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

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2.6

OPERATORS

IN NUMERICAL

ANALYSIS

can

define products (composition)

the three operators. In particular, A

is given by

f; =A(f1+1 -

ft)

11+2

- 2/1+1 +

f;.

Similarly,

f; =

V(f;

li-l)

= Ii -

2li-l

Ii-a.

note

that

(2.13)

f; = f;+J - 21i +

Ii-I.

(2.14)

shifting

and

averaging

operators

f; = f;+1 - f; -

(f;

Ii-I)

V)li

(j2 =

A-V.

'---v-'

'-,,-'

=Afi

=V!i

This shows thatthe three operators are related.

is convenientto introduce

the

shifting

and

averaging

operators,respectively

and

p" as

E/(x)

I(x

+h),

p,1(x) =

+~)

~)].

(2.15)

Note that for any positive integer

n, En

I(x)

I(x

+nh). We generalize this to

any real

number

I(x)

I(x

ah).

All

the

other finite-difference operators

can

be written in terms

(2.16)

A=E-1,

V=1-E-

(j=E

2_E-

1/2,

p,=~(El/2+E-l/2)

(2.17)

The

first two equations

(2.17)

can

be rewritten as

E=l

+A,

E = (1 -

V)-l

(2.18)

can

obtaina usefulformulafor

the

shiftingoperatorwhenit acts on polynomials

degree n or less. First note that

1 - V

= (1 -

V)(l

+ V +

...

+ V

) .

But V

armihilates all polynomials

degree n

less

(see Problem 2.33).

Therefore, for such polynomials, we have

1 = (1 -

V)(l

+ V +

. + V

) ,

(2.20)

72 2.

OPERATOR

ALGEBRA

which

shows

that E = (1 -

V)-I

= 1 +V +...+V

Now

let n -+

and

obtain

E = (1 -

V)-I

= LV

(2.19)

k~O

for

polynomials

any degree

and-hy

Taylor

expansion-for

any

(well-behaved)

function.

2.6.1.

Example.

Numerical interpolation illustrates the use of the

formulas

derived

above.

Suppose

that

are

given a

table

of the valuesof a

function,

andwe

want

the

valueof the

function

foranx located

between

two

entries.

cannot

usethe

table

directly,

butwe mayuse thefollowing

procedure.

Assume

that

the

values

ofthe

function

are

givenfor

Xl,

X2,

...

xi,

...

and

we are

interested

in thevalueof the

function

forx such

that

xi < x < xi+ 1.This

corresponds

Ii+r. where0

< r < 1.Wehave

r r ( r(r -

2 )

li+r=Eli=(1+~)/i=

1+r~+--2-~

+...

j;.

practice,

the

infinite

sumis

truncated

after

finite

number

terms.

onlytwotenusare

kept,

we have

li+r "" (1 +

r~)li

= Ii +r(fi+1 -

Ii)

r)/i

rli+l.

(2.21)

In particular. for r =

Equatiou (2.21) yields

li+l/2

t(fi

li+I).

which states the

reasonable

result

that

thevalueatthe

midpoint

approximately

equalto the

average

ofthe

values

attheend

points.

the third term of the series in Equation (2.20) is also retained. theu

[

r(r -

I) 2] r(r - I) 2

li+r""

+r~+

--2-~

Ii =Ii

+r~/i

--2-~

. r(r -

= Ii +

r(fi+l

/i)

--2-(1i+2

2li+l

+Ii) (2.22)

~-~(I-~

r~-I)

= 2 Ii +r(2 - r)li+1 +

--2-11+2'

For r =

that is. at the midpoint betweeu Xi and Xi+1. Equation (2.22) yields

1i+1/2 ""

~Ii

lli+l

!/i+2.

which

turns

outtohea

better

approximation

than

Equation

(2.21).

However,

involves

not

onlythetwo

points

either

sideofx butalsoa

relatively

distant

point,

Xi+2,

wewereto

retain terms up to

for k > 2. theu Ii+r would be giveu in terms of h. Ii+j.

