We’ve seen that in quantum mechanics, the state of an
electron in some potential is given by a wave function _{}, and physical variables are represented by operators on this
wave function, such as the momentum in the *x*-direction
_{} The Schrödinger wave
equation is a *linear* equation, which
means that if _{} and _{} are solutions, then so
is _{}, where *c*_{1}, *c*_{2} are
arbitrary complex numbers.

This linearity of the sets of possible solutions is true
generally in quantum mechanics, as is the representation of physical variables
by operators on the wave functions. The
mathematical structure this describes, the linear set of possible states and
sets of operators on those states, is in fact a *linear algebra* of operators acting on a *vector space*. From now on,
this is the language we’ll be using most of the time. To clarify, we’ll give some definitions.

The prototypical vector space is of course the set of real
vectors in ordinary three-dimensional space, these vectors can be represented
by trios of real numbers _{} measuring the
components in the *x*, *y* and *z* directions respectively.

*The basic properties
of these vectors are*:

• any vector
multiplied by a number is another vector in the space, _{};

• the sum of two
vectors is another vector in the space, that given by just adding the
corresponding components together:_{}

These two properties together are referred to as “*closure*”: adding vectors and multiplying
them by numbers cannot get you out of the space.

• A further property
is that there is a unique null vector _{} and each vector has an
additive inverse _{} which added to the
original vector gives the null vector.

Mathematicians have generalized the definition of a vector space: a general vector space has the properties we’ve listed above for three-dimensional real vectors, but the operations of addition and multiplication by a number are generalized to more abstract operations between more general entities. The operators are, however, restricted to being commutative and associative.

Notice that the list of necessary properties for a general
vector space does not include that the vectors have a magnitude—that would be
an additional requirement, giving what is called a *normed* vector space. More
about that later.

To go from the familiar three-dimensional vector space to
the vector spaces relevant to quantum mechanics, first the real numbers
(components of the vector and possible multiplying factors) are to be generalized
to complex numbers, and second the three-component vector goes an *n* component vector. The consequent *n*-dimensional complex space is sufficient to describe the quantum
mechanics of angular momentum, an important subject. But to describe the wave function of a
particle in a box requires an *infinite*
dimensional space, one dimension for each Fourier component, and to describe
the wave function for a particle on an infinite line requires the set of all
normalizable continuous differentiable functions on that line. Fortunately, all these generalizations are to
finite or infinite sets of complex numbers, so the mathematicians’ vector space
requirements of commutativity and associativity are always trivially
satisfied.

We use Dirac’s notation for vectors, _{} and call them “kets”,
so, in his language, if _{}belong to the space, so does _{} for arbitrary complex
constants *c*_{1}, *c*_{2}. Since our vectors are made up of complex
numbers, multiplying any vector by zero gives the null vector, and the additive
inverse is given by reversing the signs of all the numbers in the vector.

Clearly, the set of
solutions of Schrödinger’s equation for an electron in a potential satisfies
the requirements for a vector space: _{}is just a complex number at each point in space, so only
complex numbers are involved in forming _{}, and commutativity, associativity, etc., follow at once.

The vectors _{} are *linearly
independent* if

_{}

implies

_{}

*A vector space is n-dimensional if the maximum number
of linearly independent vectors in the space is n. *

Such a space is often called *V*^{ n}(*C*),
or ** V^{ n}**(

Now, vector spaces with finite dimension *n* are
clearly insufficient for describing functions of a continuous variable *x*.
But they are well worth reviewing here: as
we’ve mentioned, they are fine for describing quantized angular momentum, and
they serve as a natural introduction to the infinite-dimensional spaces needed
to describe spatial wavefunctions.

A set of *n* linearly independent vectors in *n*-dimensional
space is a *basis*—any vector can be written *in a unique way* as a
sum over a basis:

_{}

You can check the uniqueness by taking the difference between two supposedly distinct sums: it will be a linear relation between independent vectors, a contradiction.

