Chapter 3 - Physics of Quantum Information

Introduction

It was a great realization that information is physical and that a (classical) Turing machine is not the end of the story of computation. The physical system in which the information is stored and manipulated is important and qubits are quite different from bits.

In this chapter, some background in quantum mechanics is provided. Not all of this chapter will be directly relevant to our discussion, but it is included for the sake of completeness of our understanding of how quantum mechanics from a textbook is related to quantum computing. The connection is, as of yet, clear but the story seems incomplete from a physicists perspective and for the subject of error prevention methods, some of this chapter will be vital. In particular, the section(s) concerning the density matrix. Not only is this vital, but not usually covered in most quantum mechanics classes, either undergraduate or graduate.

It is also worth emphasizing that this chapter is primarily aimed at physicists and for those others which are interested in the background physics. It is not necessary for much of what follows.

Schrodinger's Equation

A common starting point in quantum mechanics is Schrodinger's equation. This equation is not derived, or justified here, but is given in a general form:

	$H\left\vert \Psi \right\rangle =i\hbar {\frac {\partial }{\partial t}}\left\vert \Psi \right\rangle ,\,\!$	(3.1)

where $H\,\!$ is the Hamiltonian, $\hbar \,\!$ is Planck's constant (divided by $2\pi \,\!$), and $t\,\!$ is time. The Hamiltonian contains what is known about the system's evolution. Most of the time in these notes, we let $\hbar =1\,\!$.

This equation is (formally) solved by taking the time derivative to be an ordinary derivative (we assume no explicit time dependence for $H\,\!$), so

	$H\left\vert \Psi \right\rangle =i{\frac {d\left\vert \Psi \right\rangle }{dt}}.\,\!$	(3.2)

This means that

	$-iHdt={\frac {d\left\vert \Psi \right\rangle }{\left\vert \Psi \right\rangle }},\,\!$	(3.3)

so

	${\begin{aligned}\ln \left\vert \Psi \right\rangle &=-iHt+C,\\\Rightarrow \left\vert \Psi (t)\right\rangle &=e^{-iHt}\left\vert \Psi (0)\right\rangle .\end{aligned}}\,\!$	(3.4)

Now if $H\,\!$ is Hermitian, and it is, then the matrix

	$U=e^{-iHt}\,\!$	(3.5)

is unitary. (See Appendix C - Vectors and Linear Algebra, in particular the section entitled Unitary Matrices.) Any transformation on a closed system can be described by a unitary transformation and any unitary transformation can be obtained by the exponentiation of Hermitian matrix.

Aside, for inclusion later in an appendix about Group theory.

It may seem strange to exponentiate a matrix. However, you can define a function of a matrix according to its Taylor expansion. The details of this are primarily unimportant here, but just to show how it goes, it is written out.

The Taylor expansion of an exponential is the following:

	$e^{x}=\sum _{n=0}^{\infty }{\frac {x^{n}}{n!}}\,\!$	(3.6)

and this can be used to exponentiate a matrix by letting the matrix replace $x\,\!$ in the equation. This can also be used to prove that

	$e^{ix}=\cos x+i\sin x.\,\!$	(3.7)

End Aside

The end result and important point is that the evolution of a quantum state is, in general, given by a unitary matrix

	$\left\vert \Psi (t)\right\rangle =U\left\vert \Psi (0)\right\rangle .\,\!$	(3.8)

So our objective in quantum information processing is to create a unitary evolution, and eventual measurement, which will produce a particular outcome.

Density Matrix for Pure States

Now let us consider the object (a density matrix, or density operator, of rank one)

	$\rho =\left\vert \psi \right\rangle \left\langle \psi \right\vert ,$	(3.9)

which is just the outter product of two vectors. For example, if

	$\left\langle \psi \right\vert =(0,0,1),$	(3.10)

then

$\left\langle \psi \mid \psi \right\rangle =\left\langle \psi \right\vert (\left\langle \psi \right\vert )^{\dagger }=1.$

However,

	$\left\vert \psi \right\rangle \left\langle \psi \right\vert =\left({\begin{array}{ccc}0&0&0\\0&0&0\\0&0&1\end{array}}\right).$	(3.11)

