Quantum Computation Introduction
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It is a brief presentation on quantum computation, which is created as I have investigation on guided study with my instructor Professor Sen Yang at CUHK
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Quantum Computation Introduction
- 1. Guided Study Presentation: Introduction to Quantum Computation Msc Student: Alan Leung Shek Lun (1155053380) Advisor: Professor Sen Yang
- 2. Outline of Presentation and my motivation Classical Computation Vs Quantum Computation Quantum Mechanics Linear Algebra: Dirac Notation Quantum Gate and Quantum Circuit Super dense Coding
- 7. Bit And Qubit • Bit • Carries information • Representation: voltage, magnetic domain • Qubit • Simplest quantum system • Representation: atoms, photons It is the fundamental unit for quantum computation.
- 10. Linear Algebra Vs Dirac Notation Ket : v Norm : v = v v Unit vector parallel to v : v v v
- 11. Ket Vector
- 12. Bra Vector (in the dual space)
- 13. Inner Product Y 2 = Y Y Y Y = a* n n å y n æ èç ö ø÷ ak k å y k æ èç ö ø÷ = a* n an y n y n = an 2
- 14. Working on Qubit • Ket state = vector • It obeys all usual rules for vectors 1qubit basis, 0 = 1 0 æ è ç ö ø ÷ , 1 = 0 1 æ è ç ö ø ÷ 1qubit superposition, y = a 0 + b 1 = a b æ è ç ö ø ÷2 y = 2 1 2 a 0 + b 1( )æ èç ö ø÷ = 2 a / 2 b / 2 æ è ç ö ø ÷ due to normalization condition
- 15. Famous Thought Experiment: Schrodinger’s Cat Linear combination (superposition)of quantum states: Half live + half death simultaneously Normalization condition: Summing up all the probabilities: 𝛼 2 + 𝛽 2 = 1
- 16. Linear Algebra: Dirac Bra-Ket Notation 0 is orhonormal to 1 0 1 = 1 0 æ è ç ö ø ÷ 0 1 æ è ç ö ø ÷ = 0 1 0 = 0 1 æ è ç ö ø ÷ 1 0 æ è ç ö ø ÷ = 0 0 is orthogonal to 1 , 0 and 1 are normalized to 1 0 0 = 1 0( ) 1 0 æ è ç ö ø ÷ = 1 11 = 0 1( ) 0 1 æ è ç ö ø ÷ = 1
- 17. 1 2 0 + 1( )® 0 with probability 1 2 ®1 with probability 1 2 due to normalization condition pr(0) + pr(1) =1
- 18. General Expression for Inner product y = a 0 + b 1 y = a* 0 + b* 1 y y = y y = a* 0 + b* 1( ) a 0 + b 1( )= = a* a 0 0 +a* b 0 1 + b* a 1 0 + b* b 11 = a* a + b* b = a 2 + b 2 = 1