Since all vectors in the space can be written as linear sums over the elements of the basis, the sum of multiples of any two vectors has the form:

_{}

The vector spaces of relevance in quantum mechanics also have an operation associating a number with a pair of vectors, a generalization of the dot product of two ordinary three-dimensional vectors,

_{}

Following Dirac, we write the inner product of two ket
vectors _{} as _{}. Dirac refers to this
_{} form as a “bracket”
made up of a “bra” and a “ket”. This
means that each ket vector _{} has an associated bra _{} For the case of a
real *n*-dimensional vector, _{} are identical—but we
require for the more general case that

_{}

where ^{*} denotes complex conjugate. This implies
that for a ket _{}the bra will be _{}. (Actually, bras are
usually written as rows, kets as columns, so that the inner product follows the
standard rules for matrix multiplication.)
Evidently for the *n*-dimensional
complex vector_{} is real and positive except for the null vector:

_{}

For the more general
inner product spaces considered later we require_{} to be positive, except for the null vector. (These requirements do restrict the classes
of vector spaces we are considering—no Lorentz metric, for example—but they are
all satisfied by the spaces relevant to nonrelativistic quantum mechanics.)

The *norm* of _{} is then defined by

_{}

If _{} is a member of *V*^{
n}(*C*), so is _{}, for any complex number *a*.

We require the inner product operation to commute with multiplication by a number, so

_{}

The complex conjugate of the right hand side is _{} For consistency, the
bra corresponding to the ket _{} must therefore be _{}—in any case obvious from the definition of the bra in *n* complex dimensions given above.

It follows that if

_{}

To have something better resembling the standard dot product
of ordinary three vectors, we need _{} that is, we need to
construct an *orthonormal basis* in the space. There is a straightforward procedure for
doing this called the Gram-Schmidt process. We begin with a linearly
independent set of basis vectors, _{} .

We first normalize _{} by dividing it by its
norm. Call the normalized vector _{} Now _{} cannot be parallel to _{} because the original
basis was of linearly independent vectors, but _{} in general has a
nonzero component parallel to _{} equal to _{} since _{} is normalized. Therefore, the vector _{} is perpendicular to _{} as is easily verified.
It is also easy to compute the norm of
this vector, and divide by it to get _{} the second member of
the orthonormal basis. Next, we take _{} and subtract off its
components in the directions_{} and _{} normalize the
remainder, and so on.

In an *n*-dimensional space, having constructed an
orthonormal basis with members _{} any vector _{} can be written as a
column vector,

_{}

The corresponding bra is _{}, which we write as a row vector with the elements complex
conjugated, _{}. This operation,
going from columns to rows and taking the complex conjugate, is called taking
the *adjoint*, and can also be applied to matrices, as we shall see
shortly.

The reason for representing the bra as a row is that the inner product of two vectors is then given by standard matrix multiplication:

_{}

(Of course, this only works with an orthonormal base.)

The Schwartz inequality is the generalization to any inner
product space of the result _{} (or _{}) for ordinary three-dimensional vectors. The equality sign in that result only holds
when the vectors are parallel. To
generalize to higher dimensions, one might just note that two vectors are in a
two-dimensional subspace, but an illuminating way of understanding the
inequality is to write the vector _{} as a sum of two
components, one parallel to _{}and one perpendicular to _{}. The component
parallel to _{} is just _{}, so the component perpendicular to _{} is the vector _{}. Substituting this
expression into _{} , the inequality
follows.

This same point can be made in a general inner product
space: if _{} are two vectors, then

_{}

is the component of _{} perpendicular to _{}, as is easily checked by taking its inner product with _{}.

Then

_{}

A *linear operator* *A*
takes any vector in a linear vector space to a vector in that space, _{} and satisfies

_{}

with *c*_{1},
*c*_{2} arbitrary complex
constants.

