Chapter 4 - Entanglement

Introduction

Quantum entanglement is the most uniquely quantum mechanical property of quantum systems. It is also believed to be responsible for the advantages that quantum computing systems have over classical computing systems. Entangled states puzzled the founders of quantum theory, including Einstein. For these reasons, 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, with few exceptions we still do not know how to tell if the systems being described by that matrix are entangled or not. Also, although there are ways with which to quantify the entanglement in a system of particles, these quantities are notoriously difficult to calculate. Here again, with few exceptions, we do not know how to calculate the amount of entanglement in a system analytically.

What we do understand and what we can explain is the entanglement between bipartite systems that are describable by pure quantum states. In this chapter, we will first talk about the problem of entanglement in quantum mechanics in general. We will then move on to its relevancy in quantum computing, starting with pure states of qubits. Extensions and generalizations will be discussed in later sections.

EPR Paradox

Before diving into entanglement in the context of quantum computing and information, it would be prudent to discuss the now-famous EPR paradox. It was first proposed in a paper by Einstein, Polosky, and Rosen in 1935. The original paper does include a fairly simple mathematical explanation of the paradox---it is, however, not really necessary as the thought experiment is quite easily understood conceptually with (mostly) words. A slightly simplified version of the experiment will be given here.

Suppose a neutral pi meson, which has no spin, is at rest. It then decays into an electron and a positron, necessarily going in opposite directions. The wave function can now be written as

	$\left\vert \Psi _{-}\right\rangle ={\frac {1}{\sqrt {2}}}(\left\vert \uparrow \downarrow \right\rangle -\left\vert \downarrow \uparrow \right\rangle ).$	(4.1)

As can be seen, we now have a system of two particles that have a correlated spin---one being up and the other being down---with an equal probability for each configuration being the outcome of a measurement. The system is said to be entangled, as a measurement on one will guarantee that the other particle is in the correlated state. In other words, it cannot be written as $\left\vert \psi _{1}\right\rangle \left\vert \psi _{2}\right\rangle$.

Now what is the significance here? It all depends on what interpretation of quantum mechanics is being used. The orthodox position says that the wave function is the complete representation of the system. When the measurement occurs, the wave function collapses, changing the system.

But, in the context of EPR, how can this be? Imagine the entangled electron-positron pair are at opposite ends of the galaxy when one of them is measured. The conservation of angular momentum says that the other particle all of the way on the other side of the galaxy must instantly be the opposite spin as the measured particle. EPR argued that this is a violation of locality, which says an effect cannot travel faster than the speed of light. If the very action of measurement on one particle is what caused the other particle to realize the opposite spin, then locality has been violated. Therefore, the measurement could not have caused the collapse of the wave function.

EPR concluded that this proves that quantum mechanics is incomplete---that the wave function is missing some information. There was no "spooky-action-at-a-distance," there must be some underlying property that is absent from the wave function. Einstein rejected the notion that a measurement caused this quasi-mystical collapse of the wave function---the particles do not care if they are being watched or not.

Bell's Theorem

The peculiarities of the EPR paradox were convincing enough to drive many to examine possible "hidden variable theories." The basic idea is that there exists a quantity, often denoted by $\lambda \,\!$, that must be included in the wave function to completely describe the system. J.S. Bell very elegantly showed in 1964 that this is not the case, using the very thought experiment (although slightly modified) that EPR proposed.

Suppose we have another pion at rest about to decay with detectors oriented equidistant and on opposite sides, ready to measure the spin of the electron and positron. Further suppose that, unlike the previous scenario, these detectors can be rotated in order to detect the spin in the direction of unit vectors ${\vec {a}}\,\!$ and ${\vec {b}}$ for the electron and positron respectively.

