying theta can vary from 0 to pi. Now why did I do this? Well, for now just think of it as an arbitrary choice and additionally I know that e^ i5 which can be written as cos 5 plus iota sin 5. This is the oiler expansion for a complex unit complex number and this angle e the^ i5 which is essentially occurring between k0 and k1. So this quantity is known as the relative phase. And what we can see is I have one angle variable that goes from 0 to pi and another angle variable that is going from 0 to 2 pi. And with these two parameters what I can see is this actually defines a point on the unit sphere a sphere of unit radius. So these parameters are identical to the angle variables in the spherical polar coordinates. And therefore each cubid state is defined using these uh defined using these parameters corresponds to a unique point on a unit sphere. So therefore I can write my complete single cubit state as k is equal to cos thetax 2 k 0 plus e ^ i sin thetax 2 k 1. So every single point that I'm representing on the unit sphere is what I call the block sphere. So this was a concept that was introduced by Felix block uh essentially for representing optical uh states. We are utilizing the same thing to uh talk about the uh single cubid states as well. Mind you this is only for a single cubid state. Okay. So uh yeah so we are talking yes so this is how the single state s will appear. I have theta. Now you can see that when theta goes to pi 180° it will go from k0 all the way to k 1 because uh yeah notice r is cos theta by 2. So when theta goes to pi cos theta by 2 will tend to zero because cos p<unk> by2 becomes zero and sin p<unk> by2 will become 1. So quickly going back here r1 will become 1 and r0 will become zero. Therefore when I go all the way from here theta=0 to pi I will go from k0 to k 1. So this is what is known as the block sphere. This is the block sphere representation of a single cubid state. So as an example by our definition the kit zero state is always going to be on the north pole of this uh block sphere. The k1 state is always going to be in the south pole. The k plus state is going to be along the equator on the x-axis. So this is why we call the hadamat basis as also the x basis. So these are like three examples. And if I want to do a face rotation, you I mean uh we know that the angle phi corresponding to this line over here is actually the yeah I could have gone to full screen but yes angle corresponding to this line over here is phi equal to0. So if I want to change the phi I will essentially rotate about this equator or if I'm elsewhere if I'm say over here then I will rotate it about that particular latitude. So this is the idea of the uh block sphere and this is how we represent the single cubid state. So yeah. All right. So now that I've spoken about the block sphere, I will take some questions for a couple of minutes. Uh okay, I got a private question uh talking about something. Uh so I will just say this is from Nasim Kuchi. Uh I do not want to comment on your question. I'm sorry. So I will not uh read it out the chat as well. Uh so what other question do we have? We have okay what is the meaning of the square bracket and the curved bracket? Okay fine. So when I uh have two square brackets it means I'm including both these values in the interval. These are ways of defining intervals. When I have two square brackets it means 0 is also included and pi is also included in this interval. Whereas over here when I come to 2 pi it is the same as having pH equal to 0. So when I have the square bracket on this side, I'm including zero and 2 pi is not included. Okay. So that is the idea of uh that Artika K is saying that she cannot see the presentation. Please check the uh uh setup on your side because since people are asking me questions from the slides, I would assume that people are able to see the slides. So I have one question from Sdashas asking can mixed states be represented. I will come to this next week if we have time. Okay. So uh so Suda please check your audio connection again. I think I'm audible to most people here. Uh so we have seen ah okay so this is a nice question from Sujel. We have seen that K0 and K 1 are orthogonal but here we are getting an angle between them as 180. Can I please answer this question? Yes. So yes as you are saying the angle between them seems to be 180. But remember this state is actually corresponding to theta 2 here. Okay. So when I'm having 180° on the sphere I'm actually only 90° apart in the space of my choice. So remember that it's theta by 2 here. It's cos theta* 2 k 0 plus e^ i sin theta by 2 k 1. So the I'm already scaling my angle by a factor of half here by a factor of two. So since I'm doing that or rather a factor of half not a factor of two. So since I'm doing that going from 0 to pi is actually me going from 0 to p<unk> by2. Now why did I do this? I did this so that I can embed all my states onto a single sphere. This choice is deliberate. I could have had a different representation of these states too. Okay. So, uh the point is this and another thing that you may have guessed is every antipodal point. So, I will write this down on the blocks here. I will write this down on my slides. [Music] So, so the first thing we are talking about today is the block sphere. I don't know why I capitalized the the block sphere. So over here I know my general state kai can be written as cos theta by 2 k 0 plus e ^ i5 the relative is sin theta by get one and I know that theta belongs to 0 to pi 5 belongs to 0 to 2. Now you can ask what happens if I take five to beyond 2 pi. I will essentially just keep going round and round and round. That is why this is an open interval. Whenever I go to 2 pi, I will come back to zero. So whenever I go to another 2 pi, I will come back to zero. Okay. So that's the idea here. And uh yeah, if I want the sphere diagram, I will go back to the slides. I will not draw here. So now every point. So the key takeaway points here are every point on the block sphere is a unique single cubid state. and second antipodal points. So what are antipodal points? Antipodal points are essentially I think it's a single word it's fine. So antip-odal points are points on the opposite ends of the sphere where I can draw a diameter through the body of the sphere to the south like to the other pole. So it can be the north pole and the south pole or this point and this point or this point at this point or any point here and the opposite point over there. So these are the uh points on the block sphere. So these antipodal points will always correspond to orthonormal states correspond to orthonormal states. So as you can see as you can see to every point so I will write this here to every single point on the block sphere. There is only one point that corresponds to the orthonormal state. And this also makes sense because orthonormal state because first and foremost what we need to remember is in this we have neglected the global phase. So over here no global face is considered. So if I remove the global face then since this is a two-dimensional space I can either go along the axis get zero or go along the axis get one because of that for every vector I am going to have only the other dimension along which I'm going to have uh an orthonormal vector so for every single point that I pick there is only going to be another point that corresponds to the orthonormal state. So now I hope like people have given the quizzes now. So this was one of the quiz questions where I had asked how many orthonormal bases can exist for a single cubid state. You can see that for every point I can choose another point. So for every pair of points I have an orthonormal basis because those are two orthonormal vectors. So I can choose infinitely many orthonormal bases for a single cubit quantum state. So that is the idea here. Okay. [Music] So Okay. So Dan has asked uh my universe affirmed request has not been accepted. What to do sir? Please wait for some time. I'm sure they will process it and get back to you because as you can understand like thousands of people must have applied for it. So I guess there is a processing time for that. So please uh be patient. Uh okay. So there is something about the quiz. I will come to this in a minute. Uh okay. So there is another question from VS Pratik about the significance of the block sphere. The significance of the block sphere is just this. I have now a geometric representation of the single cubid states. So I can talk about different operations in terms of manipulations of this sphere. That is that. So the clock sphere does not tell me anything physical about the uh system. In fact, as you can see, we have started from the abstract representation. I have started from some generic quantum state representation and I've just manipulated the parameter. So, I'm not talking about anything physical here. This is a purely mathematical purely geometric representation of the uh block sphere. So, I have another comment from Kik asking every point of the block sphere should represent different polarization states. Of course, because the polarization of a single photon can be treated as the state of a cubid. So since that is the case, every point here corresponds to a different polarization. You are right. And that was originally the reason why the block sphere itself was the scheme was implemented. Okay. So yeah, do I have any more questions? Don't seem to see any questions on the YouTube side. So okay so Vashnavi is asking what about the negative states? Uh okay. So I need to clarify here. There are no like these are all points on the sphere like positive and negative only corresponds to their real line. Like beyond that even if I go to complex numbers and stuff there is nothing called a like complex numbers are not well ordered. I cannot write two complex numbers and say zed 1 is greater than or lesser than zed 2. I can talk about mod zed 1 or mod zed 2 but I cannot. So let me quickly make a comment here. So this is just you know comments. So the first thing is complex numbers are not well ordered. So what do I mean by this? I cannot take two numbers zed 1 and zed 2 and define always like sometimes I may be able to define but I cannot always define for any two zed 1 and zed 2 correspond like belonging to C. I cannot say zed 1 greater than zed2 or zed 1 less than zed 2. I cannot always define this. Cannot always be defined. Okay. So there is there are no negative states on this sphere. Yes, there are states where one of the coefficients is negative but this is a vector. Remember this is a complex vector and yeah so burbanjan singh I have addressed your question before. So essentially what I do here is uh the actual representation is theta* 2. So when theta goes from 0 to<unk> by2 what I'm actually doing is sorry 0 to pi what I'm doing is moving from 0 to p<unk> by2. So that is why like they are actually 90° apart but on my sphere they look 180° apart because I've essentially doubled the angle so that um okay yeah so that seems to be uh the question any reason why we so Shahik has asked any reason why we take r0 theta by 2 and not cos theta. The reason is I essentially want to achieve orthogonality within the sphere itself. I want all my points to lie on the sphere and at the same time I want to implement orthogonality as well. So that is the reason this there is no physical motivation for this. This is just a mathematical uh representation. Well, I just saw a comment from Mani saying that we cannot show minus get one. Okay, I think right let me quickly see this is K one. The vector facing downwards is K1. This point here is K one. Uh now if you're asking about a minus sign applied to K one then that's a global phase. I can ignore it. Remember that is where we started this whole argument from. Like this is the global phase. I can always and I will always ignore this. Okay. So another question from Rajes asking is relative face always associated with K 1 and not with K zero? Yes, as a matter of convention we do that. The relative face is always