Quantum Computing

Chapter 1

Brief overview

1.1 Linear Alzebra

1. Tensor Product: A Tensor product is a method of multiplying two tensors (matrices), given by the general (pneumonic) form: (It is represented as ) ⊗

a11 a12 a13 a14 ... a11M a12M a13M a14M ... a a a a ... a M a M a M a M ...  21 22 23 24   21 22 23 24  a31 a32 a33 a34 ... a31M a32M a33M a34M ... a a a a ... M = a M a M a M a M ...  41 42 43 44  ⊗  41 42 43 44   ......   ......       ......   ......       ......   ......          where, M is any matrix. If the matrices on the LHS have the dimensions (D D ) and (D D )respectively, 1r × 1c 2r × 2c then the result matrix, on the RHS will have the dimensions: [(D + D ) (D + D )] DRAFT COPY1r 2r × 1c 2c It is really important to note that the above is NOT a formal deﬁnition for a tensor product, but it is just a pneumonic form. The result in this case seems to have the same dimensions as the original matrix in the LHS (ﬁrst matrix). Only when we expand the RHS by putting various values of M,we will get the matrix of the dimensions (as given above).

1.2 Basic Quantum mechanics

We shall look at fundamentals of quantum mechanics in greater detail in the next chapter. Some important definitions are: 1. State of a system:1 The state of a quantum system is a vector in the infinite dimensional complex vector space known as the Hilbert Space. The state of a classical system is represented by its position and momentum in a phase space. But this is not possible in Quantum Mechanics because of the built in concept of the Heisenberg’s Uncertainty principle. So, if we close in into one definite value of position for a position, the possible values that its momentum can take is infinite. So, we can only close on a given area, and say , with some probability that the particle lies within that area. So, the (classical) state of the particle lies anywhere

1Adeeperpictureofthisisgiveninthequantummechanicssection.Fornow,thisdescriptionwoulddo.

9 1.3. QUBIT - BASIC UNIT OF QUANTUM INFORMATION CHAPTER 1. BRIEF OVERVIEW

in that continuous area that we define. So, the state of the particle has infinite position and momentum coordinates. Hence, it is represented as a vector in an infinite dimensional complex vector space (Hilbert Space). The state of a N-independent particle system where the individual particle wave functions are ψ , ψ , ψ ... ψ , the combined state of the N particle system is given by: | 1 | 2 | 3 | N ψ = ψ ψ ψ ...... ψ | | 1⊗| 2⊗| 3 ⊗| N

In general, if ψ1 , ψ2 , ψ3 , ..... are vectors in N1, N2, N2, ..... dimensional complex vector spaces respectively, then| ψ| is a| vector in a (N + N + N + .....) dimensional complex vector space. | 1 2 2 2. Dirac Bra-c-Ket Notation: The dirac ket notation is a well known and frequently used here. In this notation, every column vector −→ψ is represented as: ψ which is also called a Ket vector. Similarly, a row vector is represented as ψ which is called a Bra| vector. Hence the name: BracKet notation. | 3. Local and non Local processes: By “local ”process between two particles , it is meant that influ- ences between the particles must travel in such a way that they pass through space continuously; i.e. the simultaneous disappearance of some quantity in one place cannot be balanced by its appearance somewhere else if that quantity didn’t travel, in some sense, across the space in between. In particular, this influence cannot travel faster than light, in order to preserve relativity theory. 4. Canonical Commutation Relations: The some observables in Quantum mechanics do not commute. That is, they have a non zero commutator. The commutator for a pair of operators is defined as:

[A, B]=AB BA ——— Commutator − The basic commutator relations in quantum mechanics are:

[xi,pj]=i

[xi,xj]=0

[pi,pj]=0

1.3 Qubit - basic unit of quantum information

A ’bit’, a classical two state system, represents the smallest unit of information. A classical bit is represented by 1 or 0. It can be thought of as ’true’ and ’false’ or any two complementary quantities whose union is the universe, and intersection is the null set. There is a profound reason to why the smallest unit of information DRAFTis a 1 or 0, or ’true’ or ’false’. This is because any logical querry COPY can be split into a series of ’yes’ or ’no’ questions. That is, with a series of ’yes’ or ’no’ answers to questions, we can perform any logical query. This is why a ’bit’ (which can be thought of as a most general strcture of storing information) is a 1 or a 0. A quantum bit, is just an example for a two state quantum system. A qubit can can also be an electron (with spin up and down), an ammonia molecule, etc. In quantum mechanics, a two state quantum system does not mean that it has only two states. This is what distinguishes a qubit from a classical bit. The difference comes due to a very important quantum mechanical phenomena known as interferrence.Justlikehowa classical bit can take a value 0 or 1, a quantum bit can take values 0 or 1 , and any value produced by the interferrence between the states 0 and 1 (like α 0 + β 1 ). Since,| there| are infinite such superpositions (where each of the states have 0 |and 1 given| by some| probability| amplitude α and β respectively), a qubit can exist in infinite states. If each state can store a unit of information, then the qubit can hold infinite units of information.

A Qubit, like any other two level quantum system, is (conventionally) represented by its state:

ψ = α + β | |↑ |↓ ψ = α + + β | | |− ψ = α 0 + β 1 | | | where α 2 + β 2 =1. | | | |

GO TO FIRST PAGE 10 ↑ CHAPTER 1. BRIEF OVERVIEW 1.4. MULTIPLE QUBITS

This can be misleading, since it gives us the feeling that a classical bit can at most carry 2 units of information whereas a quantum bit can carry inﬁnitely many units. But, if a measurement is done on the state of the Qubit, it collapses into one of the eigen states of the measurement. So, if measurement of ψ gives ’a’, then after this measurement, the state of the Qubit will remain a (the eigen state of the measurement| corresponding to the eigen value a). This new state a now, will got give| the same measurement results as a . In fact it will not respond to any other measurement.| Why this happens Postulates| of Quantum Mechanics. Hence, only a single→ unit of information can be retrived from a Qubit.

1.4 Multiple Qubits

Any two state system (like the electron which has a spin) can be represented by a Qubit. But what about representing the state of two electrons (which are independent of each other), using a qubit? Such a representation, we saw in the beginning of the chapter, was possible. If the state of one electron is ψ and state | 1 of the other electron is ψ2 , then the system of two independent electrons can be collectively represented by the state ψ ,where: | | ψ = ψ ψ | | 1⊗| 2 Since, we take a direct product of the two states, we may represent the now state, which is the two qubit state, as: (using the convention: i j ij ) | ⊗| ≡| ψ = α 00 + α 01 + α 10 + α 11 | 00| 01| 10| 11| where : α 2 + α 2 + α 2 + α 2 =1 | 00| | 01| | 10| | 11| 2 and αij is the probability of the first qubit being in state i , and the second Qubit being in state j .If we want| | only one of them, we must sum over the other. If we| want the probability of system being in| state 2 i only, we must sum αij over all j’s. | | | 2 2 So, the probability of measuring the first qubit to be 0 is = α00 + α01 , and obviously, the measurement 2 | 2| | | α00 + α01 will collapse the state of the qubit into ψ = | | | | . | α 2 + α 2 | 00| | 01| Therefore, we can say that in ths 2 qubit system, we can retrieve 2 units of information. It certainly DRAFTcan deliver more information than a single qubit system, but COPY there are some difficulties too. In general, we need to carry the measurement process twice to determine the information stored in both the qubits. So, earlier we were carrying out the measurement only once, and now we need to do it twice. Can it be better? Can we get away in one measurement itself? In other words, can we store some amount of information about one qubit in another, such that we can guess both the qubits, by measuring only one of them? The answer is yes. We can do such a trick that can, with certainty, retrieve the information stored in one qubit, by measuring the other. Such a two qubit state is called a Bell State or the EPR pair.

