Chapter 5 - Quantum Information: Basic Principles and Simple Examples

Introduction

Quantum computers may be a few years away. However, some quantum information processing tasks which can be performed now are quite interesting and informative. In this chapter we examine a few of these. In addition, we add related material which provides constaints on our ability to manipulate quantum information. Until fairly recently these constraints have been considered a problem, or just the way things are, and were not considered further. Now these attitudes have changed and these constraints have been found useful. These two cosntraints, the uncertainty principle and the no-cloning theorem, are presented here because they are very important to the effectiveness of the BB84 protocol as well as other quantum key distribution (QKD) protocols.

No Cloning!

Quantum mechanics is quite different from classical mechanics and some of these differences are more striking that others. One very striking example of their differences is the so-called ''No Cloning Theorem'' primarily attributed to Wootters and Zurek \cite{Nocloning}. (See also \cite{NocloningPTrev} and References therein.)

The general statement of the theorem is as follows. No Cloning Theorem: No quantum operation exists which can duplicate perfectly an arbitrary quantum state.

Put another way, there is no universal quantum copying machine which will take in a state and put out two copies of the same state for any input state. There is a very simple way in which to see this is true even if there are deeper reasons behind it. Following Neilsen and Chuang, let us consider an easy way to see this and postpone a more thorough discussion. The difference between quantum information and classical information is striking in this theorem since one may take any piece of paper and put it on a copier and get two which are nearly identical. Alternatively, you could take a picture.

Suppose there was a quantum copying machine. This machine would take some state $\left\vert \psi _{1}\right\rangle \,\!$ an arbitrary input state, (also possibly some ancillary quantum system $\left\vert \phi _{0}\right\rangle \,\!$) and produce two copies of that state. Here is how we would write this, using $U\,\!$ as our unitary (this is how states transform) which performs the copying:

	$U\left\vert \psi _{1}\right\rangle \otimes \left\vert \phi _{0}\right\rangle =\left\vert \psi _{1}\right\rangle \otimes \left\vert \psi \right\rangle .\,\!$	(5.1)

This, of course, must be true for any $\left\vert \psi \right\rangle \,\!$. So it must also be true for $\left\vert \psi _{2}\right\rangle \,\!$, meaning

	$U\left\vert \psi _{2}\right\rangle \otimes \left\vert \phi _{0}\right\rangle =\left\vert \psi _{2}\right\rangle \otimes \left\vert \psi \right\rangle .\,\!$	(5.2)

Now, we can multiply the right-hand side of Eq.~(\ref{eq:clone1}) with the Hermitian conjugate of the right-hand side of Eq.~(\ref{eq:clone2}) and similarly the left-hand sides of these two equations. This gives

	$(\left\langle \psi _{2}\right\vert \otimes \left\langle \phi _{0}\right\vert U^{\dagger })(U\left\vert \psi _{1}\right\rangle \otimes \left\vert \phi _{0}\right\rangle )=(\left\langle \psi _{2}\right\vert \otimes \left\langle \psi _{2}\right\vert )(\left\vert \psi _{1}\right\rangle \otimes \left\vert \phi _{1}\right\rangle ).\,\!$	(5.3)

This gives

$\left\langle \psi _{2}\mid \psi _{1}\right\rangle =(\left\langle \psi _{2}\mid \psi _{1}\right\rangle )^{2}.\,\!$

The only way a number to be equal to its square is if it is zero or one. This means that the two states are the same, or they are orthogonal. This certainly excludes arbitrary states, but we wanted a copier which would work with any states. Therefore, it is just not possible to have a cloning, or copying, machine for quantum states.

Uncertainty Principle

The uncertainty principle is something which is considered very important to all of quantum theory. There is no close classical analog. It is introduced here for lack of a better place to put it. However, it will be used later quite often and is central to many quantum phenomena.

Consider two observables ${\mathcal {O}}_{1}\,\!$ and ${\mathcal {O}}_{2}\,\!$. The variance of these can be written as

	$\sigma _{i}^{2}=\left\langle \Psi \right\vert ({\mathcal {O}}_{i}-\langle {\mathcal {O}}_{i}\rangle )^{2}\left\vert \Psi \right\rangle ,\,\!$	(5.4)

for $i=1,2\,\!$. The uncertainty principle is

	$\sigma _{1}^{2}\sigma _{2}^{2}\geq \left({\frac {1}{2i}}\langle [{\mathcal {O}}_{1},{\mathcal {O}}_{2}]\rangle \right)^{2}.\,\!$	(5.5)

Quantum Dense Coding

Quantum dense coding is an interesting communication protocol to begin our discussion of quantum information processing. It shows how information is obtained from a quantum state and it uses entangled states to do this. It is also a very simple protocol.