When the electron and positron pair strikes the detectors, a spin up ($+1\,\!$) or spin down ($-1\,\!$) is registered. The product of the results is then examined. If they are oriented parallel, where ${\vec {a}}={\vec {b}}\,\!$, then the result will be -1. If anti-parallel, the result is then +1. The averages are, obviously,

	${\begin{aligned}P({\vec {a}},{\vec {a}})&=-1,\\P({\vec {a}},-{\vec {a}})&=+1.\end{aligned}}\,\!$	(4.2)

Quantum mechanics tells us that for arbitrary vectors,

	$P({\vec {a}},{\vec {b}})=-{\vec {a}}\cdot {\vec {b}}.\,\!$	(4.3)

We can now introduce the hidden variable, $\lambda \,\!$. This can represent any possible amount of variables that complete the description of the system and allow for locality. We then define some functions, $A({\vec {a}},\lambda )$ and $B({\vec {b}},\lambda )\,\!,$ that will give the results for the measurement (either +1 or -1) for the electron and positron respectively.

The locality assumption tells us that the orientation of one detector will not affect the outcome of the measurement of the other detector; one can imagine a scenario where the orientation is chosen at a time too late for any information to be transferred slower than light. It must also be true that, when the detectors are parallel, the results must be

	$A({\vec {a}},\lambda )=-B({\vec {a}},\lambda )\,\!$	(4.4)

due to the conservation of angular momentum. Let us also define a probability density, $\rho (\lambda ),\,\!$ for the hidden variable. Since we know nothing of $\lambda \,\!$, this can be anything as long as it is non-negative and normalizable ($\int \rho (\lambda )d(\lambda )=1\,\!$). We can now look at the product of the measurements,

	$P({\vec {a}},{\vec {b}})=\int \rho (\lambda )A({\vec {a}},\lambda )B({\vec {b}},\lambda )d\lambda .\,\!$	(4.5)

We know from Eq.(4.4) that this can be rewritten:

	$P({\vec {a}},{\vec {b}})=-\int \rho (\lambda )A({\vec {a}},\lambda )A({\vec {b}},\lambda )d\lambda .\,\!$	(4.6)

Now for the clever part. Introducing another unit vector, ${\vec {c}}$, and noting that $[A({\vec {b}},\lambda )]^{2}=1,\,\!$

	${\begin{aligned}P({\vec {a}},{\vec {b}})-P({\vec {a}},{\vec {c}})&=-\int \rho (\lambda )[A({\vec {a}},\lambda )A({\vec {b}},\lambda )-A({\vec {a}},\lambda )A({\vec {c}},\lambda )]d\lambda \\&=-\int \rho (\lambda )[1-A({\vec {b}},\lambda )A({\vec {c}},\lambda )]A({\vec {a}},\lambda )A({\vec {b}},\lambda )d\lambda \end{aligned}}\,\!$	(4.7)

Recognizing some inequalities,

	${\begin{aligned}-1\leq A({\vec {a}},\lambda )A({\vec {b}},\lambda )\leq 1,\\\rho (\lambda )[1-A({\vec {b}},\lambda )A({\vec {c}},\lambda )]\geq 0,\end{aligned}}\,\!$	(4.8)

we get to a remarkable result,

	${\begin{aligned}|P({\vec {a}},{\vec {b}})-P({\vec {a}},{\vec {c}})|&\leq \int \rho (\lambda )[1-A({\vec {b}},\lambda )A({\vec {c}},\lambda )]d\lambda \\&\leq 1+P({\vec {b}},{\vec {c}}).\end{aligned}}\,\!$	(4.9)

The last form is known as the Bell inequality. This inequality is true for any local hidden variable theory.

What does this mean? Let us define ${\vec {a}}\,\!$ and ${\vec {b}}\,\!$ to be orthogonal and ${\vec {c}}\,\!$ to make a $45^{\circ }\,\!$ angle with both of them. Using quantum mechanics (Equation(4.3)),

${\begin{aligned}P({\vec {a}},{\vec {b}})=0\\P({\vec {a}},{\vec {c}})=P({\vec {b}},{\vec {c}})=-{\frac {1}{\sqrt {2}}}.\end{aligned}}\,\!$

Inserting the values into the Bell inequality (Equation (4.9)),

${\begin{aligned}\left|-{\frac {1}{\sqrt {2}}}\right|&\leq 1+\left(-{\frac {1}{\sqrt {2}}}\right)\\{\frac {1}{\sqrt {2}}}&\leq {\frac {{\sqrt {2}}-1}{\sqrt {2}}}\\1&\leq {\sqrt {2}}-1\end{aligned}}\,\!$

Since ${\sqrt {2}}-1\approx .41,\,\!$ the inequality is violated!

