View source for 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 is 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 any 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:
{{Equation|<math>
H \left\vert \Psi\right\rangle = i\hbar\frac{\partial}{\partial t}\left\vert \Psi\right\rangle,
\,\!</math>|3.1}}<br />
where <math>H\,\!</math> is the Hamiltonian, <!-- \index{Hamiltonian} -->
<math>\hbar\,\!</math> is Planck's constant 
<!-- \index{Planck's constant} --> 
(divided by <math>2\pi\,\!</math>), and <math>t\,\!</math> is time.  The Hamiltonian contains what
is known about the system's evolution.  
Most of the time in these notes, we let <math>\hbar = 1\,\!</math>.  


This equation is (formally) solved by taking the time derivative to be
an ordinary derivative (we assume no explicit time dependence for
<math>H \,\!</math>), so 
{{Equation|<math>
H \left\vert \Psi\right\rangle = i\frac{d \left\vert \Psi\right\rangle}{dt}.
\,\!</math>|3.2}}<br />
This means that 
{{Equation|<math>
-iHdt =  \frac{d \left\vert \Psi\right\rangle}{\left\vert \Psi\right\rangle},
\,\!</math>|3.3}}<br />
so
{{Equation|<math>\begin{align}
 \ln \left\vert \Psi\right\rangle &= -iHt + C, \\
\Rightarrow\left\vert \Psi(t)\right\rangle &= e^{-iHt}\left\vert \Psi(0)\right\rangle.  
\end{align}
\,\!</math>|3.4}}
Now if <math>H\,\!</math> is Hermitian (it is), then the matrix 
{{Equation|<math>
U =  e^{-iHt}
\,\!</math>|3.5}}<br />
is unitary.  <!-- \index{unitary matrix}-->
(If this is unclear, see [[Appendix C - Vectors and Linear Algebra]], in particular the section entitled [[Appendix C - Vectors and Linear Algebra#Unitary Matrices|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
{{Equation|<math>
\left\vert \Psi(t)\right\rangle = U\left\vert \Psi(0)\right\rangle.  
\,\!</math>|3.6}}<br />
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====

<div id="expmatrix"> ''Aside: a note about the exponentiation of a matrix.''</div>
  
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:
{{Equation|<math>
e^x = \sum_{n=0}^\infty \frac{x^n}{n!}
\,\!</math>|3.7}}<br />
and this can be used to exponentiate a matrix by letting the matrix
replace <math>x\,\!</math> in the equation.  This can also be used to prove that 
{{Equation|<math>
e^{ix}=\cos x +i\sin x.  
\,\!</math>|3.8}}<br />
''End Aside''

===Density Matrix for Pure States===

Now let us consider the object (a ''density matrix, or 
density operator, of rank one'') <!-- \index{density matrix}\index{density
matrix!pure state} -->
{{Equation|<math>
\rho = \left\vert\psi\right\rangle \left\langle \psi\right\vert,
</math>|3.9}}<br />
which is just the outer product of two vectors.  (See Appendix C.2.4, [[Appendix C - Vectors and Linear Algebra#Outer Product|Outer Product]].) 

Since <math>\left\vert \psi\right\rangle = \left\vert \psi(t)\right\rangle\,\!</math>, <math>\rho=\rho(t)\,\!</math> is also true.  If we
differentiate this with respect to <math>t\,\!</math>, we discover<!-- \index{Schr\"odinger Equation!
  for density matrix} -->
{{Equation|<math>\begin{align}
\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{align}
</math>|3.10}} 
This is merely the Schrodinger equation for a density matrix with the solution
{{Equation|<math>
\rho(t) = U\rho(0)U^\dagger.
</math>|3.11}}<br />
This follows from <math>\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\,\!</math>. 