The idea is to send two bits of information to one person from another person. The sender is commonly referred to as Alice and the receiver Bob. Bob will get two bits of information, but Alice will only send one qubit. Two bits can only be in one of four possible states, $00,01,10,\,\!$ or $11\,\!$.

We begin with the following Bell state,

	$\left\vert \phi _{+}\right\rangle ={\frac {1}{\sqrt {2}}}(\left\vert 00\right\rangle +\left\vert 11\right\rangle ).\,\!$	(5.6)

Suppose Alice has the first of these two qubits and Bob the second. We might then write the state in the following way to be even more explicit:

	$\left\vert \phi _{+}\right\rangle ={\frac {1}{\sqrt {2}}}(\left\vert 0\right\rangle _{A}\left\vert 0\right\rangle _{B}+\left\vert 1\right\rangle _{A}\left\vert 1\right\rangle _{B}),\,\!$	(5.7)

Then Alice will decide whether to act with $\mathbb {I} ,X,Z\,\!$, or $Y(=ZX)\,\!$ on her qubit. She will then send that qubit to Bob, who will act on the two qubits and measure. He will then have two bits of information from Alice who will have only sent one qubit to him. So for one qubit sent, and some shared entanglement, they can transfer two bits of information.

This is the specific protocol. Alice decides which she wants to send, $00,01,10,\,\!$ or $11\,\!$, i.e., two bits of information to Bob. She then acts on $\left\vert \phi _{+}\right\rangle \,\!$ with $\mathbb {I} ,X,Z\,\!$, or $iY(=ZX)\,\!$ respectively to do this, giving

	${\begin{aligned}\left\vert \Psi _{0}\right\rangle &=\mathbb {I} _{A}\left\vert \phi _{+}\right\rangle ={\frac {1}{\sqrt {2}}}(\left\vert 0\right\rangle _{A}\left\vert 0\right\rangle _{B}+\left\vert 1\right\rangle _{A}\left\vert 1\right\rangle _{B})\\\left\vert \Psi _{1}\right\rangle &=X_{A}\left\vert \phi _{+}\right\rangle ={\frac {1}{\sqrt {2}}}(\left\vert 1\right\rangle _{A}\left\vert 0\right\rangle _{B}+\left\vert 0\right\rangle _{A}\left\vert 1\right\rangle _{B})\\\left\vert \Psi _{2}\right\rangle &=Z_{A}\left\vert \phi _{+}\right\rangle ={\frac {1}{\sqrt {2}}}(\left\vert 0\right\rangle _{A}\left\vert 0\right\rangle _{B}-\left\vert 1\right\rangle _{A}\left\vert 1\right\rangle _{B})\\\left\vert \Psi _{3}\right\rangle &=iY_{A}\left\vert \phi _{+}\right\rangle ={\frac {1}{\sqrt {2}}}(-\left\vert 1\right\rangle _{A}\left\vert 0\right\rangle _{B}+\left\vert 0\right\rangle _{A}\left\vert 1\right\rangle _{B})\end{aligned}}\,\!$	(5.8)

Now, Alice sends her qubit to Bob. Having both qubits, Bob now acts with a CNOT on the two qubits to obtain the following:

	${\begin{aligned}CNOT_{12}\left\vert \Psi _{0}\right\rangle &={\frac {1}{\sqrt {2}}}(\left\vert 0\right\rangle _{A}\left\vert 0\right\rangle _{B}+\left\vert 1\right\rangle _{A}\left\vert 0\right\rangle _{B})={\frac {1}{\sqrt {2}}}(\left\vert 0\right\rangle _{A}+\left\vert 1\right\rangle _{A})\left\vert 0\right\rangle _{B}\\CNOT_{12}\left\vert \Psi _{1}\right\rangle &={\frac {1}{\sqrt {2}}}(\left\vert 1\right\rangle _{A}\left\vert 1\right\rangle _{B}+\left\vert 0\right\rangle _{A}\left\vert 1\right\rangle _{B})={\frac {1}{\sqrt {2}}}(\left\vert 1\right\rangle _{A}+\left\vert 0\right\rangle _{A})\left\vert 1\right\rangle _{B}\\CNOT_{12}\left\vert \Psi _{2}\right\rangle &={\frac {1}{\sqrt {2}}}(\left\vert 0\right\rangle _{A}\left\vert 0\right\rangle _{B}-\left\vert 1\right\rangle _{A}\left\vert 1\right\rangle _{B})={\frac {1}{\sqrt {2}}}(\left\vert 1\right\rangle _{A}-\left\vert 0\right\rangle _{A})\left\vert 0\right\rangle _{B}\\CNOT_{12}\left\vert \Psi _{3}\right\rangle &={\frac {1}{\sqrt {2}}}(-\left\vert 1\right\rangle _{A}\left\vert 0\right\rangle _{B}+\left\vert 0\right\rangle _{A}\left\vert 1\right\rangle _{B})={\frac {1}{\sqrt {2}}}(\left\vert 0\right\rangle _{A}-\left\vert 1\right\rangle _{A})\left\vert 1\right\rangle _{B}\end{aligned}}\,\!$	(5.9)

(Recall a CNOT acts on two qubits and flips the second one if the first is in the one state and does nothing if it is in the zero state.)

Now Bob acts with a Hadamard gate on the first qubit (the one Alice sent). Recall the Hadamard gate, $H\,\!$, acts in the following way: $H\left\vert 0\right\rangle ={\tfrac {1}{\sqrt {2}}}(\left\vert 0\right\rangle +\left\vert 1\right\rangle )\,\!$, $H\left\vert 1\right\rangle ={\tfrac {1}{\sqrt {2}}}(\left\vert 0\right\rangle -\left\vert 1\right\rangle )\,\!$, and $H^{2}=\mathbb {I} \,\!$. Then

	${\begin{aligned}H_{1}CNOT_{12}\left\vert \Psi _{0}\right\rangle &=\left\vert 00\right\rangle \\H_{1}CNOT_{12}\left\vert \Psi _{1}\right\rangle &=\left\vert 01\right\rangle \\H_{1}CNOT_{12}\left\vert \Psi _{2}\right\rangle &=\left\vert 10\right\rangle \\H_{1}CNOT_{12}\left\vert \Psi _{3}\right\rangle &=\left\vert 11\right\rangle \end{aligned}}\,\!$	(5.10)

Therefore, if Alice has sent $\left\vert \Psi _{0}\right\rangle \,\!$, then Bob will be able to recover and find $\left\vert 00\right\rangle \,\!$ and similarly for the other three possibilities. So Alice, by sending one qubit, has transmitted two classical bits of information.

Teleporting a Quantum State

Teleportation is an interesting protocol for several reasons. First and foremost, for our considerations is that it is a simple protocol which is easily described using what we already know.

Here is the scenario. Suppose Alice wants to send Bob a state $\left\vert \psi \right\rangle \,\!$ and let us further suppose that Alice doesn't know the state. We can write the state in the form

	$\left\vert \psi \right\rangle =\alpha _{0}\left\vert 0\right\rangle +\alpha _{1}\left\vert 1\right\rangle .\,\!$	(5.11)

Now if Alice wants to specify the state exactly to Bob using classical information, it would be impossible. This is because Alice would need to know the numbers $\alpha _{0}\,\!$ and $\alpha _{1}\,\!$ exactly and they may be irrational numbers of arbitrary precision. (You may say that we never need to know anything to arbitrary precision, but that somewhat misses the point. It's not, in principle, possible to do it classically. There is no way for Alice to send Bob an arbitrary number of digits in a finite time. This is if she knows the state. We are saying that she does not need to know it. Also, she can't know find out what the state is through measurment since a measurement in the $\left\vert 0\right\rangle -\left\vert 1\right\rangle \,\!$ basis will return one or the other with some probability. It will not give her the coefficients.) However, she can teleport the state exactly using the following protocol.

Alice wants to send the state $\left\vert \psi \right\rangle \,\!$ to Bob and suppose that they share a state between them

	$\left\vert \phi _{+}\right\rangle ={\frac {1}{\sqrt {2}}}(\left\vert 00\right\rangle +\left\vert 11\right\rangle ),\,\!$	(5.12)

where, as before, the first qubit is in Alice's possession and the second qubit is in Bob's possession. So we could write it as

	$\left\vert \phi _{+}\right\rangle ={\frac {1}{\sqrt {2}}}(\left\vert 0\right\rangle _{A}\left\vert 0\right\rangle _{B}+\left\vert 1\right\rangle _{A}\left\vert 1\right\rangle _{B}).\,\!$	(5.13)

However, it will be understood that the first particle is Alice's and the second Bob's.