- 19. y = a 0 + b 1 We cannot determine the value of a and b. The quantum state is not directly observable. For 0 ,with probability a 2 For 1 ,with probability b 2 Measurement will distrub the quantum state and make the superposition of states collapse to one of these states. Quantum ® Classical
- 20. 2 qubits basis (Tensor Product) 00 = 0 Ä 0 = 1 0 æ è ç ö ø ÷ Ä 1 0 æ è ç ö ø ÷ = 1´1 1´ 0 0 ´1 0 ´ 0 æ è ç ç ç ç ö ø ÷ ÷ ÷ ÷ = 1 0 0 0 æ è ç ç ç ç ö ø ÷ ÷ ÷ ÷ Similarly, 01 = 0 Ä 0 = 0 1 0 0 æ è ç ç ç ç ö ø ÷ ÷ ÷ ÷ 10 = 0 Ä 0 = 0 0 1 0 æ è ç ç ç ç ö ø ÷ ÷ ÷ ÷ 11 = 0 Ä 0 = 0 0 0 1 æ è ç ç ç ç ö ø ÷ ÷ ÷ ÷
- 21. In general description for tensor product Suppose we have 2 qubits in the system xy : a b é ë ê ù û úÄ c d é ë ê ù û ú = ac 00 + ad 01 + bd 10 + bc 11 = ac ad bd bc é ë ê ê ê ê ù û ú ú ú ú
- 22. Φ■ ψ■ + Φ■ ψ● + Φ● ψ■ + Φ● ψ● (Φ■ + Φ●)(ψ■ + ψ●)
- 23. Φ■ ψ■ + Φ● ψ●
- 24. Logic Gate
- 27. Pauli-X gate 0 ® 1 1 ® 0 C 0 = 0 1 1 0 é ë ê ù û ú 1 0 é ë ê ù û ú = 0 1 é ë ê ù û ú C 1 = 0 1 1 0 é ë ê ù û ú 0 1 é ë ê ù û ú = 1 0 é ë ê ù û ú
- 28. Pauli-X gate: It is own inverse • Doing Quantum Wire C2 = CC = 0 1 1 0 é ë ê ù û ú 0 1 1 0 é ë ê ù û ú = 1 0 0 1 é ë ê ù û ú = I y ® C y ® C2 y = y
- 29. Pauli-Y Gate 0 ® i 1 1 ® -i 0 Y = 0 -i i 0 é ë ê ù û ú Y2 = YY = 0 -i i 0 é ë ê ù û ú 0 -i i 0 é ë ê ù û ú = 1 0 0 1 é ë ê ù û ú = I
- 30. Pauli Z Gate 0 ® 0 1 ® - 1 Z = 1 0 0 -1 é ë ê ù û ú Z2 = ZZ = 1 0 0 -1 é ë ê ù û ú 1 0 0 -1 é ë ê ù û ú = 1 0 0 1 é ë ê ù û ú = I
- 31. One of the most important quantum gate
- 33. • After operation, same state • Can’t distinguish the state after one more operation • We need a quantum gate suit our needs Not a legitimate quantum gate
- 34. What does it mean by a matrix is unitary? General Single-Qubit Gate: any unitary matrix, U U y = U a b é ë ê ê ù û ú ú = a ' b ' é ë ê ê ù û ú ú = y ' U = X,H, etc. U† U = I,U† = (UT )* a b c d é ë ê ù û ú † = a* c* b* d* é ë ê ê ù û ú ú
- 35. They are Unitary C† C = 0 1 1 0 é ë ê ù û ú 0 1 1 0 é ë ê ù û ú = 1 0 0 1 é ë ê ù û ú = I H† H = 1 2 1 1 1 -1 é ë ê ù û ú 1 2 1 1 1 -1 é ë ê ù û ú = 1 0 0 1 é ë ê ù û ú = I
- 36. What is the usage of unitary matrix? • If the matrix is unitary, , it implies that the length preserved after the operation U y = y for all y Proof: U y 2 = (Uy )j * j å (Uy )j = U jk * y k * j,k,l å U jl yl = Ukj † U jl y k * j,k,l å yl = dkl y k * k,l å y l = y k * k å y k = y 2 Conclusion: Let M be a matrix. If M is unitary, then U y = y for all y .
