PHYS20672 Summary 5

Definition of a vector space

1. A vector space $\set{V}$ is a set of objects (vectors, denoted $\ket a$, $\ket b$, etc.) that is closed under two operations:
• vector addition: $$\ket a + \ket b = \ket b+\ket a\in\set{V}\qquad\text{(commutativity)}$$ $$\ket a + (\ket b + \ket c) = (\ket a+\ket b)+\ket c\qquad\text{(associativity)}$$
• multiplication by scalars (usually real or complex numbers) $\lambda$, $\mu$, etc., which satisfies: $$\lambda(\ket a + \ket b) = \lambda\ket a+\lambda\ket b$$ $$\lambda(\mu\ket a) = (\lambda\mu)\ket a$$ $$(\lambda+\mu)\ket a = \lambda\ket a + \mu\ket a$$ The set of scalars is assumed to contain a unit scalar, denoted by $1$, with the property $1\ket a=\ket a$.
2. Every vector space contains a (unique) null or zero vector $\ket0$ such that for all $\ket a$ in $\set{V}$, $$\ket a + \ket0=\ket a.$$
3. For every vector $\ket a$ in $\set{V}$ there is an additive inverse, temporarily denoted $\ket{{-}a}$, such that $$\ket a+\ket{{-}a} = \ket0.$$ When the scalars are real or complex numbers, the axioms above imply that the additive inverse can be constructed as $\ket{{-}a}=(-1)\ket a$. Thus, we can drop the pedantic notation $\ket{{-}a}$ in favour of $-\ket a$.

Shankar 1.1, Riley 8.1

Linear independence

1. A set of vectors $\{\ket{a_i},\; i=1, 2, \dots, N\}$ is said to be linearly independent if the equation $$\sum_{i=1}^N\lambda_i\ket{a_i}=\ket0$$ is satisfied only if every $\lambda_i=0$. Otherwise the vectors are linearly dependent.
2. If in a vector space there is a set of $N$ linearly independent vectors but no set of $N+1$ linearly independent vectors, the space is said to be $N$-dimensional. In this course we denote the dimensionality by a superscript: $\set{V}^N$.
3. For an $N$-dimensional vector space, a set of $N$ linearly independent vectors is called a basis; the base vectors $\ket{a_i}$ are said to span $\set{V}^N$. Given a basis $\basis{a_i}$, any vector $\ket b$ can be expressed as a linear combination of the $\ket{a_i}$: $$ \ket b = \sum_{i=1}^Nb_i\ket{a_i},$$ where the expansion coefficients (or components) $b_i$ are unique for a given $\ket b$ and choice of basis.

Shankar 1.1, Riley 8.1.1

Representations

1. Given a basis $\basis{a_i}$, the list of components $b_i$ specifies $\ket b$ completely, and is said to be a representation of the abstract vector $\ket b$: $$\ket b\longrightarrow\colvec bn.$$ The arrow "$\longrightarrow$" can be pronounced "is represented by". This is just like representing a Cartesian 3-vector $\vec b$ by a list of its Cartesian components, $\displaystyle\begin{pmatrix}b_x\\b_y\\b_z\end{pmatrix}$; here the notation $\vec b$ (like $\ket b$) denotes the vector without making reference to the coordinate system, whereas the components $b_x,b_y,b_z$ depend on the choice of axes, i.e., on the choice of basis vectors.
2. Note that the basis vectors $\ket{a_i}$ have a particularly simple representation: $$\ket{a_1}\longrightarrow\begin{pmatrix}1\\0\\\vdots\\0\end{pmatrix},\quad \ket{a_2}\longrightarrow\begin{pmatrix}0\\1\\\vdots\\0\end{pmatrix},\quad\dots,\quad \ket{a_N}\longrightarrow\begin{pmatrix}0\\0\\\vdots\\1\end{pmatrix}.$$
3. In a given representation, if $\ket w=\lambda\ket b + \mu\ket c$ then $w_i = \lambda b_i+\mu c_i$; i.e., $$\ket w \longrightarrow \begin{pmatrix} \lambda b_1+\mu c_1\\\lambda b_2+\mu c_2\\\vdots\\\lambda b_N+\mu c_N \end{pmatrix}.$$

