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 to progress our understanding of how quantum mechanics from a textbook is related to quantum computing. The connection is clear, but the story seems incomplete from a physicists perspective. 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, it often not covered in quantum mechanics classes, both undergraduate and graduate.

It is also worth emphasizing that this chapter is primarily aimed at physicists and for those others who are interested in the background physics. However, 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 (it is), then the matrix

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

is unitary. (If this is unclear, 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 a Hermitian matrix.

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

	$\left\vert \Psi (t)\right\rangle =U\left\vert \Psi (0)\right\rangle .\,\!$	(3.6)

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

Exponentiating a Matrix

Aside: a note about the exponentiation of a matrix.

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 for demonstration purposes, it is written out.

The Taylor expansion of an exponential is the following:

	$e^{x}=\sum _{n=0}^{\infty }{\frac {x^{n}}{n!}}\,\!$	(3.7)

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

	$e^{ix}=\cos x+i\sin x.\,\!$	(3.8)

End Aside

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 outer product of two vectors. (See Appendix C, Sec. C.2.4.)

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

	${\begin{aligned}{\frac {\partial \rho }{\partial t}}&=\left({\frac {\partial \left\vert \psi \right\rangle }{\partial t}}\right)\left\langle \psi \right\vert +\left\vert \psi \right\rangle \left({\frac {\partial \left\langle \psi \right\vert }{\partial t}}\right)\\&=(-iH)\rho +\rho (iH)=-i[H,\rho ],\end{aligned}}$	(3.10)

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

	$\rho (t)=U\rho (0)U^{\dagger }.$	(3.11)

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

	$\left\vert 0\right\rangle =\left({\begin{array}{c}1\\0\end{array}}\right),\;\;\;{\mbox{and}}\;\;\;\left\vert 1\right\rangle =\left({\begin{array}{c}0\\1\end{array}}\right).$	(3.12)

A superposition of these two states is

	$\left\vert \psi \right\rangle =\alpha _{0}\left\vert 0\right\rangle +\alpha _{1}\left\vert 1\right\rangle =\left({\begin{array}{c}\alpha _{0}\\\alpha _{1}\end{array}}\right),$	(3.13)

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

	$\rho _{p}=\left\vert \psi \right\rangle \left\langle \psi \right\vert =\left({\begin{array}{cc}|\alpha _{0}|^{2}&\alpha _{0}\alpha _{1}^{*}\\\alpha _{0}^{*}\alpha _{1}&|\alpha _{1}|^{2}\end{array}}\right).$	(3.14)

Note that the superposition in Eq.(3.13) can be obtained from any pure state by a unitary transformation. Here, the trace of the density matrix is an important quantity; it is

	${\mbox{Tr}}(\rho _{p})=|\alpha _{0}|^{2}+|\alpha _{1}|^{2}=1.\,\!$	(3.15)

Notice also that the determinant of this matrix is zero, indicating that it has a zero eigenvalue:

	$\det(\rho _{p})=|\alpha _{0}|^{2}|\alpha _{1}|^{2}-\alpha _{0}\alpha _{1}^{*}\alpha _{0}^{*}\alpha _{1}=0.$	(3.16)

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

	${\begin{aligned}\det(U(\left\vert 0\right\rangle \left\langle 0\right\vert )U^{\dagger })&=\det(U(\left\vert 0\right\rangle \left\langle 0\right\vert )U^{-1})\\&=\det(U)\det(\left\vert 0\right\rangle \left\langle 0\right\vert ){\frac {1}{\det(U)}}\\&=\det(\left\vert 0\right\rangle \left\langle 0\right\vert )=0.\end{aligned}}$	(3.17)

This is a characteristic of a pure state and for two-state systems, it is a necessary and sufficient condition for the density operator to represent a pure state of the system.

Measurements Revisited

If the state of a quantum system is described by

	$\left\vert \psi \right\rangle =\alpha _{0}\left\vert 0\right\rangle +\alpha _{1}\left\vert 1\right\rangle ,$	(3.18)

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

	$\left\vert \psi \right\rangle =U\left\vert 0\right\rangle .$	(3.19)

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,

	${\begin{aligned}\left\vert \psi \right\rangle &=U\left\vert 0\right\rangle \\&=\exp(-i{\vec {n}}\cdot {\vec {\sigma }}\theta )\left\vert 0\right\rangle \\&=(\mathbb {I} \cos(\theta )-i{\vec {n}}\cdot {\vec {\sigma }}\sin(\theta ))\left\vert 0\right\rangle ,\end{aligned}}$	(3.20)

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

	${\begin{aligned}\exp(-i{\vec {n}}\cdot {\vec {\sigma }}\theta )&=\left({\begin{array}{cc}1&0\\0&1\end{array}}\right)\cos(\theta )\\&\;\;\;+(-i)\left[n_{1}\left({\begin{array}{cc}0&1\\1&0\end{array}}\right)+n_{2}\left({\begin{array}{cc}0&-i\\i&0\end{array}}\right)+n_{3}\left({\begin{array}{cc}1&0\\0&-1\end{array}}\right)\right]\sin(\theta )\\&=\left({\begin{array}{cc}\cos(\theta )-in_{3}\sin(\theta )&(-in_{1}-n_{2})\sin(\theta )\\(-in_{1}+n_{2})\sin(\theta )&\cos(\theta )+in_{3}\sin(\theta )\end{array}}\right).\end{aligned}}$	(3.21)