This means that quantum mechanics is incompatible with any local hidden variable theory. The EPR paradox had stronger implications than the authors realized; if local realism is held, then quantum mechanics is incorrect. This has been repeatedly disproved experimentally. Thus no local hidden variable theory can resolve the "spooky-action-at-a-distance."

Entangled Pure States

Let us consider two quantum systems, one called A and the other B. Let us suppose the joint state of the entire system comprised of A and B is a pure state. If the subsystems are independent and have never interacted, then the state of the composite system of the two particles can be written as

	$\left\vert \Phi \right\rangle =\left\vert \psi \right\rangle _{A}\otimes \left\vert \phi \right\rangle _{B},$	(4.10)

where $\left\vert \psi \right\rangle _{A}\,\!$ is the state of particle A and $\left\vert \phi \right\rangle _{B}\,\!$ is the state of subsystem B. This tensor product structure is sometimes stated as a postulate of quantum mechanics as in Nielsen and Chuang's book. In this case the two particles are not correlated in any way---they are said to be unentangled or separable. When a pure state cannot be written in this form it is said to be entangled.

For example, the most general form for a pure state of two qubits is given by Eq. (2.30).

$\left\vert {\psi }\right\rangle =\alpha _{00}\left\vert {00}\right\rangle +\alpha _{01}\left\vert {01}\right\rangle +\alpha _{10}\left\vert {10}\right\rangle +\alpha _{11}\left\vert {11}\right\rangle .$

Examples are given below of states that are entangled and thus cannot be written in the form of Eq. (4.10).

For two particles (or systems) to become entangled, they must first interact with each other. This entanglement cannot increase (usually) by acting on an individual even both subsystems 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. For example, local unitary operations on individual particles can be written as

	$L=U_{A}\otimes U_{B},\,\!$	(4.11)

so that

	$L\left\vert \Phi \right\rangle =U_{A}\left\vert \psi \right\rangle _{A}\otimes U_{B}\left\vert \phi \right\rangle _{B}.$	(4.12)

Local unitary transformations will not change the entanglement of a system. Furthermore, local measurements or local measurements combined with unitary transformations cannot, on average, increase the entanglement between subsystems.

For later use, it is relevant to note that the density matrix for the composite system in Eq.(4.10) is

	${\begin{aligned}\rho _{ss}&=\left\vert \Phi \right\rangle \left\langle \Phi \right\vert \\&=(\left\vert \psi \right\rangle _{A}\otimes \left\vert \phi \right\rangle _{B})(_{A}\left\langle \psi \right\vert \otimes _{B}\!\!\left\langle \phi \right\vert )\\&=\left\vert \psi \right\rangle _{A}\left\langle \psi \right\vert \otimes \left\vert \phi \right\rangle _{B}\left\langle \phi \right\vert \\&=\rho _{A}\otimes \rho _{B},\end{aligned}}\,\!$	(4.13)

where $\rho _{A}=\left\vert \psi \right\rangle _{A}\left\langle \psi \right\vert \,\!$ and $\rho _{B}=\left\vert \phi \right\rangle _{B}\left\langle \phi \right\vert \,\!$ ---the density operator of a product state is the 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 is known as the "maximally entangled state" of two qubits. The "maximally" will be explained below. These four different versions are called Bell states and are

