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    Chapter 3

    Basic Postulates of Quantum

    Mechanics

    3.1 Basic Postulates of Quantum Mechanics

    We have introduced two concepts: (i) the state of a particle or a quan-

    tum mechanical system is described by a wave function or state function

    (r, t) which satisfies Schrodinger equation; (ii) the dynamical variables

    like momentum p and energy E are operators.

    Now important properties of a physical system are quantities like p, E

    which can be measured or observed; such quantities are called observables.

    There must be a means of predicting the values of observables from thestate function and the procedure for doing this is given by following set of

    postulates:

    (1) To every observable there corresponds an operator A.(2) The possible result of a measurement of an observable is one of the

    eigenvalues an of A given by the equation

    Aun = anun (3.1)

    where A is an operator and an is eigenvalue corresponding to eigen-function un.

    (3) A measurement of A on a system in an eigenstate certainly leadsto the result an, the eigenvalue.

    (4) The average value of a large number of measurements of an ob-

    servable on a system described by an arbitrary state is given

    by

    A a a = Adr (3.2)provided that

    dr = 1, and there exist suitable boundary conditions.

    Thus for example, the average value of momentum in x-direction for a state

    33

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    34 Quantum Mechanics

    (x, t) is given by

    p = (x) i

    x(x)dx. (3.3)Applying to the special case of A = x, we get the average value of position

    x =

    (x)x(x)dx

    =

    xdx =

    x|(x)|2dx. (3.4)

    This is consistent with our interpretation of (x), that |(x)|2 determinesthe probability density P(x) of the particle in space, since

    |(x)

    |2 appears

    as the weighing factor appropriate to x in the calculation of the average

    position.

    It is convenient to introduce a compact notation for the matrix element

    (|A) or |A| to mean Adr. Furthermore, the integration maynot always be over space and it may be necessary to imply integration over

    other continuous or discontinuous variables.

    3.2 Formal Properties of Quantum Mechanical Operators

    The quantum mechanical operators (observables) possess certain properties

    which are important and we discuss some of them briefly.

    (a) They are linear, i.e. if (Cn are numbers)

    =n

    Cnn (3.5)

    then we have A = n

    Cn An (3.6)(b) They obey the laws of association and distribution. Thus if A, B

    and C are three operators, we haveA( B C) = ( AB)C , (3.7)

    A(

    B +

    C) =

    A

    B +

    A

    C . (3.8)

    (c) An observable corresponds to a hermitian operator. We define thehermitian conjugate or adjoint A of an operator A by the equation

    (|A) = (|A) = ( A|) (3.9)

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    Basic Postulates of Quantum Mechanics 35

    i.e.

    Adr =

    (

    A)dr .

    An operator is said to hermitian if A = A, i.e.(|A) = (|A) = ( A|) . (3.10)

    Theorems:

    (i) The eigenvalues of a hermitian operator are real.

    (ii) The eigenfunctions of a hermitian operator corresponding to dif-

    ferent eigenvalues are orthogonal:

    We have eigenvalue equation Aun = anun (3.11)and then

    um|Aun = anum|un = anun|um. (3.12)But from Eq (3.9):

    um

    |Aun

    =

    um

    |Aun = un|Aum= amun|um. (3.13)Hence from Eqs (3.12) and (3.13):

    (am an)un|um = 0.Therefore if

    i) m = n, an = an, since un|un = 0 (3.14)

    ii) m= n,

    un

    |um

    = unumdr = 0. (3.15)

    If the eigenfunctions are normalized,

    un|un =

    unundr = 1. (3.16)

    Hence Eqs (3.15) and (3.16) can be written in a compact form :

    un|um =

    unumdr = mn (3.17)

    where mn = 0 m = n ,mn = 1 m = n.

    (3.18)

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    36 Quantum Mechanics

    mn is called the Kronecker delta. The eigenfunctions are then said to be

    orthonormal.

    (d) The eigenfunctions un of a hermitian operator (observable) form a

    complete orthonormal set so that any arbitrary state function can be

    expanded in terms of them, i.e.

    =n

    Cnun . (3.19)

    This is what we mean by a complete set. The Eq. (3.19) is called to the

    superposition principle, a basic ingredient of quantum mechanics.

    Now in analogy with vector analysis where we express a vector in terms

    of basis vectors i,j, k, the uns are called basis vectors and Cn the corre-sponding coordinates. It follows thatumdr = (um|)

    =n

    Cn(um|un)

    =n

    Cnmn

    = Cm , (3.20)where Cm is related to the probability of finding the system described by

    state in an eigenstate um. Thus

    |Cm|2 = |

    umdr|2 (3.21)

    gives the probability of operator

    A having the eigenvalue am, when the

    system is described by a state . To see this we note that the average value

    of the operator A in state is givena =

    Adr

    =m

    n

    CmCn

    um Aundr

    =m

    n

    CmCnan

    umundr

    = m n CmCnanmn=m

    |Cm|2am , (3.22)

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    Basic Postulates of Quantum Mechanics 37

    The weighing factor |Cm|2 above gives the probability of finding the Eigen-value am.

    (e) The operators do not necessarily obey a commutative law, i.e. two

    operators A and B need not give AB = B A. If AB equals B A, the operatorsare said to commute, i.e.

    [ A, B] = AB B A = 0 (3.23)[ A, B] is called the commutator of two operators.