- 37. Representation of a Matrix Element M y † = y M† for all y Proof: M y( )j † = M y( )j * = M jk * y k * k å = y k * Mkj † k å = y M† ( )j Conclusion: Let M be a matrix. Then M y 2 = y M† M y for all y
- 38. The converse proof • If the length preserved after the operation, it implies that the matrix is unitary. M y = y for all y Þ M is unitary. Proof: M† M( )jj = ej M† M ej = M ej 2 = ej 2 = 1 M† M( )jk = ej M† M ek = M ej + eiq ek( ) 2 = 0 ( j ¹ k) q is the phase factor
- 39. Another Single qubit quantum gates • Phase shift gate • Unitary • Length preserving eiq 0 0 eif é ë ê ê ù û ú ú 0 ® eiq 0 1 ® eif 1 y = a 0 + b 1 ® y ' = aeiq 0 + beif 1 Pro( 0 )= a 2 , Pro( 1 )= b 2 Probability is not affected by the phase shift
- 40. Rotation Matrix • A unitary matrix • Preserve length • => a valid quantum gate
- 41. The controlled not (CNOT) gate: 2 qubits gate Control Qubit Target Qubit xy ® x x Å y
- 43. Universal Quantum Gate • All operations may be derived from a series of given gates or unitary operation
- 44. Procedure for quantum computation
- 46. Procedure 1. Eve sends the half of the bell pair corresponding to Alice and Bob 2. Alice sends info to Bob: 00,01,10,11 3. Alice encodes the message Eve Alice Bob
- 47. Procedure 4. Alice sends her qubit to bob. Bob will have both qubits 5. Bob does a decoding operation Eve Alice Bob
- 48. Eve’s Action: H gate on 1st qubit and then CNOT gate on both qubits 00 ® 00 + 10 2 ® 00 + 11 2 Eve
- 49. Eve’s Action: H gate on 1st qubit and then CNOT gate on both qubits HA 0 A = 0 A + 1 A 2 , H 1 A = 0 A - 1 A 2 HA 00 = HA 0 A Ä HA 0 B = 0 A + 1 A 2 Ä HA 0 B = 0 A 0 B + 1 A 0 B 2 = 00 + 10 2 HA 01 = HA 0 A Ä HA 1 B = 0 A + 1 A 2 Ä HA 1 B = 0 A 1 B + 1 A 1 B 2 = 01 + 11 2 HA 10 = HA 1 A Ä HA 0 B = 0 A - 1 A 2 Ä HA 0 B = 0 A 0 B - 1 A 0 B 2 = 00 - 10 2 HA 11 = HA 1 A Ä HA 1 B = 0 A - 1 A 2 Ä HA 1 B = 0 A 1 B - 1 A 1 B 2 = 01 - 11 2
- 50. Eve’s Action: H gate on 1st qubit and then CNOT gate on both qubits 00 ® 00 + 10 2 ® 00 + 11 2 01 ® 01 + 11 2 ® 01 + 10 2 10 ® 00 - 10 2 ® 00 - 11 2 11 ® 01 - 11 2 ® 01 - 10 2 Input H gate CNOT gate
- 51. Alice‘s Action: Encode the 1st qubit If it is 00 , then apply I gate on the 1st qubit If it is 01 , then apply X gate on the 1st qubit If it is 10 , then apply Z gate on the 1st qubit If it is 10 , then apply Z gate followed by X gate on the 1st qubit I gate : 00 + 11 ® 00 + 11 X gate: 00 + 11 ® 10 + 01 Z gate: 00 + 11 ® 00 - 11 XZ gate : 00 + 11 ® 10 - 01
- 52. Bob’s Action on the First Qubit 00 + 11 2 ® 00 + 10 2 ® 00 01 + 10 2 ® 01 + 11 2 ® 01 00 - 11 2 ® 00 - 10 2 ® 10 01 - 10 2 ® 01 - 11 2 ® 11 HA 00 + 10 2 = 1 2 00 + 10 2 + 00 - 10 2 æ è ç ö ø ÷ = 00 HA 01 + 11 2 = 1 2 01 + 11 2 + 01 - 11 2 æ è ç ö ø ÷ = 01 HA 00 - 10 2 = 1 2 00 + 10 2 - 00 - 10 2 æ è ç ö ø ÷ = 10 HA 01 - 11 2 = 1 2 01 + 11 2 - 01 - 11 2 æ è ç ö ø ÷ = 11 H gateCNOT gate Output
- 53. Advantage of Super dense Coding • When actual message will be sent, half of what’s has been arrived. • Double the bandwidth • Half of number of qubits will be needed to reconstruct a classical message • We can transmit at double speed until the pre-delivered qubits run out.