00 + 11 The Bell State is given by : ψ = | | | √2 1 Here, the first Qubit being is measured to be 0 with the Probability (changing the state to : ψ = 00 ) 2 | | 1 and 1 with Probability (Changing the state to : ψ = 11 ) 2 | | Hence, P(measuring the first qubit to be 0) = P(measuring the second qubit to be 0) , also state after the measurement of the first qubit to be 0 = state after the measurement of the second qubit to be 0. Similarly, P(measuring the first qubit to be 1) = P(measuring the second qubit to be 1) , also sate after the measurement of the first qubit to be 1 = state after the measurement of the second qubit to be 1. Therefore the first qubit always gives the same result as the measurement of the second qubit. Hence, we can say that: By knowing the result of measurement to the first qubit, we can tell by certainity,

GO TO FIRST PAGE 11 ↑ 1.5. QUANTUM GATES CHAPTER 1. BRIEF OVERVIEW

the result of measurement of the second qubit. Also we can say ”The two states are perfectly correlated”. In the language of quantum mechanics, the two states are ”Entangled”.

It is also easy to note that this property cannot be satisﬁed by any arbitrary state. Hence, we need to uniquely classify these states.2 Its allpications and signiﬁcance become more prominent as we procees in the later sections.

1.5 Quantum Gates

Just as we have Classical Gates that operate on classical bit(s), we also have their quantum analogues. To Start with, consider a simple classical gate, the NOT gate:

We can now think of its quantum mechanical analogue: Consider the Gate G that is a ”quantum mechanical NOT gate”. G ”ﬂips” the state of a qubit.

G :(α 0 + β 1 ) (α 1 + β 0 ) | | → | |

We can now try to see how this process of “ﬂipping” the qubit is carrier out. We know that, in the case of two level systems, (or Spin Half) the state 0 can be changed to 1 and vice-versa using the ladder operators | | given by S and S+. The action of these operators is given by: −

S+ = Sx + iSy

S = Sx iSy − − S 0 = 1 —— Raising operator DRAFT+| | COPY S 1 = 0 —— Raising operator +| Simillarly: S 1 = 0 —— Lowering operator −| | S 0 = 0 —— Lowering operator −| Therefore we can say: textbfG (S+ + S ) ≡ − (S+ + S )(α 0 + β 1 )=α(S 0 + S 0 )+β(S 1 + S 1 ) − | | −| −| −| −| α( 1 + 0) + β( 0 + 0) ⇒ | | α 1 + β 0 ⇒ | | Hence, G(α 0 + β 1 )=α 1 + β 0 | | | | A Very important aspect to take note of is that: the state 0 is very diﬀerent from 0. The latter means a null vector. It represents void. While the former represents some| state in which a particle is present in. The state 0 does not mean void. |

2Istillcannotgetwhatissospecial,italmostseemsliketheyaretwoidenticalquantumstates.Itisliketakingtwoclassical bits 0 and 0 and saying that the result to measurement of 0 is = to the result of measurement of the other 0 state. What property of QM is being used?

GO TO FIRST PAGE 12 ↑ CHAPTER 1. BRIEF OVERVIEW 1.6. BLOCH SPHERE

Since the state of the qubit is a ket vector, we can also have a column vector representation for it: α ψ = | β now the Gate G can be deﬁned as:

α β G = β α By looking at this property, we can ”guess” the matrix form of G,tobe:

01 G = . We can verify that this is the σ + iσ or S + iS operator, as computed above. 10 x y x y

1.5.1 Other single qubit gates:

There are many single qubit gates. A major requirement for a quantum gate operator is that it must be Unitary. This has 2 consequences:

1. The conservation of probability.

2. Since the inverse of an unitary matrix is also an unitary matrix, each single bit quantum gate can be ”undone” by some other single bit quantum gate. So, the input can be obtained by performing some operation on the output. Therefore there is no ”loss of information”, unlike the classical case where the gates are not invertible.

Let us consider a Z Gate that ”leaves 0 unchanged, and ﬂips the state of 1 to 1 . | | −|

Z (α 0 + β 1 )=(α 0 β 1 ) | | | − | 10 We can also guess that: Z = . This is similar to the σ or S operator (pauli spin matrix). 0 1 z z − Let us now consider yet another important single qubit gate, the Hadamard Gate. This gate ”trans- forms the states 0 to a states between 0 and 1 ”. This gate can be equated to an operation which is DRAFT| | | COPY π 0 + 1 0 1 reﬂection of the qubit vector about the line θ = . It changes 0 to | | and changes 1 to | −| . 8 | √2 | √2

0 + 1 0 1 H(α 0 + β 1 )=(α| | + β | −| ) | | √2 √2

1 11 The Matrix form of H can be guessed as: H = √2 1 1 −

1.6 Bloch Sphere

The state of any two state system is represented by a point in a two dimensional complex vector space. Since, this is a two dimensional complex vector space, each dimension (like we have x, y, and z dimensions in the cartesian frame) is complex. Each complex number needs two real numbers to represent itself. So, the state

GO TO FIRST PAGE 13 ↑ 1.6. BLOCH SPHERE CHAPTER 1. BRIEF OVERVIEW

can now be described by four real quantities, rather that two complex quantities. Hence, we have: The state in the two dimensional complex vector space is: ψ = α 0 + β 1 | | | Since, α and β are complex, each of them can be described by two

real quantities: α =(rα,iα) and β =(rβ,iβ) The normalization condition: α 2 + β 2 = 1 , now translates to the condition on the | | | | four real quantities as: r 2 + i 2 + r 2 + i 2 =1. (1.1) | α| | α| | β| | β| Just like how the equation x2 + y2 + z2 = 1 represents the surface of a sphere, placed in a three dimensional space, the above equation (1.1) represents the surface of the sphere kept in a four dimensional space. Now, this is the motivation for us to try to describe the state of a two state system as a point on the sphere. A four dimensional space is still a bizarre object for us to imagine. So, we need to try and remove one degree of freedom here so that we get our usual two dimensional sphere in a three dimensional space. The two dimensional sphere hence obtained is known as the Bloch Sphere. For this purpose, we need to work out the above process in the polar form. So, we shall have:

iφα iφβ The complex numbers: α = rαe and β = rβe Therefore, the state of the system, ψ = r eiφα + r eiφβ | α β So, till this point, we have been working with a sphere kept in a four dimensional space, as there are four real quantities in the equation of the state. Now, we need to remove one degree of freedom, that is eliminate one real quantity from the equation of the state. For doing so, we need to recollect a very important feature of quantum mechanics that any quantum mechanical system is invariant under rotation by a overall phase. The following sentence can be realized if we go back to our basic state of the quantum mechanical system. The state is represented by ψ which is actually a probability amplitude. But what we can measure is the probability density, denoted| by ψ 2. So, even if there is an overall phase factor, like γ, in the probability amplitude, it will not aﬀect our measurement,| | and for all values of γ, the system will be identical3. So, let us multiply the system by any an overall phase angle. The choice can be decided by us, to suit our requirements.

