Appendix C - Vectors and Linear Algebra

Introduction

This appendix introduces some aspects of linear algebra and complex algebra which will be helpful for the course. In addition, Dirac notation is introduced and explained.

Vectors

Here we review some facts about real vectors before discussing the representation and complex analogues used in quantum mechanics.

Real Vectors

The simple definition of a vector - an object which has magnitude and direction - is helpful to keep in mind even when dealing with complex and/or abstract vectors as we will here. In three dimensional space, a vector is often written as

${\vec {v}}=v_{x}{\hat {x}}+v_{y}{\hat {y}}+v_{z}{\hat {z}},$

where the hat (${\hat {\cdot }}\,\!$) denotes a unit vector. The components $v_{i}\,\!$, $i=x,y,z\,\!$ are just numbers. These unit vectors are also called basis vectors because any vector (in real three-dimensional space) can be written in terms of them. In some sense they are basic components of any vector. Other basis vectors could be used though. Some other common choices are those the are used for spherical coordinates and those used for cylindrical coordinates. When dealing with more abstract and/or complex vectors, it is often helpful to ask what one would do for an ordinary three-dimensional vector. For example, properties of unit vectors, dot products, etc. in three-dimensions are similar to the analogous constructions in more dimensions.

The inner product, or dot product for two real three-dimensional vectors

${\vec {v}}=v_{x}{\hat {x}}+v_{y}{\hat {y}}+v_{z}{\hat {z}},\;\;{\vec {w}}=w_{x}{\hat {x}}+w_{y}{\hat {y}}+w_{z}{\hat {z}},$

can be computed as follows

${\vec {v}}\cdot {\vec {w}}=v_{x}w_{x}+v_{y}w_{y}+v_{z}w_{z}.$

For the inner product of ${\vec {v}}\,\!$ with itself, we get the square of the magnitude of ${\vec {v}}\,\!$ denoted $|{\vec {v}}|^{2}\,\!$,

$|{\vec {v}}|^{2}={\vec {v}}\cdot {\vec {v}}=v_{x}v_{x}+v_{y}v_{y}+v_{z}v_{z}=v_{x}^{2}+v_{y}^{2}+v_{z}^{2}.$

If we want a unit vector in the direction of ${\vec {v}}\,\!$, then we divide by its magnitude to get a unit vector,

${\hat {v}}={\frac {\vec {v}}{|{\vec {v}}|}}.$

Now, of course, ${\hat {v}}\cdot {\hat {v}}=1\,\!$ as can easily be checked.

There are several ways to represent a vector. The ones we will use most often are column and row vector notations. So, for example, we could write the vector above as

${\vec {v}}=\left({\begin{array}{c}v_{x}\\v_{y}\\v_{z}\end{array}}\right).$

In this case, our unit vectors are represented by the following

${\hat {x}}=\left({\begin{array}{c}1\\0\\0\end{array}}\right),\;\;{\hat {y}}=\left({\begin{array}{c}0\\1\\0\end{array}}\right),\;\;{\hat {z}}=\left({\begin{array}{c}0\\0\\1\end{array}}\right).\,\!$

We next turn to the subject of complex vectors along with the relevant notation. We will see how to compute the inner product later, but some other definitions will be required.

Complex Vectors

For complex vectors in quantum mechanics, Dirac notation is most often used. This notation uses a $\left\vert \cdot \right\rangle \,\!$, called a ket, for a vector, so our vector ${\vec {v}}\,\!$ would be

$\left\vert v\right\rangle =\left({\begin{array}{c}v_{x}\\v_{y}\\v_{z}\end{array}}\right).$

For qubits, i.e. two-state quantum systems, will often be written as complex vectors

	${\begin{aligned}\left\vert \psi \right\rangle &=\left({\begin{array}{c}\alpha \\\beta \end{array}}\right)\\&=\alpha \left\vert 0\right\rangle +\beta \left\vert 1\right\rangle ,\end{aligned}}$	(C.1)

where

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

are the basis vectors. The two numbers $\alpha \,\!$ and $\beta \,\!$ are complex numbers, so the vector is said to be a complex vector.

Linear Algebra: Matrices

There are many aspects of linear algebra that are quite useful in quantum mechanics. We will discuss (briefly) several of these here, but first we will provide some definitions and properties which will be useful as well as fixing notation. Some familiarity with matrices will be assumed, but many basic defintions are also included.

Let us denote some $m\times n\,\!$ matrix by $A\,\!$. The set of all $m\times n\,\!$ matrices with real entries is $M(n\times m,\mathbb {R} )\,\!$. Such matrices are said to be real since they have all real entries. Similarly, the set of $m\times n\,\!$ complex matrices is $M(m\times n,\mathbb {C} )\,\!$. For the set of set of square $n\times n\,\!$ complex matrices, we simply write $M(n,\mathbb {C} )\,\!$.