	${\begin{aligned}\left\vert \psi _{+}\right\rangle &={\frac {1}{\sqrt {2}}}(\left\vert 01\right\rangle +\left\vert 10\right\rangle ),\\\left\vert \psi _{-}\right\rangle &={\frac {1}{\sqrt {2}}}(\left\vert 01\right\rangle -\left\vert 10\right\rangle ),\\\left\vert \phi _{+}\right\rangle &={\frac {1}{\sqrt {2}}}(\left\vert 00\right\rangle +\left\vert 11\right\rangle ),\\\left\vert \phi _{-}\right\rangle &={\frac {1}{\sqrt {2}}}(\left\vert 00\right\rangle -\left\vert 11\right\rangle ).\end{aligned}}$	(4.14)

This is an orthonormal set of states that are all able to be obtained from each other by acting on one particle alone or both individual particles with unitary transformations (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 $\left\vert \phi _{+}\right\rangle \,\!$. Similarly, $\sigma _{1}\otimes \mathbb {I} \,\!$ acting on $\left\vert \psi _{+}\right\rangle \,\!$ yields $\left\vert \phi _{+}\right\rangle \,\!$, and so on.

These states certainly cannot be written in the form

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

What if they could? Let $\left\vert \psi \right\rangle _{A}=\alpha _{0}\left\vert 0\right\rangle +\alpha _{1}\left\vert 1\right\rangle \,\!$ and $\left\vert \phi \right\rangle _{B}=\beta _{0}\left\vert 0\right\rangle +\beta _{1}\left\vert 1\right\rangle \,\!$. Notice that the general form is

	${\begin{aligned}\left\vert \psi \right\rangle _{A}\otimes \left\vert \phi \right\rangle _{B}&=(\alpha _{0}\left\vert 0\right\rangle +\alpha _{1}\left\vert 1\right\rangle )\otimes (\beta _{0}\left\vert 0\right\rangle +\beta _{1}\left\vert 1\right\rangle ),\\&=\alpha _{0}\beta _{0}\left\vert 00\right\rangle +\alpha _{1}\beta _{0}\left\vert 10\right\rangle +\alpha _{0}\beta _{1}\left\vert 01\right\rangle +\alpha _{1}\beta _{1}\left\vert 11\right\rangle ,\end{aligned}}$	(4.15)

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---thus, they cannot be written as a tensor product of two 1-particle states. So, for any 2-particle state,

	$\left\vert \Psi \right\rangle =\alpha _{00}\left\vert 00\right\rangle +\alpha _{01}\left\vert 01\right\rangle +\alpha _{10}\left\vert 10\right\rangle +\alpha _{11}\left\vert 11\right\rangle ,$	(4.16)

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

Entangled Mixed States

The state in Eq.(4.1) 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

	$\rho _{s}=\sum _{i}p_{i}\rho _{A}^{(i)}\otimes \rho _{B}^{(i)},$	(4.17)

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.(4.17).

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.(4.10). In other words a pure state is entangled if it cannot be written as the product of two states of the individual subsystems.

Reduced Density Operators and the Partial Trace

The Bell states are maximally entangled states. To understand this, one may consider the fact that these states are pure states but information about the individual particles in the system is lacking. In this section, a more precise meaning of this statement is given.

Let us first consider useful tool, the partial trace. The partial trace is the trace over one of the subsystems (particle states) of a composite system. Let us suppose that the density matrix for a composite system is given by

	$\rho _{ss}=\rho _{A}\otimes \rho _{B}.\,\!$	(4.18)

The partial trace is the trace over one of the subsystems. For example, the trace over subsystem $B\,\!$ is

	${\mbox{Tr}}_{B}(\rho _{ss})=\rho _{A}{\mbox{Tr}}(\rho _{B})=\rho _{A},\,\!$	(4.19)

since ${\mbox{Tr}}(A\otimes B)={\mbox{Tr}}(A){\mbox{Tr}}(B)\,\!$ and the trace of a density matrix is one. The matrix $\rho _{A}\,\!$ is called the reduced density operator, or reduced density matrix. However, this is a special case. The density matrix for a composite system of two (or more) subsystems cannot be written in this form except in very special circumstances---when the two subsystems have never interacted and there are no correlations between them.