iφ Since the choice of γ can be artitrary, let γ = e− α , so that, we can eliminate one real variable. iφ iφ iφ iφ iφ Therefore, the state of the system, e− α ( ψ )=r e α (e− α )+r e β (e− α ) | α β Now, the LHS remains as ψ since the system is invariant under overall phase change. | Let: φ = φ φ . Then, on simplification we have: ψ = r 0 + r eiφ DRAFTβ − β COPY| α| β Now, in the above expression, the second term of the RHS has a complex coefficient. We can write this in the complex form (x + iy): ψ = r 0 +(x + iy) 1 where x = r cosθ and y = r sinθ ⇒| α| | β β Now, let us apply the normalization condition: r 2 + x 2 + y 2 =1 | α| | | | | Therefore, we now get the equation of the Bloch sphere. Hence, we got the equation of the Bloch sphere. We now need to find how the state of a system can be represented on the bloch sphere. For this, let us go to spherical coridinates. Let us map x, y and rα on to a sphere with unit radius. Let us now make the transformations: x = cosφ sinθ y=sinφ sinθ

rα = cosθ Therefore, the state now becomes: ψ = cosθ 0 +sinθ(cosφ +sinφ) 1 | | | We can write this as: ψ = cosθ 0 +(eiφ)sinθ 1 | | | 3In fact, this probability amplitude is the reason for interference.

GO TO FIRST PAGE 14 ↑ CHAPTER 1. BRIEF OVERVIEW 1.6. BLOCH SPHERE

Now, by convention, in spherical polar coridinates, θ goes from 0 to 2π.Buthere,ifweputθ = 0, we get π ψ = 0 and on putting θ = , we get ψ =(eiφ) 1 . | | 2 | | π θ So, we see that θ =0toθ = covers the entire sphere. So, we modify our θ by changing it to . Now θ 2 2 π =0toθ = covers the entire sphere. So, the ﬁnal equation of the state of the system in a Bloch sphere 2 is: θ θ ψ = cos 0 +(eiφ)sin 1 (1.2) | 2 | 2 | So, the above equation represents the single qubit in a Bloch Sphere. Now, what about multiple two state systems, or multiple qubits ? Let us take a two qubit system represented by its state ψ = ψ1 ψ2 .The state space of ψ is now a vector in the four dimensional complex vector space because| each| ⊗ of the| states are vectors in a| two dimensional Hilbert space. Following the exact similar argument as above, we can see that the state ψ can be represented as a vector on a seven dimensional sphere, kept in a eight dimensional space. This seems| confusing because in the above case, we claimed that a two state system (represented as a vector in 2) can be represented on a two dimensional surface. Extending the same argument, one should say that a compositionH of two state system (represented as a vector in 2 2) should be represented as a vector on a four dimensional surface (need not be a sphere, but still it mustH × beH some four dimensional surface). Hence, we see that the surface used to represent ψ has more dimensions than expected. We know that a seven dimensional surface has more points than a four| dimensional one. We can now conclude that the seven dimensional surface can represent more states than a four dimensional one. Therefore, there is some missing information that cannot be represented on the four dimensional surface. The seven dimensional surface carries more information than a four dimensional one. This means that when we take a tensor product of two 2-state systems, the new system formed contains information about each of the individual systems as well as some excess information that cannot be attributed to any single one of them. We arrived at a four dimensional surface from the assumption that all compositions of two 2-state systems can be represented as a tensor product of some two 2-state systems. From the fact that the sphere which represents all the compositions has more states, we can say that our old assumption that all compositions of two 2-state systems can be represented as a tensor product, fails. So, there are some states that are composed of two 2-state systems, but cannot be expresses as a tensor product. These states have information (properties) that are not related to any one of the 2-state systems. This information is lost when the qubits are separated. Hence, this information is due to the tie, or (in more sophisticated words) entanglement of the two 2-state systems. It is quite clear that all the states do not have this property. States that have this property are called entangled states. The Bell State DRAFT00 + 11 COPY 00 = | | . B √2

1.6.1 Generalizing Quantum Gates - Universal Gates

Now, since gates are unitary operators, that act on the state of the qubit, we can think of them as rotation operators on the Bloch Sphere. Now each gate performs some rotation. But any rotation, in general can be broken into a sequence of standard rotations, about the x-y plane, y-z plane and the x-z plane. Since, any rotation can be broken into a sequence of standard rotations, we can drawn the same analogy and say that any Quantum gate can be represented by a sequence of standard gates that act on the state of the qubit ψ , and produce the same answer as the original gate. | Consider a rotation operator U. We can decompose U into several basic rotation operators: β γ γ δ i cos sin i iα e 2 0 2 2 e 2 0 So, U = e  β  γ γ  δ  i  sin cos  i  0 e− 2  − 2 2  0 e− 2           

In other words, these are ”universal” operators. Similarly, we can consider universal quantum gates, that,

GO TO FIRST PAGE 15 ↑ 1.6. BLOCH SPHERE CHAPTER 1. BRIEF OVERVIEW

when manipulated appropriately, can mimic any other quantum gate. Let us consider a simple classical universal gate, the NAND gate. The XOR Gate is not a universal gate because it fails to change the parity of the bits.4:

Figure 1.1: A classical univaersal gate. NAND gate.

Let us now consider a Universal Quantum Gate: A slight modiﬁcation of the NOT gate, The controlled NOT DRAFTgate or the CNOT Gate. COPY

Figure 1.2: A CNOT Gate

A CNOT Gate is a ”two input NOT gate”. It takes two inputs: a DATA and CONTROL, ﬂips DATA if CONTROL = 1. Classically, this is analogous to the XOR Gate. leaves DATA unchanged if CONTROL = 0.