We will also refer to the set of matrix elements, $a_{ij}\,\!$ where the first index ($i\,\!$ in this case) labels the row and the second $(j)\,\!$ labels the column. Thus the element $a_{23}\,\!$ is the element in the second row and third column. A comma is inserted if there is some ambiguity. For example, in a large matrix the element in the 2nd row and 12th column is written as $a_{2,12}\,\!$ to distinguish between the 21st row and 2nd column.

Complex Conjugate

The complex conjugate of a matrix is the matrix with each element replaced by its complex conjugate. In other words, to take the complex conjugate of a matrix, one takes the complex conjugate of each entry in the matrix. We denote the complex conjugate with a ``star, e.g. $A^{*}\,\!$. For example,

	${\begin{aligned}A^{*}&=&\left({\begin{array}{ccc}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{array}}\right)^{*}\\&=&\left({\begin{array}{ccc}a_{11}^{*}&a_{12}^{*}&a_{13}^{*}\\a_{21}^{*}&a_{22}^{*}&a_{23}^{*}\\a_{31}^{*}&a_{32}^{*}&a_{33}^{*}\end{array}}\right).\end{aligned}}\,\!$	(C.2)

(Notice that the notation for a matrix is a capital letter, whereas the entries are numbers, so they are represented by lower case letters.)

Transpose

The transpose of a matrix is the same set of elements but the first row becomes the first column, the second row becomes the second column, etc. Thus the rows and columns are interchanged. For example, for a square $3\times 3\,\!$ matrix, the transpose is given by

	${\begin{aligned}A^{T}&=&\left({\begin{array}{ccc}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{array}}\right)^{T}\\&=&\left({\begin{array}{ccc}a_{11}&a_{21}&a_{31}\\a_{12}&a_{22}&a_{32}\\a_{13}&a_{23}&a_{33}\end{array}}\right).\end{aligned}}\,\!$	(C.3)

Hermitian Conjugate

The complex conjugate and transpose is called the Hermitian conjugate, or simply the dagger of a matrix. It is called the dagger, because the symbol used to denote it is a dagger ($\dagger \,\!$), viz.

	$(A^{T})^{*}=(A^{*})^{T}\equiv A^{\dagger }.\,\!$	(C.4)

For our $3\times 3\,\!$ example,

$A^{\dagger }=\left({\begin{array}{ccc}a_{11}^{*}&a_{21}^{*}&a_{31}^{*}\\a_{12}^{*}&a_{22}^{*}&a_{32}^{*}\\a_{13}^{*}&a_{23}^{*}&a_{33}^{*}\end{array}}\right).\,\!$

If a matrix is its own Hermitian conjugate, i.e. $A^{\dagger }=A\,\!$, the we call it a Hermitian matrix. (Clearly this is only possible for square matrices.) Hermitian matrices are very important in quantum mechanics since their eigenvalues are real. (See Sec.(Eigenvalues and Eigenvectors).)

Index Notation

Very often we write the product of two matrices $A\,\!$ and $B\,\!$ simply as $AB\,\!$ and let $C=AB\,\!$. However, it is also quite useful to write this in component form. In this case, if these are $n\times n\,\!$ matrices

$c_{ik}=\sum _{j=1}^{n}a_{ij}b_{jk}.\,\!$

This says that the element in the $i^{\mbox{th}}\,\!$ row and $j^{\mbox{th}}\,\!$ column of the matrix $C\,\!$ is the sum $\sum _{1}^{n}a_{ij}b_{jk}\,\!$. The transpose of $C\,\!$ has elements

$c_{ki}=\sum _{j=1}^{n}a_{kj}b_{ji}.\,\!$

Now if we were to transpose $A\,\!$ and $B\,\!$ as well, this would read

$c_{ki}=\sum _{j=1}^{n}(a_{jk})^{T}(b_{ij})^{T}=\sum _{1}^{n}b_{ij}^{T}a_{jk}^{T}.\,\!$

This gives us a way of seeing the general rule that

$C^{T}=B^{T}A^{T}.\,\!$

It follows that

$C^{\dagger }=B^{\dagger }A^{\dagger }.\,\!$

The Trace

The trace of a matrix is the sum of the diagonal elements and is denoted ${\mbox{Tr}}\,\!$. So for example, the trace of an $n\times n\,\!$ matrix $A\,\!$ is

${\mbox{Tr}}(A)=\sum _{i=1}^{n}a_{ii}\,\!$

Some useful properties of the trace are the following:

1. ${\mbox{Tr}}(AB)={\mbox{Tr}}(BA)\,\!$
2. ${\mbox{Tr}}(A+B)={\mbox{Tr}}(A)+{\mbox{Tr}}(B)\,\!$

Using the first of these results,

${\mbox{Tr}}(UAU^{-1})={\mbox{Tr}}(U^{-1}UA)={\mbox{Tr}}(A).\,\!$

This relation is used so often that we state it here explicitly.