The *identity operator I* is (obviously!) defined by:

_{}

For an *n*-dimensional vector space with an orthonormal
basis _{} since any vector in
the space can be expressed as a sum _{}, the linear operator is completely determined by its action
on the basis vectors—this is all we need to know. It’s easy to find an
expression for the identity operator in terms of bras and kets.

Taking the inner product of both sides of the equation _{} with the bra _{} gives _{} so

_{}

Since this is true for any vector in the space, it follows that that the identity operator is just

_{}

This is an important result: it will reappear in many disguises.

To analyze the action of a general linear operator *A*, we just need to know how it acts on
each basis vector. Beginning with _{} this must be some sum
over the basis vectors, and since they are orthonormal, the component in the _{} direction must be just
_{}

That is,

_{}

So if the linear operator *A* acting on _{} gives _{}, that is,_{}the linearity tells us that

_{}

where in the fourth step we just inserted the identity operator.

Since the _{}’s are all orthogonal, the coefficient of a particular _{} on the left-hand side
of the equation must be identical with the coefficient of the same _{} on the right-hand
side. That is, _{}

Therefore the operator *A*
is simply equivalent to *matrix
multiplication*:

_{}

Evidently, then, applying
two linear operators one after the other is equivalent to successive matrix
multiplication—and, therefore, *since
matrices do not in general commute, nor do linear operators*. (Of course, if
we hope to represent quantum variables as linear operators on a vector space,
this has to be true—the momentum operator _{}certainly doesn’t commute with *x*!)

It is important to note that a linear operator applied
successively to the members of an orthonormal basis might give a new set of
vectors which *no longer span the entire
space*. To give an example, the
linear operator _{} applied to any vector
in the space picks out the vector’s component in the _{} direction. It’s called a *projection operator*. The operator _{} projects a vector into
its components in the subspace spanned by the vectors_{} and _{}, and so on—if we extend the sum to be over the whole basis,
we recover the identity operator.

*Exercise*: prove
that the _{} matrix representation
of the projection operator _{} has all elements zero
except the first two diagonal elements, which are equal to one.

There can be no *inverse*
operator to a nontrivial projection operator, since the information about
components of the vector perpendicular to the projected subspace is lost.

As we’ve discussed, if a ket _{} in the *n*-dimensional space is written as a
column vector with *n* (complex) components,
the corresponding bra is a *row* vector
having as elements the complex conjugates of the ket elements. _{} then follows
automatically from standard matrix multiplication rules, and on multiplying _{} by a complex number *a *to get _{} (meaning that each
element in the column of numbers is multiplied by *a*) the corresponding bra goes to _{}

But suppose that instead of multiplying a ket by a number,
we operate on it with a linear operator. What generates the parallel
transformation among the bras? In other
words, if _{}what operator sends the bra _{} to _{}? It must be a linear
operator, because *A* is linear, that
is, if under *A* _{} and _{} then under *A*
_{} Consequently, under
the parallel *bra *transformation we
must have _{}—the bra transformation is necessarily also linear. Recalling that the bra is an *n*-element *row* vector, the most general linear transformation sending it to
another bra is an _{} matrix operating on
the bra from the *right*.

This bra operator is called the *adjoint* of *A*, written _{} That is, the ket _{} has corresponding bra _{} In an orthonormal
basis, using the notation _{} to denote the bra _{} corresponding to the
ket _{} say,

_{}

So *the adjoint operator is the transpose complex
conjugate*.

*Important*: for a *product*
of two operators (prove this!),

_{}

**An operator equal to its adjoint _{} is called Hermitian**. As we shall find in the next lecture,
Hermitian operators are of central importance in quantum mechanics. An operator equal to

_{}

The definition of adjoint naturally extends to *vectors
and numbers*: the adjoint of a ket is the corresponding bra, the adjoint of
a number is its complex conjugate. This
is useful to bear in mind when taking the adjoint of an operator which may be
partially constructed of vectors and numbers, such as projection-type
operators. The adjoint of a product of
matrices, vectors and numbers is the product of the adjoints *in reverse
order*. (Of course, for numbers the
order doesn’t matter.)

