Aaron Moore amoore91@tamu.edu
Department of Computer Science and Engineering
Department of Mathematics
Texas A&M University Math 482
April 28, 2013
Abstract
Quantum computers offer an intriguing new model for computation. It can be shown that quantum computers can offer superpolynomial speed up over the fastest classical algorithms. While the technology to create and manipulate quantum states has not yet been realized, the quantum circuit model allows us to test the theoretical limits of these future machines.
This paper will discuss a short intro to quantum computing, the quantum circuit model, and introduce the protocol for quantum teleportation.
1 Quantum Bits
Classical computers are based on computation using bits which are either 0 or 1. While this system is widely used, a model based on quantum mechanics offers the potential for dramatic speed up over the fastest classical algorithms.
To develop this system we need a quantum analog of the bit.
Consider a monochromatic light source. We can polarize the light so that it forms waves in either a horizontal or vertical plane. We can denote each of these clearly distinguishable states |↔i and |li . If we pass each of these waves through a properly oriented calcite crystal, the horizontally polarized ( |↔i ) beam will pass straight through while the vertically polarized ( |li ) beam will be deflected.
We can use this calcite crystal then as a basis to measure the polarization of an unknown light. If we then rotate the polarizer of our light source by an angle
θ the crystal will record the two components of the light cosθ |↔i + sinθ |li , or expressed as a column vector ( cosθ, sinθ ) T .
Now consider what would happen if we slowed down our light source to emit a single photon. We can still polarize this photon at an angle θ from the |↔i
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direction but what will happen if we pass it through the calcite crystal? The single photon can not split to register both of its horizontal and vertical components. The question is will it register as a 0 or a 1, a |↔i or a |li ? It turns out, if the photon has polarization a |↔i + b |li it will register |↔i with probability a
2 and |li with probability b
2
. Furthermore, the coefficients of the polarization need not be real numbers. We use this idea of polarization to define a quantum bit.
Definition 1.
where | a | 2 + | b | 2
A quantum bit, also qubit or quantum state, q ∈
C
2 = ( a, b ) T
= 1. We will usually denote q in its component form a | 0 i + b | 1 i where | 0 i = e
1
= (1 , 0) T and | 1 i = e
2
= (0 , 1) T
2 Single Qubit Operations
In order for our new quantum system to be useful, it needs to manipulate and measure quantum states while conforming to the laws of quantum mechanics.
In the single qubit cases these operations are relatively easy to categorize. First we will need to recall a few concept from linear algebra.
Definition 2.
If A ∈ is A
∗
= A T
C nxm let A
∗ denote the conjugate transpose of A . That
Definition 3.
A square matrix U ∈
C nxn is called unitary if denote the set of unitary nxn matrices SU ( n )
U
∗
U = I . We
Remark 1.
Unitary matrices are the complex analog of orthogonal matrices.
Let U ∈ SU ( n ) and x, y ∈
C n then h U x, U y i = h x, y i .
As a consequence
| U x | = | x |
If you recall that our model of quantum computation defines qubits as vectors, then it is only natural that transformations be defined as matrices; furthermore, since qubits also have the additional restraint of being unit length, it is clear that these quantum gates need to be unitary.
If you recall, to determine the action of a linear transformation on a vector, it is enough to know its effects on the basis vectors. This is the reason for the ket( | 0 i , | 1 i ) notation. As you will see, this notation makes working with complex systems much easier, but first we will introduce some of the most common gates.
Definition 4.
Common gates in vector form and using ket notation
1. The Hadamard gate denoted H
H =
1
√
2
2. The Not gate denoted X
1 1
1 − 1
H | 0 i =
H | 1 i =
√
2
( | 0 i + | 1 i )
√
2
( | 0 i − | 1 i )
X =
0 1
1 0
X | 0 i = | 1 i
X | 1 i = | 0 i
2

3. The Z gate
Z =
1 0
0 − 1
4. An arbitrary real rotation R
R
θ
= cosθ sinθ
− sinθ cosθ
Z
Z
|
|
1
0 i i =
=
| 0
−| i
1 i
R
θ
| 0 i = cosθ | 0 i + sinθ | 1 i
R
θ
| 1 i = − sinθ | 0 i + cosθ | 1 i
The other operation we need is measurement. If you recall from the physical system, we could not measure the individual components of a state. Because the single particle could not split, measurement always revealed either 0 or 1.
Somehow the act of measuring changes the state. Our system must reflect the physical.
Definition 5.
Let q be the quantum state q = a | 0 i + b | 1 i with a, b ∈ after measurement
C
2 then q → | 0 i with probability | a | 2 q → | 1 i with probability | b | 2
Graphically, we denote operations on quantum bits using quantum circuits.
We start with a very simple example.
Example 1.
| 0 i H
Solution.
This diagram is read from left to right. The input to the circuit is the state q = | 0 i which we assume can be easily prepared. The state then travels along the wire until it encounters a Hadamard gate denoted by a box with an H around it. At this point q → H | 0 i = √
2
( | 0 i + | 1 i ). The final symbol on the right is a measurement. Measurement of the state q = √
2
( | 0 i + | 1 i ) will reveal | 0 i and | 1 i with equal probability 1 / 2
While this was a very simple circuit, it is actually quite important. On a classical computer, it is impossible to implement truly random numbers. Classical computers will always produce the same ouput when given the same input.
Our first quantum circuit, however, has already extended beyond the limits of classical computers by giving us a truly random bit!
3 Multiple qubits
In order to develop more complex systems, we must find a way to combine qubits into more complicated systems. Consider two qubits in general states a = a
0
| 0 i + a
1
| 1 i and b = b
0
| 0 i + b
1
| 1 i . What would be the outcome if we measured the entire system?