placed here because that makes life easier for us. Okay. [Music] So yeah with that I will put an end to the uh block sphere uh representation and I will move to the next topic for today. Yeah. So at this rate I think we can actually look at a couple more interesting things. So yes so the other thing that I wanted to address is the following. Sorry. So yeah, I think I will again this is kind of a retrading of what professor Babapad did last week but I'm doing it just to you know keep things like I want to give an introduction from my perspective as well because since I will be dealing with some more sessions uh uh going forward I think I will you know set up my own uh notation. Okay. So yeah, so let me say I have a two cubit system represented by state capital which is some small side chronological product or texer product with state kai. Now these two single cubid states can be represented as vectors on their respective block spheres. Now that we know what block spheres are, we can talk about this. So s1 is some cos theta_1 k 0 plus e ^ i1 sin theta 1 by 2 k 1 and k is cos theta_2 theta_2 theta theta_2 by 2 k 0 plus ^ i 2 sin theta 2x2. So I have a total of four parameters. Uh what happened to Dr. Gahubadai? He's busy this week. So I will be handling uh the sessions this week. So uh yes so I need and oh yeah one more thing which I can mention right here. Theta 1 and 51 are two real numbers. So what we have seen is once we know to get rid of the global phase and we already have the constraint that mod a square + mod b² is equal to 1 for a single cubid state. You can see that I can actually uniquely specify a single cubid state using only two real numbers just two real numbers theta 1 and 51 that's it. So while yes the vectors themselves belong to the vector space C cross C, I can have a unique mapping from there to the block sphere. Okay. So yeah. So there is yeah now I can also mention that I will just continue and I will say every valid single cubit state vector in B cross C as a corresponding point on the block. And secondly, the above mapping is many to one. So there are many vectors here which correspond to the same point. What are the many vectors? Those are the many vectors that are you know the uh vectors that differ by a global phase. Okay. uh is many 2 1 okay it's now going back to yes so now there are these four parameters now I can write kai tensor kai as this cos theta 1x2 plus theta 2x2 get 0 0 e ^ i 5x2 theta 1x X2 sin theta 2x2 get 0 1 e ^ i 52 sin theta 1x2 cos theta 2x2 get 1 0 and e ^ i 51 + 52 sin theta 1x2 sin theta 2x2 get 1. Now you can take this vector and take its inner product and find out that it is equal to one. So these states are also unit vectors. This is something that you need to remember that even when you're considering two cubit states ultimately they are also unit vectors. In fact any cubid state any n cubit state it is a unit vector in its corresponding space. Now consider the following vectors I'm obtaining from this vector. So this is k capital side. This is k capital side. This one over here. Okay, here I cannot select it here. So now I define three more vectors where I do the following. Notice what I am doing. I'm essentially swapping out the uh what I'm doing here is I'm swapping out the uh basis vectors. So notice Yeah. So see I have 0 0 1 1 0 and 1 one. Here what I've done is I have swapped what have I swapped [Music] here? Wait what have I swapped between this and this? Sin theta 1 + theta by 2. Yes. So over here I have swapped 01 with 1 0 and 1 0 with 0 1. Okay. So sati I'm seeing your question. Please uh like stay patient maybe later like maybe towards the end of the course. Uh uh so yes here what I've done is I've swapped the 0 1 and 1 0 vectors. As you can see the coefficient here is sin theta 1x2 cos theta_2x2 but previously it was cos theta 1x2 sin theta 2x2. So I have exchanged those two. Next what have I done? I have done the following. I have exchanged these two like I have essentially done this is cos theta 1x2 sin theta 2x2 so this one remains correct but the last two I've kind of interchanged them 1 0 has gone to 1 1 and 1 has gone to 1 0 and finally what the other thing that I've done is instead of writing this as e ^ 51 + 52 I'm just writing it as 53 some other angle It is no longer 51 + 52. It is just you know some arbitrary angle 53. The point is these are also unit vectors. I just took this and performed some manipulations to create these vectors and these are also unit vectors. But you know what the problem with these unit vectors are? I can no longer separate them like this. So these are all state vectors that are not separable. No, they are not like see remember. Yeah. So this is a question from Sedatma are all these states fight and circle. No they are not. Fight and circle is just this one. Now using fight and cirky I've just done some manipulations and obtained three different vectors. I'm just you know just defining three different vectors and without proof I'm stating here none of these can be separated into two single cubid states. So s can be factored so to speak like this. Si dash can also be factored like essentially oh yeah s dash can be factored. So this one c prime can be factored. What happens is I will essentially swap theta 1 theta_2 and theta_2 theta_1. So I will get s2. So from Rajes uh okay so I'm seeing a bunch of questions please yes hold on one of them has even stated the answer so I will come to that in a minute so yes s and s dash can be separated but s dash and criple dash are not separable so the two cubit states which can be factored into these two single cubit states are called separable states And the ones that are not well what are they? So let us take S double dash for example. So over here what I have done is I have interchanged 1 0 with 1 1 just for reference. Go back here. Look at these coefficients. I have e ^ i 51 sin theta 1 by 2 + cos theta_2x2. Over here I have the same term occurring at 111. So essentially what I can say is I've turned 1 0 into 1 1 and 1 1 into 1 0. So this is achieved by the following transformation. The transformation is this 0 0 goes to 0 0 1 goes to 0 1 1 0 goes to 1 1 goes to 1 0. Now we know this is the C not gate. This has been introduced. So I can just just like I defined my uh single cubit state single cubit transformations I can also define my two cubit transformations. How do I define my two cubit transformations? Let me move on to the presentation. So new slide. That six transactions. So if I have by transformation m then m can be written as a i j k i bra j s some sum over i comma j j belonging to 0 1. Okay. So what does this mean? This is equal to some a 0 0. Oh no, no, no, no, no. This is just this is going to give me four terms. I'm sorry. I should write IGK. Yeah, apologies. Or you know what? We already have the notation for that. Please remember from last week when I J belongs to 0 1 2 3. So what do I mean here? A I J K I project. So here get 0 corresponds to get 0 0 get 1 corresponds to get 01. corresponds to 1 Z. So this was explained last week. So please refer to uh last week's slides and video lectures. Uh the uh uh just a reminder the recordings of these sessions are available in terms of the YouTube uh not in terms are available on YouTube as uh recordings. So you can always go back uh and view them for reference. Okay. So get three is get one one. Okay. Similarly for bra. So so I can write this as a 0 0. Mind you this zero here is get 0 0. Okay. Plus a 01. I'm not going to write the whole thing. Just going to go plus all the way till a 33 A3. Okay. So there are going to be 16 uh such terms. Four on one side, four on the other. It's a 4 + 4 matrix. And the condition for this to be a valid transformation again it has to be a unitary transformation. So let me just state that here m is unitary which implies m dagger m is equal to m m dagger is equal to i. Okay this is the identity matrix. Okay. So again remember since m dagger and m are equal to mm dagger it automatically implies that you know these are uh uh square matrices you can also say that from this definition. So yeah that is that. So going back to my presentation. Yes. So after the definition of the C not gate I will take a break. So by using this I can actually define the matrix and the matrix has this form. So again this is something that we know. So now this is C12. What do I mean by C12? The first cubit determines what happens to the second cubit. When the first cubit is zero, the second cubit remains undisturbed. When the first cubit is one, an X gate is applied to the second cubit. So this C knot or CX is the gate. So this is the control not operation also known as the control X operation because well you know I will come to that uh later. Uh so you know what now I will take some questions and before that I will also take a clip of water. Okay, some more questions. Yeah. So, so Swabankar Mandi has asked every point in block sphere is a single cubit state. Does it confirm uh that the sum of probabilities is equal to one? Yes, it is one of the corary relations. Like see sum of probabilities equal to one comes from normalization and we are considering only normalized state vectors here. So it is not that block spheres are telling us that the sum of probabilities is going to be one. We kind of assume that to go with like the fact that we are taking uh only normalized vectors is kind of it's it's for granted. Okay. So we are not uh taking considering any other kind of uh state vector. Okay. So yeah that yeah once again let me just make this clear professor Tangapad is not in town he's out of town he's away on business that is why I'm handling this week's sessions so by their magnitudes why is it not always true yes so you are right we can only compare are complex numbers by their magnitudes. If I give you two general complex numbers, let me go back to my presentation and uh to my slides here. Yeah, the PPD here. See, I've not written mod zed. I've written zed 1 greater than zed 2 or zed 1 less than zed. This I cannot always see. Although in some cases I can. In some cases I can like essentially if I have uh say the real num real value to be fully zero then I can you know if it's a purely imaginary number I can say if it is greater than or less than if it is a purely real number I can say if it is greater than or less than but otherwise I cannot like if I have a complex number and let us say I'm taking a scalar multiple of the complex number then I can assign a notion of greater or lesser. So typically this whole greater than or lesser than is a one-dimensional concept. If I can arrange my numbers on a straight line then along that straight line I can say yes number one is greater than number two and vice versa or you know vice versa but complex numbers there on a plane. So I cannot always do that. Okay. So yeah that is that uh back to this. Yes. Yes. So is asking sir are two cubits represented by two block spheres? Well I just showed that it cannot always be represented by two block spheres. Up until this say s and c prime can be represented on two block spheres because I can factor them. But we are finding that there are states where such factoring is not possible where I cannot separate them. When they are not separable they become what are known as entangled states. So any quantum state that is not separable into single cubid states is an entangled state. So of course by direct definition to have such a state I need multiple cubits or you know multiple quantum systems. when I have multiple quantum systems, multiple you know cubid systems, then it is possible for me to have integral states in uh like you know in uh that space. Okay. So yes any other question? Okay. So if the states are not separable does it mean that we cannot break them? Of course it means that we cannot. Okay. What does it imply in quantum computing that we are not able to transfer information in these states? No. The answer is actually the opposite. If I have entangled states then I can transfer some kinds of information in a way that is