Consider our two-state system, 
{{Equation|<math>
\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).
</math>|3.12}}<br />
Recall that the arbitrary superposition of these states is shown by 
{{Equation|<math>
\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),
</math>|3.13}}<br />
where <math>\alpha_0\,\!</math> and <math>\alpha_1\,\!</math> are complex numbers such that 
<math>|\alpha_0|^2 + |\alpha_1|^2 = 1\,\!</math>.  The corresponding 
''pure state'' (i.e. rank one) ''density matrix'' is given by 
{{Equation|<math>
\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).
</math>|3.14}}<br />
Note that the superposition in Eq.[[#eq3.13|(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
{{Equation|<math>
\mbox{Tr}(\rho_p) = |\alpha_0|^2 + |\alpha_1|^2 = 1.  
\,\!</math>|3.15}}<br />
Notice also that the determinant of this matrix is zero, indicating that it has a zero eigenvalue:
{{Equation|<math>
\det(\rho_p) = |\alpha_0|^2|\alpha_1|^2 - \alpha_0 \alpha_1^*\alpha_0^*
\alpha_1 = 0.
</math>|3.16}}<br />
To see this another way, note that the density operator of rank one can be written as <math>U(\left\vert 0\right\rangle\left\langle0\right\vert)U^\dagger\,\!</math>, so that the determinant is 
{{Equation|<math>\begin{align}
\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\vert0\right\rangle\left\langle 0\right\vert)\frac{1}{\det(U)} \\
                            &= \det(\left\vert 0\right\rangle \left\langle 0\right\vert) = 0.
\end{align}
</math>|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
{{Equation|<math> 
\left\vert \psi\right\rangle = \alpha_0\left\vert 0\right\rangle + \alpha_1\left\vert 1\right\rangle, 
</math>|3.18}}<br />
then the probability of finding it in the state <math>\left\vert 0\right\rangle\,\!</math> when measured in
the computational basis is <math>|\alpha_0|^2\,\!</math>.  However, this is a
particular superposition that could be written as 
{{Equation|<math>
\left\vert \psi\right\rangle = U \left\vert 0\right\rangle.  
</math>|3.19}}<br />
In the section entitled [[#Schrodinger's Equation|Schrodinger's Equation]] it was shown that this matrix <math>U\,\!</math> results
from the exponentiation of a Hermitian matrix. Recall from the section entitled [[Chapter 2 - Qubits and Collections of Qubits#The Pauli Matrices|The Pauli Matrices]] that any <math>2\times 2\,\!</math>
Hermitian matrix can be written in terms of the Pauli matrices.  To make this explicit using standard conventions, 
{{Equation|<math>\begin{align}
\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{align}
</math>|3.20}}
where <math>\vec{n}\,\!</math> is a unit vector, <math>|\vec{n}|=1\,\!</math> and <math>\vec{n}\cdot\vec{\sigma} =
n_1\sigma_1+n_2\sigma_2+n_3\sigma_3\,\!</math>. (For a proof of this, see [[Appendix C - Vectors and Linear Algebra#Transformations of a Qubit|Section C.5.1]].)  One can write this matrix out explicitly, 
{{Equation|<math>\begin{align}
 \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{align}
</math>|3.21}}
Notice this is a  ''special unitary matrix.''  (See [[Appendix C - Vectors and Linear Algebra]], in particular the subsection [[Appendix C - Vectors and Linear Algebra#Unitary Matrices|Unitary Matrices]].)