Again $\left\vert \psi \right\rangle =\left\vert \psi (t)\right\rangle \,\!$, so $\rho =\rho (t)\,\!$. If we differentiate this with respect to $t\,\!$,

	${\begin{aligned}{\frac {\partial \rho }{\partial t}}&=\left({\frac {\partial \left\vert \psi \right\rangle }{\partial t}}\right)\left\langle \psi \right\vert +\left\vert \psi \right\rangle \left({\frac {\partial \left\langle \psi \right\vert }{\partial t}}\right)\\&=(-iH)\rho +\rho (iH)=-i[H,\rho ],\end{aligned}}$	(3.12)

which is the Schrodinger equation for the density matrix, with solution,

	$\rho (t)=U\rho (0)U^{\dagger }.$	(3.13)

This follows from $\left\vert \psi (t)\right\rangle \left\langle \psi (t)\right\vert =U\left\vert \psi (0)\right\rangle \left\langle \psi (0)\right\vert U^{\dagger }\,\!$.

Consider our two-state system

	$\left\vert 0\right\rangle =\left({\begin{array}{c}1\\0\end{array}}\right),\;\;\;{\mbox{and}}\;\;\;\left\vert 1\right\rangle =\left({\begin{array}{c}0\\1\end{array}}\right).$	(3.14)

A superposition of these two states is

	$\left\vert \psi \right\rangle =\alpha _{0}\left\vert 0\right\rangle +\alpha _{1}\left\vert 1\right\rangle =\left({\begin{array}{c}\alpha _{0}\\\alpha _{1}\end{array}}\right),$	(3.15)

where $\alpha _{0}\,\!$ and $\alpha _{1}\,\!$ are complex numbers such that $|\alpha _{0}|^{2}+|\alpha _{1}|^{2}=1\,\!$. The corresponding pure state, (i.e. rank one) density matrix is given by

	$\rho _{p}=\left\vert \psi \right\rangle \left\langle \psi \right\vert =\left({\begin{array}{cc}|\alpha _{0}|^{2}&\alpha _{0}\alpha _{1}^{*}\\\alpha _{0}^{*}\alpha _{1}&|\alpha _{1}|^{2}\end{array}}\right).$	(3.16)

Note that the superposition in Eq.(3.15) can be obtained from any pure state by a unitary transformation. Here, the trace of the density matrix is an important quantity; it is

	${\mbox{Tr}}(\rho _{p})=|\alpha _{0}|^{2}+|\alpha _{1}|^{2}=1.\,\!$	(3.17)

Notice also that the determinant of this matrix is zero:

	$\det(\rho _{p})=|\alpha _{0}|^{2}|\alpha _{1}|^{2}-\alpha _{0}\alpha _{1}^{*}\alpha _{0}^{*}\alpha _{1}=0.$	(3.18)

To see this another way, note that the density operator of rank one can be written as $U(\left\vert 0\right\rangle \left\langle 0\right\vert )U^{\dagger }\,\!$, so that the determinant is

	${\begin{aligned}\det(U(\left\vert 0\right\rangle \left\langle 0\right\vert )U^{\dagger })&=\det(U(\left\vert 0\right\rangle \left\langle 0\right\vert )U^{-1})\\&=\det(U)\det(\left\vert 0\right\rangle \left\langle 0\right\vert ){\frac {1}{\det(U)}}\\&=\det(\left\vert 0\right\rangle \left\langle 0\right\vert )=0.\end{aligned}}$	(3.19)

Measurements Revisited

If the state of a quantum system is described by

	$\left\vert \psi \right\rangle =\alpha _{0}\left\vert 0\right\rangle +\alpha _{1}\left\vert 1\right\rangle ,$	(3.20)

the probability of finding it in the state $\left\vert 0\right\rangle \,\!$ when measured in the computational basis is $|\alpha _{0}|^{2}\,\!$. However, this is a particular superposition which could be written as

	$\left\vert \psi \right\rangle =U\left\vert 0\right\rangle .$	(3.21)

In the section entitled Schrodinger’s Equation it was shown that this matrix $U\,\!$ results from the exponentiation of a Hermitian matrix and from the section entitled The Pauli Matrices any $2\times 2\,\!$ Hermitian matrix can be written in terms of the Pauli matrices. To make this explicit using standard conventions,

