Multi Qubit Quantum Gates

Category: Quantum Computing

Tags: gatesmatrixquantumqubitssuperposition

Entities: CCNOT gateCNOT gatecontrolled U gateHadamard gateSWAP gateToffoli gate

Transcript

00:00

Everybody let us talk about multicubit quantum gates. To discuss multicubit quantum gates we consider an igon basis set with four cat vectors k 0 0 k 0 1 k 1 0 and k 1.

So therefore this system

00:17

will represent two cubits and four on states. So it is a dimensional dimension 4 quantum system and the standard basis vectors in this uh basis set are

00:32

represented by these column vectors. The first gate that we will discuss is C not gate.

We are putting C not in as a sub as a subscript of U. So U represents any unitary operation.

00:48

Now this uh K 00 going to K 0 this actually represents the the operation of the C not gate. Now a C not gate will have two cubits.

One is control cubit and the other is target

01:04

cubit. If the control cubit is zero the target cubit will the target cubit will have no action at the output.

But when control cubit is one, the target cubit will be flipped.

01:22

We will get the flipped version of the target cubit at the output. So that's exactly is being shown here.

So in this 000 combination, the most significant bit represents the control cubit and the least significant uh the least

01:37

significant bit is the target cubit. So that means the control cubit in cat 000 the control cubit is zero.

So when the control cubit is zero the target cubit will remain as it is. So that's what exactly is shown here.

Again in the

01:54

second combination the control cubit is zero and the target cubit is one. Since the control cubit is zero so no change in the target cubit.

But let us come to the third combination. When the control cubit is one so definitely the target cubit will get flipped.

So therefore the

02:11

output will be k11. Similarly in the last case when the control cubit is one the target cubit will be flipped and we get an output of get 1 zero.

This represents the

02:27

explicit form of the C not gate. This these this is the representation of the C not gate in terms of the outer product of the respective uh states.

And this is the representation of the explicit form

02:42

using the identity and the polyexcate. Because when you open this outer product ultimately when you solve this equation you will get this matrix and this matrix uh actually this one this is one only.

So you will get this

02:58

matrix. You look at the last part of the matrix, you will see that the last part is actually a polyex gate that represents the flipping operation when the control cubit is one.

In this case, we have only one control

03:13

cubit but we can have more than one uh control lines. So if we have two control lines that means it will be a controlled control not gate.

So this gate will have three inputs two control cubits and one target cubit. The target cubit flips

03:29

when both control cubits are one. So when both will be one only then the target cubit will flip.

The explicit form of the CC not gate is given by this equation. You expand this equation you will get the unititary matrix for uh the

03:45

CC node gate. The CC n gate is also known as the tfoli gate.

Very commonly and very famously it is known as the gate. that this is the symbol of the tophole gate where two control lines are there and one target cubit is there.

04:02

Next we will discuss the swap gate and the fracking gate. The swap gate acts on a tensor product state as so what is the tensor product state?

The tensor product state is the cat s1 s2. So there are two cubits one is k1 and the other is s2.

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Once you apply the swap gate on this set of tensor product state, so what will happen? The state will swap itself.

The swip will swap with each other. That means uh the output will be k1.

04:38

So this is the symbol of the swap gate. So you will see that after the application of the swap gate, you will see that the two uh cq bits are swapped.

The two cubit values are swept and this is the explicit form. Once you solve

04:54

this explicit form, you will find out you will find the solve means when you when you try to put it in the uh in the matrix form. So let me show how uh how can you expand this.

For example, you

05:09

have the first uh term it is k 000 bra 0. So this can be solved as so cat 0 0 is represented by 1 0 0 0 and bra 0

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will be represented as 1 0 0 0. So once you solve this matrix and you add all the other matrices you will see that ultimately uh the matrix that the output that will come uh is written here as the

05:39

swap gate matrix. So that's the unitary operator used for the swap gate.

Now we can also add a control line here. Control adding the control line means you will when the control cubit is one only then the swap will take place.

Again this is the uh

05:57

explicit form. Now we can have the controlled U gate.

Now if you look at the C not gate here uh the last quadrant of the C not gate basically is a polyex gate. So we can generalize the controlled uh we can generalize this

06:14

part and we can put any unitary operator. So that is why uh a gate is there which is controlled u gate.

So control U gate maybe a polyex gate or any other gate you want to uh any other functionality that you want to implement

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that can be done with the that's just an expression of the generalized control u gate the next gate we will I will talk about is the head gate gate is used to make and then unmake superpositions so headed gate can be

06:51

used to generate the equal superposition states of the cubits. Now the matrix form representation of the head gate is uh given by this matrix 2 +2 matrix.

So this is a single cubit gate. Now once you apply uh [clears throat] header gate

07:08

on uh k0. So if you solve this expression you will get you will obtain K 0 uh K 0 plus K 1 um 1 / under root the normalizing factor is 1 / under two

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and this uh particular state is also known as K plus once you apply gate on uh cubit one so you will get uh here a negative sign so K 0 minus K 1

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you will get so you have to you Keep in mind whenever you apply the headward gate on cat zero there will be a plus sign. Whenever you apply headam gate on one there will be a negative sign.

And this overall application of the headward gate on one is termed as

07:55

the uh get negative cubit. Right?

So the application of H on K0 is K plus application of H on K 1 is K minus.

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Now let us we we have to actually extend the use of headom gate on multiple cubits. So let us extend the application of the Hgate to two cubits.

Now this is represented by so if you represent how

08:26

can you represent the application of Hgate on two cubits. So this is the way how we represent it.