An operator is *unitary* if _{} This implies first
that *U* operating on any vector gives a vector having the same norm,
since the new norm _{}. Furthermore, inner
products are preserved, _{} *Therefore, under a
unitary transformation the original orthonormal basis in the space must go to
another orthonormal basis*.

Conversely, any transformation that takes one orthonormal
basis into another one is a unitary transformation. To see this, suppose that a linear
transformation *A* sends the members of
the orthonormal basis _{}to the different orthonormal set _{}, so _{}etc. Then the vector _{} will go to _{}having the same norm, _{}. A matrix elememt _{}, but also _{}. That is, _{} for *arbitrary* kets _{}. This is only
possible if _{}, so *A* is unitary.

A unitary operation amounts to a rotation (possibly combined
with a reflection) in the space. Evidently,
since _{} the adjoint _{} rotates the basis
back—it is the inverse operation, and so _{} also, that is, *U* and _{} commute.

We review in this section the *determinant* of a
matrix, a function closely related to the operator properties of the matrix.

Let’s start with _{} matrices:

_{}

The determinant of this matrix is defined by:

_{}

Writing the two rows of the matrix as vectors:

_{}

(*R*
denotes row), _{} is just the *area*
(with appropriate sign) of the parallelogram having the two row vectors as
adjacent sides:

This is zero if the two vectors are parallel (linearly
dependent) and is not changed by adding any multiple of _{} to _{} (because the new
parallelogram has the same base and the same height as the original—check this
by drawing).

Let’s go on to the more interesting case of _{}matrices:

_{}

The determinant of *A*
is defined as

_{}

where _{} if any two are equal,
+1 if *ijk* = 123, 231 or 312 (that is to say, an even permutation of 123)
and –1 if *ijk* is an odd permutation of 123. Repeated suffixes, of course, imply summation
here.

Writing this out explicitly,

_{}

Just as in two dimensions, it’s worth looking at this expression in terms of vectors representing the rows of the matrix

_{}

so

_{}

This is the volume of the parallelopiped formed by the three vectors being adjacent sides (meeting at one corner, the origin).

This parallelepiped volume will of course be zero if the
three vectors lie in a plane, and it is not changed if a multiple of one of the
vectors is added to another of the vectors.
That is to say, *the determinant of
a matrix is not changed if a multiple of one row is added to another row*. This is because the determinant is linear in
the elements of a single row,

_{}

and the last term is zero because two rows are identical—so the triple vector product vanishes.

A more general way of stating this, applicable to larger
determinants, is that for a determinant with two identical rows, the symmetry
of the two rows, together with the *antisymmetry* of _{} ensures that the terms
in the sum all cancel in pairs.

Since the determinant is not altered by adding some multiple of one row to another, if the rows are linearly dependent, one row could be made identically zero by adding the right multiples of the other rows. Since every term in the expression for the determinant has one element from each row, the determinant would then be identically zero. For the three-dimensional case, the linear dependence of the rows means the corresponding vectors lie in a plane, and the parallelepiped is flat.

The algebraic argument generalizes easily to _{} determinants: they are
*identically zero if the rows are linearly
dependent*.

The generalization from _{} determinants is that _{}becomes:

_{}

where *ijk…p* is summed over all permutations of 132…*n*,
and the _{} symbol is zero if any
two of its suffixes are equal, +1 for an even permutation and -1 for
an odd permutation. (*Note*: any
permutation can be written as a product of swaps of neighbors. Such a representation is in general not
unique, but for a given permutation, all such representations will have either
an odd number of elements or an even number.)

An important theorem is that for a product of two matrices *A*,
*B* the determinant of the product
is the product of the determinants, _{} This can be verified
by brute force for _{} matrices, and a proof
in the general case can be found in any book on mathematical physics (for
example, Byron and Fuller).