actually not possible in the classical case. Unfortunately we are not doing. Let me see if I have some time maybe I will see if I can do some you know quum uh I I just let us see how today's session goes. Okay. So yes a couple of repeat questions then what do we have can we design any circuit like half adder in this we are going to see that yes that is the case just uh Amitra please hang in there I will come to that that is something I definitely want to see uh want to show today uh matrix form is C not is no C not can never be represented as a 2 +2 matrix. C not gate is always going to be represented as a 4 + 4 matrix because it acts on the two cubit basis vectors. Okay. Uh okay so Shara is asking okay so if professor gabad slides are not there fine I will I will see to this after today's session okay so yeah uh m00 and 10 which theory depends to okay I'm unable to soangi sharma I'm unable to understand your question if you can you know repeat you know I will I will try to answer it if you can repeat it you know in a more clear fashion from the YouTube side I have a question which is asking why is the C not get in matrix form like this well because this is how I'm defining the transformation and from this if I start constructing the matrix elements this is what I will get so this corresponds to only like four like matrix elements so like this is essentially As you can see shuffling the order of my comput standard basis 0 0 0 1 1 10 1 10 1 10 1 10 1 10 1 10 1 10 1 10 1 10 1 10 1 0 1 1 I apply the transformation I get the same four vectors but in a different order I get 0 0 0 1 1 1 0. So because of this I can say with guarantee that this matrix is going to have only four non-zero elements and the four non-zero elements are going to be arranged in this fashion. Okay. So what other questions do we have? Okay. So, Tapas G is saying that all session slides are already uploaded on the GitHub repository. So, Shahika please go there and check again. Uh why we cannot separate them for Prabhak or will we cannot like you you're welcome to try. So there does not exist like essentially what I'm saying is I cannot find two states like this such that their tensor product will be equal to either side double dash or cle dash that is all that I'm saying it is just not possible to separate these into two single cubit states that is just a mathematical construct okay so Yeah, that is a good question. So any two cubits with the C not gate are entangled. Uh well uh I will come to this in just a minute and yeah that's a good question. So I will come to that in just a minute. Yes. So this is the form of the uh C not. However, what we need to remember here is the following. So the notion is this is one form of the matrix. But what happens? So first this is a real symmetric matrix. As you can see since it is symmetric, it is also herial and I know the product of CX with CX is identity. So this is also a unitary transformation which is wonderful because CX is CX dagger. So CX dagger CX or CX CX dagger is equal to identity. So you know over here yes this circle denotes matrix multiplication. Okay. So whenever I put a circle here I'm talking about matrix multiplication. When I put a cross inside the circle I'm talking about tensor product multiplication. Okay. So uh yes. So now what happens when I have let us say I I hope you guys remember the tensor product notation. Let me you know what open up that as well. Just one second. Yeah. So it was also from last week. Yes. Yes. So yeah. So this is the tensor product of two matrices. If I have matrix A and B then that tensor product is given like this. Now look at this and look at not this but this. So essentially look at this transformation here and this transformation here. So this is a 0 0 * a matrix B. This is a 01 * a matrix B. A 1 0 * a matrix B. A11 * a matrix B. Now the thing is these two matrices are all zeros. Since these two matrices are all zeros, I can say a 01 and a 1 0 are both zeros because in that case whatever number is here it is going to be 0. But a 0 0 * b is the identity matrix and a 1 0 * a1 * b is the is this matrix. This is essentially the poly x. Okay. So that is what I have written here. Now you can see that there exists no matrix A or B that can give me this solution. So the CX gate is fundamentally a two cubit gate. It cannot be factored into two single cubit states. Okay. So the CX gate has to be defined like this. It cannot be defined in terms of two single cubit operations. This is a fundamentally double cubit operation. Okay. So now the numbers of course in my notation C12 determines which cubid controls the transformation. In this case the first cubid controls the transformation or the second cubid. Now what about C21? What will that look like? So let me go to my Poweroint. Yes. So we know M is C one where it's controlling the first cubit is controlling the transformation of the second. Okay. So in this case I know which is equal to 1 0 [Music] 1 0 1 1 0. This is also a standard matrix representation. When I put a zero here, it means that every element, so we know that this is a 4 + 4 matrix. So it is to be understood that there is a 2 +2 block of zeros here and a 2 +2 block of zeros here. This is just a shorthand notation of writing the matrix transformation. So this is ultimately a 4 + 4 matrix. Now what about C? So what about my C bit? So I know C2 is now I'm writing the full thing. Okay. So yeah, this is the C not gate and uh yes but what will happen to C not 21 like the second cubit controls what happens to the first. So in that case how do I go about defining it? So first I know 0 0 0 nothing is going to happen to it because the second pivot is zero. So 0 0 goes to 0 0. I'm essentially defining the actions. Uh okay so I have a question from uh Sam who is asking I have taken a 0 and a 01 as equal to 1. No, I have not taken it equal to as anything. I'm merely stating that this these are the matrix equations and they have no solutions. Like I have not taken them to be any values at all. See, I'm just scalar multiplying this matrix by this matrix. When I'm multiplying by a scalar, the same number gets multiplied to every single element. And I cannot have a scalar a complex number no matter what that number is that can you know suddenly change a zero to a one 0 * any number is 0 is your you know uh like I mean it's the zero element like any multiplication there is only going to lead me to zero like it's the kernel of my multiplication group so to speak well not the multiplication group well it is you understand what I uh so yeah I went back to my slides. No. Yes. So 0 0 goes to 0 0. 0 1 on the other hand will go to 1 1 and 1 0 since the second cubit is zero nothing will happen to it is going to be 1 0 and 1 1 is going to go to 01. So this matrix is going to have the matrix elements as. So this implies C not 21 is going to have the following matrix elements. 0 0 0 + 0 1 1 + 1 0 1 1 [Music] 0 + 1 1 0 Am I writing this [Music] correctly? Yes, I am writing this correctly because this is also a symmetric matrix. Yes. So these are the matrix elements. Now again just to remind you uh 0 0 is given by the column vector 1 followed by three zeros. 01 is given by the column vector 0 1 0 0 and 1 0 it's 0 0 1 0 and 1 1 it's 0 0 1. So using this we can construct the matrix elements by taking the k and the bra vectors. Yes. So that is given by the sum of uh the uh uh sorry the yes sum of these individual matrix products which is going to give me this matrix representation. This is also the cube. So typically when no index is given when I just write C not and I show you two cubits it is always assumed that the first cubit is in control otherwise it will always be mentioned which cubit is acting as the control cubit and the target cubit. So that is always specified. Okay. So please keep that in mind. So C not gate doesn't always mean this matrix representation. Depending on which gates I'm applying the transformation to I can have a different matrix representation. Now remember this because next week this will become important. Now the point is it is possible to perform the controlled version of any single cubit unitary transformation. How? I am not going to explain how I'm definitely not going to explain because you know it is uh kind of uh uh difficult. So, so okay. So, I have I'm seeing a couple of questions. Please uh hang in there. I will want to say like go on for another five minutes and then come back to the questions. So it is possible to perform the control version of any unitary transformation. What does this mean? So I will again go back to this PowerPoint again new slide and talk about control view. Control view Now this is somewhat important for what we are going to do in the next week. Okay. So what is the control Q? So CU or let me say yeah let me call it CU this is a controlled unitary gate. where the definition is if the first or what I'm supposed to say is the controlling cubit. That is what I mean when I say the first. Okay, it's not always the first. uh what is the difference between a unitary transformation and a unitary gate? Nothing they are the same. So you know this in state M0 then do nothing to the second. Okay. However, if the control incubate get one then apply the transformation or gate or what I'll just write because it's a smaller word. U to the second cubit. So this is what a controlled unitary transformation is. And how do I apply this? Uh so CU acting on K 0 is going to give me K 0. CU the first cubit is the control cubit. Okay. always like in this case remember that 01 is the state 01 CU on 10 is going to be K 1 tensor product Q * K0 and C U 1 is 1 tensor product with U okay so this is the unitary transformation for such a unitary transformation the matrix element is are going to look like this zero I'm writing it again in the block angle Another sound U 0 0 U 0 1 U 1 0 U 1 where U is the unitary transformation that I want to apply. surface right here. This quantity right here this is the unitary transformation that I wish to apply. So this is my view. So it is possible to construct such unitary transformations. Okay. So now let me talk about what shall I talk? Yes. The next part is change of basis. So let me call this the basis change rule in quantum computing. So now I will address a question that was asked a few minutes ago. But before I start this, some of you may have the question that you know maybe are not asking uh it out of you know some politeness which I appreciate but nonetheless let me address this issue. Why am I uh kind of you know doing the topics why are we covering the topics like this? The idea was the following. So in the first couple of weeks what we essentially did was kind of set up the language for quer computing so that because the truth is this nobody can kind of you know complete a uh uh like you know nobody can give an exhaustive introduction or an exhaustive uh summary of quantum computing in any capacity in a short four-week course like nobody is going to be able to do that because this is a a rapidly evolving uh field b there is plenty of fundamentals to cover. So you know it's kind of difficult. So what we have done is we have kind of spent like half of the course talking about the basics where the materials are all present and are given to you and the recordings of the lectures are also there. So the idea is you can use the content from the first two weeks to kind of explore any other quantum computing course or textbook or material that you have be reflect be Mc Becman be Nilson Chuang or you know some other online course somewhere else. So that should give you the basics to understand and deal with any other quantum computing literature or material that you may come across in your you know in your future endeavors. So that was the point of the first two weeks of the course. In the last two weeks, what we are going to do is we have a particular agenda for this course. So we are going to pursue that. So that is why in this third week I'm taking care of like things. I'm essentially moving back through like the