...•

li+k>

and

the

result

wouldbe

accurate

than

(2.22).

Thus,

the

information we have

about

the

behavior

of a

function

distant

points;

the

better

we can

approximate

it at

(Xj,Xi+l).

The

foregoing

analysis wasbasedon

forward

interpolation.

Wemaywanttouse

back-

wardinterpolation,

where

fi-r

issoughtfor 0 < r < 1.In sucha caseweusethe

backward

difference

operator

(r(r

- I) 2 )

Ii-r

(E-

)rIi =(1 -

V)r

Ii = 1 -

--2-V

+... h-

III

differentiation

and

integration

operators

2.6

OPERATORS

NUMERICAL

ANALYSIS

2.6.2. Example. Let us check the conclusionmade in Example2.6.1 using a calculator

and a specific function, say sinx. A calculator gives sin(O.I)

= 0.0998334,

sin(O.2)

0.1986693,sin(0.3) = 0.2955202.Supposethat wewantto findsin(O.l5)by interpolation.

UsingEquation(2.21)with

r =

we obtain

sin(0.15) "" ![sin(O.I)

+ sin(0.2)] = 0.1492514.

On theother hand, using (2.22)with r = iyields

sin(0.15) ""

isin(O.I) +isin(0.2) - ksin(0.3) = 0.1494995.

Thevalueof sin(O.l5) obtainedusing a calculatoris 0.1494381.

is clearthat (2.22)gives

a betterestimate than (2.21).

III

2.6.2 Differentiation and Integration Operators

The two

most

importantoperations

mathematicalphysics

can

be writtettitttemns

the finite-difference operators. Define

the

differetttiatiott operator 0

and

the

mtegration operator J by

Of(x)

f'(x),

Jf(x)

es x

f(t)

dt.

(2.23)

Assuming that

exists, we

note

that

f(x)

[f'(x)].

This

shows that

the operatiott

antidifferentiation:

f(x)

F(x),

where F is any primitive

Ott

the

otherhand,

AF(x)

F(x

+h)

F(x)

Jf(x).

These two equations

and the fact that

J and 0

commute

(reader, verify!) show that

"exact"

relation

between

shiffing

and

differentiation

operators

Ao-

= J

JO = OJ= A = E

-1.

Usingthe

Taylor

expansion,

we can

write

(2.24)

(2.26)

2.6.3. Example. Let us calculate cos(O.I), considering

cosx

to be

(dldx)(sinx)

and

usingthevaluesgivenin

Example

2.6.2. Using

Equation

(2.26) to second

order,

we get

74 2.

OPERATOR

ALGEBRA

Thisgives

cos(O.I) '" 1[-0.295520 +4(0.198669) - 3(0.099833)] = 0.998284.

"exact"

relation

between

integration

and

difference

operators

comparison.

the

value

obtained

directly

from

calculator

is 0.995004.

The operator J is a special case of a more general operator defined by

Jaf(x)

= x

f(t)

F(x

ah)

F(x)

= (E

-l)F(x)

= (E

_1)D-

f(x),

a-1

(1+a)a-1

J = (E

-1)D-

h--

= h-'-c:--,--'---:-,--

a InE 1n(1 +

where we nsed Equation (2.26).

iii

(2.27)

(2.28)

2.6.3 NumericalIntegration

Suppose that we are interested in the numerical value of

f(x)

dx.

Let

sa a

and

b, and divide the interval [a, b] into N equal parts, each of length

h = (b -

a)jN.

The method commonly used in calcnlating integrals is to find

the integral

J:,;+ah

f(x)

dx,

where a is a suitable number, and then add all such

integrals. More specifically,

f(x)dx+

...

f(x)dx,

XQ xo+ah

xo+(M-l)ah

where M is a suitably chosen number. In fact, since xN =

+N h, we have

Ma = N. We next employ Equation (2.27) to get

Jafo

Jafa

+...+

Jaf(M-l)a

= J

(~l

fka)

k~O

where fka sa

f(xo

kah).

