1. Quantum Computation and Quantum Information – Lecture 2
Part 1 of CS406 – Research Directions in Computing
Dr. Rajagopal Nagarajan
Assistant: Nick Papanikolaou

2. Lecture 2 Topics Physical systems on the atomic scale
State vectors and basis states; Qubits
Systems of many qubits
Quantum Measurement
Entanglement
Quantum gates
Quantum coin-flipping and teleportation

3. Quantum physics and Nature
There exists a vast array of minute objects on the atomic scale: electrons, protons, neutrons, photons, quarks, neutrinos, …
Quantum mechanics is a system of laws that describes the behaviour of such objects
With computer chips getting smaller and smaller, by 2020 we will store 1 bit of data on objects of that size!

4. Quantum physics and Nature (2)
Atom-sized objects behave in unusual ways; their “state” is generally unknown at any given time, and changes if you try to observe it!
Several properties of these systems can be manipulated and measured.

6. Qubits
A qubit is any quantum system with exactly two degrees of freedom; we use them to represent binary ‘0’ and ‘1’
Hydrogen atom:
Spin-1/2 electron:
Ground state:
Excited state:
Spin-down (-ħ/2) state:
Spin-up (+ħ/2) state:

7. Qubits (2)
In general, the state of a qubit is a combination, or superposition, of two basis states
The rest state and the excited state are the basis states of the hydrogen atom
The spin-up and spin-down states are basis states for the spin-1/2 particle

8. The State Vector
The state of a quantum system is described by a state vector, written |yñ
If the basis states for a qubit are written |0ñ and |1ñ, then the state vector for the qubit is
|yñ = a |0ñ + b |1ñ
where a and b are complex numbers with
a2 + b2 = 1

9. Basis States
Instead of |0ñ and |1ñ we can use any other basis states, as long as we can distinguish clearly between the two.
Mathematically, basis states must be given by orthogonal vectors.
|1ñ
|0ñ
The inner product of the two vectors must be 0:
á0 | 1ñ = 0

10. Basis states (2)
For example, we could use the basis {|+ñ, |-ñ} to describe the state of a qubit:
|1ñ
Now: |yñ = g |+ñ + d |-ñ
orthogonality: á+ | -ñ = 0
|-ñ
|+ñ
|0ñ

11. Systems of many qubits
If we know the individual states of the electrons in the system below:
|y1ñ = 0 |0ñ + 1 |1ñ = |1ñ
|y2ñ = 1 |0ñ + 0 |1ñ = |0ñ
|y3ñ = 0 |0ñ + 1 |1ñ = |1ñ 1 2 3
... then what is the overall state of the three-particle system?

12. Systems of many qubits (2)
The state of a composite quantum system, when all the component states are known, is their tensor product:
|yñ = |y1ñ Ä |y2ñ Ä |y3ñ
This is the “outer product” of vectors
Note that this is different from the inner product áf½cñ

13. Systems of many qubits (3)
We have
|yñ = |y1ñ Ä |y2ñ Ä |y3ñ
= (0 |0ñ + 1 |1ñ) Ä (1 |0ñ + 0 |1ñ) Ä (0 |0ñ + 1 |1ñ)
= |1ñ Ä |0ñ Ä |1ñ
By convention, we write |1ñ Ä |0ñ Ä |1ñ as |101ñ

14. Quantum Measurement
To extract any information out of a quantum system, you have to perform a physical measurement
By measuring a quantum system:
you automatically change its state, the very state you’re trying to measure
you obtain, in general, a random result, which may be different from the original state

15. Quantum Measurement (2)
When you try to measure a qubit
|yñ = a |0ñ + b |1ñ
... you will never be able to obtain the values of a and b.
A measurement has to be made with respect to a particular basis.

16. Quantum Measurement (3)
If you measure with respect to the {|0ñ, |1ñ} basis:
if |yñ = |0ñ the answer will be |0ñ with probability 100%
if |yñ = |1ñ the answer will be |1ñ with probability 100%
in all other cases (e.g. a2=b2=0.5), the result will be probabilistic.
After measurement, the value of |yñ will change permanently to the result obtained.

17. Quantum Measurement (4)
If you measure with respect to a different basis, things are worse!
Measuring |yñ = a |0ñ + b |1ñ with respect to {|+ñ, |-ñ} will give one of the results |+ñ and |-ñ with particular probabilities.
Also, the value of |yñ will change permanently to the result obtained.

18. Quantum Measurement, Formally
Formally, when you measure
|yñ = a |0ñ + b |1ñ
with respect to {|0ñ, |1ñ} you will get:
result |0ñ with probability |a|2
result |1ñ with probability |b|2
If you use a different measurement basis, the result will be one of the basis states, with different probabilities

19. Measuring many qubits
We want to know the possible outcomes of measuring the two qubit state:
|yñ = (a |0ñ + b |1ñ) Ä (g |0ñ + d |1ñ) =
= ag |00ñ + ad |01ñ + bg |10ñ + bd |11ñ
prob. |ag|2 + |ad|2
prob. |bg|2 + |bd|2
the first measurement will reduce |yñ to one of these smaller states