first uh uh two or uh yeah the first two weeks and kind of I'm filling in the gaps and you know making things whole. Okay. So now what do I mean by the basis change? Basis change rule means this. Now first we have what is called the let me talk about some different basis. So first we have what is called the standard or the computational basis or the z basis. So first I will define what these bases are. Okay. So or should I you know what I will not do that it will cause more confusion. I will simply state what the basis is. So this is my computational basis which I define as B. Let me call this B S which I define as this Z0 comma K. So this is my standard basis or computational basis or the Z basis. Why this is called the Z basis? Just you know take it for what it is. Okay. Uh well I can instead of showing that I can show you this. So let me quickly yeah I can actually explain it without explaining it. Uh so yes notice K1 is in this direction. Sorry K0 was in this direction. K1 is in this direction. So these two vectors together form the Z axis. That is why they are called the Z basis. Okay. So that is quite simply the reason why it's called the Z basis because these two remember in a single cubid basis I will always have my orthonormal states lying on antipodal points. They will be on opposite poles of the sphere which implies those two vectors will form a straight line. It will form a body diameter for the sphere. Since it forms a body diameter, whatever axis that forms for the sphere is what I shall consider to be the axis of the basis. So these are called Z basis because the two vectors are aligned along the Z axis of the sphere. Okay. And I'll be another bit of water. Okay. So the second basis that we know is the hadabard basis. or the X bases and now you can see what's happening. So I will call this BH. So this is K plus comma K minus where K plus or minus is defined as K 0 plus or minus get 1 divided by <unk>2. I hope this plus or minus notation is familiar to some of you if not all of you. Essentially if it's plus take the plus. If it's minus take the minus. Okay. So that's what it is. So this is the second basis that we know. So again why do I call this the X basis? Going back to this, this is K plus and K minus has to be on the other side. So the K plus K minus basis forms the X-axis of my screen. Similarly the Y basis will form the Y-axis and you can I hope you can imagine the rest. Okay. So we have this now lucky. Okay. So if I'm given two bases uh two bases let us say if uh well let me create a new slide. Say we are given a basis that is say some P which is equal to some vector V_sub_1 comma V_sub_2. Now how do I go from? So essentially the question is how to transform PS to how to do this? Essentially the idea is I have my standard basis. You know what I will make a quick change of notation. rather than say S and H I will now use the axis notation. It's a PZ Bx. So let us say I am starting from PZ. How do I go from BZ to P? This B that I've defined here. Let me call this some BB. Okay. How do I do this? What is the transformation that takes me from here to here? So uh I really wish uh I had um yeah I would have really liked to have asked this question to you and seen the answers but you know what it's fine. So see now let me say there is an operation u that takes me to this. Okay, let me say that there is a transformation U that will convert BZ into BV. Essentially, what I'm saying is U * K0 is get V1 and U * K 1 is to get V_sub_2. I hope you can see why I'm writing like this because now V_sub_1 and V_sub_2 are orthonormal because I' I've told that this is a basis. So it goes without saying but nonetheless I will say it. These two vectors are orthogonal. Okay. Yeah. These are orthogonal. Uh yes. So there exists some transformation you that does this. Uh I'm seeing some questions. Please hang in there. I will come to them in around 2 minutes. So the transformation that will take me from here to here is nothing but u is equal to V1 0 + V2 1. This is it. This is the transformation. Okay. So if I apply you can directly see this u * k0 is going to be vub1 0 0 + v_sub_2 k v2 1 0. I know this vanishes. Therefore the result is equal to get v1 and this is equal to 1. And similarly I can say u k1 is equal to v2. I hope you can see this. I've essentially just multiplied by k0 to this. Okay. And this is the advantage of the direct bracket notation. I can define you know just some uh arbitrary abstract vectors and perform my matrix multiplications. Okay. So this is my basis change. Now how do I know that I've given the correct answer? Well that is why I defined the X basis here. Now we know K plus is this. So let us see if our transformations are indeed correct. Okay. So consider pz to px. Now what is the transformation u? U is given by k + bra 0 plus k get minus bra 1. Okay. So this will be 1x <unk>2 get 0 plus get 1 square roo<unk> of 2 I'm sorry get 1 T 0 plus 1 /<unk> 2 0 sorry jet 0 bra 1 minus bra 1 jet 1 now you can remember way back from the first uh week's content that first week's lectures that this is indeed the Adam matrix Sorry. You know what? I put it in order. Plus 0 - 1. Okay. So this is the basis change rule for the uh basis change rule for the single cubit basis. Yeah. So this is all for the single cubit basis. Okay. So now what I will say in just one line is this rule generalizes hard states as well. Do not worry. I'm not going to leave it hanging here. I will come back to this after the break. So, so what we shall do now is I think we can take a 15 minute break. Yeah, sure. The time is [Music] 11 someone. Yes, 11:26. So we shall reconvene at 11:40. Okay. So those of you who need to take a break, please uh uh go ahead and take a break. In the meantime, I will of course address the questions from the chat. And if you have any other questions like you know even from the previous weeks and so on, please feel free to post them. So I think we can yeah we can start our break now for 15 minutes. Okay. So the break has begun. Uh let me now address some of the questions are in the same axis oppositeely. Are they parallel then? Okay. So you know what? Yeah. Math physics TV has asked a rather nice question. So I will only say that see 0ero and one are along the same axis in uh the uh block sphere. So does it make them parallel? The answer is no because I hope you may like because the block sphere is not a vector space. I hope you can see that the block sphere is just you know a manifold. It is just a geometric structure. Okay. So what do I mean to say is so yeah see this is uh my block sphere. This is K 0. This is K 1. And I know this is 0 + 1 by <unk>2. Can you add these two vectors to arrive at this vector along the x-axis? You cannot. So this is not my vector space. This is just a manifold representation. Okay. So the notion of parallelism or things like that do not carry over here. That is why I explicitly wrote over here all the way up here that every valid single cubit state vector in C cross C that is my vector space that has not changed at all. Okay. corresponds to a point on the block sphere. The block sphere is a manifold. Okay. So on YouTube I have another question from Abritk. What is the no cloning theorem? If we have enough time I will address it today. Okay. If we have enough time I shall address it. Uh so yes uh or you know maybe in the final week I will take some time for it. Okay. So kindly be patient. Uh now yes back to uh the WebEx questions. Let us see if uh we have anything there. What was the last question that I saw? Explain separable and non-separable states in regard to computing. I will be doing that today. So please hang in there. Clicks are already shared. So yeah, these are announcements. Okay. So, C not essentially changes the probability of the basis K. So, the photon must change polarization. Okay. So, this is a question from Kanik. What I will say is photons are not altogether a good example for understanding quantum computing precisely because operations like C not cannot be readily applied. Okay. So do not try to understand all of quantum computing using photons because photonic quantum computing has its limits. At least photonic quantum computing where I leverage the polarization of the photon. It has several strong limits. I will not say anything more than that. But you can understand the basic ideas especially single cubid transformations. You can understand them very well in terms of photon polarization. Okay. So yeah. So this is a question that I've seen. I so sam had asked a question like roughly half an hour ago uh talking about k 0 k 11 one are always equal to one and are always equal to zero I don't understand in what context you have said this if you can repeat your question towards the end I will come back to it uh Uh okay. Yes. So this is something that I will take some time to address during the break. Uh or you know what Induma your question I will actually pick up once the break is over. Okay. This will be the first thing that I shall uh talk about way down in the comments part. Okay. So, uh right so what is the difference between a unitary transformation and gate? Well, [Music] uh I've already told you there is no difference. So, Sneha has asked is Kit 10 uh 11 1 0. No, it is not. Take the corresponding matrices, multiply them and see. you will see that it's a different uh like it's actually a matrix and it's not the zero matrix. Okay. So give the difference between a unitary transformation and a controlled unitary transformation. Uh well the control unitary transformation is a subset of all unitary transformations. So when one cubit decides what transformation is being applied to the second cubit that is a controlled unitary transformation. So this is from Burli. So this question so essentially a unitary transformation is the whole general thing. Any transformation that I apply to my cubit vectors like cubid state vectors is a unitary transformation. If that transformation corresponds to a case where the value of my first uh cubid state determines on what happens to the second then you know I uh consider that to be a control unitary transformation. Okay. So yeah so K first has asked how is K01111 uh okay that was for this one that is for when the second cubit is in control okay so essentially the second cubit is one so the x gate is applied to the first cubit so that was the only case in which I mentioned so in this case this is what I have so c12 is not the same as c21 please understand that Okay. [Music] So can I please explain how I arrived at this matrix? So at 1111 I was doing this. So yeah I mean just I've given the vectors here. So the bra form of them is the transpose conjugate like I really don't want to uh you know what I will take why not I will do this. I will do this. So I'll just quickly set a new slide here. Therefore, CN to 1 is equal to 1 0 0 multiplied by 1 0 0 0 plus what is what is my second vector that is 01 1. So 01 is going to be 0 0 0 1 0 1 1. Second vector is third vector is 1 0 1 0. 