The Gate operation is represented ast: CNOT:( A + B ) A, B A | | →| ⊕ The action of this Gate can be described explicitly:( DAT A, CONT ROL DAT A,CONTROL ) 00 00 | →| |01→|11 |10→|10 |11→|01 | →|

4This is not clear to me

GO TO FIRST PAGE 16 ↑ CHAPTER 1. BRIEF OVERVIEW 1.7. IMPORTANT CONVENTIONS

In a Quantum circuit, the CNOT gate is represented as:

Figure 1.3: Circuit representation of a CNOT gate. The top wire has the control bit and the bottom has the Target or the Data bit.

We can now look at the matrix representation of this Gate.

1000 0100 CNOT = 0001 0010    

1.7 Important conventions

α 0 + β 1 —————– X NOT gate———— α 1 + β 0 | | | | α 0 + β 1 —————– Z Zgate—————– α 0 - β 1 | | | | 0 + 1 0 1 α 0 + β 1 —————– H Hadamard gate—— α| | + β | −| DRAFT| | √2 COPY√2 A most remarkable or unique feature that we notice about a quantum gate, is that it operates on single bits, which are like a superposition of 2 probabilistic classical bits. The unitary(existence of inverse) nature of these operators enables the input Qubit to be easily retrieved.

1.8 Measurement Basis

Measurement is an operation that is performed on system, to determine with certainty, the state of the system. Measurement also has a meaning in classical bits. Each classical bit had a even probability of it being 0 or 1. When a measurement is done, we know by certainty what value that bit has. The classical bit is 0 with probability one half or 1 with probability one half. Both these choices forever exclude each other. Whereas in the quantum case, the diﬀerence comes here. We have superposition states. For example, a qubit can be 0 with probability one half and 1 with probability one half. Note the usage of and and or. The meaning| of measuring some property| (operator) of a state ψ is nothing but ﬁnding the eigen values of that operator (corresponding to the property). The allowed| values of any property (or, the result of measurement of any property) are limited to the eigen values of the operator representing this property.But it is not so straight forward: what if we are trying to measure the property corresponding to the operator A, and the given state ψ is not an eigen state of A? | GO TO FIRST PAGE 17 ↑ 1.8. MEASUREMENT BASIS CHAPTER 1. BRIEF OVERVIEW

For example, till now we took ψ = α 0 + β 1 . This means we took the ket vectors 0 and 1 as our basis | | | | | states. These vectors are nothing but the eigen vectors of the σz or the Sz operator. So, we were in the Sz eigen basis. That is, the operator Sz is diagonal in this basis. Now, if we measure a property corresponding to the Sz operator, then the measurements results can have only two allowed values: 1 and -1, because these are the eigen values of Sz. But what if we want to measure a property corresponding to the Sx operator ? Here, 0 and 1 are certainly not the eigen vectors of the S operator (they better not be because if they | | x are the eigen vectors of Sx also, that would mean Sx and Sz commute). Here, we make use of a key property of the eigen basis. They form a complete basis. So, any state ψ can be expressed as a superposition of the | Sx eigen states, and the corresponding eigen values are the result of the measurement. So, to start with, we have a state ψ in the S basis. Now when we want to measure a property corresponding to S ,we | z x ﬁnd the eigen states of Sx, and write ψ as a superposition of those eigen states. Now the coeﬃcients of the eigen states are the measurement results,| and the eigen states are those to which ψ shall collapse after measurement. |

So, whichever property (operator) we are measuring, we must expand ψ in the eigen basis of that operator (correspond to the property), and that ﬁnd the coeﬃcients of those states| in the expansion of ψ . |

1.8.1 Quantum Circuits

A quantum circuit, like any other, is read from left to right. A line in the circuit represents a wire. A wire here may just denote the path of the photon (or passage of time, etc. There is no physical path or channel.) The input to a circuit is the qubit (represented by ψ , which by convention, is assumed to be in the computational basis state). | An important feature not allowed in quantum circuits is loops. There is no feedback from one part of the circuit to another. Thsee feedback forms are of 2 general types: several wires joined together FANIN , and a copy of the qubit in one wire, going to multiple wires FANOUT. These are not allowed. Refer to ﬁgure. As DRAFTan interesting consequence of this, the copying of a qubit is an COPY impossible task.

Figure 1.4: FANIN and FANOUT characteristics:

Also, we can generalize the representation of a controlled not gate which takes N qubits. The generalized representation is shown in the ﬁgure:

GO TO FIRST PAGE 18 ↑ CHAPTER 1. BRIEF OVERVIEW 1.8. MEASUREMENT BASIS

Figure 1.5: Generalized CNOT gate

Another important aspect of quantum circuits are meters used to measure the quantum bits.5 As we have already discussed, the measurement will collapse the qubit into a probabilistic classical bit (distinguished by drawing a double arrow line)6. ψ (= α 0 + β 1 ), upon measurement, will change to a classical bit that gives result 0 with probability α| 2and 1| with| probability β 2. A representation of the quantum meter is shown below: | | | | DRAFT COPY Figure 1.6: Representation of a Quantum Bit measuring device:

1.8.2 Quantum Copying or Cloning circuits

One of the key features of quantum computation is that it disallows the replication or copying of Qubits. We can see why this happens. Let us take up the issue of replicating or the copying of a classical bit. To accomplish this classically, one could do the following:

5measurement here means to identify the qubit. measuring a state α 0 + β 1 means to extract out α and β. | | 6The subtle difference here is that a probabilistic classical bit is fundamentally different from a quantum bit. This is because aclassicalhassomefiniteprobabilityofcarryingonly one piece of information, and it gives that upon measurement. Qubit on the other hand carries infinite information, but when measured, it has some probability of collapsing on to one piece

GO TO FIRST PAGE 19 ↑ 1.8. MEASUREMENT BASIS CHAPTER 1. BRIEF OVERVIEW

Figure 1.7: Classical method of copying a bit. Here stands for the XOR operation. Here, we take an arbitrary bit x and perform an XOR operation with that⊕ bit and y, where y is another bit that is constantly = 0.

Let us now try the following with the qubit also: Here, we blindly replace the classical XOR with the quantum XOR or . A valid question would be : have we copied all the information stored in ψ .The answer obviously is⊕ NO. It is obvious because, we agreed that an inﬁnite amount of information is| stored in ψ . So, all of it cannot possibly be copied. This theorem is called the no cloning theorem.Thereisa short| proof of this theorem. Suppose we have two states x and y , such that there exists a machine U which copies the state x into y ( y is called the target| state ),| then the system of these two qubits is in the state: x y .| So, we have:| | | ⊗|

U( x y )=(x x ) (1.3) | ⊗| | ⊗| Here U is copying the value of a state into another. We claim that U is some unitary, universal operator that can clone any quantum state. Let us not take two arbitrary quantum states ψ and φ . Let the target state be represented as s . According to the above claim, we have: | | | U( ψ s )=(ψ ψ ) (1.4) | ⊗| | ⊗| similarly, U( φ s )=(φ φ ) (1.5) | ⊗| | ⊗| DRAFTWe can take an inner product of equations (1.4) and (1.5). To COPY take an inner product of the equations means to take the inner products on the two RHS to form a new RHS and inner product of two LHS to form a new LHS. The inner product of any two matrices (applies to even column vectors - since they too are matrices) is nothing but the product of the hermitian conjugate of the ﬁrst with the second matrix. The LHS of equation (1.4) is of the form A x . This is nothing but a product A B. The hermitian conjugate | · of (A B)isB† A†. The LHS will then be: · ·