Shankar 1.1, Riley 8.3

Scalar product

1. The scalar product (if defined for a particular vector space $\set{V}$) is a generalization of the dot product $\vec a\cdot\vec b$ defined for Cartesian vectors. It is a scalar function of two vectors $\ket a,\ket b\in\set{V}$, written $\braket ab$, with the following properties:
   • If $\ket w=\lambda\ket b+\mu\ket c$, then $$\braket aw =\lambda\braket ab+\mu\braket ac;$$ i.e., the function $\braket aw$ is linear in the second argument.
   • $\braket ab=\conj{\braket ba}$. Here we are assuming a complex vector space; for a real vector space, the condition would be simply $\braket ab=\braket ba$.
   • In this course we also require $$\braket aa\ge 0,$$ where $\braket aa=0$ if and only if $\ket a=\ket0$.
   The first two properties imply that if $\ket w=\lambda\ket b+\mu\ket c$, then $$\braket wa=\conj\lambda\braket ba+\conj\mu\braket ca,$$ i.e., the scalar product is "conjugate linear" (or "antilinear") in its first argument.
2. A vector space with a scalar product is called an inner product space; "inner product" is just another name for the scalar product.
3. Once an inner product is defined, the norm (or "length") of a vector $\ket b$ can be defined as $\norm b\equiv\sqrt{\!\braket bb}\ge0$.
4. If $\norm b=1$, we say that $\ket b$ is normalized to unity; we could also say that $\ket b$ is a unit vector.
5. Two important inequalities that follow from the properties of the scalar product are:
   • Schwarz's inequality: $$\abs{\braket ab} \le \norm a\norm b,$$ which is the generalization of $\abs{\vec a\cdot\vec b}=ab\abs{\cos\theta}\le ab$ for Cartesian vectors.
   • The triangle inequality: if $\ket c=\ket a+\ket b$, then $$\norm c\le\norm a+\norm b.$$
   For the proofs, see Examples 6.51.

Shankar 1.2, Riley 8.1.2, 8.1.3

Orthonormal bases

1. Given the existence of a scalar product, the orthogonality of two vectors $\ket a,\ket b$ can be defined by the condition $\braket ab=0$. A set of vectors $\kete i \in \set{V}^N$ that have unit norm and are mutually orthogonal is called an orthonormal set. The vectors of an orthonormal set satisfy $$\braket{e_i}{e_j} = \delta_{ij} = \begin{cases} 1\quad\text{if $i=j$}\\0\quad\text{if $i\ne j$.} \end{cases}$$ If the orthonormal set contains $N$ unit vectors $\kete i$, they form an orthonormal basis for $\set{V}^N$.
2. The components $b_i$ of a vector $\ket b$ with respect to an orthonormal basis $\basis{e_i}$ are given by the simple expressions $b_i=\braket{e_i}b$, analogous to $b_x = \vec i\cdot\vec b$, etc., for the components of a Cartesian vector. The scalar product $\braket bc$ also takes a simple form, $$\braket bc=\sum_i\conj{b_i}c_i=\rowvec bn\colvec cn,$$ where the product of row and column vectors follows the usual rule for matrix multiplication. (Note that these simple formulas for $b_i$ and $\braket bc$ do not hold for a general, non-orthonormal basis.)
3. Bras and kets. The symbols $\ket a,\ket b,\dots$ we have used for abstract vectors have been called kets by Dirac. Their representations (see above) are column vectors. For each ket $\ket b$ there is a corresponding bra vector, $\bra b$, which is represented by the row vector $(\conj{b_1}, \conj{b_2}, \dots,\conj{b_N})$ in the preceding formula for the scalar product. The set of bra vectors is also a vector space, which is distinct from the space $\set{V}^N$ that the kets $\ket b$ belong to; the vector space of bras is said to be dual to $\mathbb V^N$. The existence of a distinction between these two vector spaces should be clear from the representation of kets by column vectors and bras by row vectors. Similarly, the one-to-one correspondence between kets $\ket b$ and bras $\bra b$ should also be apparent from this representation: $$\ket b\longrightarrow\colvec bn\equiv\vec b,\qquad \bra b\longrightarrow\rowvec bn\equiv\vec b^\dagger;$$ i.e., the row vector representing a bra is the Hermitian conjugate (conjugate transpose) of the column vector representing the ket, and vice versa. The scalar product $\braket bc$ can be regarded as a combination of the bra $\bra b$ with the ket $\ket c$ to form the "bra[c]ket" $\braket bc$; this explains the silly (but useful) names.
4. Gram-Schmidt orthogonalization is a general method for constructing an orthonormal basis $\basis{e_i}$ from a given non-orthonormal basis $\basis{a_i}$ for $\set{V}^N$. We...
   • Let our first unit vector be $\kete 1=\ket{a_1}/\norm{a_1}$.
   • Let $\kete 2 = C_2(\ket{a_2}-\kete 1\braket{e_1}{a_2})$, so that $\braket{e_1}{e_2}=0$. (You should check this!) Choose $C_2$ so that $\kete 2$ is normalized to unity: $$\braket{e_2}{e_2} = \abs{C_2}^2\left(\braket{a_2}{a_2}-\abs{\braket{a_2}{e_1}}^2\right)=1,$$ which fixes the value of $\abs{C_2}$; the phase can be chosen arbitrarily.
   • On the $j$th step, let $$\kete j=C_j\left(\ket{a_j}-\sum_{i=1}^{j-1}\kete i\braket{e_i}{a_j}\right),$$ so that $\braket{e_i}{e_j}=0$ for $i\lt j$. Again, we can find the value of $\abs{C_j}$ that makes $\braket{e_j}{e_j}=1$.
   At each step, $\ket{a_j}$ is linearly independent of the unit vectors $\kete i$ with $i\lt j$, as the latter were constructed using only the vectors $\ket{a_i}$ with $i\lt j$. The process therefore terminates on step $j=N$, after the construction of $\kete N$. The existence of the Gram-Schmidt process proves that an orthonormal basis can be found for any finite-dimensional inner-product space.