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a
0
| 0 i + a
1
| 1 i b
0
| 0 i + b
1
| 1 i
Individually, a will measure 0 with probability | a
0
| 2 and 1 with probability | a
1
| 2 ; b will have similar probabilities. Since these two bits are measured independently, their joint probabilities are simply the product of the individual outcomes. Therefore, the overall measurement will reveal P (00) = | a
0 b
P (01) = | a
0 b
1
| 2
, P (10) = | a
1 b
0
| 2
, and P (11) = | a
1 b
1
| 2 where P ( x
0
| 2
) denotes the
, probability of viewing x. We need a way to reflect these joint probabilities while still retaining the use of the single qubit gates defined earlier.
Definition 6.
Let a ∈
C
2 m and b ∈
C
2 n
. The tensor product a ⊗ b ∈
C
2 m + n takes two elements from smaller spaces and combines them into a larger one in the following way.
| 0 i ⊗ | 0 i =
1
0
⊗
1
0
:= | 00 i := (1 , 0 , 0 , 0)
T
| 0 i ⊗ | 1 i =
1
0
⊗
0
1
:= | 01 i := (0 , 1 , 0 , 0)
T
| 1 i ⊗ | 0 i =
0
1
⊗
1
0
:= | 10 i := (0 , 0 , 1 , 0)
T
| 1 i ⊗ | 1 i =
0
1
⊗
0
1
:= | 11 i := (0 , 0 , 0 , 1)
T
The tensor product satisfies the following properties. Let V ∈ elements v
0 and v
1
, W ∈
C
2 n with elements w
0 and w
1
. Also let c ∈
1. ( v
0
+ v
1
) ⊗ w
0
= v
0
⊗ w
0
+ v
1
⊗ w
0
2.
v
0
3.
c ( v
⊗
0
( w
0
+ w
1
) = v
⊗ w
0
) = ( cv
0
)
0
⊗
⊗ w w
0
0
=
+ v
0 v
0
⊗ w
1
⊗ ( cw
0
)
C
2 m
C
.
with
Example 2.
In order to fully see the power of this system we will walk through an example.
| 0 i H
| 0 i X
To read this diagram, again we start from the left. The system begins with input
| 0 i ⊗ | 0 i which we assume can be easily produced. Both states then travel down the wires at the same speed. A Hadamard gate is then applied to the top qubit producing state H | 0 i⊗| 0 i = √
2
( | 0 i + | 1 i ) ⊗| 0 i . Next, a not (X) gate is applied to the second qubit producing the state √
2
( | 0 i + | 1 i ) ⊗ X | 0 i = √
2
( | 0 i + | 1 i ) ⊗ | 1 i
Thus before the measurements, the system is in state
√
2
( | 0 i ⊗ | 1 i + | 1 i ⊗ | 1 i ) = √
2
√
2
( | 0 i + | 1 i ) ⊗ | 1 i =
| 01 i + | 11 i . The measurements will then reveal 01 and 11 with equal probability 1/2.
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We will introduce one more important gate needed for an important protocol.
Unlike all the previous gates, the controlled-not gate acts on two qubits.
Definition 7.
The controlled not gate with source on the top bit and target on the second applies a not gate to the second bit only if the first bit it | 1 i .
•
It has the following action on the basis
| 00 i → | 00 i
| 01 i → | 01 i
| 10 i → | 11 i
| 11 i → | 10 i
We can similarly define the reversed operation with the target on the top bit.
•
It has the following action on the basis, it flips the top bit when the bottom bit is | 1 i .
| 00 i → | 00 i
| 01 i → | 11 i
| 10 i → | 10 i
| 11 i → | 01 i
Example 3.
While the controlled not gate seems simple, it is important to see the action when combined with other gates.
| 0 i
| 0 i
H
•
|
As with Example 2, after the Hadamard gate the system is in state
1 i ) ⊗ | 0 i =
√
2
( | 0 i +
√
2
( | 0 i ⊗ | 0 i + | 1 i ⊗ | 0 i ). Then, after the controlled not gate is applied, we have the state √
2
( | 0 i ⊗ | 0 i + | 1 i ⊗ | 1 i ) = √
2
( | 00 i + | 11 i ). Notice how the second bit was flipped only in the component where the first bit was | 1 i .
We must now make an important observation about the measurement. Recall that measurement collapses the superposition of a single qubit. That means if a measurement of the first bit reveals | 0 i , all components with a 1 in the first bit are eliminated (and vice-versa) and the state is renormalized to keep unit length. That means in this problem, if the measurement reveals a | 0 i then the second bit will become | 0 i and if it measures | 1 i the second qubit will contain only | 1 i . Somehow, measuring the first qubit changes the value of the second.
We say that these qubits are entangled.
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4 Quantum Teleportation
We will now present an extended example which will serve to demonstrate a practical application of quantum computing that relies on many of the previous examples. Suppose that Alice has a state a | 0 i + b | 1 i that she wants to communicate to Bob. If Alice and Bob also share an entangled pair, like in example
3, they can each act locally on the bits they own to transfer the state over any distance! After following the below circuit, Alice will successfully communicate her state to Bob. Note that Alice owns the top two bits while Bob has the bottom one. For a detailed walkthrough see the attached document.
a | 0 i + b | 1 i
| 0 i
H
•
| 0 i
• H
Corrections
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5 Sources
1. Klappenecker, Andreas. Quantum Algorithms. faculty.cs.tamu.edu/klappi/csce640f12/qcircuit.pdf. 2004. Web
2. Nielsen, Michael, and Isaac Chuang. Quantum Computation and Quantum Information. 1st ed. Cabridge, UK: Cambridge University Press,
2000. Print.
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