0 0 1 0 multiply multiply 0= 0 plus the fourth term is the opposite of this 1* 016. So the first one essentially what I will be doing is take this multiply with all this take this multiply with. So this is going to be my first row. So this will give me a 4 + 4 matrix that is 1 0 0 0 So it's going to be this plus over here as you can see I'm going to get a1 where this one and this one gets multiplied. So this corresponds to the second row. So everything else is going to be zeros 0 0 0 or zeros over here. The third row, third pel 1 0. Finally, the fourth one is going to be fourth row, second column. fourth row. Second but this is going to give me 1 0 0 0 1 0 0 1 0 0 1 0 0. So this is how I arrived at the matrix. Okay. Yes. Now for a question that everybody uh has been asking. So what about py? So what is y? What is my y basis is equal to what? So let us quickly go to the block sphere diagram and ask ph is zero along this direction. Let me first figure out uh what the theta and phi values corresponding to this are because once I know the theta and phi values then I can you know just find out what the state is because this is a vector on the block sphere. If I know what theta and phi are then I know what the vector is. So over here I know that theta is equal to<unk> by2. So that is something that I already know because when I come to the xy plane theta has to be by 2. Okay. So in my y basis I know theta is equal to 5 by2. Okay. Now what about phi? Remember phi is measured starting from the x-axis in the counterclockwise direction and the clockwise direction. So from here and one for the circle is 2 pi. So if I go up to the y-axis then I'm going 90°. So five is also equal to by 2. So, one of the vectors over here is so one of the vectors here is going to be uh just second. Okay. Yeah. So, what was I saying? Yes. So, the first vector that I'm going to have is the following. It is cos theta by 2. So that is yeah cos by 4 because theta is<unk> by2 so theta* 2 is<unk> by 4 so<unk> by 4 get 0 plus e ^ i by 2 sin by 4 get Yeah, we are almost at the end of the break. So, I'll still take a couple more questions before I resume. So, this will give me a state vector that is 1 upon roo<unk> get 0 plus I * 1 upon <unk>2. So, I'll just write it as I by <unk>2 get 1. So, this is the first vector. Okay. So, I hope you can see that. Now what about my second thing? So second point is over here. So here again theta is equal to<unk> by 4. But phi is equal to 90. Another 90 and another 90 270° or 3<unk> by2. Okay. So my second vector is so this is vector 1. My second vector is again cos by 4 get 0 plus e ^ i 3 by 2 sin by 4 get 1. So this will come out to be again 1x <unk>2 get 0 minus i by <unk>2 get one. So it's as straightforward as this. So these are my two basis vectors for uh my y-axis. This is the vector that I shall call K I and this is the vector I shall call minus K get minus I just for notation because I have a plus I here I have a minus I here in the superp position. Okay. So I will just write this as K I comma minus K minus I not minus K I K minus I. the minus I is within the gate. Okay. So this is my y basis. I hope this is clear to everyone. So what is the use of basis change? We are going to see that in a minute. Okay. So okay, please tell the difference between global and relative phase. I have told this several times before. I will say it just one more time. I'm not going to go into the weeds of it. A global phase is a phase factor that I can remove as common between the two cubits. So I can just remove that and I can dispose off of it. This relative face is something that I cannot throw away because the relative face as you can see is what allows me to define the different orthonormal bases. Like even here if I look at the uh adamat basis what is the difference between them? The difference between them is a plus becomes a minus that is also a relative phase. That is a relative phase of 180° because e ^ i is minus one. Okay. So that so for any unknown basis how to find u I just told you how to do that. This is the method. Take your basis vectors, multiply by bra 0, expand this and write your matrix for hadaband. I tried it and showed it that this method works. For any other basis vectors, take your two basis vectors here, define your transformation and calculate this matrix. Just calculate that matrix. Uh yes, I see that you have some doubts with the quiz. Uh I have also seen your question before. I will address them later because yes there are a couple of mistakes. Do not worry if there is a mistake in the question that is kind of everybody is given marks for that. So the corrected versions we will share with you later. Okay. So please kindly bear with us on that. Uh yes, I will address the advantages of superp position and entanglement. Just give me some time. [Music] Uh okay. So yes on slide six. So I suppose you're asking this. Yes. How to write CU matrix if the second cubit is the control cit. Uh try it yourself you know call it a homework. So Anabu has asked how to do CU if the second cubit is the control cubit. Just write your elements write the inner products and just try you know that's the best way to do it. So consider it a homework from my side. Okay. So yeah with that I've come to the end of the messages. Do I have anything more on YouTube? Uh, no, nothing more. All right. So, yes, the first thing that I said that I will address is after the break is over. So, now the break is over officially. So, what is the thing that I wanted to address? I said I will I will be addressing something. What did I say? I will be addressing I will back the thing here. You know what the person who asked that question if you're still around please ask the question again. I mean yeah. So the last question that I have not answered is yes. So how do I move from like you know so this is BY. Now how to move from say BZ to BY. So let me just put up a partition here. at the desk. How to go from B X to BY or alternatively from say from BV to some you know PW. Why does it have to be X to Y? Now why does it have to be you know uh just to fix things? How do I go from any two different bases? The answer is let me say I have basis one b1 if it's given by v_sub_1 v2 and v_sub_2 that is given by u1 and u2 you know what I've used u for transformations so let me say I'm using this using w here yeah why W1 to W2. How do I do the basis change? The answer is quite simple. Say U takes get V1 to give me get W1. Just assume this that you get V2 get W2. This will give you The inner product sorry W1 V1 plus W2 V2 that is it. Oh yes, two cubit PHs this is something I'm coming to this I'm coming to this and I'm coming to this. So yes I mean this is your question is kind of the reason why I'm doing all of this. I mean also like you know to cover the syllabus into context. Yeah we have we have plenty of time. Okay. So this is how we define the transformation. Just take the vectors, take this outer product between the two, take their sum. Whatever matrix you have will be the matrix that will take the transformation. So how to go from bx to py? So, so if I'm talking about transforming from current so if I'm talking about bx y then the transformation u is going to be k I let me say U plus is going to give me get I and U minus is going to give me get minus I. This is my transformation. Therefore, U is equal to get I get plus minus I get minus. Okay. [Music] So, yeah. You know what? I will expand this thing just for this case. So remember that there is a 1x roo<unk>2 factor here and there is also a 1x <unk>2 factor here. So when I take K and B I will end up getting by two here times 0 0 sorry 0 0 plus 0 1 plus iota 1 0 plus iota 1 Um that is this is this plus another half types 0 0 plus wait am I doing this correctly yes I'm doing this correctly now remember that this is 0 - 1 by <unk>2 and this is 0 - iota 1x<unk>2 okay so I'm going to have a minus sign here not a plus sign 01us iota * 1 2 plus iota. So this is the second term. So just one second. Yeah. So this implies wait. So this implies u as you can see this term and this term are going to get cancelled. This term and this term are going to get cancelled. So just cancel them out. 2 is therefore going to be half and half is going to become 1. It's going to be 0 0 7 plus iota * 1 1 and this in its matrix form can be written as 1 0 0 I as simple as that. Over here all multiplications are normal multiplications. They are not conical products. They're not tensor products. Tensor products will appear if and only if this symbol appears. We do not have that here. And therefore we are not here. Oh. Okay, I think that is not supposed to be the case. Yes. Yeah. Okay. So, yeah, this is the matrix that will take me from BX to BY. Similarly, you can work out the other things. I've worked out one example here. Please try and work out the others. So how do we decide when to do tensor products and when to do normal product? Well, that comes from the definition of the two entities. See here this is a single cubid transformation. Like see if there is a tensor product it will be mentioned like it's not something that you choose to do. Remember the two products are different things. They give us different objects like K0 tensor K0 is different from K0 bra zero. I hope you understand that there the two K zeros are coming from two different vector spaces. Here they are essentially coming from the same vector spaces. These are two different bases for the same vector space. Okay, please understand that. Okay, so is there anything else that I would like to comment here right now? Nothing. is now but let us go to the why did I introduce basis change I will quickly highlight what happens to an end cubit system okay so again to Krishna Gupta's question it is just you know be patient I will I will tell you See tensor products define something else. So you will use tensor products only when you want to define those objects. Tensor products are used to define multi-ubit operations. Tensor products are used to define other things. Okay. So please uh this just hang. So yeah, uh let me just have something Yeah. So question is why study basis change? What is the whole point of my basis change? So the answer to that is oh you know what before I begin this last thing this is change for multi-ubit systems So uh let me say now here I'm talking about in cubit systems. I'm not only talking about two cubit systems. I'm talking about general n cubit systems because gradually I need to start introducing that notation as well. So here the idea is get zero corresponds to 0 tensor 0 tensor up to n * 0. So this is n times. So this is a n cubit state. Okay, which I'm equivalently writing as 0 0 0 or I shall simply call this as 0 n. Okay. So get one is 0 0 0 1. Get two 1 0. Okay. Three is 0 0 0 1 1. Similarly, I keep essentially moving up in the binary classification and I will say 2 ^ n minus one which is 1 one 1. So starting from all zeros to all ones I'll just say this is one and you know what let's not do 1 and 0 n looks nice one n does not look nice. So why are we writing it in this form? Because this is my this entire thing this is my multi cubit standard basis I have just standard basis. So for a single cubit I had k 0 and k 1 as my standard basis. Here I have all the n bit binary streams as my standard basis vectors. Okay. So yes now I will actually take a couple of minutes to now Krishna I will answer your question. Uh Sufyan I will come to your question in a minute. Uh so and Kavitak Mari has actually asked a nice question which I can use to kind of uh uh kick off this whole discussion. So what does it mean by two vectors coming from two different spaces? Because here both spaces are complex. So how is it different from uh one another? Remember and always remember this. Never forget this. The vector space corresponds to a cubit. Okay. So a physical cubit. So a cubit is a physical object. So yes, so this is a physical object. The point is mathematically the two vector spaces are equivalent. But the vectors in vector space one corresponds to the physical state of object one. The vectors in vector space 2 correspond to the physical states of object two. Here this is say I will say this is cubit one you know let me say cubit number one cubit number one this is cubit number n minus one like in between there are cubit 2 cubit 3 and so on this is cubit number n minus one and this is cubit So that is the ordering that I am maintaining here. So to generate the basis of this particular vector space I am taking the tensor product of the individual uh yes connect you are writing. I am writing the tensor product of the individual uh basis elements computational basis elements for each vector. Okay. So that is the n cubit basis. For a two cubit basis I will just be taking two of these. But as you can see in that case I will stop at three. But for n cubits how many vectors how many boolean strings am I going to have? I will have 2 ^ n strings. So that goes from all zero to all one which is 2 ^ n minus one. Like the decimal value is 2 ^ n minus one. Okay. So that is why I'm writing it like this. So