- 54. Reference • Nielsen, M. A., & Chuang, I. L. (2000). Quantum Computation and Quantum Information. Quantum, 546, 1231. • Griffiths, D. J. (2016). Introduction to Quantum Mechanics. Cambridge University Press. • Kurzgesagt. “Quantum Computers Explained – Limits of Human Technology.” YouTube, YouTube, 8 Dec. 2015, www.youtube.com/watch?v=JhHMJCUmq28&t=77s. • Quantum Computing CFD. (2017). Research Overview. [online] Available at: http://www.qcfd.pitt.edu/research-overview [Accessed 15 Dec. 2017]. • Radical, T. (2017). Quantum Computing: The Qubits. [online] RADICALNEWS. Available at: http://radicalnews.in/education/122/quantum-computing- the-qubits [Accessed 15 Dec. 2017].
Editor's Notes
- This outline is my motivation for studying the quantum computation for this early stage. I am fascinated by some science fiction that we can teleport anything even object in macroscopic scale to an far distant part.
- A classical computer is made up of various component doing different tasks
- The computer chip containing these three things
- A transistor is a switch allowing or block ing the way for information flow through wich is made up of bit 0 / 1 https://gfycat.com/gifs/search/quantum+computer https://www.google.com.sg/search?q=quantum+computation+gif&rlz=1C5CHFA_enHK691HK691&tbm=isch&source=iu&ictx=1&fir=6OxNRNmj0CKtKM%253A%252CjyH_KpikqttDdM%252C_&usg=__XIdDMw3dgzI0xeR6_8VPhsjIRsc%3D&sa=X&ved=0ahUKEwiWlNGO24nYAhVKUbwKHUvcDisQ9QEIODAE#imgrc=Vg7yFKkBvqqysM:
- The main difference between the two due to the ability of a quantum state to represent many possible classical state at the same time. For classical bit, 0/1 Quantum bit is in supposition not only in real domain, but also can be represented by complex number if we have N qubit then there are 2^N possible state in the superposition. Theoretically, if a quantum computer with 30 qubits, there would be 1,000,0000,000 one billion possible state. Quantum with 300 qubits would roughly have the possible states as the total number of atoms in the universe.
- A bit can be one of the two states 0 /1 It can be represented by a transistor switch set to off or on Or by an arrow pointing up or down. It is only represented by 2 points only. A qubit, has many possible states. The states can be represented by an arrow pointing to a location on a sphere (Bloch Sphere), .i.e. any point on a 3D sphere. The north pole to equivalent to state 0, the south pole is state 1. The other locations are quantum super positions of 0 and 1. Classically, we may interpret the state as simultaneously 1 and 0. Every possible state could be stored, and processed in parallel with all the others
- the state is the vector and it is exactly the same as the column vector with component alpha and beta, i.e., the amplitude for 0 and 1 state respectively we can write the name as psi to represent the state. So the state of a qubit obeys all the rule of a vector, for example, it can be multiplied by a scalar however, if we multiply it by 2, it has to be the same probability amplitude, because of the normalisation condition
- the constraint is the normalization constants, the sum of the square of the probability amplitude must be equal to 1 think of 0 and 1 is orthonormal (orthogonal and normalized vector) we can use the classical physics to describe the everyday life object, exact description is ensure. While quantum computer, we cannot describe the state for sure, but only the probability we can not prepare two states at the same time
- 0 and 1 indicate the state, called the ket computational basis states, information 0 and 1 quantum state of a qubit is in fact a vector in a 2 dimensional complex vector space a linear combination of these two state = superposition the coefficient can be real or complex =probability amplitude why we take this complex description let us know build up a sort of model of computation so in particular states, this mathematical model can help us more easily to realize the quantum mechanics In general, we can think of the state of a qubit as just the being a normalized vector in the two dimensional complex vector space what does a quantum state mean? How should we interpret it ?How should we
- The implication is that you can store the infinite amount of classical information in qubit originally, it is in superposition state, and after the measurement it becomes the state 0 with 1/ root 2 and the state 1 with 1/root 2
- What if someone give me a unknown state, can i determine the value of the component? No, the quantum state isn’t directly observable which is a fundamental constraint, we can only get partial information the process is measurement, in the computational basis given the qubit, upon measurement will give me classical bit apparatus microprocess interact with the qubit measurement and give the outcome 0 / 1, In quantum computation we have the quantum state, then we are manipulating using the quantum gate, and then we do the measurement to read out the result The fact is the the measurement will disturb the system and collapse the quantum bit the classical bit the alpha and beta are now gone, there is no trace to it, no more information you can extract there is hidden information you can’t get full information about them