Now, Alice also has $\left\vert \psi \right\rangle \,\!$, so together we can write the total system's state as

	$\left\vert \psi \right\rangle \otimes \left\vert \phi _{+}\right\rangle =(\alpha _{0}\left\vert 0\right\rangle +\alpha _{1}\left\vert 1\right\rangle )\otimes {\frac {1}{\sqrt {2}}}(\left\vert 00\right\rangle +\left\vert 11\right\rangle ).\,\!$	(5.14)

Now Alice can act on the two qubits in her possession with a CNOT gate. This produces:

	$CNOT_{12}=\alpha _{0}\left\vert 0\right\rangle \otimes {\frac {1}{\sqrt {2}}}(\left\vert 00\right\rangle +\left\vert 11\right\rangle )+\alpha _{1}\left\vert 1\right\rangle \otimes {\frac {1}{\sqrt {2}}}(\left\vert 10\right\rangle +\left\vert 01\right\rangle ).\,\!$	(5.15)

Now she applies a Hadamard gate to the first qubit to get

	${\begin{aligned}H_{1}CNOT_{12}&={\frac {1}{2}}\alpha _{0}(\left\vert 0\right\rangle +\left\vert 1\right\rangle )\otimes (\left\vert 00\right\rangle +\left\vert 11\right\rangle )+{\frac {1}{2}}\alpha _{1}(\left\vert 0\right\rangle -\left\vert 1\right\rangle )\otimes (\left\vert 10\right\rangle +\left\vert 01\right\rangle )\\&={\frac {1}{2}}[\left\vert 00\right\rangle (\alpha _{0}\left\vert 0\right\rangle +\alpha _{1}\left\vert 1\right\rangle )+\left\vert 01\right\rangle (\alpha _{0}\left\vert 1\right\rangle +\alpha _{1}\left\vert 0\right\rangle )\\&\;\;\;\;+\left\vert 10\right\rangle (\alpha _{0}\left\vert 0\right\rangle -\alpha _{1}\left\vert 1\right\rangle )+\left\vert 11\right\rangle (\alpha _{0}\left\vert 1\right\rangle -\alpha _{1}\left\vert 0\right\rangle )],\end{aligned}}\,\!$	(5.16)

where the second equality follows from rearranging terms. Now Alice measures the first two qubits which are in her possession. If she gets the result $\left\vert 00\right\rangle \,\!$ she tells Bob that he has the state. If she gets the result $\left\vert 01\right\rangle \,\!$, then she tells Bob to apply an $X\,\!$ operation to the state in his possession and then he'll have $\left\vert \psi \right\rangle \,\!$. If she gets $\left\vert 10\right\rangle \,\!$, she tells him to apply a Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle Z\,\!} operation, and if she gets $\left\vert 11\right\rangle \,\!$ she tells him to apply $iY=ZX\,\!$. In each case, he will obtain the desired state Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left\vert \psi\right\rangle\,\!} thus ''teleporting'' the state $\left\vert \psi \right\rangle \,\!$ from Alice to Bob.

QKD: BB84

The objective of cryptography is to encode information in such a way that only the sender and receiver can read the information transmitted between them. Many ingenious codes have been developed, but none are complete secure with the exception of the one-time key pad. This is a randomly generated key that allows the sender to encode information, the receiver to decode the information, and allows no others to decifer it since it is randomly generated and used only once.

Quantum cryptography provides a way in which to generate a shared random key. The amazing thing is that, in the ideal case, it can be proven secure even though it is transmitted over a public channel. Since this method of cryptography works by the generation of a key, it is often called quantum key distribution (QKD). The simplest protocol for generating such a key is called the BB84 protocol after Bennett and Brassard, its inventors. (See \cite{cryptorev:02} and references therein for a nice history of quantum cryptography.)

Quantum cryptography is now an active field \cite{cryptorev:02} with systems being developed for commercial use. (Some such systems already exist.) However, most (if not all) protocols which enable secret communication using quantum states are based on the very simple ideas contained in the BB84 protocol. It is therefore instructive, as well as quite interesting from a quantum information prespective, to introduce this protocol here and show how it works.