when I want to get a multi-cubit basis then from single cubid basis then I will take the tensor product. Okay. Then and only then I will take the tensor product. I will take the tensor product when I want to construct a multi cubit transformation from a single cubit transformation or a multicubid state vector from a single cubid state vector. Okay. So no physical object. Therefore, the n cubit standard basis is generated by the in single cubit. Let me mention this here. Tensor product. Yeah, I will not be using chronical product too much. But understand that they are both equivalent. Tensor product of the n single cubid standard basis. So essentially d n for n cubit basis of zed is going to be defined by p1. Oh no this is not right. Uh yes you know what I will just write it like this. Get 0 comma 1 tensor product. Get 0 comma 1 answered. So on and so forth get zero. Get one. Have I done? That was wrong. Please disregard. Yes. So this is performed n times. Okay. So this is the n cubit standard basis. Okay. Now how do I talk about basis change? Now essentially for any basis Now remember as one of you also pointed out correctly in the chat the number of basis vectors is going to be equal to the dimensional of the dimensionality of the vector space. So there is going to be 2 ^ n basis vectors no matter what they are 2 ^ n distinct orthonormal basis vectors. So for any basis BB that is equal to V_sub_1 V2 you can see where this is going V3 V2 ^ N minus one the transformation is given by so U that takes X PZ to P V is again given as U is equal to get okay B1 0 plus oh no I should have written B 0 here I'm sorry V 0 B1 I apologize because I need 2 ^ n total vector. So it has to go from 0 to 2 ^ n minus one. So here it is p 0. I realize that the moment I wrote this. Okay. Plus v_sub_1 k1 plus v_sub_2 k2 plus so forth. B 2 ^ n - 1 get 2 ^ n - 1 sorry 2 ^ n - 1. So this is the transformation. Okay. This is the transformation that takes me from uh any uh from my n cubit computational basis standard basis to the any generic n cubit basis. So this is the basis change route for n cubits. Now again I have a quick question how do I define two? Yeah. Now what about DZ? How do I do this? Can anyone like quickly give me a response in the chat? Let me see if I can do this. Uh the logistic map be the same. Uh I do not understand your question. question please like maybe wait whatever you are trying to ask maybe I will you know answer it later on yes thank you so this was from swap mill yes because see ud dagger is the reverse of u so ud dagger will take me from bb to bz and what is ud dagger it is just the dagger of these individual elements because a plus b yes thank you for you know multiple people writing is yeah, it's going to be this v 0 + 1 v1 plus I'm going to leave the third one out 2 ^ n - 1 v 2 ^ n -1 so it's as simple as this whatever applies to n cubits also applies to one cubit so go back and do the same thing over there okay so now I can tell you if I need to move from say BW to PV. I can do this. This is equivalent to moving from BW to B Z to BB. I can always do this because I just need to find out the transformation from BZ to BW. Take the dagger of that and then take the transformation from BZ to BV and take the matrix multiplication of both. Because if I start from here and apply U dagger I will go to BZ and then if I apply the next operation say a B I will go to BV. So I can also do this this kind of a basis change. Now one question that I would like to ask you and I will give you around a minute to respond since you are able to respond. What are my unitary transformations doing at the end of the day? the various unitary transformations that I'm seeing what are they doing and now we come to the question can someone please like based on your observation so far what are what is my generic unitary transformation doing rotating sting No. Okay. Surprisingly. Yeah. So Susant finally has hit upon the correct answer. The rest of you well you are saying something that is essentially correct but well it's not what I was looking for. See ultimately what we are doing every unitary transformation every quantum gate operation is a basis change that is what. So why study basis change? The answer is because this is not a proper sentence because you know because it's a consumption but you know what I don't care about it any unitary transformation any transformation at all only performs a basis sorry performs a basis change. That is it. All of quantum computing the algorithm that you saw yesterday the you know the basis rotation is essentially a basis change. Okay. So everything is ultimately just a basis change. Any unitary transformation applied on a single cubit, multiple cubits, however we look at it, it is only performing a basis change. And somehow by just changing the basis over and over again by choosing some appropriate basis change transformations, we are able to do computation. That is the beauty of quantum computing. That is where like essentially every quantum algorithm this further implies every quantum circuit algorithm is only performing coming. This is six only performing let me say a sequence sequence of basis changes. And we finally measure the state. Extract or Z output. Okay. So now I will take some questions. So Utin is asking a gate like X gate does not perform basis change. Who said it does not? Remember what does my X gate do? So you know what? Yeah, I will write it here in X * K0 takes me to K 1 and X * K 1 takes me to K 0. The basis permutation is also a basis change because originally my basis was K01. Now my basis is K 1 Z. Even that is a different basis. Yes, it is a very small change. It's a minor change but it is a change nonetheless. No. Now this measurement is not a basis change. That is why I mentioned this separately. All my gate transformation that is why I said quantum circuit here. Yes, let me write that in red over here measurement is not a basis but everything before measurement is definitely a basis sheet. Okay. So ah yes now Ramon Shaha has asked uh when we are changing bases our quantum state does not change for this I will just give you a brief comment there are two different ways of looking at quantum computing what I'm talking about is the second way the first way is you talk about the state itself changing and giving the output alternatively You can think about your state being a constant and your base is constantly changing and then you are saying that yeah finally I will apply a measurement and I will get my desired output. Both approaches are equivalent. When yes when you are changing the basis the quantum state does not change. When you don't want to change your uh basis then your quantum state will start changing. So you either change your basis or you change your quantum state. They are both equivalent. Okay. uh in uh mathematics and uh mechanics you will uh uh encounter this thing called active and passive rotations. So active rotation yes so Banu is talking about the stringer and Eisenberg picture this is essentially that but I don't want to go into the details okay so just remember it like this whatever I have mentioned here is very self-consistent so please you know for those of you who are not familiar with quantum mechanics just remember it like this it's perfectly fine okay so let me quickly finish writing what I was writing Here this is also a basis change. And what is the basis change that I'm doing here? The basis change is a permutation. This is a B sorry this is a basis permutation. Okay. I'm changing the order of my basis. What happens to the probabilities during uh change of basis? They change. They do change. Which is why like I mean I don't just set up a quantum computer and immediately perform a measurement. Do I? I perform a then I perform a measurement. I will get a different outcome with a different probability. But that is what I consider to be the output of my quantum algorithm. So yes, the probability does change if I like however you consider it. If it is a basis change or if it is you know a vector transformation they are both one and the same. Uh no like see what is happening is happening. So Rajes has asked does this mean that we can't change the state and the basis at the same time? Well when I'm changing my state I can think of it as me changing my state or I can think of it as me changing my basis. They're both one and the same. The same thing I'm describing using two different language but a gate is applied to orthonormal basis. Yes. So rigid has asked but gate is applied to orthonormal basis. Does the resultant need to be another orthonormal basis? Yes, of course. This is something that professor Ghabadha has pointed out in the beginning. See consider any nubid state. Consider a state s. Let me say that U acting on S gives me KI or you know what how do I say this? Ah okay I will I will define it like this. Consider a basis. Consider a basis that is v_sub_1 v_sub_2. I'm explaining this with single cubit basis but this applies to multi cubit as well. Okay. Now let us say u v_sub_1 is equal to some w1 and u v2 is equal to some w2. Now what is the inner product between W2 and W1? Oh yeah, this is what I was supposed to address. Yes, I finally remembered. So uh so what is this going to be? This is going to be U V1. This is U V2 is W2. which implies p W2 is the transposed conjugate of the left hand side. So that is going to be equal to p v2 u dagger. So p v2 u dagger. Now I know u dagger u is identity because u is a unitary transformation. I should also mention that u is unit. Okay. So this implies this is equal to v_sub_2 v_sub_1 which is equal to zero because this was originally a basis pv. So yes a unitary transformation when applied to any basis will give me another unitary transformation. So when I apply another unitary transformation to it I will get another orthonormal basis. That is the whole point of quantum computing. It will the probabilities will say like see again to uh JSha we are dealing with normalized states and any unitary transformation will keep a normalized state normalized that is not going to change. Okay. So yeah [Music] uh yes so I hope this part is clear and this entire thing applies to the incubit case as well applies to the n cubit case as well. So there there will be many such combinations and you can rest assured that for every combination it is going to hold. I saw another important question. Yes, of course. Like every slide of every day will be uploaded. So do not worry. Uh so uh someone named RS is asking how this topic is used in quantum computing. This topic is quantum computing. Like we don't have to use it anywhere like I'm essentially moving into the mathematics of quantum computing. Now all of quantum computing any quantum algorithm is ultimately just performing different basis changes. Okay. So uh okay so Krishna Gupta has once again asked as we measure is there a state change also? Yes there is a state change when you measure but again that is not a basis change. Remember basis changes are reversible as I have pointed out here. If u takes me from zed to v then u dagger will take me from v to zed. But in measurement such a thing does not happen. Okay. So uh okay so Khan you have asked me a question what is the duration you can ask this question to professor vish tomorrow. Okay. So yes, now I've spoken enough about basis change. Now I will come to the last part which is essentially you know I will do a little bit of physics and or before I start doing the little bit of physics I will address the notion of entanglement. Let us consider two cubit entanglement. Okay. So let us come to the matter of two cubit impact. So Yeah, sure. Now question is this [Music] uh let me quickly switch over to my PDF. Now see you can think of it like this. Uh okay. So is it possible for two cubit to be entangled or more? I will come to this in just a second. So see this particular change I can take K0 to K1 by applying the X gate. H I can take K 1 to K X by also applying some K or I can take K0 to K plus by applying the gate. So in a