The Determinant

For a square matrix, the determinant is quite a useful thing. For example, an $n\times n\,\!$ matrix is invertible if and only if its determinant is not zero. So let us define the determinant and give some properties and examples.

The determinant of a $2\times 2\,\!$ matrix

	$N=\left({\begin{array}{cc}a&b\\c&d\end{array}}\right),\,\!$	(C.5)

is given by

	$\det(N)=ad-bc.\,\!$	(C.6)

Higher-order determinants can be written in terms of smaller ones in the standard way.

The determinant of a matrix $A\,\!$ can be also be written in terms of its components as

	$\det(A)=\sum _{i,j,k,l,...}\epsilon _{ijkl...}a_{1i}a_{2j}a_{3k}a_{4l}...,\,\!$	(C.7)

where the symbol

	$\epsilon _{ijkl...}={\begin{cases}+1,\;{\mbox{if }}\;ijkl...=1234...({\mbox{in order, or any even number of permutations}}),\\-1,\;{\mbox{if }}\;ijkl...=2134...({\mbox{or any odd number of permutations}}),\\\;\;\;0,\;{\mbox{otherwise}},\;({\mbox{meaning any index is repeated}}).\end{cases}}\,\!$	(C.8)

Let us consider the example of the $3\times 3\,\!$ matrix $A\,\!$ given above. The determinant can be calculated by

$\det(A)=\sum _{i,j,k}\epsilon _{ijk}a_{1i}a_{2j}a_{3k},\,\!$

where, explicitly,

	$\epsilon _{ijk}={\begin{cases}+1,\;{\mbox{if }}\;ijk=123,231,\;{\mbox{or}}\;312,({\mbox{These are even permutations of }}123),\\-1,\;{\mbox{if }}\;ijk=213,132,\;{\mbox{or}}\;321({\mbox{These are odd permuations of }}123),\\\;\;\;0,\;{\mbox{otherwise}},\;({\mbox{meaning any index is repeated}}).\end{cases}}\,\!$	(C.9)

so that

	${\begin{aligned}\det(A)&=&\epsilon _{123}a_{11}a_{22}a_{33}+\epsilon _{132}a_{11}a_{23}a_{32}+\epsilon _{231}a_{12}a_{23}a_{31}\\&&+\epsilon _{213}a_{12}a_{21}a_{33}+\epsilon _{312}a_{13}a_{21}a_{32}+\epsilon _{213}a_{13}a_{21}a_{32}.\end{aligned}}\,\!$	(C.10)

Now given the values of $\epsilon _{ijk}\,\!$ in Eq.~(\ref{eq:3depsilon}), this is

$\det(A)=a_{11}a_{22}a_{33}-a_{11}a_{23}a_{32}+a_{12}a_{23}a_{31}-a_{12}a_{21}a_{33}+a_{13}a_{21}a_{32}-a_{13}a_{21}a_{32}.\,\!$

The determinant has several properties which are useful to know. A few are listed here.

1. The determinant of the transpose of a matrix is the same as the determinant of the matrix itself: $\det(A)=\det(A^{T}).\,\!$
2. The determinant of a product is the product of determinants: $\det(AB)=\det(A)\det(B).\,\!$

From this last property, another specific property can be derived. Suppose we take the determinant of the product of a matrix and its inverse we find

$\det(UU^{-1})=\det(U)\det(U^{-1})=\det(\mathbb {I} )=1,\,\!$

since the determinant of the identity is one. This implies that

$\det(U^{-1})={\frac {1}{\det(U)}}.\,\!$

The Inverse of a Matrix

The inverse of a square matrix $A\,\!$ is another matrix, denoted $A^{-1}\,\!$ such that

$AA^{-1}=A^{-1}A=\mathbb {I} ,\,\!$

where $\mathbb {I} \,\!$ is the identity matrix consisting of zeroes everywhere except the diagonal which has ones. For example the $3\times 3\,\!$ identity matrix is

$\mathbb {I} _{3}=\left({\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}}\right).\,\!$

It is important to note that a matrix is invertible if and only if its determinant is nonzero. Thus one only needsd to calculate the determinant to see if a matrix has an inverse or not.

Unitary Matrices

A unitary matrix $U\,\!$ is one whose inverse is also its Hermitian conjugate, $U^{\dagger }=U^{-1}\,\!$ so that

$U^{\dagger }U=UU^{\dagger }=\mathbb {I} .\,\!$

If the unitary matrix also has determinant one, it is said to be a special unitary matrix. The set of $n\times n\,\!$ unitary matrices is denoted U$(n)\,\!$ and the set of special unitary matrices is denoted SU$(n)\,\!$.