To see that any state <math>\left\vert \psi\right\rangle\,\!</math> for arbitrary coefficients
<math>\alpha_0\,\!</math>, <math>\alpha_1\,\!</math> can be obtained by choosing <math>\vec{n}\,\!</math> and <math>\theta\,\!</math>
appropriately, the state <math>\left\vert 0\right\rangle\,\!</math> can be chosen as a starting point.  
Then, 
{{Equation|<math>\begin{align}
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{align}
</math>|3.22}}
For example, choosing <math>\theta=0\,\!</math> gives the original state; choosing
<math>\vec{n} = (0,1,0)\,\!</math> and <math>\theta = \pi/2\,\!</math> gives <math>\left\vert 1\right\rangle\,\!</math>; and choosing
<math>\vec{n} = (0,1,0)\,\!</math> and <math>\theta = \pi/4\,\!</math> gives an equal superposition.  
In general, when the system is in the state  <math>\left\vert \psi\right\rangle = U\left\vert 0\right\rangle\,\!</math>,
the probability of finding the state <math>\left\vert 0 \right\rangle \,\!</math> when a measurement is made in the computational basis is given by 
{{Equation|<math>\begin{align}
|\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{align}
</math>|3.23}}
and the probability of finding the state <math>\left\vert 1\right\rangle\,\!</math> if the initial state is <math>\left\vert 0\right\rangle\,\!</math>, is 
{{Equation|<math>\begin{align}
|\left\langle 1\right\vert U\left\vert 0\right\rangle|^2 &= |(-in_1+n_2)\sin(\theta)|^2   \\
                    &= (n_1^2+n_2^2)\sin^2(\theta).   
\end{align}
</math>|3.24}}
Notice that the probabilities add up to one if <math>\vec{n}\,\!</math> is a unit vector.  

What this shows is that there is a transformation that takes the state
<math>\left\vert 0\right\rangle\,\!</math>, which has probability <math>1\,\!</math> of being in the state <math>\left\vert 0\right\rangle\,\!</math> and
probability <math>0\,\!</math> of being in the state <math>\left\vert 1\right\rangle\,\!</math>, and transforms it
(using a <nowiki>"rotation''</nowiki>) 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 that(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 <math>D\,\!</math> dimensions, a ''mixed state density operator'' 
(or density matrix, see [[Appendix E - Density Operator: Extensions|Appendix E]]) <!-- \index{density matrix} \index{density operator} --> is a matrix that is used to
describe a more general state of a quantum system.  This can be written as 
{{Equation|<math>
\rho_D = \sum_i a_i \rho_i
\,\! </math>|3.25}}<br />
where <math>a_i\geq 0\,\!</math>, <math>{\sum}_i a_i=1\,\!</math>, and the <math>\rho_i\,\!</math> are pure states.  There is also a generalization of the Bloch sphere which is described in [[Appendix E - Density Operator: Extensions|Appendix E]].  

Mixed state <!-- \index{density matrix!mixed state} --> density operators 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 operator. 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 and is commonly used in quantum statistical mechanics.  Another set of physical systems that are described by density operators will be given elsewhere.  

====General Properties====

In general, a density operator has the following properties:
{{Equation|<math>\begin{align}
\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{align}
\,\!</math>|3.26}}
If, in addition, it is a pure state density operator, then 
{{Equation|<math>
\rho^2 = \rho.
\,\!</math>|3.27}}<br />
The second property in Eq.[[#eq3.26|(3.26)]] 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 operator for a two-state system is represented by a rank two matrix, <math>\rho_m\,\!</math>, which can be written as 
{{Equation|<math>
\rho_m = \left[a_1\rho_1 + a_2\rho_2\right],
</math>|3.28}}<br />
where <math>\rho_1 = \left\vert\psi_1\right\rangle\left\langle \psi_1\right\vert\,\!</math>, <math>\rho_2 = \left \vert \psi_2\right\rangle\left\langle \psi_2\right\vert \,\!</math> 
and <math>a_1 + a_2=1\,\!</math>.  The <math>a_i\,\!</math> are probabilities and must positive and sum to one.
(Note, if <math>\left\vert \psi_1\right\rangle =\left\vert \psi_2\right\rangle\,\!</math>, or if one <math>a_i\,\!</math> or one 
<math>\left\vert \psi_i\right\rangle\,\!</math> is zero, then this reduces to a pure state.)  In this mixture, 
the probability of finding the state <math>\left\vert \psi_1\right\rangle\,\!</math> is <math>a_1\,\!</math>
and the probability of finding the state <math>\left\vert \psi_2\right\rangle\,\!</math> is <math>a_2\,\!</math>.