It can also be proved that if the *rows* are linearly independent, the determinant cannot be zero.

(*Here’s a proof*:
take an _{} matrix with the *n* row vectors linearly independent. Now consider the components of those vectors
in the *n* – 1 dimensional subspace perpendicular to (1, 0, … ,0). These *n*
vectors, each with only *n* – 1
components, must be linearly dependent, since there are more of them than the
dimension of the space. So we can take some combination of the rows below the
first row and subtract it from the first row to leave the first row (*a*, 0, 0, … ,0), and *a* cannot be zero since we have a matrix with *n* linearly independent rows. We can then subtract multiples of this
first row from the other rows to get a determinant having zeros in the first
column below the first row. Now look at the *n*
– 1 by *n* – 1 determinant to be
multiplied by *a*.

Its rows must be linearly independent since those of the original matrix were. Now proceed by induction.)

To return to three dimensions, it is clear from the form of

_{}

that we could equally have taken the *columns *of *A*
as three vectors, _{} in an obvious notation,
_{}, and linear dependence among the *columns* will also
ensure the vanishing of the determinant—so, in fact, linear dependence of the
columns ensures linear dependence of the rows.

This, too, generalizes to _{}: in the definition of determinant _{}, the row suffix is fixed and the column suffix goes over all
permissible permutations, with the appropriate sign—but the same terms would be
generated by having the *column*
suffixes kept in numerical order and allowing the row suffix to undergo the
permutations.

It
is perhaps worth mentioning how the inverse of a _{} matrix operator can be
understood in terms of vectors. For a
set of linearly independent vectors _{}, a reciprocal set _{} can be defined by

_{}

and the obvious cyclic definitions for the other two reciprocal vectors. We see immediately that

_{}

from which it follows that the inverse matrix to

_{}.

(These reciprocal vectors are important in *x*-ray crystallography, for example. If a crystalline lattice has certain atoms at
positions _{}, where *n*_{1}, *n*_{2}, *n*_{3}
are integers, the reciprocal vectors are the set of normals to possible planes
of the atoms, and these planes of atoms are the important elements in the
diffractive *x*-ray scattering.)

If an operator *A*
operating on a ket _{} gives a multiple of
the same ket,

_{}

then _{} is said to be an *eigenket
*(or, just as often, *eigenvector*, or *eigenstate*!) of *A* with *eigenvalue _{}*.

Eigenkets and eigenvalues are of central importance in quantum mechanics: dynamical variables are operators, a physical measurement of a dynamical variable yields an eigenvalue of the operator, and forces the system into an eigenket.

In this section, we shall show how to find the eigenvalues
and corresponding eigenkets for an operator *A*. We’ll use the notation _{} for the set of
eigenkets _{} with corresponding
eigenvalues *a _{i}*. (Obviously, in the eigenvalue equation here
the suffix

The first step in solving _{} is to find the allowed
eigenvalues *a _{i}*.

Writing the equation in matrix form:

_{}

This equation is actually telling us that the *columns* of the matrix _{}are linearly dependent!
To see this, write the matrix as a row vector each element of which is
one of its columns, and the equation becomes

_{}

which is to say

_{}

the columns of the matrix are indeed a *linearly dependent* set.

We know that means the determinant of the matrix _{} is zero,

_{}

Evaluating the determinant using _{} gives an *n*^{th}
order polynomial in_{} sometimes called the *characteristic
polynomial*. Any polynomial can be
written in terms of its roots:

_{}

where the *a _{i}*’s,
the roots of the polynomial, and

Notice that the diagonal term in the determinant _{} generates the leading
two orders in the polynomial _{}, (and some lower order terms too). Equating the coefficient of _{} here with that in _{},

_{}

Putting _{} in both the
determinantal and the polynomial representations (in other words, equating the_{}-independent terms),

_{}

So we can find both the sum and the product of the
eigenvalues directly from the determinant, and for a _{} matrix this is enough
to solve the problem.