Unitary matrices are particularly important in quantum mechanics because they describe the evolution, or change, of quantum states. They are able to do this because unitary matrices have the property that rows and columns, viewed as vectors, are orthonormal. (To see this, an example is provided below.) This means that when they act on a basis vector of the form (one 1, in say the $j$th spot, and zeroes everywhere else)

	$\left\vert j\right\rangle =\left({\begin{array}{c}0\\0\\\vdots \\1\\\vdots \\0\end{array}}\right),\,\!$	(C.11)

the result is a normalized complex vector. Acting on a set of orthonormal vectors of the form given in Eq.(C.11), will produce another orthonormal set.

Let us consider the example of a $2\times 2\,\!$ unitary matrix,

	$U=\left({\begin{array}{cc}a&b\\c&d\end{array}}\right).\,\!$	(C.12)

The inverse of this matrix is the Hermitian conjugate, so the inverse is given by

	$U^{-1}=U^{\dagger }=\left({\begin{array}{cc}a^{*}&c^{*}\\b^{*}&d^{*}\end{array}}\right),\,\!$	(C.13)

provided that the matrix $U\,\!$ satisfies the constraints

	${\begin{aligned}|a|^{2}+|b|^{2}=1,\;&\;ac^{*}+bd^{*}=0\\ca^{*}+db^{*}=0,\;&\;|c|^{2}+|d|^{2}=1,\end{aligned}}\,\!$	(C.14)

and

	${\begin{aligned}|a|^{2}+|c|^{2}=1,\;&\;ba^{*}+dc^{*}=0\\b^{*}a+d^{*}c=0,\;&\;|b|^{2}+|d|^{2}=1.\end{aligned}}\,\!$	(C.15)

Looking at each row as a vector, the constraints in Eq.(C.14) are the orthonormality conditions for the vectors forming the rows. Similarly, the constraints in Eq.(C.15) are the orthonormality conditions for the vectors forming the columns when viewed as vectors.

More Dirac Notation

Now that we have a definition of Hermitian conjugate, we consider the case for a $1\times n\,\!$ matrix, i.e. a vector. In Dirac notation, we had

$\left\vert \psi \right\rangle =\left({\begin{array}{c}\alpha \\\beta \end{array}}\right),\,\!$

So the Hermitian conjugate comes up so often that we use the following notation for vectors,

$\left\langle \psi \right\vert =(\left\vert \psi \right\rangle )^{\dagger }=\left({\begin{array}{c}\alpha \\\beta \end{array}}\right)^{\dagger }=\left(\alpha ^{*},\;\beta ^{*}\right).\,\!$

This is a row vector. Let us consider a second complex vector

$\left\vert \phi \right\rangle =\left({\begin{array}{c}\gamma \\\delta \end{array}}\right).\,\!$

The inner product between $\left\vert \psi \right\rangle \,\!$ and $\left\vert \phi \right\rangle \,\!$ is computed as follows:

	${\begin{aligned}\left\vert \phi \right\rangle \left\langle \psi \right\vert &\equiv (\left\vert \phi \right\rangle )^{\dagger }\left\vert \psi \right\rangle \\&=(\gamma ^{*},\delta ^{*})\left({\begin{array}{c}\alpha \\\beta \end{array}}\right)\\&=\gamma ^{*}\alpha +\delta ^{*}\beta .\end{aligned}}$	(C16)

If these two vectors are orthogonal, then their inner product is zero, $\left\langle \phi \mid \psi \right\rangle =0\,\!$. The inner product of $\left\vert \psi \right\rangle \,\!$ with itself is

$\left\langle \psi \mid \psi \right\rangle =|\alpha |^{2}+|\beta |^{2}.\,\!$

If this vector is normalized then $\left\langle \psi \mid \psi \right\rangle =1\,\!$.

More generally, we will consider vectors in $N\,\!$ dimensions. In this case we write the vector in terms of a set of basis vectors $\{\left\vert i\right\rangle \}\,\!$, where $i=0,1,2,...N-1\,\!$. This is an ordered set of vectors which are just labeled by integers. If the set is orthogonal, then

$\left\langle i\mid j\right\rangle =0,\;\;{\mbox{for all }}i\neq j,\,\!$

and if they are normalized, then

$\left\langle i\mid i\right\rangle =1,\;\;{\mbox{for all }}i.\,\!$

If both of these are true, i.e., the entire set is orthonormal, we can write,

$\left\langle i\mid j\right\rangle =\delta _{ij},\,\!$

where the symbol $\delta _{ij}\,\!$ is called the Kronecker delta and is defined by

