Hassani S. Mathematical Physics: A Modern Introduction to Its Foundations

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4.4

SPECTRAL

DECOMPOSITION

121

The characteristic polynomial is det(H - A1) = (A+ 2)2(A -

2)2

Thus, Al =

-2

with

multiplicity m1 = 2, and).2 = 2

With

multiplicity m2 = 2. Tofindthe

eigenvectors,

first

look

at the matrix equation (H+21) la} = 0, or

-1-;

-1+;

-I-.i)

GI)

I+!

"'2

=0.

o "'3

2 4

Thisis a systemof

linear

equations

whose

"solution"

Wehavetwo

arbitrary

parameters,

soweexpecttwolinearly

independent

solutions.

Forthe

two choices

=2, cz=0 and

= 0,

Cl2

= 2, we obtain, respectively,

and

which happen to he orthogonal. We simply normalize them to ohtain

and

Similarly, the second eigenvalue equation, (H- 21)

la} = 0, gives rise to the conditions

"'3

= -!(l+ ;)("'1

+"'2)

and "'4 =

-!(l-

;)("'1 - "'2), whichprodnce the orthonormal

vectors

and

le4} = I", (

.).

2...,2 -

1-;

Theunitary

matrix

that

diagonaIizes

Hcanbe

constructed

fromthesecolunm

vectors

using the remarks hefore Example 3.4.4, which implythat

we simply put the vectors lei}

together

columns,

theresulting

matrix

is U

2 0

0 2 0

= _

2-,/2

1+; 1+;

-1-;

1-;

-1+; -1+;

122 4.

SPECTRAL

DECOMPOSITION

and the unitary

matrix

will be

2 0

1+ i

-1-

U=(Ut)t=_,

2,f'i

2 0

-I+i

-1-

0 2

-I+i

1 + i

can

easily

check

that U diagonalizes H, i.e.,

that

UHU

is diagonal.

application

diagonalizalion

electromagnetism

4.4.12.

Example.

In somephysicalapplications the abilityto diagonaIize matricescan

be very useful. As a simple but illustrative example, let us consider the motion

a charged

particle in a constantmagnetic field pointing in the zdirection.

The

equation

motion for

suchaparticleis

(eX

m--.! = qv x B =

qdet

dt 0

which in

component

form becomes

-=--V

dt m

-=0.

Ignoring the uniform motion in the z direction, we need to solve the first two coupled

equations, whichin matrix form becomes

)

( 0

dt v

-;;;

-1

(4.10)

where we have introduced a factor

i to

render

the

matrix

hermitian, and defined

(j)

qB/m.

the 2 x 2

matrix

were

diagonal, we

would

get

two

uncoupled equations,

which

could

solve easily. Diagonalizing

the

matrix. involves finding a

matrix

such

that

wecoulddo sucha diagonalization, wewouldmuttipty(4.10)by Rto get'

which

can

bewritten as

(vi)

_ .

(1'1

-d

-leV

t v

where

5The fact that R is independent of t is crucial in this step. This fact, in tum, is a consequence of the independence from t of

the original 2 x 2 matrix.

4.4 SPECTRAL DECDMPDSITIDN 123

Wethenwouldhavea

pairof

uncoupled

equations

that

havev' = v'

e-iwtttt

andv' = v'

e-iw!LZ

asa solutionset in whichv' andv'

XOx

YOy

'Ox

areintegration constants.

To find R, we need the normalizedeigenvectors

(~i

b).

But these are obtainedin

preciselythesame

fashion

asin

Example

4.4.4. Thereis,

however,

arbitrariness

in the

solutions due to the choice in numbering the eigenvalues.

we choose the normalized

eigenvectors

(-i)

le2) =

-J'i

1 '

thenfromcommentsatthe endof Section 3.3, we get

-1

t 1

=R=-J'il

With

this

choiceof R,wehave

t t

1 (-i

R=(R)

-J'i'

so that1'1 = 1 =

-1'2.

HavingfoundRt, we can write

(

Vx)

= Rt

) =

-J'i

-i)

(Vbxe~i"")

1 v' e

lwf

(4.11)

thex andy

components

of velocityatt = 0 areVOx andVDy'respectively, then

(

vox)

= Rt

) ,

Substituting iu (4.11), weobtain

(

)

= R

(vox)

(-iVo

VDy..,fi

vOx

-i)

«-iVox

voy)e~iwt)

= (

vOx

cos

VDy

sin

)

(ivox+voy)elwt

-vOxsinwt+voyCOSW(

simultaneous

diagonalization

defined

This gives the velocity

asa

function

time. Antidifferentiating once with respect to time

yields the position vector. II

many situations

physical interest, it is desirable to know whether two

operators are simultaneously diagonalizable.