Shankar 1.2, 1.3.1, Riley 8.1.2

Linear operators

1. A linear operator $\op A$ is a vector-valued function of a vector, written $\op A\ket v$, which has the property of linearity: $$\op A(\lambda\ket a+\mu\ket b) = \lambda(\op A\ket a)+\mu(\op A\ket b)$$ where $\lambda,\mu$ are scalars and $\ket a,\ket b$ are vectors in $\set{V}$. In this course we suppose that $\op A\ket a\in\set{V}$ if $\ket a\in\set{V}$.
2. Simple examples of operators are
   • the identity operator (or unit operator) $\op 1$ which has no effect when applied to a vector: $\op1\ket b=\ket b$ for all $\ket b\in\set{V}$.
   • the zero operator $\op0$ which gives the null vector $\ket0$ when applied to any vector: $\op0\ket b=\ket0$ for all $\ket b\in\set{V}$.
3. The sum of two operators, $\op A+\op B$, is defined by its action $$(\op A + \op B)\ket b = \op A\ket b+\op B\ket b$$ for all $\ket b\in\set{V}$. Multiplication of an operator $\op A$ by a scalar $\lambda$ is defined by $$(\lambda\op A)\ket b = \lambda(\op A\ket b).$$
4. The product of two operators, $\op A\op B$ is defined by $$(\op A\op B)\ket b = \op A(\op B\ket b);$$ i.e., $\op B$ is applied first, then $\op A$. From this definition you should be able to show that the product is associative: $(\op A\op B)\op C = \op A(\op B\op C)$. Note, however, that the operator product need not be commutative: $\op A\op B\ne\op B\op A$, in general.
5. If there exists an operator $\op B$ such that $\op B\op A=\op1$, then $\op B\equiv\op A^{-1}$ is called the inverse of $\op A$. Not all operators have an inverse!
6. More examples of operators:
   • The outer product of two vectors $\ket a$ and $\ket c$, written $\ketbra ca$, is defined to be an operator whose action on vector $\ket b$ is given by $$(\ketbra ca)\ket b = \ket c\braket ab = (\braket ab)\ket c;$$ the factor $\braket ab$ is a scalar, of course.
   • Given an orthonormal basis $\basis{e_i}$ for a vector space, one can define projection operators $\op P_i=\ketbra{e_i}{e_i}$ that extract from a vector $\ket b=\sum_jb_j\kete j$ the part that is "parallel" to $\kete i$: $$\op P_i\ket b = \ketbra{e_i}{e_i}\left({\textstyle\sum_jb_j\kete j}\right) = b_i\kete i.$$ Note that all information about $b_j$ for $j\ne i$ is lost in the process of projection, so $\op P_i$ is an example of an operator that has no inverse. By summing both sides of the last equation with respect to $i$ one obtains the result $$\left({\textstyle\sum_i\op P_i}\right)\ket b = \sum_ib_i\kete i = \ket b;$$ that is, we have $\sum_i\op P_i =\op1$, or $$\sum_i\outer{e_i}{e_i}=\op1.$$ This simple (but extremely useful) formula expresses the fact that the orthonormal vectors $\basis{e_i}$ span $\set{V}$. It is called the completeness relation or a resolution of unity. Some further properties of projection operators are $$\op P_i\op P_j = \begin{cases} \op P_i\quad&\text{if $j=i$}\\ \op0\quad&\text{if $j\ne i$.} \end{cases}$$