====Description of Open Quantum Systems: An Example====

One example of the utility of a density matrix is the following
statistical problem.  Let us consider a collection of electrons in a box, where their
spin is a two-state system being either up or down when measured.  If
a subset of these electrons is prepared in the state <nowiki>''up''</nowiki> before
being put in the box and the rest <nowiki>''down,''</nowiki> then the description of
the system of particles is given by 
{{Equation|<math>
\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,
</math>|3.29}}<br />
where the fraction of  <nowiki>''up''</nowiki> particles is <math>a_u\,\!</math> and the fraction of <nowiki>''down''</nowiki> is <math>a_d\,\!</math>.  Our system is described by this density matrix---if a particle is chosen at random from the box and measured, the state of the particle is <math>\left\vert \uparrow\right\rangle\,\!</math> with probability <math>a_u\,\!</math>
and <math>\left\vert \downarrow\right\rangle\,\!</math> with probability <math>a_d\,\!</math>.  This is known as the "statistical
interpretation" of the density operator. 


There is another example that is more relevant for our purposes. Let us consider another two-state system.
If there is some probability <math>p\,\!</math> for an error to occur, let us say it is a unitary operator <math>U_e\,\!</math>, then the density matrix for the
system is 
{{Equation|<math>
\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.  
</math>|3.30}}<br />
This is the same form as Eq.[[#eq3.29|(3.29)]].  

Note that in each 
case the probabilities associated with the density matrix <math>p,1-p\,\!</math>, and
<math>a_u,a_d\,\!</math>, (generally, <math>a_i\,\!</math>) are classical probabilities;
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 <math>\left\vert \psi\right\rangle = \alpha_0 \left\vert 0\right\rangle + \alpha_1\left\vert 1\right\rangle\,\!</math>
given by the square of the moduli of the coefficients.  This is an
important distinction!  The state
<math>\left\vert \psi\right\rangle\,\!</math> can be taken to the state <math>\left\vert 0\right\rangle\,\!</math> with a unitary
transformation.  If this is done, then a measurement of the state can give the 
result <math>\left\vert 0\right\rangle\,\!</math> with no probability for obtaining
<math>\left\vert 1\right\rangle\,\!</math>.  However, for nonzero <math>\left\vert \psi\right\rangle\,\!</math> and a non-identity
operator <math>U\,\!</math>, the matrix <math>\rho_e\,\!</math> has rank two and thus can never have
probability <math>1\,\!</math> for a measurement result of either of the two states, <math>\left\vert 0\right\rangle\,\!</math> or <math>\left\vert 1\right\rangle\,\!</math>.  
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 <math>1\,\!</math>.
For the mixed state density operator this is not possible.  

For example, the state 
{{Equation|<math>
\rho = \left(\begin{array}{cc} 
              1/2 & 0 \\ 
                0 & 1/2 \end{array}\right),
</math>|3.31}}<br />
is the one for which we have the least amount of knowledge, since either result of a two-state measurement has equally probable outcomes.  This is called the
maximally mixed state. <!-- \index{maximally mixed state! two qubits}-->  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. If one
eigenvalue is zero, as it is in the case of a pures state, then there is a definite best guess.  

Note, however, that these outcomes are basis dependent.  Changing the basis by a unitary transformation from the state <math>\left\vert \psi\right\rangle\,\!</math> with <math>\alpha_0 = \alpha_1 = 1/\sqrt{2}\,\!</math>, changes the probability of obtaining the 0/1 result from 1/2 and 1/2 to 1 and 0.  There is no possibility of a basis change on the maximally mixed state which will give probabilities 1 and 0.  The two probabilities are always 1/2 and 1/2.  