	$\delta _{ij}={\begin{cases}1,&{\mbox{if }}i=j,\\0,&{\mbox{if }}i\neq j.\end{cases}}$	(C.17)

Now consider $(N+1)\,\!$-dimensional vectors by letting two such vectors be expressed in the same basis as

$\left\vert \Psi \right\rangle =\sum _{i=0}^{N}\alpha _{i}\left\vert i\right\rangle ,\,\!$

and

$\left\vert \Phi \right\rangle =\sum _{j=0}^{N}\beta _{j}\left\vert j\right\rangle .\,\!$

Then the inner product is

	${\begin{aligned}\left\langle \Psi \mid \Phi \right\rangle &=\left(\sum _{i=0}^{N}\alpha _{i}\left\vert i\right\rangle \right)^{\dagger }\left(\sum _{j=0}^{N}\beta _{j}\left\vert j\right\rangle \right)\\&=\sum _{ij}\alpha _{i}^{*}\beta _{j}\left\langle i\mid j\right\rangle \\&=\sum _{ij}\alpha _{i}^{*}\beta _{j}\delta _{ij}\\&=\sum _{i}\alpha _{i}^{*}\beta _{i},\end{aligned}}$	(C.18)

where we have used the fact that the delta function is zero unless $i=j\,\!$ to get the last equality. For the inner product of a vector with itself, we get

$\left\langle \Psi \mid \Psi \right\rangle =\sum _{i}\alpha _{i}^{*}\alpha _{i}=\sum _{i}|\alpha _{i}|^{2}.\,\!$

This immediately gives us a very important property of the inner product. It tells us that in general,

$\left\langle \Phi \mid \Phi \right\rangle \geq 0,\;\;{\mbox{and}}\;\;\left\langle \Phi \mid \Phi \right\rangle =0\Leftrightarrow \left\vert \Phi \right\rangle =0.\,\!$

(Just in case you don't know, the symbol $\Leftrightarrow \,\!$ means "if and only iff" sometimes written as "iff.")

We could also expand a vector in a different basis. Let us suppose that the set $\{\left\vert e_{k}\right\rangle \}\,\!$ is an orthonormal basis ($\left\langle e_{k}\mid e_{l}\right\rangle =\delta _{kl}\,\!$) which is different from the one considered earlier. We could expand our vector $\left\vert \Psi \right\rangle \,\!$ in terms of our new basis by expanding our new basis in terms of our old basis. Let us first expand the $\left\vert e_{k}\right\rangle \,\!$ in terms of the $\left\vert j\right\rangle \,\!$:

	$\left\vert e_{k}\right\rangle =\sum _{j}\left\vert j\right\rangle \left\langle j\mid e_{k}\right\rangle ,$	(C.19)

so that

	${\begin{aligned}\left\vert \Psi \right\rangle &=\sum _{j}\alpha _{j}\left\vert j\right\rangle \\&=\sum _{j}\sum _{k}\alpha _{j}\left\vert e_{k}\right\rangle \left\langle e_{k}\mid j\right\rangle \\&=\sum _{k}\alpha _{k}^{\prime }\left\vert e_{k}\right\rangle ,\end{aligned}}$	(C.20)

where

	$\alpha _{k}^{\prime }=\sum _{j}\alpha _{j}\left\langle e_{k}\mid j\right\rangle .$	(C.21)

Notice that the insertion of $\sum _{k}\left\vert e_{k}\right\rangle \left\langle e_{k}\right\vert \,\!$ didn't do anything to our original vector. It is the same vector, just in a different basis. Therefore, this is effectively the identity operator

$\mathbb {I} =\sum _{k}\left\vert e_{k}\right\rangle \left\langle e_{k}\right\vert .\,\!$

This is an important and quite useful relation. Now, to interpret Eq.(C.19), we can draw a close analogy with three-dimensional real vectors. The inner product $\left\vert e_{k}\right\rangle \left\langle j\right\vert \,\!$ can be interpreted as the projection of one vector onto another. This provides the part of $\left\vert j\right\rangle \,\!$ along $\left\vert e_{k}\right\rangle \,\!$.