For anything bigger, the method is to solve the polynomial
equation _{}to find the set of eigenvalues, then use them to calculate the
corresponding eigenvectors. This is done
one at a time.

Labeling the first eigenvalue found as *a*_{1}, the corresponding equation for the components *v _{i} *of the eigenvector

_{}

This looks like *n*
equations for the *n* numbers *v _{i}*, but it isn’t: remember
the rows are linearly dependent, so there are only

the ratios of the vector components *v*_{1},
…, *v _{n}*, then finally the
eigenvector is normalized. The process
is then repeated for each eignevalue.
(Extra care is needed if the polynomial has coincident roots—we’ll
discuss that case later.)

For a *Hermitian
*matrix, it is easy to establish that the eigenvalues are always *real*. (*Note*:
A basic postulate of Quantum Mechanics, discussed in the next lecture, is that
physical observables are represented by Hermitian operators.) Taking (in this section) *A* to be hermitian, _{}and labeling the eigenkets by the eigenvalue, that is,

_{}

the inner product
with the bra _{} gives _{} But the inner product of the *adjoint* equation (remembering _{})

_{}

with _{} gives _{} so _{}, and all the eigenvalues must be real.

They certainly don’t have to all be different—for example,
the unit matrix *I *is Hermitian, and
all its eigenvalues are of course 1. But let’s first consider the case where
they *are* all different.

It’s easy to show that the eigenkets belonging to *different
*eigenvalues are *orthogonal*.

If

_{}

take the adjoint of the first equation and then the inner
product with _{}, and compare it with the inner product of the second
equation with _{}:

_{}

so _{} unless the eigenvalues
are equal. (If they *are* equal, they are referred to as *degenerate* eigenvalues.)

Let’s first consider the nondegenerate case: *A* has all eigenvalues distinct. The eigenkets of *A*, appropriately normalized, form an orthonormal basis in the
space.

Write

_{}

Now

_{}

so

_{}

Note also that, obviously, *V* is unitary:

_{}

We have established, then, that for a Hermitian matrix with
distinct eigenvalues (nondegenerate case), the unitary matrix *V* having columns identical to the
normalized eigenkets of *A* *diagonalizes* *A*, that is, _{}is diagonal.
Furthermore, its (diagonal) elements equal the corresponding eigenvalues
of *A*.

Another way of saying this is that the unitary matrix *V* is the transformation from the
original orthonormal basis in ths space to the basis formed of the normalized
eigenkets of *A*.

We’ll now move on to the general case: what if some of the
eigenvalues of *A* are the same? In this case, any linear combination of them
is also an eigenvector with the same eigenvalue. *Assuming
they form a basis in the subspace*, the Gram Schmidt procedure can be used
to make it orthonormal, and so part of an orthonormal basis of the whole space.

However, we have not actually established that the
eigenvectors *do* form a basis in a
degenerate subspace. Could it be that
(to take the simplest case) the two eigenvectors for the single eigenvalue turn
out to be parallel? This is actually the
case for some 2×2 matrices—for example, _{}, we need to prove it is *not*
true for Hermitian matrices, and nor are the analogous statements for
higher-dimensional degenerate subspaces.

A clear presentation is given in Byron and Fuller, section 4.7. We follow it here. The procedure is by induction from the 2×2 case. The general 2×2 Hermitian matrix has the form

_{}

where *a*, *c* are real. It is easy to check that if the eigenvalues
are degenerate, this matrix becomes a real multiple of the identity, and so
trivially has two orthonormal eigenvectors. Since we already know that if the eigenvalues
of a 2×2 Hermitian matrix are distinct it can be diagonalized by the unitary
transformation formed from its orthonormal eigenvectors, we have established
that *any* 2×2 Hermitian matrix can be
so diagonalized.