	${\begin{aligned}\left\vert \psi \right\rangle &=U\left\vert 0\right\rangle \\&=\exp(-i{\vec {n}}\cdot {\vec {\sigma }}\theta )\left\vert 0\right\rangle \\&=(\mathbb {I} \cos(\theta )-i{\vec {n}}\cdot {\vec {\sigma }}\sin(\theta ))\left\vert 0\right\rangle ,\end{aligned}}$	(3.22)

where ${\vec {n}}\,\!$ is a unit vector, $|{\vec {n}}|=1\,\!$ and ${\vec {n}}\cdot {\vec {\sigma }}=n_{1}\sigma _{1}+n_{2}\sigma _{2}+n_{3}\sigma _{3}\,\!$. One can write this matrix out explicitly

	${\begin{aligned}\exp(-i{\vec {n}}\cdot {\vec {\sigma }}\theta )&=\left({\begin{array}{cc}1&0\\0&1\end{array}}\right)\cos(\theta )\\&\;\;\;+(-i)\left[n_{1}\left({\begin{array}{cc}0&1\\1&0\end{array}}\right)+n_{2}\left({\begin{array}{cc}0&-i\\i&0\end{array}}\right)+n_{3}\left({\begin{array}{cc}1&0\\0&-1\end{array}}\right)\right]\sin(\theta )\\&=\left({\begin{array}{cc}\cos(\theta )-in_{3}\sin(\theta )&(-in_{1}-n_{2})\sin(\theta )\\(-in_{1}+n_{2})\sin(\theta )&\cos(\theta )+in_{3}\sin(\theta )\end{array}}\right).\end{aligned}}$	(3.23)

Notice this is a special unitary matrix. (See Appendix C - Vectors and Linear Algebra, in particular the subsection Unitary Matrices.)

To see that any state $\left\vert \psi \right\rangle \,\!$ for arbitrary coefficients $\alpha _{0}\,\!$, $\alpha _{1}\,\!$ can be obtained by choosing ${\vec {n}}\,\!$ and $\theta \,\!$ appropriately, the state $\left\vert 0\right\rangle \,\!$ can be chosen as a starting point. Then

	${\begin{aligned}U\left\vert 0\right\rangle &=\left({\begin{array}{cc}\cos(\theta )-in_{3}\sin(\theta )&(-in_{1}-n_{2})\sin(\theta )\\(-in_{1}+n_{2})\sin(\theta )&\cos(\theta )+in_{3}\sin(\theta )\end{array}}\right)\left({\begin{array}{c}1\\0\end{array}}\right)\\&=\left({\begin{array}{c}\cos(\theta )-in_{3}\sin(\theta )\\(-in_{1}+n_{2})\sin(\theta )\end{array}}\right).\end{aligned}}$	(3.24)

For example, choosing $\theta =0\,\!$ gives the original state; choosing ${\vec {n}}=(0,1,0)\,\!$ and $\theta =\pi /2\,\!$ gives $\left\vert 1\right\rangle \,\!$; and choosing ${\vec {n}}=(0,1,0)\,\!$ and $\theta =\pi /4\,\!$ gives an equal superposition. In general, when the system is in the state $\left\vert \psi \right\rangle =U\left\vert 0\right\rangle \,\!$, the probability of finding the state $\left\vert 0\right\rangle \,\!$ when a measurement is made in the computational basis is given by

	${\begin{aligned}|\left\langle 0\right\vert U\left\vert 0\right\rangle |^{2}&=|\cos(\theta )-in_{3}\sin(\theta )|^{2}\\&=\cos ^{2}(\theta )+n_{3}^{2}\sin ^{2}(\theta ),\end{aligned}}$	(3.25)

and the probability of finding $\left\vert 1\right\rangle \,\!$ is

	${\begin{aligned}|\left\langle 1\right\vert U\left\vert 1\right\rangle |^{2}&=|(-in_{1}+n_{2})\sin(\theta )|^{2}\\&=(n_{1}^{2}+n_{2}^{2})\sin ^{2}(\theta ).\end{aligned}}$	(3.26)

Notice the probabilities add up to one if ${\vec {n}}\,\!$ is a unit vector.