Transformations

Suppose we have two different orthogonal bases, $\{e_{k}\}\,\!$, $\{j\}\,\!$. The numbers $\left\langle e_{k}\mid j\right\rangle \,\!$ for all the different $k\,\!$ and $j\,\!$ are often referred to as matrix elements since the set forms a matrix with $k\,\!$ labelling the rows, and $j\,\!$ labelling the columns. Therefore, we can write the transformation from one basis to another with a matrix transformation. Let $M\,\!$ be the matrix with elements $m_{kj}=\left\langle e_{k}\mid j\right\rangle \,\!$. Then the transformation from one basis to another, written in terms of the coefficients of $\left\vert \Psi \right\rangle \,\!$, is

	$A^{\prime }=MA,$	(C.22)

where

$A^{\prime }=\left({\begin{array}{c}\alpha _{1}^{\prime }\\\alpha _{2}^{\prime }\\\vdots \\\alpha _{n}^{\prime }\end{array}}\right),\;\;{\mbox{ and }}\;\;A=\left({\begin{array}{c}\alpha _{1}\\\alpha _{2}\\\vdots \\\alpha _{n}\end{array}}\right).\,\!$

This sort of transformation is a change of basis. However, most often when one vector is transformed to another, the transformation is represented by a matrix. Such transformations can either be represented by the matrix equation, like Eq.~(\ref{eq:matrixeq}), or the components

	$\alpha _{k}^{\prime }=\sum _{j}\alpha _{j}\left\langle e_{k}\mid j\right\rangle =\sum _{j}m_{kj}\alpha _{j}.$	(C.23)

For a general transformation matrix $T\,\!$, acting on a vector, the matrix elements in a particular basis $\left\vert i\right\rangle \,\!$ are

$t_{ij}=\left\langle i\right\vert (T)\left\vert j\right\rangle ,\,\!$

just as elements of a vector can be found using

$\left\langle i\mid \Psi \right\rangle =\alpha _{i}.\,\!$

A similarity transformation of an $n\times n\,\!$ matrix $A\,\!$ by an invertible matrix $S\,\!$ is $SAS^{-1}\,\!$. There are (at least) two important things to note about similarity transformations,

1. Similarity transformations leave determinants unchanged. (We say the determinant is invariant under similarity transformations.) This is because $\det(SAS^{-1})=\det(S)\det(A)\det(S^{-1})=\det(S)\det(A){\frac {1}{\det(S)}}=\det(A).\,\!$
2. Simultaneous similarity transformations of matrices in an equation will leave the equation unchanged. Let $A^{\prime }=SAS^{-1}\,\!$, $B^{\prime }=SBS^{-1}\,\!$, and $C^{\prime }=SCS^{-1}\,\!$. If $AB=C\,\!$, then $A^{\prime }B^{\prime }=C^{\prime }\,\!$ since $A^{\prime }B^{\prime }=SAS^{-1}SBS^{-1}=SABS^{-1}=SCS^{-1}=C^{\prime }\,\!$. The two matrices $A^{\prime }\,\!$ and $A\,\!$ are said to be similar.

Eigenvalues and Eigenvectors

A matrix can always be diagonalized. By this, it is meant that for every complex matrix $M\,\!$ there is a diagonal matrix $D\,\!$ such that

	$M=UDV,\,\!$	(C.24)

where $U\,\!$ and $V\,\!$ are unitary matrices. The entries of the diagonal matrix $D\,\!$ are called the singular values of the matrix $M\,\!$. However, the singular values are not always easy to find.

For the special case that the matrix $M\,\!$ is Hermitian ($M^{\dagger }=M\,\!$), the matrix $M\,\!$ can be written as

	$M=UDU^{\dagger },$	(C.25)

where $U\,\!$ is unitary ($U^{-1}=U^{\dagger }\,\!$). In this case the elements of the matrix $D\,\!$ are called eigenvalues. Very often eigenvalues are introduced as solutions to the equation

$M\left\vert v\right\rangle =\lambda \left\vert v\right\rangle \,\!$

where $\left\vert v\right\rangle \,\!$ a vector called an eigenvector.

To find the eigenvalues and eigenvectors of a matrix $M\,\!$, we follow a standard procedure which is to calculate the following

	$\det(\lambda \mathbb {I} -M)=0,$	(C.26)

and solve for $\lambda \,\!$. The different solutions for $\lambda \,\!$ is the set of eigenvalues and this set is called the spectrum. Let the different eigenvalues be denoted by $\lambda _{i}\,\!$, $i=1,2,...,n\,\!$ fo an $n\times n\,\!$ vector. If two eigenvalues are equal, we say the spectrum is degenerate. To find the eigenvectors, which correspond to different eigenvalues, the equation

$M\left\vert v\right\rangle =\lambda _{i}\left\vert v\right\rangle \,\!$

must be solved for each value of $i\,\!$. Notice that this equations holds even if we multiply both sides by some complex number. This implies that an eigenvector can always be scaled. Usually they are normalized to obtain an orthonormal set. As we will see by example, degenerate eigenvalues require some care.