For

instance, if there exists a basis

a Hilbert space

a quantum-mechanical system consisting

simultaneous

eigenvectors

two operators, then one can measure those two operators at the

same time.

particular, they are not restricted by an uncertainty relation.

4.4.13. Definition.

Two

operators

aresaidtobesimultaneouslydiagonalizable if

they can be written interms

a/the

same set a/projection operators, as in Theorem

4.4.6.

124 4.

SPECTRAL

OECOMPOSITION

This definitionis consistentwith the matrixrepresentation

the two operators,

becauseif we take the orthonormalbasis

B = (I

ej)

) discussedrightafterTheorem

4.4.6, we obtaindiagonalmatricesfor bothoperators.

What

are the conditionsunder

which two operators can be simultaneously diagonalized? Clearly, a necessary

condition is

that

thetwo

operators

commute.

Thisisan

immediate

consequence of

the orthogonalityof the projectionoperators, whichtrivially implies PiPj = P

jPi

for all i and

is also apparentin the matrix representation

the operators: Any

two diagonal matrices commute.

What

about sufficiency? Is the commutativity

of the two operators sufficient for them to be simultaneously diagonalizable? To

answer this question, we need the following lemma:

4.4.14.

Lemma.

operatorTcommutes with a normal operator A ifand only if

T commutes with all the projection operators

Proof

The

"if"

part is trivial. To prove the "only

if"

part, suppose AT = TA,

and let

Ix} be any vector in one of the eigenspaces

A, say Mi- Then we have

A(T

Ix}) =T(A Ix}) =

T(A.j

Ix}) =Aj(T Ix»);i.e., T Ix)

isinMj,orMj

isinvariant

under T. Since Mj is arbitrary, T leaves all eigenspaces invariant.

In particular, it

leaves M

t,the orthogonal complement

M j (the direct sum

all the remaining

eigenspaces), invariant. By Theorems 4.2.3 and 4.2.5, TPj

= PjT; and this holds

forallj.

necessary

and

sufficient

condition

for

simultaneous

diagonalizability

4.4.15.

Theorem.

A necessary

and

sufficient condition

for

two normal operators

and

B to be simultaneously diagonalizable is [A, B] = O.

Proof As

claimed

above,

the

"necessity"

trivial.

Toprovethe

"sufficiency,"

let

A =

Lj~l

AjPj and B =

Lf~l

I-'kQk, where {Aj}and {Pj} are eigenvalues and

projections

A, and

{M}

and {Qk! are those of B. Assume [A, B] =

Then

by Lemma 4.4.14,

AQk

QkA.

Since

commutes with A, it must commute

with the latter's projectionoperators: P

jQk

QkP

i- Tocomplete the proof, define

Rjk

PjQk,

and note that

Rjk

(PjQk)t

Qlpj

QkPj

PjQk

Rjk,

(Rjk)2

(PjQd

PjQkPjQk

PjPjQkQk

PjQk

Rjk.

Therefore, R

are hermitian projection operators. Furthermore,

r r

LRjk=LPjQk=Qk.

j=l

-----

Similarly,

Lf=l

jQk

= Pj

Lf=l

= Pi- We can now write A

and B as

r r s

A =

LAjPj

LLAjRjk,

j=l

j~l

k~l

, , r

B =

LI-'kQk

LLI-'kRjk.

k~l

k~lj=l

spectral

decomposition

ofa

Pauli

spin

matrix

4.5

FUNCTIONS

OPERATORS

125

By definition, they are simultaneously diagonalizable. D

4.4.16. Example. Let us findthe spectraldecompositionof thePaotispinmatrix

(

-i)

U2 = i 0 .

Theeigenvalues and

eigenvectors

havebeenfound in

Example

4.4.4. These

are

and

The

subspaces

JY[J..

j areone-dimensional;

therefore,

(1)

1 1

PI =

1eJ)

(ell =

-J'i

-i)

=:2 i

1(1)

1(1

P2 = !e2)

(e21

= - .

-2

-z

-I

Wecan checkthatPI +P2 =

(6?)

and

-i)

1 '

-i)

2-'

III

What

restrictions are to be imposed on the most general operator T to make

it diagonalizable? We saw in Chapter

2 that T can be written in terms

its so-

called Cartesian components as T

= H +iH' where both H and H' are hermitian

and can therefore be decomposed according to Theorem 4.4.6. Can we conclude

that T is also decomposable? No. Because the projection operators used in the

decomposition

may

not be the same as those used for H'. However,

H and

H' are simultaneously diagonalizable such that

and

H' =

I>~Pko

k~1

(4.12)

then T

L:k~1

()'k +

iA~)Pk.

follows that T has a spectral decomposition,

and therefore is diagonalizable. Theorem 4.4.15 now implies that H and

H' must

commute. Since,

H =

(T +Tt) and H' =

(T-

Tt),

we have [H,

H']

= 0

and

ouly

[T,Tt] = 0; i.e., T is normal. We thus get

back

to the condition with which

we started the whole discussion

spectral decomposition in this section.