Put another way, independent of basis (unitary transformations),
one always has a probability greater than zero of measuring
<math>\left\vert \uparrow\right\rangle\,\!</math> and probability greater than zero of measuring
<math>\left\vert \downarrow\right\rangle\,\!</math> when the state is represented by a mixed state density operator. Thus the  state described by the density matrix is
a ''mixed state'' <!-- \index{mixed state density matrix}--> in the sense
that it can be considered a statistical mixture of the  two states
<math>\left\vert \uparrow\right\rangle\,\!</math> and <math>\left\vert \downarrow\right\rangle\,\!</math>.  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 or its corresponding diagonal matrix.  Let <math>\rho =
U\rho_d U^\dagger\,\!</math>.  It can now be seen, 
{{Equation|<math>\begin{align}
\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{align}
</math>|3.32}}

====Two-State Example: Bloch Sphere====

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

A very convenient representation of two state density matrices, one
that can written in the so-called Bloch sphere <!-- \index{Bloch sphere}-->
representation given the fact that the density matrix is Hermitian, 
{{Equation|<math>
\rho_2 = \frac{1}{2}(\mathbb{I} + \vec{n}\cdot\vec{\sigma}),
</math>|3.33}}<br />
where, for the density matrix to be positive <math>|\vec{n}| \leq 1\,\!</math>, and the
<math>\sigma_i\,\!</math> are the Pauli matrices <!-- \index{Pauli matrices}-->
{{Equation|<math>
\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).
</math>|3.34}}<br />
The matrix entries on the RHS of this equation are the [[Chapter 2 - Qubits and Collections of Qubits#The Pauli Matrices|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
{{Equation|<math>
\lambda_\pm = \frac{1\pm|\vec{n}|}{2}.
</math>|3.35}}<br />
When <math>|\vec{n}| = 1\,\!</math>, the state is pure, i.e., that the matrix 
has rank one since it has one eigenvalue one and one zero.  If <math>|\vec{n}|
< 1\,\!</math>, 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 (<math>\vec{n}\cdot \vec{n} =1\,\!</math>), 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 <math>\rho^2 =\rho\,\!</math> the condition that <math>\vec{n}\cdot\vec{n} =1\,\!</math> for a pure
state can also be determined.  The square in the Bloch sphere
representation yields
{{Equation|<math>
\rho_2^2 = \frac{1}{4}\left(\mathbb{I} + 2\vec{n}\cdot\vec{\sigma} + (\vec{n}\cdot\vec{\sigma})^2\right), 
</math>|3.36}}<br />
and using 
{{Equation|<math>
\sigma_i \sigma_j = \mathbb{I}\delta_{ij} + i\epsilon_{ijk}\sigma_k,
</math>|3.37}}<br />
then <math>\rho_2^2 =\rho_2\,\!</math> if and only if <math>\vec{n}\cdot\vec{n} =1\,\!</math>.  This technique is
used for higher dimensions.  See [[Appendix E - Density Operator: Extensions|Appendix E]].