What this shows is that there is a transformation that takes the state $\left\vert 0\right\rangle \,\!$, which has probability $1\,\!$ of being in the state $\left\vert 0\right\rangle \,\!$ and probability 0 of being in the state $\left\vert 1\right\rangle \,\!$ and transform it (using a "rotation'' into a state with a different (and generic) probability of each. This means that the density matrix corresponding to this system always has determinant zero, meaning (for a two-state system) it has one eigenvalue 1 and another eigenvalue 0. (The determinant is the product of the eigenvalues.)

Density Matrix for Mixed States

For a system with $D\,\!$ dimensions, a mixed state density matrix (or density operator, see Appendix \ref{app:cohvec}), is a matrix which us used to describe a more general state of a quantum system and can be written as

	$\rho _{D}=\sum _{i}a_{i}\rho _{i},\,\!$	(3.27)

where $a_{i}\geq 0\,\!$, $\sum _{i}a_{i}=1\,\!$ and the $\rho _{i}\,\!$ are pure states. There is also a generalization of the Bloch sphere which is described in Appendix {app:polvec}.

The mixed state density matrices are important in all descriptions of physical implementations of quantum information processing. For this reason, a bit of labor has gone into this section, the rest of which is devoted to the physical interpretation and properties of this description of a quantum system.

Density Matrix for a Mixed State: Two States

A mixed state density matrix (for a two-state system) is a rank two density matrix, $\rho _{m}\,\!$, which can be described by

	$\rho _{m}=\left[a_{1}\rho _{1}+a_{2}\rho _{2}\right],$	(3.28)

where $\rho _{1}=\left\vert \psi _{1}\right\rangle \left\langle \psi _{1}\right\vert \,\!$, $\rho _{2}=\left\vert \psi _{2}\right\rangle \left\langle \psi _{2}\right\vert \,\!$ and $a_{1}+a_{2}=1\,\!$. The $a_{i}\,\!$ are probabilities and must sum to one. (Note, if $\left\vert \psi \right\rangle _{1}=\left\vert \psi \right\rangle _{2}\,\!$, or if one $a_{i}\,\!$ or one $\left\vert \psi \right\rangle _{i}\,\!$ is zero, this reduces to a pure state.) For example, the probability of finding the state $\left\vert \psi _{1}\right\rangle \,\!$ is $a_{1}\,\!$ and the probability of finding the state $\left\vert \psi _{2}\right\rangle \,\!$ is $a_{2}\,\!$.

Density Matrix for the Description of Open Quantum Systems: An Example

One example of the utility of a density matrix is the following statistical problem. Let us consider the collection of two-state systems, this will be a collection of electrons in a box and their spin is a two-state system, being either up or down when measured. If a subset of these electrons was prepared in the state ''up'' before being put in the box, and the rest ''down,'' then the description of the system of particles is given by

	$\rho =a_{u}\left\vert \uparrow \right\rangle \left\langle \uparrow \right\vert +a_{d}\left\vert \downarrow \right\rangle \left\langle \downarrow \right\vert ,$	(3.29)

where the fraction of ''up'' particles is $a_{u}\,\!$ and the fraction of ''down'' is $a_{d}\,\!$. Our system is described by this density matrix because if a particle is chosen at random from the box and measured, the state of the particle is $\left\vert \uparrow \right\rangle \,\!$ with probability $a_{u}\,\!$ and $\left\vert \downarrow \right\rangle \,\!$ with probability $a_{d}\,\!$. This is known as the statistical interpretation of the density operator.

There is another example which is more relevant for our purposes. For a certain system (again a two-state system is take as an example) if there is some probability $p\,\!$ for an error to occur, let us say our example is a unitary operator $U_{e}\,\!$, then the density matrix for the system is

	$\rho _{e}=(1-p)\left\vert \psi \right\rangle \left\langle \psi \right\vert +pU_{e}\left\vert \psi \right\rangle \left\langle \psi \right\vert U_{e}^{\dagger }.$	(3.30)

This is the same form as Eq.(\ref{eq:tsdmatex1}).