4.5 Functionsof Operators

Functions

transformations were discussed in Chapter 2. With the power

spectral decomposition at our disposal, we can draw many important conclusions

about them.

126

SPECTRAL

OECOMPOSITION

First, we

note

thatifT

L;~l

A,P" then, because

orthogonality

the

P,'s

, ,

LATPi"'"

yn=

LAfPi.

i=1

;=1

Thus, any polynomial

pinT

has a spectral decomposition given by

p(T)

L;=l

p(Ai)Pi. Generalizing this to functions expandable in

power

series gives

f(T)

Lf(Ai)Pi.

i=l

(4.13)

4.5.1.

Example.

Let us investigatethe spectral decomposition of the followingunitary

(actuallyorthogonal)matrix:

-Sine)

cos a .

(

COS B -

()

sinO

Wefindthe

eigenvalues

(

CaSe - A

-SinB)

det .

=1.

-2cosOA+I=0.

SIn

v cos\] - A

yieldingAl =

and

= e

. ForAl we have

(reader,

provide

themissingsteps)

CO~9S~~il~)

(:~)=

a2 =

ial

andforA2.

(

COS

e-i()

sinB

- sinO )

("I)

cosB -

a2 =

-i"l

lez) =

Wenote that the

are one-dimensionaland spannedby [ej). Thus,

_ie-iO)

I ( e

-te +

-Ie

PI = leI) (ell =

(;)(1

Pz = lez)(ezl

(I.)

Clearly,PI + P

= 1, and

-i8

i9 1

(e-i()

e PI +e Pz = 2

ie-i8

-i)

O=!(l

-i)

I '

D·

iO)

if)

= U.

III

we takethenatural log

this

equation

anduse

Equation

(4.13), we obtain

InU = In(e-iO)Pl + In(eiO)pz=

-iOPI

+ ieP

-OPI

+ OPz)

iH, (4.14)

where

-fjPl

+8P2 isahennitian

operator

becausefj isrealandPIandP2

are

hermitian.

InvertingEquation(4.14)givesU= e

" , where

H =

O(-PI

+Pz) = 0

(~i

~).

The

square

root

operator

plagued

multivaluedness.

the

real

numbers,

have

only

two-valued

ness!

4.5

FUNCTIONS

OPERATORS

127

The example above showsthat the unitary 2 x 2 matrix Ucan be written as

an exponentialof an anti-hermitianoperator. This is a generalresult.

Infact, we

havethe following theorem,whoseproof is left as an exercisefor thereader (see

Problem4.22).

4.5.2.Theorem. A unitary operatorUon afinite-dimensionalcomplexinnerprod-

uctspacecanbewrittenas U =e

whereH is

hermitian.

Furthermore,

unitary

matrix can be broughtto diagonalform by a unitary transformation matrix.

The last statemeutfollowsfrom Corollary4.4.10 and the fact that

for anyfunctiou

f that can be expandedin a Taylorseries.

A usefulfunctionof anoperatorisits squareroot.A natural way

to definethe

squareroot of an operatorAis

.jA

L~~l

(±A)Pi.

This clearly givesmany

candidatesfortheroot,becauseeachtermin the sumcanhaveeithertheplussign

or themious sign.

4.5.3.Definition. Thepositive squareroot

apositiveoperatorA =

L~=l

A,P,

is.;A

L~~l

APi.

The uniquenessof the spectraldecompositionimpliesthat the positivesquare

root of a positiveoperatoris unique.

4.5.4.Example. Letus

evaluate

,fA

where

A=CS3i

~).

First,we haveto spectrally decomposeA.Itscharacteristic

equation

- lOA

+ 16 = 0,

withrootsAl = 8 and

= 2. Since botheigenvaluesarepositiveandAis

hermitian,

conclude

that

Ais indeedpositive(Corollary 4.4.8). Wecanalso easily findits

normalized

eigenvectors:

Thus,

and

(-i)

1<2)

=.,fi

1 .

and

;)

-i)

1 '

can

easily

check

!hat

(,fA)2

-i)

=.,fi

-i

III

128 4.

SPECTRAL

DECOMPOSITION

Intuitively, higher

and

higher

powers

when

acting on a few vectors

the space, eventually exhaust all vectors,

and

further increase in

power

will be

a repetition

lower powers.

This

intuitive

idea

can

made

more precise by

looking at

the

projection operators. We have already

seen

that Tn =

LJ=l

AjPj,

n =

1,2,

....