Matrix representation of operators

1. If $\ket c=\op A\ket b$, projection on to vector $\kete i$ of an orthonormal basis extracts the component $c_i=\braket{e_i}{c}=\brae i(\op A\ket b)$. If we expand $\ket b$ in terms of the basis vectors, $\ket b=\sum_jb_j\kete j$, this gives $$c_i = \sum_j \brae i\op A\kete jb_j = \sum_jA_{ij}b_j\,, $$ where $A_{ij}=\brae i\op A\kete j$. We recognize this as the rule for multiplying a column vector by a matrix $\mat A$ with elements $A_{ij}$. Thus, the equation $$\ket c = \op A\ket b$$ has the representation $$\colvec cn = \matrix An\colvec bn\quad\text{or, more briefly,}\quad\vec c=\mat A\vec b.$$
2. Aside:
   • Note that column $j$ of matrix $\mat A$ gives the components of $\op A\kete j$. If the columns of $\mat A$ are not linearly independent, there will be a combination of the vectors $\op A\kete j$ that satifies $\sum_j u_j\op A\kete j=\ket0$, where not all of the $u_j$ are zero. This could also be written $\op A\ket u = \ket0$, where $\ket u=\sum_j u_j\kete j$.
   • An important consequence of this is that if the equation $\op A\ket b=\ket c$ has a solution, the solution $\ket b$ will not be unique: $\ket b+\alpha\ket u$ with $\alpha\ne0$ will be a different but equally acceptable solution. Accordingly, the operator $\op A$ has no inverse.
   • Alternatively, note that if the columns of $\mat A$ are linearly dependent, then $\det\mat A=0$, so that the inverse of the matrix doesn't exist.
3. Examples of matrix representations of operators:
   • The identity: $\brae i\op1\kete j = \delta_{ij}$. Thus, $$\op1\longrightarrow\diagmat 111\equiv\mat I$$ is the unit matrix.
   • The outer product $\ketbra ca$ has matrix elements $\braket{e_i}c\braket a{e_j}=c_i\conj{a_j}$: $$\ketbra ca\longrightarrow \begin{pmatrix} c_1\conj{a_1}&c_1\conj{a_2}&\dots&c_1\conj{a_N}\\ c_2\conj{a_1}&c_2\conj{a_2}&\dots&c_2\conj{a_N}\\ \vdots&\vdots&\ddots&\vdots\\ c_N\conj{a_1}&c_N\conj{a_2}&\dots&c_N\conj{a_N} \end{pmatrix}\equiv\vec c\,\vec a^\dagger.$$
   • The matrix representation of the projection operator $\op P_1=\kete 1\brae 1$ is therefore $$\op P_1\longrightarrow\diagmat 100\equiv\mat P_1\,.$$ Similarly, the matrix $\mat P_i$ will be zero everywhere except for a single $1$ at position $(i,i)$ on the diagonal.

Shankar 1.2, 1.3.1, Riley 8.1.2

Adjoint operators (Hermitian conjugates)

1. For a linear operator $\op A$, we define its adjoint (or Hermitian conjugate) to be the operator $\op A^\dagger$ for which $$\bra u\op A^\dagger\ket v = \conj{\bra v\op A\ket u}$$ for all $\ket u,\ket v\in\set{V}$. Taking $\ket u$ and $\ket v$ to be orthonormal basis vectors $\kete i$ and $\kete j$ shows at once that the matrix elements of $\op A^\dagger$ are $$A^\dagger{}_{ij} = \brae i\op A^\dagger\kete j =\conj{\brae j\op A\kete i} = \conj{A_{ji}}\,,$$ i.e., the matrix $\mat A^\dagger$ is the conjugate transpose (Hermitian conjugate) of the matrix $\mat A$.
2. You should be able to show that:
   • $(\op A\op B){}^\dagger=\op B^\dagger\op A^\dagger$;
   • $(\lambda\op A)^\dagger=\conj\lambda\op A^\dagger$ (where $\lambda$ is a scalar);
   • if $\op Q=\ketbra ca$ then $\op Q^\dagger=\ketbra ac$.

Hermitian operators

1. An operator $\op A$ is said to be self-adjoint (or Hermitian) if $\op A=\op A^\dagger$, so that $$\bra u\op A\ket v=\conj{\bra v\op A\ket u}$$ for all $\ket u,\ket v\in\set{V}$. From this, the matrix elements of $\op A$ satisfy $A_{ij}=\conj{A_{ji}}$: the matrix $\mat A=\mat A^\dagger$ is Hermitian.
2. Note that in a real vector space (i.e., one in which the scalars are real numbers), the matrix of a self-adjoint operator will be real/symmetric, $A_{ij}=A_{ji}\in\set{R}$.
3. Exercise:
   • Show that if $\op A=\op A^\dagger$, then $\bra u\op A^2\ket u\ge0$ for all $\ket u\in\set{V}$.
   • Hint: Write $\op A^2=\op A\op1\op A$ and make use of the completeness relation $\op1=\sum_i\ketbra{e_i}{e_i}$.