Note that in each case the probabilities associated with the density matrix $p,1-p\,\!$, and $a_{u},a_{d}\,\!$, (generally, the $a_{i}\,\!$) are classical probabilities. That is, they are associated with a classical probability distribution--the probability for error/no error and up/down. These are not probabilities associated with the superposition of the quantum state in the equation $\left\vert \psi \right\rangle =\alpha _{0}\left\vert 0\right\rangle +\alpha _{1}\left\vert 1\right\rangle \,\!$ given by the square of the moduli of the coefficients. This is an important distinction for the following reason. The state $\left\vert \psi \right\rangle \,\!$ can be taken to the state $\left\vert 0\right\rangle \,\!$ with a unitary transformation. This state is deterministic in the sense that the result $\left\vert 0\right\rangle \,\!$ will be obtained from a measurement in the computational basis since there is no probability for obtaining $\left\vert 1\right\rangle \,\!$. However, for nonzero $\left\vert \psi \right\rangle \,\!$ and a non-identity operator $U\,\!$, the matrix $\rho _{e}\,\!$ has rank two and thus can never have probability $1\,\!$ for either of the two states, $\left\vert 0\right\rangle \,\!$ or $\left\vert 1\right\rangle \,\!$. Thus, we have maximum knowledge about a pure state since there is a way to choose a measurement, perhaps after a unitary transformation, which achieves a certain result with probability one. For the mixed state density operator this is not possible. The state

	$\rho =\left({\begin{array}{cc}1/2&0\\0&1/2\end{array}}\right),$	(3.31)

for which we have the least amount of knowledge is called the maximally mixed state. The state could be either up or down with equal probability and neither is a better guess. If the two eigenvalues are not equal, then there is a better guess, or bet, as to the result of a measurement and if one eigenvalue is zero, there is a definite best guess.

To be more specific, independent of basis (unitary transformations), one always has probability greater than zero of measuring $\left\vert \uparrow \right\rangle \,\!$ and probability greater than zero of measuring $\left\vert \downarrow \right\rangle \,\!$. Thus the state described by the density matrix is a mixed state in the sense that it can be considered a statistical mixture of the two states $\left\vert \uparrow \right\rangle \,\!$ and $\left\vert \downarrow \right\rangle \,\!$. This, because classical probabilities are included separately, is significantly different from the pure state density matrix, which is a special case of all density matrices.

To see that mixtures remain after a unitary transformation on the system, note that a unitary matrix does not change the eigenvalues. This is because the eigenvalue equation is the same for a Hermitian matrix and its corresponding diagonal matrix. Let $\rho =U\rho _{d}U^{\dagger }\,\!$, then

	${\begin{aligned}\det(\rho -\lambda \mathbb {I} )&=\det(U(\rho _{d}-\lambda \mathbb {I} )U^{\dagger })\\&=\det(U)\det(\rho _{d}-\lambda \mathbb {I} )\det(U^{\dagger })\\&=\det(U)\det(\rho _{d}-\lambda \mathbb {I} )\det(U^{-1})\\&=\det(\rho _{d}-\lambda \mathbb {I} ).\end{aligned}}$	(3.32)

Open Quantum Systems

Open quantum systems are those which interact with an "environment" or have some other outside influence. Very often only closed quantum systems are considered in intoductory quantum mechanics since open systems are more difficult to treat. However, many quantum systems are, in practice, open quantum systems since they interact with some environment, or have some outside influence. When quantum systems are treated as closed, the interacting constituents are considered to be the whole system in the sense that they do not, to a good approximation, interact with anything else.

As will be shown later, errors are not restricted to unitary errors. Furthermore, errors which change the eigenvalues of a density matrix cannot arise in closed quantum systems since a closed quantum system evolved under a unitary transformation. Thus, any time an error is discovered which changes the eigenvalues, it is due to an environmental interaction. The general description of such errors, which are not restricted to unitary errors, will be postponed until Chapter {ch:noise}.