[U( ψ s )]†[U( φ s )] | ⊗| | ⊗| ( ψ s )†U†[U( φ s )] ⇒ | ⊗| | ⊗| ( ψ s )U†U( φ s ) ⇒ |⊗ | | ⊗|

In the last step, we see that (A B)† =(A† B†). Also since U is unitary, U†U = I. Hence the LHS becomes: ⊗ ⊗

( ψ s )( φ s ) ( ψ φ ) ( s s ) |⊗ | | ⊗| ⇒ | ⊗ | In deriving the above step, we have used the formula: (A B) (C D)=(A C) (B D). Also, since the target state s is normalized, s s =1. ⊗ · ⊗ · ⊗ · | | LHS: ( ψ φ ) ∴ |

GO TO FIRST PAGE 20 ↑ CHAPTER 1. BRIEF OVERVIEW 1.9. QUANTUM TELEPORTAION

Coming to the RHS and applying similar simpliﬁcations, we have:

( ψ ψ )†( φ φ ) | ⊗| | ⊗| ( ψ ψ )( φ φ ) ⇒ |⊗ | | ⊗| ( ψ φ )2 ⇒ | Therefore, we can write the new equation formed as:

( ψ φ )=(ψ φ )2 | | This equation is of the form x = x2. The solutions to this equation are x = 1 and x = 0. The first case ( ψ φ ) = 1 means that both ψ and φ are identical states. But then, this would mean that U only copies states | that are identical. That| is, U can| only copy some specific state. But this contradicts our assumption that U is an universal operator. In the other case where ( ψ φ ) = 0, we see that ψ and φ are both orthogonal states. So, the cloning operator U will only clone orthogonal| states. | | Therefore, we can conclude that there is no universal operator that can copy any arbitrary state. If it can copy a given quatum state, then the only other state that it can copy is one which is orthogonal to the given state. In the figure given below:

Figure 1.8: Repeating the same above process for a quantum case DRAFTThe process can be concisely described as: COPY [α 0 + β 1 ] 0 ——————— CNOT ——————— α 00 + β 11 | | | | | The output however, is not equal to ψ ψ because when we multiply [α 0 + β 1 ]by[α 0 + β 1 ], we get: α2 00 + αβ 01 + αβ 10 + β2 11 .| But| this certainly isn’t equal to our| result obtained| from| the| CNOT gate since| the cross| terms| 01 and| 10 are not present. But one can now say that we can copy the quantum bit if αβ = 0. But this must| mean| that at least one of them must be 0. If this is so, then the bit will no longer be a “qubit”. Hence it is impossible to copy the state of a Qubit.

1.9 Quantum Teleportaion

1.9.1 Bell States

We have encountered these Bell States earlier. They are known to be the most correlated pair. The result of the measurement on the ﬁrst qubit is the same as the result of the same measurement on the second qubit. In this case, on measuring the ﬁrst qubit of the Bell State, one obtains 2 possible answers: 0 with probability 1 1 and 1 with probability . On measurement on the second qubit one obtains 2 possible answers: 0 with 2 2

GO TO FIRST PAGE 21 ↑ 1.9. QUANTUM TELEPORTAION CHAPTER 1. BRIEF OVERVIEW

1 1 probability and 1 with probability . The measurement on the always yields the same result. These Bell 2 2 states are formed by taking the two qubit system (in the computational basis state), and pass them into a Hadamard gate followed by a CNOT gate. A simple table for this would be:

Table 1.1: Table Showing the input and output states of a Hadamard Transform, used to produce Bell States: In Process and Out

00 + 10 00 + 11 00 —— H —— | | —— CNOT —— | | B00 | √2 √2 ≡|

01 + 11 01 + 10 01 —— H —— | | —— CNOT —— | | B01 | √2 √2 ≡|

00 10 00 11 10 —— H —— | −| —— CNOT —— | −| B10 | √2 √2 ≡|

01 11 01 10 11 —— H —— | −| —— CNOT —— | −| B11 | √2 √2 ≡|

The generalized bell state is given by:

0,y +( 1)x 1,y Bx,y | − | | ≡ √2

1.9.2 EPR paradox and Bell’s inequality

We shall look at these in detail in the next chapter: Fundamentals of Quantum mechanics. We just saw that if we have the result of measurement of one qubit in the Bell state, the result of measurement on the second qubit is determined. Not only this, but even if the two qubits are (practically) inﬁnitely far away, the result of measurement os determined instantaneously, after measuring one of them. It is surprising DRAFThow this is possible because we know that information cannot COPY be transmitted at a speed greater than the velocity of light. Therefore, following this diﬃculty, EPR suggested that the two half’s of the Bell states already had some more information, that describes the state of the two partice system at any given time, and that we have not accounted for that information in our formulation of Quantum mechanics. Also, EPR called this information, hidden variables. Later on ,John Bell suggested a mechanism to test for the presence of local hidden variables. He devised an inequality which and he claimed that, if a quantum system does not satisfy, it is impossible for it to have a local hidden variable.

1.9.3 Application of Bell States: Quantum Teleportation

Consider the following problem: Bob and Alice are two friends living far apart. They together generated an EPR pair or a Bell State. Alice and Bob have the 2 halfs of the state. Now Bob is hiding, and Alice’s mission is that she must deliver a state ψ to Bob. From| looking at the problem, we can see that things are bad for Alice. She can only send classical information. She cannot copy the state ψ . Alice does not even know the state ψ , and even if she knew it, it would take inﬁnite amount of information| (and time) to describe ψ since it takes| values in a continuous space. So, the only thing left for Alice is to utilize the EPR| pair (take advantage of the property of coherence of measurement). So, brieﬂy, what Alice does is the following:

GO TO FIRST PAGE 22 ↑ CHAPTER 1. BRIEF OVERVIEW 1.9. QUANTUM TELEPORTAION

1. Alice interacts ψ with her half of the EPR pair, and gets one of the four possible results: 00, 01, 10, 11. | 2. She sends the obtained result as classical information to Bob. 3. Bob, knows that his measurement7, i.e; when he interacts ψ with his half of the EPR pair, the result will be the same. So since Bob knows the result (sent by Alice),| he decides an appropriate operation8 on his half of the EPR pair, and he gets ψ . | Note that here the ψ has been communicated from Alice to Bob, without being actually transmitted. The information was conveyed| without any transport. So, it can be called Teleported, and hence the name : Quantum Teleportation.