Unitary operators

1. An operator $\op U$ is said to be unitary if $\op U\op U^\dagger=\op U^\dagger\op U=\op1$, i.e. if $\op U^{-1}=\op U^\dagger$. The matrix representing a unitary operator therefore satisfies $\mat U\mat U^\dagger=\mat U^\dagger\mat U=\mat I$.
2. Exercise:
   • Prove that unitary operators preserve the scalar product of two vectors; that is, if $\ket{a'}=\op U\ket a$ and $\ket{b'}=\op U\ket b$, then $\braket{a'}{b'}=\braket ab$.
   • Hence show that the norm of a vector is preserved by $\op U$; i.e., $\norm{a'}=\norm a$.
3. For a real vector space, a unitary operator can be represented by a real, orthogonal matrix, $\mat O$, so that $\mat O\tp{\mat O}=\mat I$, where $\tp{\mat O}$ denotes the transpose of $\mat O$.
4. The columns of a unitary matrix can be regarded as an orthonormal set of vectors. The same is true of the rows.
   • Proof (for columns): From each of the columns $j$ of the matrix $\mat U$, construct a vector $\ket{u_j}\equiv\sum_iU_{ij}\kete i$. Then $$\braket{u_k}{u_j} = \sum_i \conj{U_{ik}}U_{ij} =\sum_iU^\dagger{}_{ki}U_{ij} = \delta_{kj}\,,$$ which proves that the vectors $\basis{u_j}$ form an orthonormal set.
   • Exercise: Prove the orthogonality of the rows.
5. We can show that a change of basis can be effected by means of a unitary matrix:
   • If we have two orthonormal bases $\basis{e_i}$ and $\basis{e'_i}$, we can use either to expand an arbitrary vector: $$\ket b=\sum_jb'_j\ket{e'_j}=\sum_kb_k\kete k.$$ Projecting both sides onto $\ket{e'_i}$ gives $$b'_i=\sum_k\braket{e'_i}{e_k}b_k\equiv \sum_kU_{ik}b_k;$$ the matrix of the transformation is $\mat U$ with elements $U_{ik}=\braket{e'_i}{e_k}$.
   • We now show that $\mat U$ is unitary. First note that $U^\dagger{}_{kj} = \conj{U_{jk}} = \conj{\braket{e'_j}{e_k}}=\braket{e_k}{e'_j}$, so that $$ (\mat U\mat U^\dagger)_{ij} =\sum_kU_{ik}U^\dagger{}_{kj}=\sum_k\braket{e'_i}{e_k}\braket{e_k}{e'_j} =\braket{e'_i}{e'_j}=\delta_{ij}.$$ Here we have used the completeness relation $\sum_k\ketbra{e_k}{e_k}=\op1$ to do the sum on $k$.
6. Aside on active and passive transformations. Just as in the case of a rotation, a given unitary transformation with matrix elements $U_{ik}$ can be interpreted in two ways. As we have just seen, the equation $b'_i=\sum_kU_{ik}b_k$ expresses the components of $\ket b$ with respect to one basis, $\basis{e'_i}$, in terms of the components of $\ket b$ with respect to another basis, $\basis{e_i}$. Here the vector $\ket b$ is the same in each case, but the "axes" have been changed; this corresponds to the passive view of rotations, met before in PHYS10672. But the same equation, $b'_i=\sum_kU_{ik}b_k$, has a different interpretation when the quantities $b'_i$ are instead the components of a different vector $\ket{b'}$ with respect to the original basis, $\basis{e_i}$. In this case $$\ket{b'}=\op U\ket b,\quad\text{i.e.,}\quad\ket{b'}\ne\ket b,$$ and the coefficients $U_{ik}$ are the matrix elements of the unitary operator $\op U$ relating the two vectors, $U_{ik}=\brae i\op U\kete k$. This corresponds to the active view of rotations, also met in PHYS10672. Both interpretations are regularly needed in physics, but it should always be clear which one is being used.
7. The transformation of the matrix elements of operators to a new representation (basis) is almost as straightforward as the transformation of the components of a vector. In the new representation with basis vectors $\basis{e'_i}$ we have $$A'_{ij} = \bra{e'_i}\op A\ket{e'_j} =\bra{e'_i}\op1\op A\op1\ket{e'_j} =\sum_{k,l}\braket{e'_i}{e_k}\brae k\op A\kete l\braket{e_l}{e'_j},$$ where we have used the completeness relation twice. Noting that $\braket{e'_i}{e_k}=U_{ik}$ and $\braket{e_l}{e'_j}=U^\dagger{}_{lj}$ we find $$\mat A' = \mat U\mat A\mat U^\dagger.$$