General Properties

In general, a density matrix has the following properties:

	${\begin{aligned}\rho =\rho ^{\dagger },&\;\;\;{\mbox{it is hermitian}},\\\rho \geq 0,\;&\;\;\;{\mbox{it is positive semi-definite}},\\{\mbox{Tr}}(\rho )=1,\;&\;\;\;{\mbox{it is normalized}}.\end{aligned}}\,\!$	(3.33)

If, in addition, it is a pure state, then

	$\rho ^{2}=\rho .\,\!$	(3.34)

Eq.(\ref{eq:dmatpos}) really means that the eigenvalues of the density matrix are greater than or equal to zero.

Two-State Example: Bloch Sphere

Since our interest is primarily in qubits, which are two-state systems, we return again to an example.

A very convenient representation of two state density matrices, one can written in the so-called Bloch sphere representation given the fact that the density matrix is Hermitian,

	$\rho _{2}={\frac {1}{2}}(\mathbb {I} +{\vec {n}}\cdot {\vec {\sigma }}),$	(3.35)

where, for the density matrix to be positive $|{\vec {n}}|\leq 1\,\!$, and the $\sigma _{i}\,\!$ are the Pauli matrices

	${\vec {\sigma }}=(\sigma _{x},\sigma _{y},\sigma _{z})=\left(\left({\begin{array}{cc}0&1\\1&0\end{array}}\right),\left({\begin{array}{cc}0&-i\\i&0\end{array}}\right),\left({\begin{array}{cc}1&0\\0&-1\end{array}}\right)\right).$	(3.36)

The matrix entries on the RHS of this equation are called the Pauli matrices. The eigenvalues are given by

	$\lambda _{\pm }={\frac {1\pm |{\vec {n}}|}{2}}.$	(3.37)

When $|{\vec {n}}|=1\,\!$, the state is pure, i.e., that the matrix has rank one since it has one eigenvalue one and one zero. If $|{\vec {n}}|<1\,\!$, the density matrix represents a mixed state since rank is greater than one--there are two non-zero eigenvalues. These leads to the following picture: the pure states lie on the surface of the sphere (${\vec {n}}\cdot {\vec {n}}=1\,\!$), and mixed states lie in the interior of the sphere with the maximally mixed state at the origin. This is supposedly due to Bloch. Hence the name Bloch sphere.

Using $\rho ^{2}=\rho \,\!$ the condition that ${\vec {n}}\cdot {\vec {n}}=1\,\!$ for a pure state can also be determined. The square in the Bloch sphere representation yields

	$\rho _{2}^{2}={\frac {1}{4}}\left(\mathbb {I} +2{\vec {n}}\cdot {\vec {\sigma }}+({\vec {n}}\cdot {\vec {\sigma }})^{2}\right),$	(3.38)

and using

	$\sigma _{i}\sigma _{j}=\mathbb {I} \delta _{ij}+i\epsilon _{ijk}\sigma _{k},$	(3.39)

then $\rho _{2}^{2}=\rho _{2}\,\!$ if and only if ${\vec {n}}\cdot {\vec {n}}=1\,\!$. This technique is used for higher dimensions.

Two density matrices $\rho _{1}=(1/2)(\mathbb {I} +{\vec {n}}\cdot {\vec {\sigma }})\,\!$ and $\rho _{2}=(1/2)(\mathbb {I} +{\vec {m}}\cdot {\vec {\sigma }})\,\!$, correspond to orthogonal states when

	${\mbox{Tr}}(\rho _{1}\rho _{2})={\frac {1}{4}}{\mbox{Tr}}{\big (}\mathbb {I} +({\vec {n}}\cdot {\vec {\sigma }})({\vec {m}}\cdot {\vec {\sigma }}){\big )}={\frac {1}{2}}(1+{\vec {n}}\cdot {\vec {m}})=0.$	(3.40)

This implies that

	${\vec {n}}\cdot {\vec {m}}=|{\vec {n}}||{\vec {m}}|\cos(\theta )=-1.$	(3.41)

Since the magnitudes must be one, the orthogonal states correspond to pure states on a surface of a sphere which are represented by antipodal points.