To carry out the induction process, we now assume any (*n* – 1) ×(*n* – 1) Hermitian matrix can be diagonalized by a unitary
transformation. We need to prove this
means it’s also true for an *n* × *n* Hermitian matrix *A*. (Recall a unitary
transformation takes one complete orthonormal basis to another. If it diagonalizes a Hermitian matrix, the
new basis is necessarily the set of orthonormalized eigenvectors. Hence, if the matrix can be diagonalized, the
eigenvectors do span the *n*-dimensional
space.)

Choose an eigenvalue *a*_{1}
of *A*, with normalized eigenvector _{} (We put in *T* for transpose, to save the awkwardness
of filling the page with a few column vectors.)
We construct a unitary operator *V*
by making this the first column, then filling in with *n* – 1 other normalized vectors to construct, with _{}, an *n*-dimensional
orthonormal basis.

Now, since _{}, the first column of the matrix *AV* will just be _{}, and the *rows* of
the matrix _{} will be _{} followed by *n* – 1 normalized vectors orthogonal to
it, so the first column of the matrix _{} will be *a*_{1} followed by zeros. It is easy to check that _{} is Hermitian, since *A* is, so its first row is also zero
beyond the first diagonal term.

This establishes that for an *n *× *n* Hermitian matrix, a
unitary transformation exists to put it in the form:

_{}

But we can now perform a second unitary transformation in
the (*n* – 1) ×(*n* – 1) subspace orthogonal to _{} (this of course leaves
_{} invariant), to
complete the full diagonalization—that is to say, the existence of the (*n* – 1) ×(*n* – 1) diagonalization, plus the argument above, guarantees the
existence of the *n* × *n* diagonalization: the induction is
complete.

As discussed above, a Hermitian matrix is diagonal in the
orthonormal basis of its set of eigenvectors: _{}, since

_{}

If we are given the matrix elements of *A* in some other orthonormal basis, to diagonalize it we need to rotate from the
initial orthonormal basis to one made up of the eigenkets of *A*.

Denoting the initial orthonormal basis in the standard fashion

_{}

the elements of the matrix are _{}.

A transformation from one orthonormal basis to another is a *unitary* transformation, as discussed above,
so we write it

_{}

Under this transformation, the matrix element

_{}

So we can find the appropriate
transformation matrix *U* by requiring that _{} be diagonal with
respect to the *original* set of basis
vectors. (Transforming the operator in
this way, leaving the vector space alone, is equivalent to rotating the vector
space and leaving the operator alone. Of
course, in a system with more than one operator, the same transformation would
have to be applied to all the operators).

In fact, just as we discussed for the nondegenerate
(distinct eigenvalues) case, the unitary matrix *U* we need is just
composed of the normalized eigenkets of the operator *A*,

_{}

And it follows as before that

_{}

(The repeated suffixes here are of course *not* summed over.)

If some of the eigenvalues are the same, the Gram Schmidt procedure may be needed to generate an orthogonal set, as mentioned earlier.

The same unitary operator *U* that diagonalizes an Hermitian matrix *A* will also diagonalize *A*^{2}, because

_{}

so

_{}

Evidently, this same process works
for any power of *A*, and formally for
any function of *A* expressible as a
power series, but of course convergence properties need to be considered, and
this becomes trickier on going from finite matrices to operators on infinite
spaces.

From the above, the set of powers of an Hermitian matrix all commute with each other, and have a common set of eigenvectors (but not the same eigenvalues, obviously). In fact it is not difficult to show that any two Hermitian matrices that commute with each other have the same set of eigenvectors (after possible Gram Schmidt rearrangements in degenerate subspaces).