This is H gate this is the tensor symbol and two. So this basically tells us you are applying the head operation on two cubits.

So headed

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gate with tensor product two. So this tensor two on the supererscript of H implies that we have to extend the header gate on two cubits.

And uh this is what I have shown the meaning of H chron 2. This is H uh

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supererscript cron 2. Uh when if you apply uh h supererscript chron 2 on get 000 state.

So this is the chron operation between two head gates right

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and this is the value you can solve this h supererscript cron 2 uh expression if you solve this part. So that will be given by this expression.

This 4 + 4 uh matrix right this 1x2 is basically you

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will get from 1x under root2 1x under root2 once you multiply the two you will get it as 1x2 and since we are applying h supererscript on 2 on k 0 state. So we have to put in the product this k0 this is this column vector is representing

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the k 00 state. So once you solve this expression you will uh see that this is the equal superposition of the two cubits.

This is the output of the head output of the header gate applied on two

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cubits. Now I have just given the second method to solve these expressions also.

This is h super scriptron 2 acting on k 00 state. So this can be separately written like this.

And what is the

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meaning of h supererscript on two is that one h is acting on one cat zero like this. So if you solve this again you will get the same uh output as was in the previous method previous case.

All right. Now let us extend.

So uh with

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the what we have seen the application with the application of the hedam gate is that once you apply hedam on k0 as I have just now told you that you have to remember it is k0 plus k 1 under root2.

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So there is a plus sign when you apply h on k0 there is a negative sign once you apply h on uh k 1. So this you have to keep in mind because we have to take care of this uh negative sign once it is applied on the uh on k one.

So we have

11:12

to be very sure about the sign of the uh numerator expression. All right.

So if this is the case let us try to put these two expressions uh in a general form. So what we what we see here is that uh we

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are applying the headmode gate on k x. So k x is representing either zero or one.

So x is a variable that gives us the uh if it is k if we are applying it on

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k 0 or we are applying it on k 1. So if we are applying it on cat zero.

So that means uh if you see here in this expression x is on the exponent part. So zero exponent will give this uh one here.

So plus sign will remain as it is.

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But when x is one when that means when you are applying gate on k one so that means here negative sign will appear. So this expression actually is the generalized form for these two uh for these two

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operations that means headmode acting on cat zero and head acting on cat one. Now let us further simplify the expression like so this is the application of the head gate on a single cubid right so we

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are we are we are just we are just trying to put a summation here. So what this summation will take care of it will take care of k 0 plus k 1 or k 0 minus k 1 that's why this summation is there

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right and this minus1 exponent x doy will take care of the uh sign right so if you will try to expand this you will see that either you will get get either you will

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get this expression or you will get this expression If you expand this expression. All right.

So this is the application of header gate generalized expression on single cubit. But let us extend the same concept to two cubits.

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So this is h supererscript drawn to acting on a two cubit system. A two cubit system is represented like this also.

This is a basis set for two cubits. These are the igon states mentioned straightway.

This is the

13:43

decimal form representation of the igon values. And this x is basically x do this for the two cubits.

X1 is one cubit and x2 is the other cubit. Now you you pick the expression from here which

14:00

which uh which is the headmode acting on one cubit. Here we are applying hedam on two cubits.

So use this expression to write two uh brackets and two expressions for different cubit that means x1 and x2. All right expression

14:18

remains the same. Now if you solve little further you can uh you can just combine this minus1 with this minus1 and you will get in the exponent x1 y1 + x2 y2.

Okay. And uh this will be a 2 cubit

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system. That means four on states will be there.

So x doy basically x doy is a dotproduct of the vector components. What are the vector components?

x1 y1 x1 y1 x2 y2 are the ve are the components

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of the vectors for x and y. Now we have to extend this expression to an n cubit system.

That means we want to know what happens when you apply the head gate on an n cubit system. Now your kx is

15:09

basically an n cubit u n cubit uh quantum system quantum state and we apply the uh edge supererscript n that means we are applying header uh on n cubit system and we have to

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basically generalize we have to expand this expression only you note the coefficient value it is coming out to be 1 / under <unk>2² I'm not writing here four because I just want to uh just want to have a

15:43

generalized expression because to this exponent two I want to put n because I'm using it for the n cubit system. So it will be 1 / under <unk>2 raised to the power m exactly the same expression comes which which gives the uh which

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which if you expand this expression you will get the header acting on the n cubit system. Now the starting point again I say starting point is the headm acting on single cubit uh whether it is

16:18

k 0 or k 1 you have to keep track of if you are applying headm on cat zero you will get a plus sign here k 0 plus k 1 so plus sign is there once you apply the header on k 1 you will get a negative sign so that you have to just take care

16:36

of and that's why these we we are actually deriving these expressions so that we should not miss this negative sign. Now in quantum computing sometimes we have to write a different equation than this

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equation number one. This is because we we write this equation because we usually start with the all zero state.

Normally the quantum system is initialized with the all zero state. So that is why we we write a simpler form of the

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equation number one. Now if we are starting with K 0 state that means the starting state is K zero.

In that case to our advantage we need not to take care of the negative sign because we know that if the uh if we are applying

17:23

the header on get zero state whether it's a 2 cubit 1 cubit or n cubit system we not we need not to worry about this sign because we will get the plus sign. So therefore this becomes easier for us.

This uh this shows the header word

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acting on K 0 that's an n cubit system and you will see that there is no uh we are not taking care of the negative sign because we know that the negative sign will not come. So this is the uh reduced

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expression if we are starting with all zero state. So this is all about the multicubid quantum systems.

Thank you very much.