Examples

Consider a $2\times 2\,\!$ Hermitian matrix

	$\sigma =\left({\begin{array}{cc}1+a&b-ic\\b+ic&1-a\end{array}}\right).$	(C.27)

To find the eigenvalues of this, we follow a standard procedure which is to calculate the following

	$\det(\sigma -\lambda \mathbb {I} )=0,$	(C.28)

and solve for $\lambda \,\!$. The eigenvalues of this matrix are given by

$\det \left({\begin{array}{cc}1+a-\lambda &b-ic\\b+ic&1-a-\lambda \end{array}}\right)=0,\,\!$

which implies the eigenvalues are

$\lambda _{\pm }=1\pm {\sqrt {a^{2}+b^{2}+c^{2}}}.\,\!$

and the eigenvectors are

$v_{1}=\left({\begin{array}{c}i\left(-a+c+{\sqrt {a^{2}+4b^{2}-2ac+c^{2}}}\right)\\2b\end{array}}\right),v_{2}=\left({\begin{array}{c}i\left(-a+c-{\sqrt {a^{2}+4b^{2}-2ac+c^{2}}}\right)\\2b\end{array}}\right).\,\!$

These expressions are useful for calculating properties of qubit states as will be seen in the text.

Now consider a $3\times 3\,\!$ matrix

$N=\left({\begin{array}{ccc}1&-i&0\\i&1&0\\0&0&1\end{array}}\right).\,\!$

First we calculate

$\det \left({\begin{array}{ccc}1-\lambda &-i&0\\i&1-\lambda &0\\0&0&1-\lambda \end{array}}\right)=(1-\lambda )[(1-\lambda )^{2}-1].\,\!$

This implies that the eigenvalues \index{eigenvalues} are

$\lambda =1,0,{\mbox{ or }}2.\,\!$

Let $\lambda _{1}=1\,\!$, $\lambda _{0}=0\,\!$, and $\lambda _{2}=2\,\!$. To find eigenvectors, we calculate

	${\begin{aligned}Nv&=\lambda v,\\\left({\begin{array}{ccc}1&-i&0\\i&1&0\\0&0&1\end{array}}\right)\left({\begin{array}{c}v_{1}\\v_{2}\\v_{3}\end{array}}\right)&=\lambda \left({\begin{array}{c}v_{1}\\v_{2}\\v_{3}\end{array}}\right).\end{aligned}}$	(2.29)

for each $\lambda \,\!$. For $\lambda =1\,\!$ we get the following equations:

	${\begin{aligned}v_{1}-iv_{2}&=v_{1},\\iv_{1}+v_{2}&=v_{2},\\v_{3}&=v_{3},\end{aligned}}$	(2.30)

so $v_{2}=0\,\!$, $v_{1}=0\,\!$, and $v_{3}\,\!$ is any non-zero number, but we choose it to normalize the vector. For $\lambda =0\,\!$,

	${\begin{aligned}v_{1}&=iv_{2},\\v_{3}&=0,\end{aligned}}$	(2.31)

and for $\lambda =2\,\!$,

	${\begin{aligned}v_{1}-iv_{2}&=2v_{1},\\iv_{1}+v_{2}&=2v_{2},\\v_{3}&=2v_{3},\end{aligned}}$	(2.32)

so that $v_{1}=-iv_{2}\,\!$. Therefore, our three eigenvectors are

$v_{0}={\frac {1}{\sqrt {2}}}\left({\begin{array}{c}i\\1\\0\end{array}}\right),\;v_{1}=\left({\begin{array}{c}0\\0\\1\end{array}}\right),\;v_{2}={\frac {1}{\sqrt {2}}}\left({\begin{array}{c}-i\\1\\0\end{array}}\right).\,\!$

The matrix

$V=(v_{0},v_{1},v_{2})=\left({\begin{array}{ccc}i/{\sqrt {2}}&0&-i/{\sqrt {2}}\\1/{\sqrt {2}}&0&1/{\sqrt {2}}\\0&1&0\end{array}}\right)\,\!$

is the matrix that diagonalizes $N\,\!$ in the following way,

$N=VDV^{\dagger },\,\!$

where

$D=\left({\begin{array}{ccc}0&0&0\\0&1&0\\0&0&2\end{array}}\right)\,\!$

or, we may write this as

$V^{\dagger }NV=D.\,\!$

This is sometimes called the eigenvalue decompostion of the matrix and is also written as,

	$N=\sum _{i}\lambda _{i}v_{i}v_{i}^{\dagger }.\,\!$	(C.33)

(ONE MORE EXAMPLE WITH DEGENERATE E-VALUES)

Tensor Products

The tensor product (also called the Kronecker product) is used extensively in quantum mechanics and throughout the course. It is commonly denoted with a $\otimes \,\!$ symbol, but this symbol is also often left out. In fact the following are commonly found in the literature as notation for the tensor product of two vectors $\left\vert \Psi \right\rangle \,\!$ and $\left\vert \Phi \right\rangle \,\!$

	${\begin{aligned}\left\vert \Psi \right\rangle \otimes \left\vert \Phi \right\rangle &=\left\vert \Psi \right\rangle \left\vert \Phi \right\rangle \\&=\left\vert \Psi \Phi \right\rangle .\end{aligned}}$	(C.34)

Each of these has its advantages and we will use all of them in different circumstances.