We thus need an expression for J

to evaluate the

integral. Such an expression can be derived elegantly, by noting that

S E

1 I

E ds = - =

-J

DinED

InE

so that

[by Equation (2.27)]

(2.29)

trapezoidal

rule

for

numerical

integration

2.6

OPERATORS

NUMERICAL

ANALYSIS

where we expanded (1+AY using the binomial infinite series.Equations (2.28)

and (2.29) givethe desired evaluationof the integral.

Let us make a few remarks before developing any commonly used rules of

approximation.First, once

h is set, the function can be evaluatedonly at

+nh,

wheren is apositiveinteger.This meansthat fn is givenonly for positiveiotegers

n. Thus, in the sum in (2.28) ka must be an integer. Since k is an integer, we

conclude that

a must be an integer. Second, since N =

for some integer M,

we must choose N to be a multiple

a. Third, if we areto be able to evaluate

Jaf(M-l)a [the last term io (2.28)], J

cannot have powers of A higher than a,

because An f(M-l)a contaics a term of the form

f(xo

l)ah

nh)

= f(XN +(n -

a)h),

whichforn > a givesf atapointbeyondtheupperlimit.Thus,inthepower-series

expansionof J

we must make sure that

no power

A beyond a is retained.

There are several specific

Ja's

commonly used in numerical integration. We

will consider these next. The trapezoidal

rnle

sets a = 1. Accordiog to the

remarks above, we therefore

retain terms up to the firstpower in the expansionof

Then

(2.29) gives JI =J =h(1 +

~A).

Substituting this in Equation (2.28),

we obtain

(

N- l )

N-l

h(1 +

~A)

fk = h

~[!k

~(fk+l

fk)]

N-l

="2

L(fk

+ fk+l) =

"2(fo+2fl

·+2fN-l

+fN).

k~O

(2.30)

Simpson's

one-third

rule

for

numerical

integration

Simpson's

three-eighths

ruie

for

numerical

integration

Simpson's one-third rule sets a = 2. Thus, we have to retain all terms

up to the A2 term. However, for

a = 2, the third power of A disappears in

Equation

(2.29), and we get an extra ''power'' of accuracy for free! Because of

this,Simpson'sone-thirdruleispopularfornumericalintegrations.Equation

(2.29)

yieldsJ2 = 2h(1 +A +

2). Substituting this in (2.28) yields

h N/2-1 h N/2-1

- L

(61

+6A +A 2)fzk = - L (flk+2 +

4flk+l

flk)

3 k=O 3 k=O

(fo

4fl

2fz

4/3

+...+

4fN-l

fN).

(2.31)

is understood,of course,thatN is aneveninteger.Thefactor tgivesthismethod

its name.

For Simpson's three-eighths rule, we set

a = 3, retain terms up to A3, and

use Equation

(2.29) to obtaio

J3 = 3h(1 +

2 +

3) =

(81 +

l2A

+6A 2 +A3).

76 2.

OPERATOR

ALGEBRA

Substitutingin (2.28),we get

N/3-t

1=-

(81+12A+6A

z+A

)f3k

8 k=O

N/3-t

= 8"

(f3k+3 +3f3k+2 +

3f3k+t

+13k).

(2.32)

III

2.6.4. Example. Letususe

Simpson's

one-third

rulewith

four

intervals

evaluate

the

familiar

integrall =

dx.

With

h =

0.25

andN =4,

Equation

(2.31)

yields

0.25

(I +4e

+2e

.5 +4e

+e) =

1.71832.

This

very

tothe

"exact"

result

e - I '"

1.71828.

2.7 Problems

2.1. ConsideralinearoperatorTonafinite-dimensional vectorspaceV.Showthat

there exists a polynomial P such that P(T) =

Hint: Takea basis B ={Iai)}!:I

aod considerthe vectors

lal)

}~O

for large enough M aod concludethat there

existsapolynomial

PI(n

suchthatPI (T) lal) =

Do thesamefor

laz),

etc.Now

take

the product of all suchpolynomials.