Two density matrices <math>\rho_1=(1/2)(\mathbb{I} +\vec{n}\cdot\vec{\sigma})\,\!</math> and 
<math>\rho_2=(1/2)(\mathbb{I} +\vec{m}\cdot\vec{\sigma})\,\!</math>, correspond to orthogonal 
states when 
{{Equation|<math>\begin{align}
\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{align}</math>|3.38}}<br />
This implies that 
{{Equation|<math>
\vec{n}\cdot\vec{m} = |\vec{n}||\vec{m}|\cos(\theta) = -1.
</math>|3.39}}<br />
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 [[#eq3.11|Eq.(3.11)]])
<center><math>
\rho(t) = U(t)\rho(0) U^\dagger(t).
\,\!</math></center>
In general an open system will evolve according to 
<center><math>
\rho = U \rho_0 U^\dagger,
\,\!</math></center>
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 [[Appendix C - Vectors and Linear Algebra#Transformations of a Qubit|Section C.5.1]] one each of the Pauli matrices to show that indeed,
{{Equation|<math>
U\vec{m}\cdot\vec{\sigma} U^\dagger = \sum_{ij} m_i R_{ij} \sigma_j.
\,\!</math>|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 
<center><math>
\rho_0 = \frac{1}{2}(\mathbb{I} + \vec{m}\cdot\vec{\sigma}),
\,\!</math></center>
as in [[#eq3.33|Eq.(3.33)]].  (Recall <math>\vec{m}\cdot\vec{\sigma} = \sum\!{}_i \; m_i \sigma_i\,\!</math>.)  Now act on <math>\rho</math> with <math>U\,\!</math> as given in [[Appendix C - Vectors and Linear Algebra#Transformations of a Qubit|Section C.5.1]] by the so-called adjoint action <math>\rho = U \rho_0 U^\dagger\,\!</math>, 
{{Equation|<math>
\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)].
\,\!</math>|3.41}}
To do this calculation explicitly, it helps (but is not necessary) to use the following identity,
{{Equation|<math>
\sum_i \epsilon_{ijk}\epsilon_{ilm} = \delta_{jl}\delta_{km} - \delta_{jm}\delta_{kl}.
\,\!</math>|3.42}}
Then, if one only considers the non-trivial part of the density operator, <math>\vec{m}\cdot\vec{\sigma}\,\!</math>, the result is 
{{Equation|<math>
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),
\,\!</math>|3.43}}
where
{{Equation|<math>
(\vec{n}\times \vec{m})\cdot\vec{\sigma} = \sum_{ijk} \epsilon_{ijk} n_im_j\sigma_k.
\,\!</math>|3.44}}
Therefore, the result of the action of <math>U\,\!</math> is to produce, from  <math>\vec{m}\,\!</math>, the vector
{{Equation|<math>
\vec{m}^\prime = (\vec{n}\cdot\vec{m})\vec{n} + (\vec{m} - (\vec{n}\cdot\vec{m})\vec{n})\cos(\theta) + (\vec{n}\times \vec{m})\sin(\theta).
\,\!</math>|3.45}}
Using [[#eq3.42|Eq.(3.42)]], this is 
{{Equation|<math>
\vec{m}^\prime = (\vec{n}\cdot\vec{m})\vec{n} + [(\vec{n} \times \vec{m}) \times \vec{n}]\cos(\theta) + (\vec{n} \times \vec{m})\sin(\theta).
\,\!</math>|3.46}}
This equation can be interpreted as follows.  We consider three possible 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 <math>(\vec{m}\cdot\vec{n})\vec{n}\,\!</math> does not change.  The parts perpendicular to <math>(\vec{m}\cdot\vec{n})\vec{n}\,\!</math> 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 <math>(\vec{m}\cdot\vec{n})\vec{n}\,\!</math>.  One such component comes from the cross product of <math>\vec{n}\,\!</math> and <math>\vec{m}\,\!</math> and the other from <math>(\vec{n}\times\vec{m})\times\vec{n}\,\!</math>.

===Expectation Values===


The expectation value <!-- \index{expectation value}--> 
of an operator <math>\mathcal{O}\,\!</math>, is given by 
{{Equation|<math>
\langle \mathcal{O} \rangle = \mbox{Tr}(\rho \mathcal{O}),
</math>|3.47}}<br />
and is the "average value" of the operator.  For a pure state 
<math>\rho_p = \left\vert\psi\right\rangle\left\langle\psi\right\vert\,\!</math>, this reduces to 
{{Equation|<math>
(\langle \mathcal{O} \rangle)_p = \left\langle\psi\right\vert \mathcal{O}\left\vert \psi\right\rangle.  
</math>|3.48}}<br />


[[Chapter 4 - Entanglement#Introduction|Continue to '''Chapter 4 - Entanglement''']]