The tensor product is also often used for operators. So several examples will be given, one which explicitly calculates the tensor product for two vectors and one which calculates it for two matrices which could represent operators. However, these are not different in the sense that a vector is a $1\times n\,\!$ or an $n\times 1\,\!$ matrix. It is also noteworthy that the two objects in the tensor product need not be of the same type. In general a tensor product of an $n\times m\,\!$ object (array) with a $p\times q\,\!$ object will produce an $np\times mq\,\!$ object.

In general, the tensor product of two objects is computed as follows. Let $A\,\!$ be an $n\times m\,\!$ and $B\,\!$ be a $p\times q\,\!$ array

	$A=\left({\begin{array}{cccc}a_{11}&a_{12}&\cdots &a_{1m}\\a_{21}&a_{22}&\cdots &a_{2m}\\\vdots &&\ddots &\\a_{n1}&a_{n2}&\cdots &a_{nm}\end{array}}\right),$	(C.35)

and similarly for $B\,\!$. Then

	$A\otimes B=\left({\begin{array}{cccc}a_{11}B&a_{12}B&\cdots &a_{1m}B\\a_{21}B&a_{22}B&\cdots &a_{2m}B\\\vdots &&\ddots &\\a_{n1}B&a_{n2}B&\cdots &a_{nm}B\end{array}}\right).$	(C.36)

Let us now consider two examples. First let $\left\vert \phi \right\rangle \,\!$ and $\left\vert \psi \right\rangle \,\!$ be as before,

$\left\vert \psi \right\rangle =\left({\begin{array}{c}\alpha \\\beta \end{array}}\right),\;\;{\mbox{and}}\;\;\left\vert \phi \right\rangle =\left({\begin{array}{c}\gamma \\\delta \end{array}}\right).\,\!$

Then

	${\begin{aligned}\left\vert \psi \right\rangle \otimes \left\vert \phi \right\rangle &=\left({\begin{array}{c}\alpha \\\beta \end{array}}\right)\otimes \left({\begin{array}{c}\gamma \\\delta \end{array}}\right)\\&=\left({\begin{array}{c}\alpha \gamma \\\alpha \delta \\\beta \gamma \\\beta \delta \end{array}}\right).\end{aligned}}$	(2.37)

Also

	${\begin{aligned}\left\vert \psi \right\rangle \otimes \left\langle \phi \right\vert &=\left\vert \psi \right\rangle \left\langle \phi \right\vert \\&=\left({\begin{array}{c}\alpha \\\beta \end{array}}\right)\otimes \left({\begin{array}{cc}\gamma ^{*}&\delta ^{*}\end{array}}\right)\\&=\left({\begin{array}{cc}\alpha \gamma ^{*}&\alpha \delta ^{*}\\\beta \gamma ^{*}&\beta \delta ^{*}\end{array}}\right).\end{aligned}}$	(2.38)

Now consider two matrices

$A=\left({\begin{array}{cc}a&b\\c&d\end{array}}\right),\;\;{\mbox{and}}\;\;B=\left({\begin{array}{cc}e&f\\g&h\end{array}}\right).\,\!$

Then

	${\begin{aligned}A\otimes B&=\left({\begin{array}{cc}a&b\\c&d\end{array}}\right),\otimes \left({\begin{array}{cc}e&f\\g&h\end{array}}\right)\\&=\left({\begin{array}{cccc}ae&af&be&bf\\ag&ah&bg&bh\\ce&cf&de&df\\cg&ch&dg&dh\end{array}}\right).\end{aligned}}$	(2.39)

Properties of Tensor Products

Some properties of tensor products which are useful are the following (with $A\,\!$, $B\,\!$, $C\,\!$, $D\,\!$ any type):

1. $(A\otimes B)(C\otimes D)=AC\otimes BD\,\!$
2. $(A\otimes B)^{T}=A^{T}\otimes B^{T}\,\!$
3. $(A\otimes B)^{*}=A^{*}\otimes B^{*}\,\!$
4. $(A\otimes B)\otimes C=A\otimes (B\otimes C)\,\!$
5. $(A+B)\otimes C=A\otimes C+B\otimes C\,\!$
6. $A\otimes (B+C)=A\otimes B+A\otimes C\,\!$
7. ${\mbox{Tr}}(A\otimes B)={\mbox{Tr}}(A){\mbox{Tr}}(B)\,\!$

(See Ref.~(\cite{Horn/JohnsonII}), Chapter 4.)