single cubit system. So first thing that we need to understand is any single cubid gate. Any single cubid gate does the following like it essentially moves my moves the vector on the block sphere. Um one point to another. That is all it does. I start from one vector. When I apply the gate, it goes to another point. That is all. If I'm thinking about a basis change, I start from one axis and I go to another axis. Which is why we refer to the single cubit transformations or even multi-cubit transformations as rotations because unitary transformations are the generalization of 3D rotations. Like when I do a three-dimensional say rotate about the x-axis by this much angle, y-axis by this much angle. The generalization of that concept is what is called a unitary transformation. That rotation can be seen as a change of basis. That rotation can be seen as a a change of axis. It can be seen as a change of state vector. All these pictures are equivalent. Okay. Now the question is how does or rather how do I ask this? So this is the first statement. Any single cubic gate moves the vector or the block sphere from one point to another. Now if okay if a single cubit if there are two cubits I'm just realizing something and well it's fine I will not tell you what it is you know when you give the quiz you will understand what I'm essentially doing is I'm giving the answers to some of the uh uh quiz questions. It's fine. It's fine. So, and we apply only single cubid transformations to that. to them then what happens? So there are two cubits and I'm only applying single cubit transformations. So wait before that sug has asked the way of finding the matrix for a particular gate. No, it is the same because I just told you every gate transformation is also a basis state. So whatever state you have, you just have to find out the orthonormal state to the but the issue the the reason why I'm talking about basis changes it is easier to visualize some of the gate transformations. Uh okay so there is another question I will quickly uh address this uh going back to so this is from uh uh Selkumar JS soil Kumar has mentioned is it possible to go from rivers from the block sphere to the vector form the answer is not always because see this is a many to one mapping many vectors correspond to a single point on the block sphere. Okay. So if I want to do the reverse it will go to infinitely many vectors because of the global phase. So what you need to understand is yes if I ignore the global phase then the reverse is possible. If I am writing my vector to be clear, if I'm confining to writing my single cubit state vector like this in this form and this form only then yes I can do the reverse. But if I'm allowing this global phase to exist, if I'm not ignoring the global phase, then the reverse process is not always possible. Okay. So okay so well Krishna I have not yet asked a question so you don't have to answer it. Uh yes uh okay yes to the end of today's end of this. So there are two cubits and I am applying only uh single cubit transformations to two cubits. Okay, I have not applied any two cubit transformation. Essentially every transformation is of the form. So this implies every transformation is of the form uh every transformation is of the form say a n insert b where a and b are single given transformations. and the single cubid transformations. Then what happens is A will cause one kind of motion on block sphere one. B will cause one kind of motion with on block sphere two. Okay. So there is going to be constant motion and ultimately my state vectors are going to be still the same. I hope you can understand that. Okay. So what I'm trying to say is the resultant state will still be separable. Yes, Suzan, you are right. That's what I said. So I can either assume and I also represented that here. I can either assume that I'm doing a basis change in which case I will use u or when I'm considering the gate transformation I will consider you diver. They are both uniquely defined. See that's why I've written these two here. So this will be my gate transformation and this will be my uh you know basis stage operation and they are both equivalent in a certain way. Okay. So now essentially what I'm trying to answer here so let me write the third point this way the two states or rather I should say the two cubit state is always separable. Okay. So, yes. So, you're essentially right. If U is the basis transformation then U dagger will be the gate transformation or if U dagger is the basis transformation then U will be the gate transformation. Yes, you're right. So they will always be separable. So remember this if I start with two cubits and if I only apply single cubit transformations to them the two cubit state will always be separable. Why? because a will perform one transformation, b will perform another transformation, a will perform like essentially if I'm starting with say k0 k0 if I'm starting with k0 tensor k0 and if I apply a tensor b to this then what will I have? I will have a get zero tensor B get zero and then if I apply some C tensor D then I will have C A get zero tensor T B0 where here I do matrix multiplication and I take the tensor product of the two. So I can still separate the resultant state as two single cubid states. So these are still separable. Okay. So yes or you should actually say the other way around. If a state is non-separable, it is entangled. Okay. So separability is the thing that we define first. We do not define entanglement first. We define separability first. And any state that is not separable is entitled. Okay. So now the question is when will I get entitlement? So how to entitle to single cubid states? How to do this? How to perform the entire now? Can I just apply the C not gate? Because that is a two cubit transformation that is that cannot be broken into a tensor B like this. I know a gate. So, can I just apply the C not gate and get my uh entanglement? The answer unfortunately is no because see what does my C not do? I take 0 0 to 0 1 0 0 1 to 0 1 sorry 0 0 to 0 0 1 to 0 1 1 0 to 1 1 1 to 1 0 these are still separable states. These are two separable states. So now what do I do here? So what I can do is the following. Yes, as some of you are pointing out, I can do hadamat C not to get entanglement. But is that the only way in which I can get entanglement? Like Rajes, Sedhart and you know yeah the two of you have pointed out that gate can be used. What is so special about the Haramad gate? any controlled gate we need. Is it only any control gate? Again, let me come back and show you. If you're talking about any control gate, I've defined any control gate here for a two cubit state. Where is it? Okay, it's here. See, again, I'm writing the states as separable states. How is entanglement happening here? See, I have 00 0 1 1 0 1 1 still not happening. So I I mean I'm still not getting any superposition. So what do I do here? There is still no superposition. But the idea is my Okay, I'm seeing some more interesting uh things. So two states have to interact. Yes. How do I make them interact? That is the physicality of things. But how do I make them interact? Yeah. It's nice to see some engagement here. So adamat and c not is one choice. But is that the only choice? That was my question. Is there no other way I can perform internal bitment control u gates? Apparently they are not helping me. What do I do? Haramad and C not. Yes, I understand. So, isamad C not the only way to get entanglement? I will write it here. I will write the question here. So, even people who are having some audio issues can understand it. this. So this C not is this my only option? Is this my only option? Yes. Is this my only option to get entanglement connected by some face vector. Okay, I'm seeing a lot of answers. Now I will answer the question instead of looking at your answers and judging which is correct. Since you know what you've typed, you will figure out what the answer is. Okay, so the point is this. Yes, of course, I'm going to need some gate like the C not gate? I am going to need a control gate. Okay, but what kind of a control gate? So the answer to that is this. So let us assume. Okay, let me assume. So this is I should have put a question mark here not a full stop. [Music] So this slide title is creating entanglement. So how do I create entitlement? So uh so Maduri Muhaband you're talking about using projection operators that can create mixing. That is not what entanglement is. Okay. So let me say I have a control. You get the only thing I need to ensure is whatever basis I use here, I should be able to represent You know what? Give me just one second. I will not write the circuit here. Yes, I will go back and ask this. So let me take yes let me take CU itself I will I will redraw that here. Yeah that way things will become easier. So this is how I represent the control U gate. Okay I have a control cubit here. When this is one I will apply U over here. Okay. So I have some state x here some state y here. If kx is equal to k 1 then what is if then apply u to k y. Okay. So this is the idea of yeah I'm also seeing some comments from the YouTube side. Well my question so to Nidia my question is how do I get a cubid to satisfy the properties of entanglement a cubid state a cubid state to satisfy those properties. So see this is how I have defined it. Cu is defined on the standard basis. CU acts depending on what exists. So whatever my controlled basis my controlled controlling cubit is basis sorry whatever basis my control cubit is in. Yes this is what I meant to say. Whatever basis my control cubit is in as long as I can turn my input into a superposition of that basis. So that is why we used That is why we used say haramad over here hadamad over here. So what happens when I apply hadamad? When I apply hadamad at this point my state becomes 0 + 1 by <unk>2. Then again these are all matrices. These are linear transformations. So all I need to do is just one second. All I have to do is yes a rotation with a C knot or if I have a CU like essentially I'm trying to address the general question. Now two operations will be performed because when it is zero I will have a controlled operation where nothing is happening and then when it is one I will have an operation where the transformation U is applied. Uh we are not talking about topholi gates yet. So please do not invoke the topholigate. I will introduce the tophollegate next week. Okay. So right now yes the answer that I'm asking I'm essentially trying to make everyone here use what has been discussed in today's session. That's it. Okay. So uh yeah. Uh is it doing? Yes. So what is going to be the output in this case? The output in this case is going to be one upon <unk>2 k 0 k y plus one upon <unk>2 k 1 u acting on y. This is going to be the state. Now how do I know that this is an entangled state? This looks entangled on the surface of it. But is this always entangled? Okay, I'm seeing some yeses. Uh, so this is my resultant state. So if I have a Yeah, for those of you who are saying that the recording video is not clear, try changing the video quality on YouTube. Maybe you have a network issue and you know the things are uh kind of getting downscaled on their own. Please change the video quality because we are streaming on uh uh full HD. So you should be getting uh the full clarity uh video. Okay. So now how do I know that the state is entangled or is this always entangled? So this is my state. The next question that I'm asking is is this always entangle? Yeah. Now let me see some answers. Please mind you, I have not told you what the state y is. Only when uy and y are different. Well again when is that different? example when u is equal to identity. Okay, I will state that here. U is not the identity gate. Okay, so thank you for pointing that out. So U not equal to I. I must be a pure state. We are only talking about pure states now. Nobody has gone to mix state=0. Okay. Huh? So let us take an example now. Okay, let me take an example and show you what is happening. So first