If two _{} Hermitian matrices *A*,
*B* commute, that is, *AB* = *BA*, and *A* has a *nondegenerate*
set of eigenvectors _{} then _{} that is, _{} is an eigenvector of *A*
with eigenvalue *a _{i}*. Since

Now suppose *A* *is*
degenerate, and consider the _{}subspace _{} spanned by the
eigenvectors _{} of *A* having
eigenvalue _{}. Applying the
argument in the paragraph above, _{} must also lie in this
subspace. Therefore, if we transform *B* with the same unitary
transformation that diagonalized *A*, *B* will not in general be
diagonal in the subspace _{}, but it will be what is termed *block diagonal*, in
that if *B* operates on any vector in _{} it gives a vector in _{}.

*B* can be written as two
diagonal blocks: one _{}, one _{}with zeroes outside these diagonal blocks, for example, for *m*
= 2, *n* = 5:

_{}

And, in fact, if there is only *one*
degenerate eigenvalue that second block will only have nonzero terms on the
diagonal:

_{}

*B* therefore operates on two
subspaces, one *m*-dimensional, one (*n*-*m*)-dimensional, *independently*—a
vector entirely in one subspace stays there.

This means we can complete the diagonalization
of *B* with a unitary operator that *only *operates on the _{} block _{}. Such an operator
will also affect the eigenvectors of *A*, but that doesn’t matter, because
all vectors in this subspace are eigenvectors of *A* with the *same*
eigenvalue, so as far as *A* is concerned, we can choose any orthonormal
basis we like—the basis vectors will still be eigenvectors.

This establishes that any two *commuting*
Hermitian matrices can be diagonalized at the same time. Obviously, this can
never be true of *noncommuting* matrices,
since all diagonal matrices commute.

Any unitary matrix can be diagonalized by a unitary
transformation. To see this, recall that
*any* matrix *M* can be written as a sum of a Hermitian matrix and an anti
Hermitian matrix,

_{}

where both *A*, *B* are Hermitian. This is the matrix analogue of writing an
arbitrary complex number as a sum of real and imaginary parts.

If *A*, *B* commute, they can be simultaneously
diagonalized (see the previous section), and therefore *M* can be diagonalized. Now,
if a *unitary* matrix is expressed in
this form _{}with *A*, *B*
Hermitian, it easily follows from_{} that *A*, *B* commute, so *any unitary matrix U can be diagonalized by a unitary transformation.* More generally, if a matrix *M* commutes with its adjoint _{}, it can be diagonalized.

(*Note*: it is *not* possible to diagonalize *M* unless both *A*, *B* are simultaneously
diagonalized. This follows from _{} being Hermitian and
antiHermitian for any unitary operator *U*,
so their off-diagonal elements cannot cancel each other, they must all be zero
if *M* has been diagonalized by *U*, in which case the two transformed
matrices _{} are diagonal,
therefore commute, and so do the original matrices *A*, *B*.)

It is worthwhile looking at a specific example, a simple rotation of one orthonormal basis into another in three dimensions. Obviously, the axis through the origin about which the basis is rotated is an eigenvector of the transformation. It’s less clear what the other two eigenvectors might be—or, equivalently, what are the eigenvectors corresponding to a two dimensional rotation of basis in a plane? The way to find out is to write down the matrix and diagonalize it.

The matrix

_{}

Note that the determinant is equal to unity. The eigenvalues are given by solving

_{}

The corresponding eigenvectors satisfy

_{}

The eigenvectors, normalized, are:

_{}

Note that, in contrast to a Hermitian matrix, the
eigenvalues of a unitary matrix do not have to be real. In fact, from _{}, sandwiched between the bra and ket of an eigenvector, we
see that any eigenvalue of a unitary matrix must have unit modulus—it’s a
complex number on the unit circle. With
hindsight, we should have realized that one eigenvalue of a two-dimensional
rotation had to be _{}, the product of two two-dimensional rotations is given be
adding the angles of rotation, and a rotation through _{} changes all signs, so
has eigenvalue -1. Note that the
eigenvector itself is independent of the angle of rotation—the rotations all
commute, so they must have common eigenvectors. Successive rotation operators
applied to the plus eigenvector add their angles, when applied to the minus
eigenvector, all angles are subtracted.