Expectation Values

The expectation value of an operator ${\mathcal {O}}\,\!$, is given by

	$\langle {\mathcal {O}}\rangle ={\mbox{Tr}}(\rho {\mathcal {O}}),$	(3.42)

and is the "average value" of the operator. For a pure state $\rho _{p}=\left\vert \psi \right\rangle \left\langle \psi \right\vert \,\!$, this reduces to

	$(\langle {\mathcal {O}}\rangle )_{p}=\left\langle \psi \right\vert {\mathcal {O}}\left\vert \psi \right\rangle .$	(3.43)

Entangled States

Quantum entanglement is the most uniquely quantum mechanical property of quantum systems. It is also believed to be responsible for all of the information-processing advantages quantum systems have over classical systems. Entangled states puzzled even Einstein and other founders of quantum theory, so it is not surprising that they have become a central part of many investigations into quantum theory and especially quantum information theory.

There are many open problems in this area of research. Some of the most basic and fundamental questions about entangled states are still unanswered. For example, given a mixed-state density matrix for a quantum system, we still do not know how to tell if the systems being described by that matrix are entangled or not. Also, with few exceptions, we do not know how to quantify the entanglement in a system of particles.

What we do understand and what we can explain, is the entanglement between systems of particles which are describable by pure quantum states. Thus this section is concerned with pure entangled states of qubits. Extentions of these ideas will be discussed in this section.

Entangled States vs. Separable States

Let us consider two particles, one will be called particle A and the other particle B which are in pure states. (It is also possible to separate a system into two subsystems and call the subsystems A and B.) If the particles are independent and have never interacted, then the state of the system of the composite system of the two particles can be written as

	$\left\vert \Phi \right\rangle =\left\vert \psi \right\rangle _{A}\otimes \left\vert \phi \right\rangle _{B},$	(3.44)

where $\left\vert \psi \right\rangle _{A}\,\!$ is the state of particle A, and $\left\vert \phi \right\rangle _{B}\,\!$ is the state of particle B. This is sometimes stated as a postulate of quantum mechanics \cite{Nielsen/Chuang:book}.

The state, Eq.~(\ref{eq:simpsep}) is not entangled, so it is called separable. More precisely it is referred to as a simply separable state. In general, a state is separable if its density matrix can be written in the form

	$\rho _{s}=\sum _{i}p_{i}\rho _{A}^{(i)}\otimes \rho _{B}^{(i)},$	(3.45)

where $\rho _{A}^{(i)}\,\!$($\rho _{B}^{(i)}\,\!$) is a valid density matrix for subsystem $A\,\!$ ($B\,\!$), and $\sum _{i}p_{i}=1,\;p_{i}\geq 0\,\!$. An entangled state is one that cannot be written in the form of Eq.~(\ref{eq:defsepstate}).

For a pure state, the situation is simpler. A pure state is entangled if and only if it cannot be written in the form of Eq.~(\ref{eq:simpsep}). In other words a pure state is entangled if it cannot be written as the product of two states of the individual subsystems.

For two particles (or systems) to become entangled, they must interact with each other. They could then be taken far apart and still share their entanglement. This entanglement cannot increase by acting on an individual particle, or even both particles separately. Only joint measurements on both, or interactions between the two, can increase entanglement. Actions on an individual particle, without involving the other, are called local actions or local operations. Local operations on individual particles can be written as

	$L=A\otimes B,$	(3.46)

so that

	$L\left\vert \Phi \right\rangle =A\left\vert \psi \right\rangle _{A}\otimes B\left\vert \phi \right\rangle _{B}.$	(3.47)

A common example, and one often used, is a local unitary transformation

	${\mathcal {U}}_{L}\left\vert \Phi \right\rangle =U_{A}\left\vert \psi \right\rangle _{A}\otimes U_{B}\left\vert \phi \right\rangle _{B}.$	(3.48)

Local unitary transformations will not change the entanglement of a system.

Notice that the density matrix for the composite system in Eq.~(\ref{eq:simpsep}) is

	${\begin{aligned}\rho _{ss}&=\left\vert \Phi \right\rangle \left\langle \Phi \right\vert \\&=(\left\vert \psi \right\rangle _{A}\otimes \left\vert \phi \right\rangle _{B})(_{A}\left\langle \psi \right\vert \otimes \sideset {_{B}}{\left\langle \phi \right\vert })\\&=\left\vert \psi \right\rangle _{A}\left\langle \psi \right\vert \otimes \left\vert \phi \right\rangle _{B}\left\langle \phi \right\vert \\&=\rho _{A}\otimes \rho _{B},\end{aligned}}$	(3.49)

where $\rho _{A}=\left\vert \psi \right\rangle _{A}\left\langle \psi \right\vert \,\!$ and $\rho _{B}=\left\vert \phi \right\rangle _{B}\left\langle \phi \right\vert \,\!$ so the density operator of a product state is a product of density operators.