For the cases where the two subsystems are entangled, there are at least two ways to calculate the partial trace. One is to write the matrix form of the state in terms of a sum of the tensor products of Pauli matrices. (See Appendix E - Density Operator: Extensions, Sec. Two-State Example: Bloch Sphere.) The other is to realize that the trace can be calculated by summing the projections onto the diagonal elements of the subsystem over which you are tracing. For example, for a general $2\times 2\,\!$ density matrix $\rho _{2}\,\!$, the trace is

	${\mbox{Tr}}(\rho _{2})={\mbox{Tr}}(\left\vert 0\right\rangle \left\langle 0\right\vert \rho _{2}+\left\vert 1\right\rangle \left\langle 1\right\vert \rho _{2})=\left\langle 0\right\vert \rho _{2}\left\vert 0\right\rangle +\left\langle 1\right\vert \rho _{2}\left\vert 1\right\rangle .$	(4.20)

For the general case, let us consider a density matrix for a bipartite system, $\rho \,\!$. Let the subsystem $A\,\!$ have Greek letters as indices and let the subsystem $B\,\!$ have Latin indices. Then

	${\begin{aligned}\rho &=\sum _{\alpha ,\beta ,i,j}\rho _{\alpha \beta ,ij}\left\vert \alpha i\right\rangle \left\langle \beta j\right\vert \\&=\sum _{\alpha ,\beta ,i,j}\rho _{\alpha \beta ,ij}\left\vert \alpha \right\rangle _{A}\otimes \left\vert i\right\rangle _{B\;A}\!\left\langle \beta \right\vert \otimes {}_{B}\!\left\langle j\right\vert \\&=\sum _{\alpha ,\beta ,i,j}\rho _{\alpha \beta ,ij}\left\vert \alpha \right\rangle _{A\;A}\!\left\langle \beta \right|\otimes \left|i\right\rangle _{B\;B}\!\left\langle j\right|.\end{aligned}}$	(4.21)

To calculate the reduced density matrix of subsystem $A\,\!$ by tracing over subsystem $B\,\!$, the trace over $B\,\!$ is taken by computing

	${\begin{aligned}\rho _{A}&={\mbox{Tr}}_{B}(\rho )\\&=\sum _{k}{}_{B}\!\left\langle k\right\vert \rho \left\vert k\right\rangle _{B}\\&=\sum _{k}\sum _{\alpha ,\beta ,i,j}{}_{B}\!\left\langle k\right\vert \left(\rho _{\alpha \beta ,ij}\left\vert \alpha \right\rangle _{A\;A}\!\left\langle \beta \right\vert \otimes \left\vert i\right\rangle _{B\;B}\!\left\langle j\right\vert \right)\left\vert k\right\rangle _{B}\\&=\sum _{k}\sum _{\alpha ,\beta ,i,j}\rho _{\alpha \beta ,ij}\left\vert \alpha \right\rangle _{A\;A}\!\left\langle \beta \right\vert \delta _{ki}\delta _{jk}\\&=\sum _{\alpha ,\beta ,k}\rho _{\alpha \beta ,kk}\left\vert \alpha \right\rangle _{A\;A}\!\left\langle \beta \right\vert \\\end{aligned}}$	(4.22)

For the partial trace of a $4\times 4\,\!$ density matrix $\rho \,\!$ over the subsystem $B\,\!$,

	${\mbox{Tr}}_{B}(\rho )={}_{B}\left\langle 0\right\vert \rho \left\vert 0\right\rangle _{B}+{}_{B}\left\langle 1\right\vert \rho \left\vert 1\right\rangle _{B},$	(4.23)

which leaves the part of the matrix corresponding to $A\,\!$ alone and projects the $B\,\!$ part onto the two diagonal elements and then adds those. Now let us calculate the partial trace of a Bell state, for example $\left\vert \psi _{+}\right\rangle \,\!$. Assuming the first state is the $A\,\!$ state and the second $B\,\!$, we see