For

various n'sonecan"solve"for Pj in terms

powers ofT. Since

there are

only

a finite

number

Pj 's,

only

a finite

number

powers

T will

suffice. In fact, we

can

explicitlyconstructthe polynomialin Tfor Pi-

thereis such

a polynomial, by Equation

(4.13) it

must

satisfy

pj(T)

L~~l

Pj(Ak)Pk,

where Pj is somepolynomial to be determined. By orthogonality

the projection

operators,

Pj O.k)

must

be zero unless k =

in which caseit

mnst

In other

words,

Pj O.k) = 8kj. Such a polynomial

can

be explicitly constructed:

Therefore,

(4.15)

and we have the following result.

4.5.5. Proposition. Every function

a normal operator

a finite-dimensional

vectorspace

canbe expressedas apolynomial. Infact.from Equations (4./3)and

(4.15),

(4.16)

4.5.6.

Example.

Let us write

.,fA

of the last exampleas a polynomialin A. Wehave

Substitutingin Equation(4.16), we obtain

v'8

.,fi .,fi

v'8

.,fA

j):j

PI (A)

+../i:i.P2(A)

-(A

-2)

-(A-

8) =

-A+-.

6 6 6 3

TheRHS is

clearly

(first-degree)

polynomial in A,andit is easy to

verify

that

it is the

matrix

..fA

obtainedinthepreviousexample. II

4.6

POLAR

DECOMPOSITION

129

4.6 PolarDecomposition

We have seen many similarities between operators and complex numbers.

For

instance, hermitian operators behave very

much

like the real numbers: they have

real

eigenvalues;

their

squares

are

positive;

every

operator

canbe

written

asH+i

H',

where both H and H' are hermitian; and so forth. Also, unitary operators can be

written as expiH, where His hermitian. So unitary operators are the analogue

complex numbers

unit magnitude such as e

•

A general complex number can

be written as

re'", Can we write an arbitrary operator in an analogous way?

The

following theorem provides the answer.

polar

decomposition

4,6.1.

Theorem.

(polardecompositiontheorem)An operatorAonafinite-dimen-

theorem

sionalcomplexinnerproductspacecanbewrittenas A =

whereRisa(unique)

positive operatorandU a unitaryoperator.

A is invertible,then Uis alsounique.

Proof

We will prove the theorem for the case where the operator is invertible.

The

proof

the general case

can

be found in books on linear algebra (such as

[Halm 58]).

The

reader may show that the operator

A is positive. Therefore, it has a

unique positive square root

R.We let V = RA

-I,

or VA = R. Then

RA-1(RA-1)t

RA-1(A-1)tRt

R(AtA)-IRt

= R(R

2)-I

R(RtR)-IRt

RR-1(Rtl-1Rt

= 1

and

and Vis indeed a unitary operator. Now choose U

= vt to get the desired decom-

position.

To prove uniqueness we note that

= U'R' implies that R =

UtU'R'

and

= RtR =

(utu'R'lt(utU'R')

R'tU'tUUtU'R'

R'tR'

= R'2. Since the

positive transformation

(or R'2l has only one positive square root, it follows

that R

= R'

A is invertible, then so is R = UtA. Therefore,

= U'R

URR-

U'RR-

U = U',

and Uis also unique.

is interestingto note that the positivedefiniteness

Rand the nonuniqueness

U are the analogue

the positivity

r and the nonuniqueness

in the

polar representation

complex numbers:

z = re

= re

(

+ 2mr)

EZ.

130 4.

SPECTRAL

DECOMPOSITION

In practice, R is found by spectrally decomposing AtA and taking its positive

square root.

Once Ris found, Ucan be calculated from the definition A

= UR.

4.6.2.

Example.

Letus findthepolardecomposition of

(-gi

We have

The

eigenvalues

and

eigenvectors

of R

are

routinely

found

tobe

I (i)

Al = 18,

1.2

= 2,

let}

= 2../2

The

projection

matrices

are

i../7\

Thus,

TofindU,we notethatdetAis nonzero.Hence,Ais invertible, whichimplies

that

Ris also

invertible.

The

inverse

of Ris

Theunitary mattixis simply

U =

AR-

(-iI5../2

3v'14\

3i-m

15../2)·

Itis leftforthe

reader

toverify

that

Uis

indeed

unitary.

4.7 Real Vector Spaces

The treatment so far in this chapterhas focused on complex inner product spaces.

The complex number system is far more complete than the

real numbers. For

example, in preparation for the proof of the spectral decomposition theorem, we

used the existence of

n roots of a polynomial of degree n over the complex field

6It is

important

topay

attention

to theorderof thetwo

operators:

OnedecomposesAtA,notAAt.