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

	${\begin{aligned}U\left\vert 0\right\rangle &=\left({\begin{array}{cc}\cos(\theta )-in_{3}\sin(\theta )&(-in_{1}-n_{2})\sin(\theta )\\(-in_{1}+n_{2})\sin(\theta )&\cos(\theta )+in_{3}\sin(\theta )\end{array}}\right)\left({\begin{array}{c}1\\0\end{array}}\right)\\&=\left({\begin{array}{c}\cos(\theta )-in_{3}\sin(\theta )\\(-in_{1}+n_{2})\sin(\theta )\end{array}}\right).\end{aligned}}$	(3.22)

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

	${\begin{aligned}|\left\langle 0\right\vert U\left\vert 0\right\rangle |^{2}&=|\cos(\theta )-in_{3}\sin(\theta )|^{2}\\&=\cos ^{2}(\theta )+n_{3}^{2}\sin ^{2}(\theta ),\end{aligned}}$	(3.23)

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

	${\begin{aligned}|\left\langle 1\right\vert U\left\vert 1\right\rangle |^{2}&=|(-in_{1}+n_{2})\sin(\theta )|^{2}\\&=(n_{1}^{2}+n_{2}^{2})\sin ^{2}(\theta ).\end{aligned}}$	(3.24)

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

	$\rho _{D}=\sum _{i}a_{i}\rho _{i},\,\!$	(3.25)

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 should go into understanding the density matrix, the rest of this section is devoted to the physical interpretation and properties of this description of a quantum system. The first description presented is called the ensemble interpretation of the density matrix. This is perhaps the easiest to understand. Another set of physical systems which are described by density matrices will be given elsewhere.

General Properties

In general, a density matrix has the following properties:

	${\begin{aligned}\rho =\rho ^{\dagger },&\;\;\;{\mbox{it is hermitian}},\\\rho \geq 0,\;&\;\;\;{\mbox{it is positive semi-definite}},\\{\mbox{Tr}}(\rho )=1,\;&\;\;\;{\mbox{it is normalized}}.\end{aligned}}\,\!$	(3.26)

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

	$\rho ^{2}=\rho .\,\!$	(3.27)

The second property in Eq.(3.28) really means that the eigenvalues of the density matrix are greater than or equal to zero.

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.) In this mixture, 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 $a_{2}\,\!$.

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.(3.31).

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)

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,

	$\rho _{2}={\frac {1}{2}}(\mathbb {I} +{\vec {n}}\cdot {\vec {\sigma }}),$	(3.33)

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

	${\vec {\sigma }}=(\sigma _{x},\sigma _{y},\sigma _{z})=\left(\left({\begin{array}{cc}0&1\\1&0\end{array}}\right),\left({\begin{array}{cc}0&-i\\i&0\end{array}}\right),\left({\begin{array}{cc}1&0\\0&-1\end{array}}\right)\right).$	(3.34)

The matrix entries on the RHS of this equation are the The Pauli matrices discussed above. It is not difficult to convince yourself that any Hermitian matrix can be written as a real linear combination of the three Pauli matrices and the identity. The eigenvalues are given by

	$\lambda _{\pm }={\frac {1\pm |{\vec {n}}|}{2}}.$	(3.35)

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

	$\rho _{2}^{2}={\frac {1}{4}}\left(\mathbb {I} +2{\vec {n}}\cdot {\vec {\sigma }}+({\vec {n}}\cdot {\vec {\sigma }})^{2}\right),$	(3.36)

and using

	$\sigma _{i}\sigma _{j}=\mathbb {I} \delta _{ij}+i\epsilon _{ijk}\sigma _{k},$	(3.37)

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

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

	${\begin{aligned}{\mbox{Tr}}(\rho _{1}\rho _{2})&={\frac {1}{4}}{\mbox{Tr}}{\big (}\mathbb {I} +({\vec {n}}\cdot {\vec {\sigma }})({\vec {m}}\cdot {\vec {\sigma }}){\big )}\\&={\frac {1}{2}}(1+{\vec {n}}\cdot {\vec {m}})=0.\end{aligned}}$	(3.38)

This implies that

	${\vec {n}}\cdot {\vec {m}}=|{\vec {n}}||{\vec {m}}|\cos(\theta )=-1.$	(3.39)

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.

Rotations of Bloch Vectors

As shown above, the solution to the Schrodinger equation for the density operator is (see Eq.(3.11))

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

In general an open system will evolve according to

$\rho =U\rho _{0}U^{\dagger },\,\!$

whether or not the time dependence is explicitly taken into account. When the density operator is represented using the Bloch vector, the vector is rotated by the unitary transformation. This is seen through an explicit calculation.