- We say that the c-ons are “independent” if knowledge of the state of one of them does not give useful information about the state of the other. Our first table has this property. If the first c-on (or cake) is square, we’re still in the dark about the shape of the second. Similarly, the shape of the second does not reveal anything useful about the shape of the first. compare the coefficient, the product of superposition of states can not give the bell state so they cannot be decomposed into different unit
- On the other hand, we say our two c-ons are entangled when information about one improves our knowledge of the other. Our second table demonstrates extreme entanglement. In that case, whenever the first c-on is circular, we know the second is circular too. And when the first c-on is square, so is the second. Knowing the shape of one, we can infer the shape of the other with certainty. for whenever systems interact, the interaction creates correlations between them. There are many ways to create entangled states. One way is to make a measurement of your (composite) system that gives you partial information. We can learn, for example, that the two systems have conspired to have the same shape, without learning exactly what shape they have. This concept will become important later.
- We have described the state of a qubit, now we want to know how the change the state of a qubit, i.e. quantum logical gate. It is a way to manipulating quantum information (the state of qubit / set of qubit), which can help information processing like quantum teleportation. Here are some logic gate that we will investigate in details.
- Quantum NOT gate is a generalisation of the classical NOT gate it take 0 state to 1 state, vice versa it can act linear to flip the state for the superposition of states In the quantum circuit representation the line means the quantum wire represent a single qubit there is a qubit state on the left initially, after passing the quantum gate, the state is flipped there is a matrix representation, 2 by 2 Pauli Matrix X it acts linearly on vector the simplest quantum circuit of all is just a wire, the quantum bit being preserved in time but in fact it is the hardest system of all the quantum state can be very fragile, very small, single atom, single photon. While because of the tininess, it can be perturbed very easily by their environment.
- Let’s analyse the simple quantum gate, two Xs give us back the original state of qubit, we are doing a quantum wire what happen to psi, we can work out the matrix multiplication
- After acting on the pure states, it gives a balanced superposition They are different states with equal probability
- After we work, we may ask what can Hadmard gate do for us? when you want to go from one place to another another, we may do by distance if we have a new device, we can go straight away to there by displacement, Hadmard gate can span the range of states which is possible for a computer to be in. Doing this kind of expansion can create the possibility to shortcut basically we are moving in a way impossible to a classical computer and do computation faster another example, we play chess and the rule is set to your favor, it can give you more flexibility.
- change in quantum state, which is represented as a column vector with component alpha and beta matrix acting on the initial state General single qubit state can be any unitary matrix U dagger times U = I identity operation Quantum gate is described as the unitary action The unitary matrix acting on the input state question we may ask Why would we expect the single qubit gate is described by the unitary matrix? What does it mean by unitary physically? What single qubit gate are there? We may investigate what we have encountered is that the x gate and hadmard gate: whether are they the unitary? What other simple qubit gate we can built up?
- Tackle the Second question:What does it mean by a matrix is unitary ? Unitary matrices perverse A unitary matrix acting on any state psi then taking the absolute value, the length preserve to the original vector sign for all psi It is the advantage as we need the input state has to be normalised to have length 1 And the output state to be a legitimate physical quantum state, which means the length has to be 1 So if the matrix is unitary, it implies that the length preserves Doing the square make the sum of multiplication with the complex conjugate, which induce two more indices Make the kronikel delta Take any matrix m M acting on psi the length preserved for all psi If and only if m is unitary This ensure that the quantum gate must be unitary, they are the only linear operation which preserve normalisation. However, it does not completely answer the question why the quantum gate should be linear in the first place, and why should we have a matrix representation. At this stage, we would accept the fact and we’ll understand it later. We are now going to prove the other direction why the statement is true.