	${\begin{aligned}{\mbox{Tr}}_{B}(\left\vert \psi _{+}\right\rangle \left\langle \psi _{+}\right\vert )&={}_{B}\left\langle 0\right\vert (\left\vert \psi _{+}\right\rangle \left\langle \psi _{+}\right\vert )\left\vert 0\right\rangle _{B}+{}_{B}\left\langle 1\right\vert (\left\vert \psi _{+}\right\rangle \left\langle \psi _{+}\right\vert )\left\vert 1\right\rangle _{B},\\&={}_{B}\left\langle 0\right\vert \left({\frac {1}{2}}(\left\vert 01\right\rangle +\left\vert 10\right\rangle )(\left\langle 01\right\vert +\left\langle 10\right\vert )\right)\left\vert 0\right\rangle _{B}\\&\;\;+{}_{B}\left\langle 1\right\vert \left({\frac {1}{2}}\left(\left\vert 01\right\rangle +\left\vert 10\right\rangle \right)(\left\langle 01\right\vert +\left\langle 10\right\vert )\right)\left\vert 1\right\rangle _{B}\\&={\frac {1}{2}}(\left\vert 1\right\rangle _{A}\left\langle 1\right\vert )+{\frac {1}{2}}(\left\vert 0\right\rangle _{A}\left\langle 0\right\vert ),\end{aligned}}$	(4.24)

which can be rewritten simply as

	${\mbox{Tr}}_{B}(\left\vert \psi _{+}\right\rangle \left\langle \psi _{+}\right\vert )=\rho _{A}=\left({\begin{array}{cc}1/2&0\\0&1/2\end{array}}\right).$	(4.25)

This is quite an interesting and significant find. The density matrix for the whole system of two qubits is in a pure state indicating maximal knowledge. However, the reduced density matrix, representing our knowledge of one of the individual particles, is completely (or maximally) mixed, indicating minimal knowledge. This means that the two particles or subsystems taken together are in a definite, pure state; yet when taken separately, they contain as little information as possible. This fact indicates entanglement.

It is important to note that for a pure state, the trace over subsystem $A\,\!$ produces the same result as the trace over subsystem $B\,\!$. In other words, for a pure state $\rho _{p}\,\!$,

	${\mbox{Tr}}_{A}(\rho _{p})={\mbox{Tr}}_{B}(\rho _{p}).\,\!$	(4.26)

How Entangled Is It?

The previous discussion showed that there is a definite notion of maximal entanglement for pure states. From the determinant condition, there is a method for identifying unentangled pure states. A question may arise: how entangled is it if it is not separable nor maximally entangled? There are now many ways of defining measures of entanglement which will be explored in a later section and an appendix. Here, a common way of measuring the entanglement for pure states is given which is based on the partial trace of a pure state.

Let us consider the extreme cases. First, we found that if the states are Bell states, then the partial trace of the bipartite system of two qubits will yield a density operator for the subsystem that is maximally entangled. Due to the fact that the trace of the density operator is always one, it should be clear that the partial trace of a separable pure state gives a pure state density operator. Notice that the purity of the partial density operator provides a candidate for a measure of entanglement. Before we discuss this further, let us note the following important result, called the Schmidt decomposition.

Schmidt Decomposition

For any pure state density operator of a bipartite system, say with constituents A and B, it can be taken as

	$\left\vert \Psi \right\rangle _{AB}=\sum _{ij}a_{ij}^{\prime }\left\vert \psi _{i}^{\prime }\right\rangle _{A}\otimes \left\vert \phi _{j}^{\prime }\right\rangle _{B}.\,\!$	(4.27)

This can also be put in the form

	$\left\vert \Psi \right\rangle _{AB}=\sum _{i}a_{i}\left\vert \psi _{i}\right\rangle _{A}\otimes \left\vert \phi _{i}\right\rangle _{B}.\,\!$	(4.28)

In this case it is clear that the reduced density operator is the same for each subsystem and thus the purity of the reduced density operator can be used as a measure of entanglement.

Extensions and Open Problems

Continue to Chapter 5 - Quantum Information: Basic Principles and Simple Examples