Eigenvalue problems

1. Revision of your existing knowledge of eigenvalue problems. Note that you are expected to be able to apply this knowledge, as you have met many examples already in PHYS10071, PHYS10672 (principal axes of symmetric tensors) and PHYS20401 (small oscillations problems).
   • The eigenvalue equation for an operator $\op A$ is the equation $$\op A\ket u=a\ket u,$$ where $a$ is a scalar, the eigenvalue, and $\ket u\ne\ket 0$ is the eigenvector. Solving an eigenvalue problem means finding $\{a_i\}$, the set of possible values for $a$, and also the corresponding eigenvectors $\basis{u_i}$.
   • Using the matrix representation for an operator on $\set{V}^N$, the eigenvalue equation becomes $\mat A\vec u=a\vec u$, or $$(\mat A - a\mat I)\vec u = \vec0\,.$$ Since $\vec u\ne\vec0$, this implies that the columns of $\mat A-a\mat I$ are linearly dependent. [If you want to check this last point, it's helpful to note that the last equation can be written $\sum_ju_j\left[\sum_i\vec e_i(A_{ij}-a\delta_{ij})\right]=\vec0$; the quantities in square brackets are the columns of $\mat A-a\mat I$, regarded as column vectors.]
   • Thus, for there to be a nontrivial (i.e., nonzero) solution of this system of $N$ linear equations, the determinant of the matrix of coefficients must vanish: $$\det(\mat A-a\mat I) = 0.$$ This is a polynomial equation of degree $N$ in $a$, and is often called the characteristic equation or (less often, these days) the secular equation. By the Fundamental Theorem of Algebra, a polynomial equation of degree $N$ has exactly $N$ roots, $a_i\in\set{C}$. If a root is repeated $m>1$ times, we say that the eigenvalue has degeneracy $m$, or that its multiplicity is $m$.
   • Once the eigenvalues are known, the eigenvectors can be found by solving the equations $(\mat A-a_i\mat I)\vec u=\vec0$ for $\vec u$. Eigenvectors can be determined only up to a multiplicative constant: i.e., if $\vec u$ is an eigenvector, then $C\vec u$ is an equally good solution of the eigenvalue equation, provided $C\ne0$, but $C\vec u$ is not regarded as being an eigenvector that is distinct from $\vec u$.
   • Distinct eigenvalues ($a_i\ne a_j$) always correspond to distinct eigenvectors. But if a particular eigenvalue is degenerate, with degeneracy $m>1$, we can say only that there may be up to $m$ linearly independent eigenvectors belonging to this eigenvalue. Thus, the total number of linearly-independent eigenvectors of an $N\times N$ matrix is always less than or equal to $N$.
   • The Fundamental Theorem of Algebra, applied to the polynomial $P(a)=\det(\mat A-a\mat I)$, also tells us that $P(a)$ can be factorized in the form $$\det(\mat A-a\mat I)=\prod_{i=1}^N(a_i-a).$$ By setting $a=0$ on each side, we obtain $$\det\mat A=\prod_{i=1}^Na_i\,{;}$$ the determinant of a matrix is equal to the product of its eigenvalues. By comparing coefficients of $a^{N-1}$ on each side, one can also show (with more effort) that $$\Tr\mat A=\sum_{i=1}^Na_i\,{;}$$ the trace of a matrix (i.e., the sum of its diagonal elements) equals the sum of its eigenvalues. Note that these last two, important results are always valid, regardless of whether there are exactly $N$ or fewer than $N$ linearly-independent eigenvectors. They can be a useful check on your calculation of the eigenvalues of a matrix.
2. Eigenvalues and eigenvectors of Hermitian operators, $\op A=\op A^\dagger$
   • Suppose that $\ket{u_i}$ and $\ket{u_k}$ are eigenvectors belonging to eigenvalues $a_i$ and $a_k$. Then, starting from $$\op A\ket{u_i}=a_i\ket{u_i},$$ we can take the scalar product with $\ket{u_k}$ to obtain $$\bra{u_k}a_i\ket{u_i} =\bra{u_k}\op A\ket{u_i} =\conj{\bra{u_i}\op A\ket{u_k}} =\conj{\bra{u_i}a_k\ket{u_k}} =\conj{a_k}\braket{u_k}{u_i},$$ which gives $$(a_i-\conj{a_k})\braket{u_k}{u_i}=0.$$ We can draw two conclusions from this:
     • By setting $k=i$ and noting that $\braket{u_i}{u_i}>0$, we find $a_i-\conj{a_i}=0$; the eigenvalues of an Hermitian operator are real numbers.
     • If $a_i\ne a_k$, the factor $(a_i-\conj{a_k})$ is nonzero. [Remember, all of the eigenvalues are real, so $\conj{a_k}=a_k$.] We deduce that the other factor is zero: $\braket{u_k}{u_i}=0$. Thus, the eigenvectors belonging to distinct eigenvalues are orthogonal.
   • Less obviously (we won't prove it), in the case of an eigenvalue with degeneracy $m$, there will be exactly $m$ linearly independent eigenvectors. (Compare this with the general case, $\op A\ne\op A^\dagger$, where the number of linearly-independent eigenvectors may be less than $m$.) The Gram-Schmidt process can be applied to these $m$ eigenvectors to obtain an orthonormal set of $m$ eigenvectors belonging to the same eigenvalue.
   • Thus, the $N$ normalized eigenvectors of an Hermitian operator $\op A$ on $\set{V}^N$ form an orthonormal basis for $\set{V}^N$, or, in the case of degeneracy, they can be chosen so that they do.