Bell States

The simplest examples of entangled states are the entangled states of two two-state systems. There are four different versions of what the ``maximally entangled state of two qubits. The ``maximally will be explained below. These four different versions are called Bell states and are

	${\begin{aligned}\left\vert \psi _{+}\right\rangle &={\frac {1}{\sqrt {2}}}(\left\vert 01\right\rangle +\left\vert 10\right\rangle ),\\\left\vert \psi _{-}\right\rangle &={\frac {1}{\sqrt {2}}}(\left\vert 01\right\rangle -\left\vert 10\right\rangle ),\\\left\vert \phi _{+}\right\rangle &={\frac {1}{\sqrt {2}}}(\left\vert 00\right\rangle +\left\vert 11\right\rangle ),\\\left\vert \phi _{-}\right\rangle &={\frac {1}{\sqrt {2}}}(\left\vert 00\right\rangle -\left\vert 11\right\rangle ).\end{aligned}}$	(3.50)

These are an orthonormal ste of states and are all able to be obtained from each other by acting on one particle alone or both individual particles, i.e. acting with local unitary transformations. For example consider the local unitary transformation $\mathbb {I} \otimes \sigma _{3}\,\!$ acting on $\left\vert \phi _{-}\right\rangle \,\!$. The result is $\left\vert \phi _{+}\right\rangle \,\!$. Acting with $\sigma _{1}\otimes \mathbb {I} \,\!$ on $\left\vert \psi _{+}\right\rangle \,\!$ gives $\left\vert \phi _{+}\right\rangle \,\!$, etc.

These states certainly cannot be written in the form

$\left\vert \Phi \right\rangle =\left\vert \psi \right\rangle _{A}\otimes \left\vert \phi \right\rangle _{B}.\,\!$

If they could, then, letting $\left\vert \psi \right\rangle _{A}=\alpha _{0}\left\vert 0\right\rangle +\alpha _{1}\left\vert 1\right\rangle \,\!$ and $\left\vert \psi \right\rangle _{B}=\beta _{0}\left\vert 0\right\rangle +\beta _{1}\left\vert 1\right\rangle \,\!$, and notice that the general form is

	${\begin{aligned}\left\vert \psi \right\rangle _{A}\otimes \left\vert \phi \right\rangle _{B}&=(\alpha _{0}\left\vert 0\right\rangle +\alpha _{1}\left\vert 1\right\rangle )\otimes (\beta _{0}\left\vert 0\right\rangle +\beta _{1}\left\vert 1\right\rangle ),\\&=\alpha _{0}\beta _{0}\left\vert 00\right\rangle +\alpha _{1}\beta _{0}\left\vert 10\right\rangle +\alpha _{0}\beta _{1}\left\vert 01\right\rangle +\alpha _{1}\beta _{1}\left\vert 11\right\rangle ,\end{aligned}}$	(3.51)

so the coefficient of $\left\vert 00\right\rangle \,\!$ times the coefficient of $\left\vert 11\right\rangle \,\!$ minus the coefficient of $\left\vert 01\right\rangle \,\!$ times the coefficient of $\left\vert 10\right\rangle \,\!$ is zero. This is not true for any of the Bell states, therefore they cannot be written as a tensor product of two 1-particle states. So, for some 2-particle state,

	$\left\vert \Psi \right\rangle =\alpha _{00}\left\vert 00\right\rangle +\alpha _{01}\left\vert 01\right\rangle +\alpha _{10}\left\vert 10\right\rangle +\alpha _{11}\left\vert 11\right\rangle ,$	(3.52)

the state is separable, or unentangled, if $\alpha _{00}\alpha _{11}-\alpha _{01}\alpha _{10}=0\,\!$. Otherwise it is entangled.