Now let us look into the process more closely: 00 + 11 let | | be the Bell state or the EPR state that Alice and Bob together created. Let the state to be √2 conveyed to Bob be ψ (the state to be teleported) = α 0 + β 1 ,whereα and β are unknown amplitudes. Now, on acting the EPR| pair, with the state ψ , we get:| | | ψ = ψ B | 0 | | 00 1 ψ0 = [α 0 ( 00 + 11 )+β 1 ( 00 + 11 )] | √2 | | | | | | Now, the ﬁrst qubit represents the message to be teleported and the second is Alice’s half of the Bell state. The last is Bob’s half of the Bell state. Alice’s now will get her qubits into a CNOT gate and obtain ψ ,where: | 1 1 ψ1 = [α 0 ( 00 + 11 )+β 1 ( 10 + 01 )] | √2 | | | | | |

Now, on sending ψ through a Hadamard gate, we get ψ where: | 1 | 2 1 0 + 1 0 1 ψ2 = α| | ( 00 + 11 )+β | −| ( 10 + 01 ) | √2 √2 | | √2 | | 1 [α( 0 + 1 )( 00 + 11 )+β( 0 1 )( 10 + 01 )] ⇒ 2 | | | | | −| | | DRAFTOn expanding and rearranging the terms, we can COPY obtain the following expressions: 1 ψ = [ 00 (α 0 + β 1 )+ 01 (α 1 + β 0 )+ 10 (α 0 β 1 )+ 11 (α 1 β 0 )] | 2 | | | | | | | | − | | | − |

So, as per the convention,

Figure 1.9: The ﬁrst two bits are Alice’s and the next is Bob’s as shown.

7measurement means any interaction with the system 8kind of inverse operation of what Alice did

GO TO FIRST PAGE 23 ↑ 1.9. QUANTUM TELEPORTAION CHAPTER 1. BRIEF OVERVIEW

if Alice performs a measurement and obtains 00, she will send it to Bob, then Bob will see which Qubit of his in the expression will also give the same result. So, he will come to the conclusion that the state he wants is with the Qubit 00 . So, we have: |

Table 1.2: Alice’s measurement and the corresponding state ψ : | Alice’s measurement result State ψ that Bob shall recover and get as the ﬁnal message from Alice: | 3 00 ψ (00) [α 0 + β 1 ] | 3 ≡ | | 01 ψ (01) [α 1 + β 0 ] | 3 ≡ | | 10 ψ (10) [α 0 β 1 ] | 3 ≡ | − | 11 ψ (11) [α 1 β 0 ] | 3 ≡ | − |

From the measurement results, Bob will apply a certain operation and retrieve the corresponding Qubit. For example, if the measurement result is 1. 00, then Bob will let ψ as it is. | 2. 01, then he will act the ψ state with an X gate9. | 3. 10, he will apply Z gate to ψ . | 4. 11 then he will ﬁrst apply X gate to ψ and then Z gate on the rest. | Therefore, the resultant operation summarizes to:

ZM1 XM2 ψ = ψ | 2 |

where M1 and M2 are the two bits of information send by Alice. These are the results of Alice’s measurement.

Summarizing all the above operations, we can not draw a circuit diagram that Bob must follow, to retrieve the state ψ : DRAFT| COPY

Figure 1.10: The top 2 lines of input are for Alice (the ﬁrst 2 qubits belong to Alice). The bottom qubit belongs to Bob. The measurement operation that Bob must follow is described in the generalized fashion as ZM1 XM2 .

9NOT gate, these conventions will be used in most of the places.

GO TO FIRST PAGE 24 ↑ CHAPTER 1. BRIEF OVERVIEW 1.10. QUANTUM ALGORITHMS

1.9.4 Resolving some ambiguities

The whole process is slightly surprising because it causes several doubts that imply that the teleportation violates the laws of Quantum Computation, that we earlier agreed upon. The following are the ambiguities:

1. This seems to imply that ψ is being copied by Bob from Alice. This is against the law of no cloning theorem that we discussed.| The subtlety that is hidden here is that, both copies of ψ never coexist because as Bob gets the result of Alice’s measurement (in odder for him to create his| ψ , Alice had already destroy his copy due to the measurement. |

2. The process of Teleportation seems to say that the information ψ is being conveyed instantly, since it does not explicitly involve passage of any data. This is misleading| because the fact that Bob must get Alice’s measurement result, which is in the form of classical information, is a very very key concept without which the teleportation is impossible. So, the speed of conveying the information is limited by the speed of light10, and does not violate the theory of relativity.

1.10 Quantum Algorithms

Now, we can look at a few like why Quantum Computation is preferred to Classical computation, can a quantum computer do all that a classical computer is capable11, etc. However, Quantum gates cannot be used directly as classical logic gates because the former are inherently reversible, while the later are DRAFTirreversible. But still we can build classical reversible gates. COPY

1.10.1 Simulating classical circuits using Quantum circuits

Any classical circuit or gate can be replaced by an equivalent reversible gate known as the Toffoli gate.A Tafolli gate is has three input bits. The last input bit, the target bit is flipped if all the other bits are 1. It leaves the first two bits unchanged and performs an XOR operation of (AND(all other bits)) with the last bit. The Taffoli gate can now be used to simulate NAND12. This gate is now reversible and has an inverse which is the Taffoli gate itself. The gate can be used as a quantum as well as a classical gate. In the quantum case, the Taffoli gate takes a state 110 and gives 111 . It simply permutes the computational basis states. | |

10speed of classical information cannot cross the velocity of light - Theory of relativity 11It would be surprising if this was not possible because we know that all classical phenomena can be explained through Quantum mechanics. 12If it can simulate a NAND gate, which is a classical universal gate, then it can simulate all classical gates.

GO TO FIRST PAGE 25 ↑ 1.11. QUANTUM PARALLELISM CHAPTER 1. BRIEF OVERVIEW

Figure 1.11: A Toffoli gate - Reversible classical gate: Truth table and circuit representations. The last bit is flipped if the first two bits are 1.

1.11 Quantum Parallelism

Parallelism has a diﬀerent meaning in the quantum case, as compared to the classical case. In the classical case, at any given unit of time, only one unit of task is in execution. While, in the quantum case, parallelism actually takes its real meaning. At any given unit or instance in time, all the tasks run in parallel. This feature in Quantum Mechanics (for the multiple qubit case) is achieved due to presence of the superposition of bits like: [αij ij ]i,j. One of the most simple operations that can be performed on two qubits is an XOR operation (denoted| by ). For all practical purposes, let us assume XOR gate to be a universal gate. So, if we can do in parallel two⊕ XOR operations, then we have showed quantum parallelism in action. Let us now claim that corresponding to any function ’f’ where f:x f(x) we can always deﬁne a unitary transformation → Uf , such that

U x, y = x, y f(x) f | | ⊕

We also impose a condition that for the quantum transformation or circuit given by Uf , the inputs should DRAFTnot be in the computational basis. COPY A rough circuit diagram of our set up is given below: (Here, since y is permanently set to 0 the value of x, 0 f(x) is equal to f(x) itself). In other words, | | ⊕ U x, 0 = x, 0 f(x) f(x) f | | ⊕ ⇒