- considering the j component of unit vector M acting on the Kth unit vector <ej|M|ek>Mjk For the diagonal part Doing with the basis, the length preserve 1 For the non-diagonal part, it becomes zero
- what does it look like? super-dense coding theta and phi on the diagonal side, there may be a phase change the superposition act linearly, there is length preserved For example, we take the initial state acted by the quantum gate with theta 0 and phi pi then we will have two different supposition of states if we apply the hadmard gate on these supposition of states, we will get the basis state, so if we measure in the computational basis, we will get 0 or 1 state this gives us a way to distinguish the states Another example:rotation matrix is a unitary matrix which preserves length=> it is a valid quantum gate Pauli Matrix sigma for electron spin X gate, Y gate, Z gate to do information processing for super dense cooling
- investigating in the multi qubit gate control qubit target qubit for 2 qubit system the 4 possible states of a 2 qubit system 00, 01, 10, 11 (alpha)00+(beta)01+(gamma)10+(delta)11 the mechanism: if the control gate is 1, then the target gate will be flipped to 0, otherwise, it will do nothing to the gate
- the action of the 00 and 01 state is to leave them alone, the action of the 10 and 11 state is to flip the target qubit the matrix representation of CNOT gate can be verify
- What they are and how they should be designed ? Classical computer AND, NOT gate composing them together => we can compute any function other logic gates like OR gate, addition, subtraction, multiplication, can be built by the elementary operations So we will say AND, NOT gate are universal gate, which means we are able to use those to compute any function Another one is the NAND they are equivalent given one set you can use that set to simulate another, each model can simulate the other For quantum computer CNOT, single-qubit gate can built up any arbitrary / unitary operation on a qubit. in principle, we pick any unitary operation, it is possible to decompose into the the elementary qubit gate. CNOT+ single-qubit gates are universal for quantum computation. It is not the only one, but a convenient gate
- First, ahead of time, Eve has two qubit in the lab, both start with the zero state, apply with the hadamard gate then do the CNOT gate one will go to one person Alice and one go off to the third party bob Alice and Bob each get half of a Bell pair. That is to say, two qubits are placed into a superposition where either both are false or both are true, and then Alice and Bob each take one of those qubits. There's a lot of flexibility in who actually makes the Bell pair. Alice can do it, Bob can do it, or an unrelated third party can do it. Regardless, what matters is that the Bell pair can be delivered ahead of time and stored for later use. Second, Alice decides what information she wants to send to Bob. She can send two bits (i.e. one of four possibilities). We'll call the possible messages 00, 01, 10, and 11. Third, Alice encodes the message by applying operations to her qubit (the one from the Bell pair). The operations are based on the message she wants to send. If she wants to send 00, she does nothing. For 01, she rotates the qubit 180° around its Z axis. For 10 she instead rotates 180° around its X axis. Otherwise the message is 11 and she rotates both 180° around the X axis and then 180° around the Z axis. Note that the 11 case is technically a rotation around the Y axis, but it's nice to split it into X and Z rotations because it makes the circuit simpler. It means Alice can just apply the X rotation if the second bit is true, and afterwards the Z rotation if the first bit is true.
- Fourth, Alice sends her qubit to Bob. So Bob will end up with both halves of the Bell pair, but Alice has operated on one of them. Fifth, Bob does a decoding operation. He conditionally-nots his qubit, conditioned on Alice's qubit. This will flip the value of his qubit in the parts of the superposition where hers is true. Then he rotates Alice's qubit by 180° around the diagonal X+Z axis (i.e. applies the Hadamard operation).
- For 3 qubit 000, 0001, the first 2 control qubits are 0, so the target qubit will be 0
- http://michaelnielsen.org/