3. Eigenvalues and eigenvectors of unitary operators, $\op U^{-1}=\op U^\dagger$. If $\op U$ is unitary and the eigenvalue equation is $$\op U\ket{v_j}=\omega_j\ket{v_j},$$ then:
   • The eigenvalues may be real or complex, but all have unit modulus, $\mod{\omega_j}=1$.
   • If $\omega_j\ne\omega_k$, then $\braket{v_j}{v_k}=0$; the eigenvectors belonging to distinct eigenvalues are orthogonal.
   • A unitary operator on $\set V^N$ has exactly $N$ linearly-independent eigenvectors $\basis{v_j}$ and they can be chosen so that they form an orthonormal basis for $\set V^N$. (As in the case of an Hermitian operator, "choice" arises only in the case of degeneracy.)
   For the proof of the first two statements, see Examples 6.58.
4. "Diagonalization" of Hermitian and unitary operators
   • Solving the eigenvalue problem for an operator $\op A$ that is either Hermitian or unitary is often called diagonalization. The reason for the name is that the matrix of $\op A$ is diagonal in the representation given by the orthonormal set of eigenvectors, $\ket{u_i}$: $$\bra{u_i}\op A\ket{u_j} = \bra{u_i}a_j\ket{u_j} = a_j\delta_{ij}\,;$$ i.e., in this representation, $$\op A\longrightarrow\diagmat{a_1}{a_2}{a_N}\equiv\mat A^\text{diag}.$$
   • The unitary transformation from basis $\basis{e_i}$ to the basis of eigenvectors, $\basis{u_j}$, is given by $$\mat A^\text{diag}=\mat T^\dagger\mat A\mat T,$$ where $T_{ij}=\braket{e_i}{u_j}$; that is, the $j$th column of $\mat T$ is the $j$th eigenvector of $\mat A$.
   • By using the eigenfunctions $\basis{u_i}$ as the basis, we can also write any Hermitian or unitary operator in a particularly simple, general form. Starting from the eigenvalue equation $$\op A\ket{u_i}=a_i\ket{u_i},$$ we take the outer product of each side with $\ket{u_i}$ (i.e., we multiply both sides on the right by $\bra{u_i}$) and then sum over $i$: $$\sum_i\op A\ketbra{u_i}{u_i} = \sum_ia_i\ketbra{u_i}{u_i}.$$ But $\sum_i\ketbra{u_i}{u_i}=\op1$ (the completeness relation for $\basis{u_i}$), so $$\op A = \sum_ia_i\ketbra{u_i}{u_i}.$$ This is called the spectral representation of the operator $\op A$.
5. Aside [not needed for the examples]: Diagonalization of commuting operators -- Operators $\op A$ and $\op B$ are said to commute if $\op A\op B=\op B\op A$.
   • Let $\ket u$ be an eigenvector of $\op B$, so that $\op B\ket u=b\ket u$. Then if $\op A$ commutes with $\op B$, we can show that $\op A\ket u$ is also an eigenvector of $\op B$, corresponding to the same eigenvalue $b$: $$\op B\left(\op A\ket u\right) = \op A\left(\op B\ket u\right) = b\op A\ket u\,,$$ where we have used $\op A\op B=\op B\op A$ in the first step and the eigenvalue equation $\op B\ket u=b\ket u$ in the second step. Two cases arise:
     • If the eigenvalue $b$ is nondegenerate, $\op A\ket u$ must be simply proportional to $\ket u$: $$\op A\ket u=a\ket u\,;$$ this shows that $\ket u$ is also an eigenvector of $\op A$; the coefficient of proportionality, $a$, is the eigenvalue.
     • Suppose instead that $b$ has degeneracy $m>1$ and that the corresponding eigenvectors are $\{\ket{u_i},i=1,\,\dots,m\}$. The vector $\op A\ket{u_i}$ is an eigenvector of $\op B$ with eigenvalue $b$, so it must be possible to express it as a linear combination of the vectors $\{\ket{u_i},i=1,\,\dots,m\}$: $$\op A\ket{u_i}=\sum_{k=1}^ma_{ki}\ket{u_k}\,;$$ the notation for the coefficients has been chosen so that $a_{ki}=\bra{u_k}\op A\ket{u_i}$. Now consider the eigenvalue problem $\op A\ket v=a\ket v$, where $\ket v$ is a vector in the $m$-dimensional subspace spanned by the vectors $\{\ket{u_i},i=1,\,\dots,m\}$. By making the expansion $\ket v=\sum_{i=1}^mv_i\ket{u_i}$, the eigenvalue problem can be expressed in matrix form: $$\sum_{j=1}^ma_{ij}\,v_j=a\,v_i\,;$$ if you are in any doubt about this, it would be a good exercise to check it. But the last equation is simply a standard $m\times m$ matrix eigenvalue problem, which can be solved to obtain the eigenvalues and eigenvectors of $\op A$ that are -- at the same time -- eigenvectors of $\op B$ with eigenvalue $b$. Thus, if they commute, the operators $\op A$ and $\op B$ can be diagonalized simultaneously.
     In fact, you have met concrete examples of this result already in your quantum mechanics course. For example, the energy eigenfunctions of the hydrogen atom can be taken to be eigenfunctions of $\op{\vec L}^2$, the square of the orbital angular momentum operator. Mathematically, this is possible because the energy operator, $\op H$, commutes with $\op{\vec L}^2$ -- as would any other rotationally symmetric operator. Taking this a step further, $\op H$ and $\op{\vec L}^2$ both commute with $\op L_z$, so the energy eigenfunctions can also be chosen to be eigenfunctions of $\op L_z$. Thus, the energy eigenfunctions can be labelled by the quantum number $l$ (which determines the eigenvalue of $\op{\vec L}^2$), the azimuthal quantum number $m$ (which determines the eigenvalue of $\op L_z$) and by a further quantum number $n_r$ which (given the value of $l$) determines the energy.