Figure 1.12: Quantum circuit that computes f(x) (since y = 0 ) simultaneously for 0 and 1 | |

GO TO FIRST PAGE 26 ↑ CHAPTER 1. BRIEF OVERVIEW 1.11. QUANTUM PARALLELISM

0,f(0) + 1,f(1) The output, in this case is:| | . Hence, in one run of the function Uf , we have computed √2 both f(0) and f(1). A single execution of Uf was able to compute f(x) for multiple values of x. Hence the name “parallelism”. This idea of parallelism can be extended to multiple qubit systems also. We can have a quick look at how this is done. So, our main objective (when generalized to n qubit case) is that, given n qubit states, we must compute the values for f(x) for all the n states in parallel, that is, in a single evaluation of Uf .Todothis, just like the previous case, let us take a (n+1) qubit system with the last qubit equal to 0 (It is similar to the y that we set to 0 in the two qubit case). We now need all permutations of the ﬁrst n| qubits (with each qubit| in the computation| basis) in the system. To produce all the permutations of n qubit system (where initially, all qubits are set to 0 ) , we require n hadamard gates. We ﬁrst set all the qubits to 0 initially. Each of the hadamard gate, acting| on a qubit produces |

0 + 1 H 0 = | | | √2 0 + 1 0 + 1 therefore: (H 0 )(H 0 )= | | | | | | √2 √2 0 + 01 + 10 + 11 (H 0 )(H 0 )= | | | | ⇒ | | √ 2 2

This can now be generalized to:

0 + 1 0 + 1 0 + 1 0 + 1 (H 0 )(H 0 )(H 0 )...... (H 0 )= | | | | | | .... | | | | | | √2 √2 √2 √2 i all permutations of n qubits i (H 0 )(H 0 )(H 0 )...... (H 0 )= ∈{ } | ⇒ | | | | √2n The above statement is in short represented as:

i n n i all permutations of n qubits H⊗ 0 ⊗ = ∈{ } | (1.6) | √2n

n DRAFTwhere H⊗ denotes the parallel action of n hadamard gates. This COPY is also called a hadamard transform to the ﬁrst n qubits. Now this (n+1) qubit system can be sent to the function Uf , and for each permutation, the function Uf will produce f(x) where x permutations of n qubits . The ﬁnal answer that will come as output from the ∈{ } function Uf is:

n n 1 Uf (H⊗ 0 ⊗ 0 )= x, f(x) | | √2n  |  x all permutations of n qubits ∈{ }   Therefore, we can compute 2n values of f(x) for diﬀerent values of x, in one single evaluation of the function f. It is clear that this parallelism could not be possible if the qubits were not in the superposition of states. In other words, we have shown that the parallelism works only in the quantum case and not in the case of probabilistic classical bits because the classical bits cannot be in superposition of states. But the story does not end here. There is a catch. The problem is that even if Uf computes all the values of f(x) in one single evaluation, and returns the result, as shown above, we are still restricted (by the nature of measurement that can be made in quantum mechanics) to only obtain one value of f(x). After this measurement, the entire output will collapse to an eigen state. So, though we have shown to compute f(x) for 2n values of x, we are allowed to look at f(x) for only one value of x. But there is a way of getting out of here, and we shall see how, in the later sections.

GO TO FIRST PAGE 27 ↑ 1.11. QUANTUM PARALLELISM CHAPTER 1. BRIEF OVERVIEW

1.11.1 Example of Quantum Parallelism - Deutsch Jozsa Algorithm

In the previous few sections, we have been talking about features that are present only on quantum computers and not in classical ones. In this section, we shall see how these features can come together and outperform the classical computer (for this problem which is going to be presented). This problem is given by Deutsch and Jozsa. The problem formulation is given as the following situation:

Alice in Amsterdam selects a number between 0 to 2n 1 and mails it in a letter to Bob in Boston. Bob takes the number x (sent by Alice) and computes f(x) and replies− with the result, which is either 0 or 1. Now Bob has promised to use a function which is of one of the two kinds; either f(x) is constant for all values of x, or f(x) is balanced, i.e; equal to 1 for exactly half of all possible values and 0 for the other half. Alice’s goal is to determine with certainty weather Bob has chosen a constant or a balanced function, corresponding with him as little as possible. How far can she succeed ?

n 1 Let us start by taking the naive classical method, which is when Alice shall query Bob at most 2 − +1 times (until she gets a different number from Bob). If she gets a different answer at any instant, she can n 1 conclude that f(x) is balanced. If she gets a particular 2 − + 1, then it means that the function is constant n 1 (this is because, if it hadn’t been constant, it must be balanced and hence it must have 2 − values of the n 1 other kind. But our function in this case has 2 − + 1 values of one kind.). So, in this solution, Alice and n 1 Bob need to communicate for at most for 2 − + 1 values of x. Let us try to do better by taking the quantum analogue of this problem. Let Alice and Bob be allowed to exchange Qubits instead of classical bits, and let f(x) now be a function that operates on classical bits. The output of the function, however the type of output remains the same. Let us now work through the process for the simple n = 2 case. So, we need Alice to choose a number between 0 and 3. Alice has 4 choices, and we need two qubits to distinctly represent 4 different quantities. So, let us suppose Alice chooses some number that is represented by the state ψ = 01 . Now, let us send each qubit through a Hadamard gate, and | 0 | obtain the new state ψ1 . The parallel action of two Hadamard gates on the state ψ0 produces ψ1 . So, we have: | | |

input state (Alice’s number): ψ = 01 | 0 | 2 now:H⊗ ψ = ψ | 0 | 1 0 + 1 0 1 ψ1 = | | | −| ⇒| √2 √2 DRAFTNow, let us apply the transformation Uf to the two qubits obtained.COPY (Uf is the quantum circuit that we discussed in the last subsection). Just for recap,

U x, y = x, y f(x) f | | ⊕ We can now send the two qubits of ψ1 into the quantum circuit Uf , as x and y respectively. Let us deﬁne the new qubit ψ , which is obtained| after acting ψ with U . So,| we get: | | 2 | 1 f 0 + 1 0 1 ψ2 = Uf | | | −| | √2 √2 0 ( 0 1 )+ 1 ( 0 1 ) On expanding the product, we get: ψ = U | | −| | | −| | 2 f 2

0 ( 0 1 ) 1 ( 0 1 ) On separating the terms: ψ = U | | −| + U | | −| (1.7) | 2 f 2 f 2 Here, both the terms which are separated out and being added, are of the