There are two ways to see this. One is to simply act with the matrices in the Euler angle parameterization in Section C.5.1 one each of the Pauli matrices to show that indeed,

	$U{\vec {m}}\cdot {\vec {\sigma }}U^{\dagger }=\sum _{ij}m_{i}R_{ij}\sigma _{j}.\,\!$	(3.40)

This is easily seen to be a standard rotation matrix. (See for example http://en.wikipedia.org/wiki/Rotation_matrix.)

Another way to do this is to take

$\rho _{0}={\frac {1}{2}}(\mathbb {I} +{\vec {m}}\cdot {\vec {\sigma }}),\,\!$

as in Eq.(3.33). (Recall ${\vec {m}}\cdot {\vec {\sigma }}=\sum \!{}_{i}\;m_{i}\sigma _{i}\,\!$.) Now act on $\rho$ with $U\,\!$ as given in Section C.5.1 by the so-called adjoint action $\rho =U\rho _{0}U^{\dagger }\,\!$,

	$\rho =[\mathbb {I} \cos(\theta /2)-i{\vec {n}}\cdot {\vec {\sigma }}\sin(\theta /2)]{\frac {1}{2}}(\mathbb {I} +{\vec {m}}\cdot {\vec {\sigma }})[\mathbb {I} \cos(\theta /2)+i{\vec {n}}\cdot {\vec {\sigma }}\sin(\theta /2)].\,\!$	(3.41)

To do this calculation explicitly, it helps (but is not necessary) to use the following identity,

	$\sum _{i}\epsilon _{ijk}\epsilon _{ilm}=\delta _{jl}\delta _{km}-\delta _{jm}\delta _{kl}.\,\!$	(3.42)

Then, if one only considers the non-trivial part of the density operator, ${\vec {m}}\cdot {\vec {\sigma }}\,\!$, the result is

	$e^{-i{\vec {n}}\cdot {\vec {\sigma }}\theta /2}{\vec {m}}\cdot {\vec {\sigma }}e^{i{\vec {n}}\cdot {\vec {\sigma }}\theta /2}={\vec {m}}\cdot {\vec {\sigma }}\cos(\theta )+({\vec {n}}\cdot {\vec {m}})({\vec {n}}\cdot {\vec {\sigma }})(1-\cos(\theta ))+({\vec {n}}\times {\vec {m}})\cdot {\vec {\sigma }}\sin(\theta ),\,\!$	(3.43)

or

	${\begin{aligned}e^{-i{\vec {n}}\cdot {\vec {\sigma }}\theta /2}{\vec {m}}\cdot {\vec {\sigma }}e^{i{\vec {n}}\cdot {\vec {\sigma }}\theta /2}&={\frac {1}{2}}{\vec {m}}\cdot {\vec {\sigma }}\cos(\theta )+{\frac {1}{2}}({\vec {n}}\cdot {\vec {m}})({\vec {n}}\cdot {\vec {\sigma }})\cos(\theta )+({\vec {n}}\cdot {\vec {m}})({\vec {n}}\cdot {\vec {\sigma }})\\&\;\;\;\;+({\vec {n}}\times {\vec {m}})\cdot {\vec {\sigma }}\sin(\theta )+{\frac {1}{2}}[({\vec {n}}\times {\vec {m}})\times {\vec {n}}]\cdot {\vec {\sigma }}\cos(\theta )\end{aligned}}\,\!$	(3.44)

where

	$({\vec {n}}\times {\vec {m}})\cdot {\vec {\sigma }}=\sum _{ijk}\epsilon _{ijk}n_{i}m_{j}\sigma _{k}.\,\!$	(3.45)

Therefore, the result of the action of $U\,\!$ is to produce, from ${\vec {m}}\,\!$, the vector

	${\vec {m}}^{\prime }={\vec {m}}\cos(\theta )+({\vec {n}}\cdot {\vec {m}}){\vec {n}}(1-\cos(\theta ))+({\vec {n}}\times {\vec {m}})\sin(\theta ).\,\!$	(3.46)

This equation can be interpreted as follows. We consider three components of the vector, the part along the axis of rotation and the two parts in the plane perpendicular to the axis of rotation. The part of the vector along the axis of rotation $({\vec {m}}\cdot {\vec {n}}){\vec {n}}\,\!$ does not change. The parts perpendicular to $({\vec {m}}\cdot {\vec {n}}){\vec {n}}\,\!$ change just like a vector rotated in a plane, but these parts are rotated in the plane perpendicular to the rotation axis and sitting at the end of the vector $({\vec {m}}\cdot {\vec {n}}){\vec {n}}\,\!$. It takes a bit of geometry and vector algebra to show this is the case.

Expectation Values

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

	$\langle {\mathcal {O}}\rangle ={\mbox{Tr}}(\rho {\mathcal {O}}),$	(3.47)

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

	$(\langle {\mathcal {O}}\rangle )_{p}=\left\langle \psi \right\vert {\mathcal {O}}\left\vert \psi \right\rangle .$	(3.48)

Continue to Chapter 4 - Entanglement