2.2. Use mathematicalinductionto showthat

[A,

Am] =

2.3. For D aodTdefinedin Exarople 1.3.4:

(a) Showthat [D, T] = 1.

(b) Calculate the lineartraosformationsD

aodT

2.4. ConsiderthreelinearoperatorsLI,

Lz,

aod

satisfyingthe commutationre-

lations [LI,

Lz]

L3,

[L3,

Ltl =

Lz,

[Lz,

L3]

=LI, aod definethe new operators

L±

= LI ± iLz.

(a) Showthat the operatorL

commuteswith

Lk,

k =

1,2,3.

(b) Showthatthe set

{L+,

L3}

isclosedundercommutation, i.e.,thecommuta-

tor of aoytwo ofthemcaobe writtenas alinear combinationofthe set.Determine

these commutators.

in terms of L+, L_, aod

L3.

2.5. Prove the rest of Proposition2.1.11.

2.6. Showthat

[[A,

B],

= 0, then for everypositiveintegerk,

B] = kAk-I[A, B].

Hint:Firstprovetherelationforlowvaluesofk; thenuse mathematicalinduction.

2.7

PROBLEMS

2.7. Show that for 0 and T defined in Example 1.3.4,

[ok,

T] = kok-

and

[Tk,

_kT

•

2.8. Evaluate the derivative of

H-

(t) in terms of the derivative of H(t).

2.9. Show that for any ct,

E lRand any H E ,c,(V), we have

2.10. Show that (U+T)(U - T) =U

- T

and only

[U, T] =

2.11. Prove that

A and B are hermitian, then i[A, BJ is also hermitian.

2.12. Find the solution to the operator differential equation

iii

= tHU(t).

Hint: Make the change of variable y = t

and use the result of Example 2.2.3.

2.13. Verify that

~H3

(dH)

H+

(dH).

dt dt dt dt

2.14. Show that

A and B commute, and ! and g are arbitrary functions, then

!(A)

and g(B) also commute.

2.15. Assuming that US, T], T] = 0 = [[5, T], 5], show that

[5, exp(tT)]

t[S,

exp(tn.

Hint: Expand the exponential and use Problem 2.6.

2.16. Prove that

exp(HI

H3)

= exp(Hj ) exp(H2) exp(H3)

exp{-!([HI,

H2J

+[HI, H3]+[H2,

H3])}

provided that HI, H2, and

commute with all the commutators. What is the

generalization to HI

+...+H

2.17. Denoting the derivative

A(t)

show that

d . .

-[A,

B] = [A, BJ+[A, B].

2.18. Prove Theorem 2.3.2. Hint:

Use

Equation (2.8) and Theorem2.1.3.

2.19. Let A(t) es exp(tH)Ao exp(

-tH),

where H and

are constant operators.

Show

thatdA/dt

= [H,A(t)]. What happens when Hcommutes with A(t)?

78 2.

OPERATOR

ALGEBRA

2.20. Let

If},

Ig) E C(a, b) with the additional property that

f(a)

= g(a) =

feb)

= g(b) =

Show that for such fuuctions, the derivative operator D is

anti-hermitian, The inner productis defined as usual:

g) es f

!*(t)g(t)

dt.

2.21.

this problem, you will go through the steps

proving the rigorous state-

Heisenberg

ment

the

Heisenberg

nncertainty

principle.

Denotethe expectation (average)

uncertainty

principle

value

an operator A in a state

11jI}

by A

avg.

Thus, A

avg

= (A) =

(1jI1

11jI).

The

uncertainty (deviation from the mean) in state

11jI)

the operator A is given by

LlA =

J«A

- AavgP) = J(1jI1 (A - A

g1P

11jI)·

(a) Show that for any two hermitian operators A and B, we have

Hint: Apply the Schwarz inequality to an appropriate pair

vectors.

(b) Using the above and the triangle inequality for complex numbers, show that

1(1jI1

[A,

11jI)1

:::

(1jI1

1j1

)

(1jI1

1j1

) .