0 1 same form: Uf x | −| (1.8) | √2 GO TO FIRST PAGE 28 ↑ CHAPTER 1. BRIEF OVERVIEW 1.11. QUANTUM PARALLELISM

0 1 Therefore, let us ﬁrst try to compute a general form for the expression: Uf x | −| | √2 0 1 x, 0 x, 1 Hence, we get: Uf x | −| Uf | −| | √2 ⇒ √2 x, 0 f(x) x, 1 f(x) On the action of Uf the expression becomes: | ⊕ −| ⊕ √2 0 f(x) 1 f(x) Which can be further simpliﬁed to: x | ⊕ −| ⊕ | √2 Assuming that we know the property of the (XOR) operation, let us consider two cases: f(x) = 0 and f(x) = 1. ⊕

Taking the ﬁrst case: f(x) = 0: 0 0 1 0 0 1 x | ⊕ −| ⊕ x | −| ⇒| √2 ⇒| √2

Similarly, taking the second case: f(x) = 1: 0 1 1 1 1 0 x | ⊕ −| ⊕ x | −| ⇒| √2 ⇒| √2 0 1 Therefore, we can summarize by giving a general form for x | −| by inspecting the above two cases, | √2 as the following: 0 1 0 1 x | −| =( 1)f(x) x | −| (1.9) | √2 − | √2 Now, since we have obtained a general form for the expression, we can substitute values of x to get the terms for the expressions for ψ2 , that we obtained earlier in (1.7). The ﬁrst and second terms of the RHS (of expression in (1.7)) can| be evaluated directly by substituting values of x (of expression in (1.9)) to be 0 and 1respectively. Therefore, we obtain:

Putting x = 0 in the first term and x = 1 in the second term, of the RHS: 0 1 0 1 ψ =( 1)f(0) 0 | −| +( 1)f(1) 1 | −| DRAFT| 2 − | 2 − COPY| 2 The above expression gives the same value if f(0) = f(1), and a different value otherwise. So, we have two different possibilities of evaluation: one is if f(0) = f(1), the other is when f(0) = f(1). Let us consider the case f(0) = f(1). So, let f(0) = f(1) = c. Hence, we get:

c 0 + 1 0 1 c ψ2 =( 1) | | | −| . Here, we can write ( 1) as | − √2 √2 − ± Similarly, for the other case, f(0) = f(1), we have the following argument. If f(0) = f(1), then ( 1)f(0) ( 1)f(1). So, we have: let ( 1)f(0) = ( 1)f(1) =c − − − − − − 0 1 0 1 Hence, the expression becomes: ψ2 = c | −| | −| . Here, we can write c as | √2 √2 ± So, we obtain two expressions for ψ . Now we can write them as: | 2 0 + 1 0 1 | | | −| f(0) = f(1) ± √2 √2  ψ = (1.10) {| 2  }  0 1 0 1  | −| | −| f(0) = f(1) ± √2 √2   GO TO FIRST PAGE  29 ↑ 1.11. QUANTUM PARALLELISM CHAPTER 1. BRIEF OVERVIEW

Now, we have obtained ψ . So, let us now act the first qubit with ψ ,toproduce ψ . So, we get: H ψ | 2 | 2 | 3 | 2 = ψ3 . Let us see the action of H on the first qubit, separately for the two cases which are (1.10) and (1.10). Let| us expand the product in (1.10) and see the action of H on the first qubit.

00 + 01 + 10 + 11 H ψ = H | | | | | 2 ± 2 We can act H with the ﬁrst qubit and leave the second one unchanged. So, we get: 1 0 + 1 0 + 1 0 1 0 1 H ψ2 = | | 0 | | 1 + | −| 0 | −| 1 | ±2 √2 | − √2 | √2 | − √2 | 1 Hence, on simplifying: H ψ2 = [ 00 01 + 10 11 + 00 10 01 + 11 ] | ±2√2 | −| | −| | −| −| | 0 1 H ψ2 = 0 | −| ⇒ | ±| √2 Similarly, on carrying out the calculations, for the other case of ψ , we get: | 2 0 1 H ψ2 = 1 | −| | ±| √2 So, we obtain two expressions for ψ just like how we got for ψ . Now we can write them as: | 3 | 2 0 + 1 0 | | f(0) = f(1) ±| √2  ψ = (1.11) | 3   0 1  1 | −| f(0) = f(1) ±| √2   The above expression (1.11) can be writtern in a more shorter and concise manner, in the following way.:

0 + 1 ψ3 = f(0) f(1) | | (1.12) | ±| ⊕ √2 Hence, by measuring the ﬁrst qubit, Alice can, by certainty can say weather f is balanced or constant. Hence, by a single run of the function, Alice was able to determine a global property of the function. Earlier we saw that though we could compute the function for lots of inputs, but we could measure only one result. Here, DRAFTwith the same constraint, we have accomplished our task. So, COPY on summarizing we have done the following steps: 1. The input state was prepared. 2. Hadamard Transform on the input state was performed.

3. The qubit state was now passed to the Uf transformation. 4. The ﬁrst qubit of resultant state was passed through the Hadamard gate.

We can now generalize the whole procedure for the multiple qubit case, that is Alice can choose any number between 0 to 2n where n can be as large. So, in the multiple qubit case, Alice’s query register can be represented by a n qubit state, all of which are initially set to zero. Alice’s input state is a (n+1) qubit state with the last qubit set to 1 . | n Alice’s input state: ψ = 0 ⊗ 1 | 0 | | Now, let us perform a Hadamard Transform on the query register and pass the last qubit (answer register, which Bob will modify and send 1 into a Hadamard gate, to get ψ1 . (this is similar to (1.8)). In the previous case, instead of the transform,| we had only one gate. From| (1.6 ), we know the result of Hadamard

GO TO FIRST PAGE 30 ↑ CHAPTER 1. BRIEF OVERVIEW 1.11. QUANTUM PARALLELISM

x Transform on N qubits, where all are initially set to 0 , gives | , and the hadamard gate on the | √2n x 0,1 n ∈{ } 0 1 single qubit 1 gives | −| . So, on their parallel action, we get: | √2 x 0 1 ψ1 = | | −| (1.13) | √2n √2 x 0,1 n ∈{ } where (x 0, 1 n) means all permutations of 0 and 1 which are of length n. ∈{ } Bob now takes the input state sent by Alice and computes the function f using the transformation Uf . Bob now sends the answer in ψ (the form of ψ resembles the equation in (1.9)): | 2 | 2 ( 1)f(x) x 0 1 ψ2 = − | | −| (1.14) | √2n √2 x 0,1 n ∈{ } Alice has now a set of Qubits in which the result of Bob’s function is stored in the amplitude of the superpo- ( 1)f(x) x 0 1 sition state: − | | −| . She now interferes the terms in superposition with a Hadamard √2n √2 x 0,1 n ∈{ } transform, to get ψ3 . To calculate the effect of the Hadamard transform, we can first see how the transform in general works with| the n-qubit state x ,x ,x ,....x : | 1 2 3 n n Hence, we need to calculate: H⊗ x ,x ,x ,....x | 1 2 3 n Let us now go one more level down, and first try to calculate the result of Hadamard transform on a single n qubit: H⊗ x | DRAFT COPY

GO TO FIRST PAGE 31 ↑