20. Measuring many qubits (2)
The second measurement will reduce |yñ to one of the four states |00ñ, |01ñ, |10ñ, |11ñ.
ag |00ñ + ad |01ñ
bg |10ñ + bd |11ñ
|00ñ
|01ñ
|10ñ
|11ñ

21. Measuring many qubits (3)
By multiplying the branches in the overall tree, we can obtain the probability of each result. So for the state
|yñ = ag |00ñ + ad |01ñ + bg |10ñ + bd |11ñ
two consecutive measurements will give
result |00ñ with probability |ag|2
result |01ñ with probability |ad|2
result |10ñ with probability |bg|2
result |11ñ with probability |bd|2

22. Entanglement
There exist states of many-qubit systems that cannot be broken down into a tensor product
E.g.: there do not exist a, b, g, d for which
m |00ñ + n |11ñ = (a |0ñ + b |1ñ) Ä (g |0ñ + d |1ñ)
These are termed entangled states.

23. The Bell states
For a two-qubit system, the four possible entangled states are named Bell states:

24. Measuring Entangled States
After measuring an entangled pair for the first time, the outcome of the second measurement is known 100% 1
|0ñ
|0ñ
0.5 1
0.5
|1ñ
|1ñ

25. Review Thus far we have seen: Now we will take a look at quantum gates
how qubits are represented
how many qubits can be combined together
what happens when you measure one or more qubits
where entangled pairs come from, and what happens when you measure them
Now we will take a look at quantum gates

26. Quantum gates
As in classical computing, a gate is an operation on a unit of data, here: a qubit
A quantum gate is represented by a matrix that may be applied to a state vector
We will talk about this in more detail next time; for now we will look at some examples of commonly used quantum gates:
the Hadamard gate (H)
the Pauli gates (I, σx, σy, σz)
the Controlled Not (CNot)

27. The Hadamard gate
The Hadamard gate acts on one qubit, and places it in a superposition of |0ñ and |1ñ :

28. The Pauli gates The Pauli gates act on one qubit, as follows:
phase shift, σz:
σz(a |0ñ + b |1ñ) = a |0ñ - b |1ñ
bit flip, σx:
σx(a |0ñ + b |1ñ) = a |1ñ + b |0ñ
phase shift and bit flip, σy:
σy(a |0ñ + b |1ñ) = a |1ñ - b |0ñ
identity, I, does not change the input

29. The Controlled Not Gate
The CNot gate acts on two qubits:
CNot( |00ñ ) = |00ñ
CNot( |01ñ ) = |01ñ
CNot( |10ñ ) = |11ñ
CNot( |11ñ ) = |10ñ

30. Quantum Coin Flipping
Quantum coin flipping is based on the following game:
Alice places a coin, head upwards in a box.
Alice and Bob then take turns to optionally turn the coin over (without looking at it).
At the end of the game, the box is opened and and Bob wins if the coin is head upwards.
In the quantum version of the game, the coin is a quantum state

31. Quantum Coin Flipping (2)
Assume that Alice can only perform a flipping operation, i.e. gate σx
Remember: σx(a |0ñ + b |1ñ) = a |1ñ + b |0ñ
There is a strategy that allows Bob to win always: he must perform Hadamard operations.
Thus Bob places the state of the coin in a superposition of “heads” and “tails”!

32. Quantum Coin Flipping (3)
Person
Action performed
State
|0ñ
Bob H
Alice σx

33. The No-cloning principle
It has been proved by Wootters and Zurek that it is impossible to clone, or duplicate, an unknown quantum state.
However, it is possible to recreate a quantum state in a different physical location through the process of quantum teleportation.

34. Quantum Teleportation: The Basics
If Alice and Bob each have a single particle from an entangled pair, then:
It is possible for Alice to teleport a qubit to Bob, using only a classical channel
The state of the original qubit will be destroyed
How?
Using the properties of entangled particles

35. Quantum Teleportation
Alice wants to teleport particle 1 to Bob
Two particles, 2 and 3, are prepared in an entangled state
Particle 2 is given to Alice, particle 3 is given to Bob

36. Quantum Teleportation (2)
In order to teleport particle 1, Alice now entangles it with her particle using the CNot and Hadamard gates:
Thus, particle 1 is “disassembled” and combined with the entangled pair
Alice measures particles 1 and 2, producing a classical outcome: 00, 01, 10 or 11.

37. Quantum Teleportation
Depending on the outcome of Alice’s measurement, Bob applies a Pauli operator to particle 3, “reincarnating” the original qubit
If outcome=00, Bob uses operator I
If outcome=01, Bob uses operator σx
If outcome=11, Bob uses operator σy
If outcome=10, Bob uses operator σz
Bob’s measurement produces the original state of particle 1.

38. Quantum Teleportation (Summary)
The basic idea is that Alice and Bob can perform a sequence of operations on their qubits to “move” the quantum state of a particle from one location to another
The actual operations are more involved than we have presented here; see the standard texts on quantum computing for details
Recommended: S. Lomonaco, “A Rosetta Stone for Quantum Computation” [see www]

39. Review
Quantum gates allow us to manipulate quantum states without measuring them
Quantum states cannot be cloned
Teleportation allows a quantum state to be recreated by exchanging only 2 bits of classical information
Quantum coin flipping is more fun than classical coin flipping!