things first I apply and then I consider C not but what I'm doing is this is my input state get zero here and get plus here wait did I say equal to 1. Oh god. Yeah. I'm sorry. I'm sorry. I'm really sorry. Let me quickly go back. I'm really sorry here. Uh thank you for pointing that out. You're not equal to identity. Yes. Uh yes. So what will be the output in this case? The output is going to be k 0 k + 1x <unk>2 k 0 + 1x <unk>2 x acting on k plus sorry k 1 x acting on k plus what is this state? The state is nothing but 1 by 2 get 0 0 + 0 1 + 1 by 2 get 1 0 + get 1 1 which is equal to 1 by 2 * get 0 0 0 + get 0 1 + get 1 0 + get 1 1. Now this is the same as get plus. This is very separate. So any y will not work here. any y will never work here. Okay, so that is the first thing that one has to remember. So what more do I need to stipulate? This is not always entangled. As you can see, I chose a y where this state was not entangled. Like u was set to be x and I chose y to be plus and you know this state is plus+. So no entanglement is occurring should always be orthogonal not again why should they be computational basis? I mean computational basis is nice. I like that. But we are talking about the general notion of entanglement. Let me try to take minus here. You know what? Wait, I don't have to z let me duplicate this slide. And over here instead of taking K get plus let me take K get minus. Then what happens? Just give me one second. Uh yeah. So in this case what what will I have? I will have minus and another minus. So this is going to give me k 0 1 minus one one. I think I should just erase all the signs. Yes. And over here I'm going to have - 1 0 - 1 1 sorry + 1 1. So this is going to give me - plus and this is equal to k minus k minus. two examples. Let me call the title of this slide and cut it out. So see even now it is not. Now let me take a generic operation. Let me say I have a uh let me say I do. Yes. So the first thing I definitely need is this has to be a superposition of my standard basis. If I'm using a CU gate that much we can understand. So condition one is the controlling cubit superposition of the Yes. So these are the conditions for creating entanglement standard basis. That is the first condition. The second condition we still don't know what it is but I know that if I apply hadamand here and if I take any vector from the hadamand basis I'm not getting an output. So the second condition is I'm going to write the second condition over here is actually the target cubit must be in a superp position. of the vectors in the control cubit session. Sorry. Control cubits not session control cubit spaces I'm sorry cubits spaces. So essentially if I were to choose any vector other than plus or minus because see the first cubit is in the haraman basis. Now essentially the first condition is uh I will write it check two slides. So this is the last thing that I will be doing today. The idea is the second the target cubit must always be a vector that is in superposition of these two vectors. Which is why when we took it to be zero or one which is a superposition of my hadaman basis vectors. My control cubit was in the hard basis and my target cubit must be in a different basis. It should be in a basis that is sorry I E target cubit must not be in the same basis as the control. So what do we do here in this circuit? Typically when we want to create entanglement, I change my control cubit to the haramat basis which is a superposition of my standard basis vectors and then I take the second cubit. So which implies this circuit right here. If I keep this, let us say I start with K0. If I do anything other than the hadamat basis here. So if so the idea is y I can write it like this + y not = 0 and minus y not = 0. both of these if both of these conditions are simultaneously satisfied then my resultant state is guaranteed to be entangled. So this is the second condition. So the idea is whatever basis I'm choosing here that basis should not be here. I can even have the standard basis. I can have any basis it should not be the basis here. The control cubit however should always be the basis that is a superposition of my standard basis vectors. my target cubit should be in any other basis than the one that I have chosen here. So that is the full general condition. So first one is needs to be in a superposition of the standard basis vectors. So this is absolutely necessary and the second condition which is also absolutely necessary is it should be in a basis that is not in the same basis as the control. I should put an apostrophe here. bits pieces but cubic spaces. So y can be 0 or one. Yes, of course it can be zero or because that is not in the hardat basis. It can be in any basis except the basis that we chose for the control equipment. So if we can have that again I'm not talking about the physicality here. I'm just talking about the mathematics here. The physics part well next week it's fine because I will let professor Vishind do his physics so that we can you know kind of pick up from that and move on. With this I will bring an end to my session. I will maybe take questions for five more minutes before ending the session for good. So any more questions from any of you and like looking at both the chats. If you have any questions please ask them. Previous slide. Yes. Go to previous slide. What if we exchanged the bases essentially put minus above and zero below? Yes, of course then we will work then things will work because if no if minus goes above and I apply hardat then no it won't because that will give me a one I will not have entanglement. Remember I need to do something here to change the basis. either I get rid of this addabad and take the minus upstairs or I keep the uh zero here and only change the basis of this. Okay. So all right. Any more questions? Maybe one or two more questions I will take. uh quantum commutation relations help us in the case of see commutation relations do apply in this case but for basics you do not need to know what they are so to Jatin Palot's question for understanding the basics you don't need any uh you know understanding of commutation relations uh so yes like further certification and career you know let me go to the end of the course and I shall talk about that. uh is it is it not essential the controlled operation we can have any other basis like zed or y see zed is not a superposition of my standard basis vectors y yes yes of course I can have the basis needs to be anything other than the z basis or the control cubit and the target cubit can be any vector other than the control cubits basis it can also be the z basis uh is that the inner product of the base is not equal to Z. Similar to having projection not equal to Z. It is the same. It is not similar. It is literally the same. So when I say this I mean that Y is some superposition of K plus and K minus. So adamad and pi in control uh yes instead of using control x you can use control y but not control zed. With control zed you will not get uh oh yeah I also have to mention that here. Uh yeah thank you for pointing that out. It is not equal to identity or zed. Okay. So yes, how we identify the target cubit is obtained by changing the basis. Again, we can there are infinitely many choices. That's what I'm trying to emphasize here. There are infinitely many uh choices to create entanglement. That is essentially what I wanted to show through this example. Okay. So uh had transform is used in coding theory. Yes. See the hard matrices apply appear everywhere. They appear in combinatric. They appear in coding theory. They appear like all over the place. So don't like you know don't be surprised if it suddenly shows up somewhere. Uh it is also it also appears in spectral analysis fer transforms and things like that which we shall be looking at uh next week because that is going to be a part of the HHL algorithm. Okay. So yeah after that yes we have moved to how do we de-entangle maybe next week like you know we have spent too much time on things here so we are at an end of uh things so what are the applications with respect to machine learning soar you why you've raised a hand uh but I don't see any questions from you so if you have any questions please type them into the chat So applications with respect to machine learning. So the problem point is neither me nor professor Gangabadhai nor professor Bish are machine learning uh people we have our own interests in you know the computational aspects of quantum computing. However next week when we talk about the HL algorithm we will give you a brief glimpse at the machine learning part of things. If you want to learn more about quantum machine learning, maybe you can contact someone on the uh uh CAD side. Uh they should be able to provide some contacts of some reference material for you. Uh can we have controlled gate? Yes, of course we can. We can. Yes. Uh maybe next time I will uh uh talk about that. So yeah asking about hands-on simulation. See that is why we gave you the uh access to Qiverse. Whatever you are seeing here you are free to construct those circuits perform measurements and you know check those things on universe yourself because the thing is we have like hundreds of people at least hundreds of people listening to this session. So to have a coordinated hands-on session becomes very difficult for us. We've tried it before. uh hands-on sessions is feasible only when the number of attendance is number of attendees is small and it is even better when we are you know having a face tof face interaction because otherwise we can't exactly see what you are doing and you know uh things like that so yes uh Raman Saha I saw your example yes that is what we do typically to get entanglement so Yes, C not gate is the target. Yes, of course. So, how is what possible later? Wait, Jason, uh, what did you ask? No, like simulation, I didn't say that we will do it afterwards. Simulation, you kind of do it on your own. If you have any doubts, please contact the people at CDK or ask us. We will let you know how things are done. That is why we kind of gave an introduction to Kuniverse. Doing both you know hands-on simulation like doing hands-on simulation and an online course is very very unwieldy. As someone who's done this like been an instructor in this course uh in the online mode like for four or five times, I can tell you it's very difficult. It's in fact impossible. So you'll have to practice by yourself. You can maybe get in touch with us after the course and we can you know have a one-on-one interaction or something like that to see if there are any issues. Changing the cubit itself and transforming bases are equivalent. Is this true only for single cubid states or any n cubit state? It is true for n cubit states. It is true for the entirety of quantum computing. Okay. So yeah that was the last question from susant. I think with that I will end uh today's session. I thank you all for listening. I will uh of course I will be in the session tomorrow but tomorrow's session will be conducted by professor Vish. So I hand over uh control to uh uh the team at CAD. So yeah thank you Joti Sharanji for your uh valuable and fruitful session. I hope the participants have gained maximum out of it and uh I could see the session was very interactive and lot of almost all the doubts of the participants were resolved. Uh thank you sir for your session. So dear participants uh this we are closing today's session and uh tomorrow we will meet at uh 10:00 a.m. again for the next part of the this course. And uh as always uh I encourage you to uh keep u looking at the GitHub page for any updates including the content of the course and the links of the quiz and also YouTube links and u also you we can interact in the Zuli forums for any doubts and any sessions. Uh before closing I would like to thank all the organizers and including uh Miss Mandana and Girish and all the others who have supported uh in this today's session. So this way uh dear participants we close today's session. Thank you all.