Functions of operators

1. Because we know how to add and multiply operators, we can build up simple polynomial expressions such as $\op A^2$, $2\op A^3+\op1$, etc., which are functions of the operator $\op A$. Perhaps it is possible to extend this idea to functions defined via a power series, $$f(\op A)=\sum_{r=0}^\infty f_r\op A^r\,,\quad\text{where}\quad\op A^0\equiv\op1.$$ To be more specific, suppose that $\op A$ is an operator on $\set V$ with eigenvalues $\{a_i\}$ and eigenvectors $\basis{u_i}$ that span $\set V$; this will certainly be the case if $\op A$ is Hermitian or unitary. To find out how $f(\op A)$ acts on a general vector $\ket b=\sum_ib_i\ket{u_i}$, it would be enough to know its action on $\ket{u_i}$: $$f(\op A)\ket{u_i} =\sum_{r=0}^\infty f_r\op A^r\ket{u_i} =\sum_{r=0}^\infty f_r\,a_i^r\ket{u_i} =f(a_i)\ket{u_i}.$$ For this to give a definite result for every $\ket{u_i}$, the series for $f(a_i)$ should converge for each $a_i$. Clearly, then, we can't say whether $f(\op A)$ is well defined without knowing the spectrum of $\op A$ (its eigenvalues) and the radius of convergence of the power series.
2. One example of a function that can be very usefully extended to operators is the exponential function, $$f(z)=\exp(z)=\sum_{r=0}^\infty\frac{z^r}{r!}\,,$$ which has no singularities in the finite complex plane (it is an entire function), so that its Taylor series about $z=0$ converges for all finite $z$. Now, if $\op A$ is an operator on $\set V^N$, its eigenvalues will be finite and $$f(\op A)=\exp(\op A)=\sum_{r=0}^\infty\frac{\op A^r}{r!}$$ will have a well-defined action on every $\ket{u_i}$: $$\exp(\op A)\ket{u_i}=\exp(a_i)\ket{u_i}\,;$$ we see that $\ket{u_i}$ is an eigenfunction of $\exp(\op A)$ with eigenvalues $e^{a_i}$. Therefore, in the representation in which $\op A$ is diagonal, $\exp(\op A)$ is also diagonal: $$\exp(\op A)\longrightarrow\diagmat{e^{a_1}}{e^{a_2}}{e^{a_N}}.$$
3. Examples (not examinable):

   • The exponential function of an operator is often met in advanced quantum mechanics. In quantum mechanics, the state of a system is represented by a ket $\ket{\psi(t)}$, which obeys the time-dependent Schroedinger equation $$i\hbar\pdby{}t\ket{\psi(t)}=\op H\ket{\psi(t)}.$$ This is a first-order differential equation whose solution is $$\ket{\psi(t)}=\exp[-i\op Ht/\hbar]\ket{\psi(0)}\equiv\op U_t\ket{\psi(t)},\quad\text{where $\op U_t\equiv\exp[-i\op Ht/\hbar]$}$$ and $\ket{\psi(0)}$ is the state of the system at some initial time $t=0$. Note that if $\ket{\psi(0)}$ happens to be an energy eigenstate with energy $E$, the time dependence of $\ket{\psi(t)}$ reduces to a scalar factor $\exp[-iEt/\hbar]$; this is a result that you have already have met for the time dependence of an energy eigenfunction.

     We can call operator $\op U_t\equiv\exp[-i\op Ht/\hbar]$ the time-development operator. It is easy to verify that the time-development operator is unitary: $$\op U_t{}^\dagger\op U_t=e^{i\op Ht/\hbar}\,e^{-i\op Ht/\hbar}=\op1.$$ Of course, it must satisfy an equation of motion that is consistent with the time-dependent Schroedinger equation for $\ket{\psi(t)}$, namely $$i\hbar\pdby{\op U_t}t=\op H\op U_t\,;$$ you can check that this holds by directly differentiating the series for $\exp[-i\op Ht/\hbar]$.

   • The exponential function also appears in the formula for the partition function of a quantum system: $$Z=\sum_ie^{-\beta E_i},$$ where $E_i$ are the energy levels (eigenvalues of the Hamiltonian, $\hat H$) and $\beta=1/(\kT)$. Thus, $$e^{-\beta E_i} = \bra{u_i}\exp(-\beta\op H)\ket{u_i},$$ where $\ket{u_i}$ is an energy eigenfunction. So the partition function is the sum of the diagonal elements (i.e., the trace) of the matrix of $\exp(-\beta\op H)$: $$Z = \sum_i\bra{u_i}\exp(-\beta\op H)\ket{u_i}=\Tr[\exp(-\beta\op H)].$$ This is quite a useful formula, because the trace is independent of the choice of basis. Even if we can't calculate the eigenvectors and eigenvalues of $\op H$ exactly, we may still be able to approximate the sum $$Z = \Tr[\exp(-\beta\op H)]=\sum_j\brae j\exp(-\beta\op H)\kete j$$ by making an appropriate choice of basis, $\basis{e_i}$.

     [Too vague. Example might be where $\op H=\op H_0+\lambda\op H_1$ and $\basis{e_i}$ are eigenvectors of $\op H_0$. Possible then to expand $Z(\lambda)$ (or, more usefully, $F(\lambda)-F(0)=-\kT\log[Z(\lambda)/Z(0)]$) in powers of $\lambda$. But a specific example would be more suitable for year 4.]