"1,

B' = B - fJ1,

where"

and fJ are real

numbers. Show that A' and B' are hermitian and [A', B'] = [A, B].

(d) Now use all the results above to show the celebrated uncertainty relation

(M)(LlB)

(1jI

I[A, B]

11jI}1.

What does this reduce to for position operator x and momentum operator p

[x, p] =

i1i?

2,22. Showthat U = exp A is unitary if and only if A is anti-hermitian,

2.23. Find

for each

the following linear operators.

(a) T :

JR2

]R2 given by

(b)

]R3

]R3 given by

T(~)

(;x~2:;~).

-x+2y+3z

]R2 given by

(X)

(xcosll-

y sin

II)

\xsmll+ycosll'

2.7

PROBLEMS

where 0 is a real number.

What

is TtT?

(d)

T :

1(;2

"""*

1(;2

given by

)

) •

a'2 1CXl +(X2

(e) T :

1(;3

"""*

1(;3

given by

2.24. Show that

P is a (hermitian) projection operator, so are (a) 1 - P and (b)

UtpU for any unitary operator U.

2.25. For the vector

(a)

Find

the associated projection matrix, Pa.

(b) Verify that P

does project an arbitrary vector in

1(;4

along la).

Pais

also a projection operator.

2.26.

Let

lal)

al =

(I,

-I)

and la2) sa 82 =

(-2,

-I).

(a) Construct (in the form

a matrix) the projection operators PI and P2 that

projectonto the directions

lal)

and

la2),respectively. Verify thatthey are indeed

projection operators.

(b) Construct (in the form

a matrix) the operator P = PI +P2 and verify

directly that

it is a projection operator.

(c)

Let

Pact on an arbitraryvector(x, y, z).

What

is the

dot

product

the resulting

vector with the vector 8

I x 82?

What

can you say about P and yourconclusion in

(b)?

2.27.

Let

p(m)

L~=llei)

(eil be a projection operator constructed out

the

firstm orthonormalvectors

the basis B = {lei) 1f:,1

"17.

Show

that

p(m)

projects

into the subspace spanned by the first

m vectors in B.

2.28.

What

is the length

the projection

the vector(3, 4,

-4)

ontoa line whose

parametricequation is

x = 21+I, Y =

-I

+3, z = 1

-I?

Hint:

Find

a unitvector

in the direction

the line and construct its projection operator.

2.29.

The

parametric equation

a line L in a coordinate system with origin 0 is

x=2/+1,

y = 1 +I,

Z=-2/+2.

80 2.

OPERATOR

ALGEBRA

A point P has coordinates (3,

-2,

1).

(a) Using the projection operators, find the length of the projection of 0 P on the

line

(b) Findthe vectorwhose beginning is P and ends on the line L andperpendicnlar

(c)

From

this vector calculate the distance from P to the line L.

2.30.

Let

the operator U :

1(;2

be given by

(al)

-i

~+.!!Z...

..(i..(i

Is U unitary?

2.31. Show that the product of two unitary operators is always unitary, but the

product of two hermitian operators is hermitian

and only

they commute.

2.32.

Let

5 be an operator that is both unitary and hermitian. Show that

(a) 5 is involutive (i.e., 52 = 1).

(b) 5 = p+ -

P-,

where p+ and

P-

are hermitian projection operators.

2,33. Show that when the forward difference operator is applied to a polynomial,

the degree

the polynomial is reducedby

(Hint: Considerx

first.) Then show

that

an+1 annihilates all polynomials of degree n or less.

2.34. Show that

and V

annihilate any polynomial of degree n.

2.35. Show that all of the finite-difference operators commute with one another.

2.36. Verify the identities

= 6E

=a,

v +a =2p,6=E -

E-

= P, _

(6/2).

2.37. By writing everything in terms of E, show that 6

= a - v = a V.

2.38. Write expressions for E

a, V, and p, in terms of 6.

2.39. Show that

sinh-I m.

2.40. Show that

- a

~a5

...

)

and derive Equation (2.27).