    (f) If two observables commute, then it is possible to find a set of func-

    tions which are simultaneously eigenfunctions of

    A and

    B. If they do not

    commute, i.e. [

    A,

    B] = 0, this cannot be done except for a state which

    has [ A, B] = 0.We now show that if un is a simultaneous eigenfunction of A and B

    corresponding to eigenvalues an and bn, then

    [ A, B]un = 0 . (3.24)Now Aun = anun , (3.25a)

    Bun = bnun . (3.25b)

    Further ABun = Abnun= bn Aun= bnanun , (3.26a)

    B

    Aun =

    Banun

    = an Bun= anbnun . (3.26b)Therefore

    ( AB B A)un = (bnan anbn)un = 0 ,i.e.

    [

    A,

    B]un = 0 . (3.27)

    The implication of this result is as follows:

    Since un form a complete set so that an arbitrary function can beexpanded in terms of them, it follows that

    [ A, B] = 0 .

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    38 Quantum Mechanics

    Since is arbitrary, [ A, B] = 0. Thus if a set of simultaneous eigenfunctionsof two observables

    A and

    B exist, then

    A and

    B commute.

    If [ A, B] = 0 and [ A, C] = 0 but [ B, C] = 0, then it is not possible to findfunctions which are simultaneously eigenfunctions of A, B and C. It is onlypossible to find eigenfunctions for A and B or for A and C. Correspondingto this, we can only make simultaneous measurements of the observables

    corresponding to the pair of operators A and B or to the pair A and C, itis not possible to measure B, C together.3.3 Continuous Spectrum and Dirac Delta Functions

    (a) Continuous Spectrum

    So far we have considered the case when the eigenvalues of an operatorA are discrete. In case eigenvalues of an operator take on continuous values,the sum in the completeness relation (3.19) takes the form of an integral

    (x) =

    C(a)ua(x)da , (3.28)

    where the label a corresponds to continuous set of eigenvalues and replaces

    the discrete label n in Eq. (3.19). For simplicity we first consider to be afunction of a single variable x only; the generalisation to three dimensions

    is straightforward.

    Now using (3.28)ua(x)(x)dx (ua |)

    =

    daC(a)(ua |ua) , (3.29)

    where a lies in the domain of integration of a. Now the orthogonalitycondition (3.18) becomes

    ua(x)ua(x)dx (ua |ua)= 0 when a = a . (3.30)

    For a = a, it does not have to vanish. In fact it must be infinitely largeat a = a because if it is finite at a = a, then the integral on right-handside of Eq. (3.29) vanishes but (x) does not vanish in general. But this

    infinity must be such thatua(x)(x)dx = C(a

    )

    da(ua |ua) , (3.31)

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    Basic Postulates of Quantum Mechanics 39

    so that in analogy with Eq. (3.20), we have

    C(a) = ua(x)(x)dx (3.32)

    with da(ua |ua) = 1 . (3.33)

    Thus (ua |ua) has the peculiar property: it is zero everywhere except ata = a [see Fig. 3.1] and at a = a it is infinitely large such that its integralis 1.

    (aa)

    ag1

    g2

    a

    Fig. 3.1 The Dirac delta function.

    Such a function is called Dirac -function and is written as

    (a a) = 0 for a = a= for a = a (3.34a)

    such that da(a a) = 1 . (3.34b)

    Thus we write

    (ua |ua) = ua(x)ua(x)dx= (a a) , (3.35)

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    40 Quantum Mechanics

    corresponding to orthonormality relation (3.17) for discrete set. Substitut-

    ing Eq. (3.38) into Eq. (3.29), we have

    ua(x)(x)dx = daC(a)(a a) .Comparing it with Eq. (3.35), we have

    daC(a)(a a) = C(a) . (3.36)

    This is the fundamental property of the -function which we require.

    The generalisation to three dimensions is obvious

    x

    r, a

    a, a

    a .

    Thus Eqs. (3.28), (3.32), (3.34), (3.35) and (3.36) respectively become

    (r) =

    C(a)ua(r)da (3.37)

    C(a) =

    ua

    (r)(r)dr (3.38)

    (a a) = 0 for a = a ,= for a = a.

    (3.39a)

    da(a a) = 1 , (3.39b)

    (ua |ua) = (a a) , (3.40)C(a)(a a)da = C(a) . (3.41)

    Here(a a) = (ax ax)(ay ay)(az az) . (3.42)

    (b) Closure Relations We have

    (r) =n

    Cnun(r) (for discrete set) (3.43a)

    =

    C(a)ua(r)da (for continuous set) . (3.43b)

    ThenCn =

    un(r

    )(r)dr , (3.44a)

    C(a) =

    ua

    (r)(r)dr . (3.44b)

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    Basic Postulates of Quantum Mechanics 41

    Substituting in Eq. (3.43a), we have for discrete set

    (r) = n un(r)(r)drun(r)=

    n

    un(r)un(r

    )

    (r)dr .

    Thus it follows that n

    un(r)un(r

    ) = (r r) . (3.45a)

    The corresponding relation for continuous set is

    ua(r)u

    a

    (r)da = (r r) . (3.45b)

    These are known as Closure Relations. These are equivalent to complete-

    ness relations, since one can write (e.g. discrete set)

    (r) =

    (r)(r r)dr

    = (r)n

    un(r)un(r)dr

    =n

    un(r

    )(r)dr

    un(r)

    =n

    Cnun(r) . (3.46)

    (c) A Simple Representation of -function Consider the function

    eia(xx)da .Now eia(xx

    ) is an oscillating function and the above integral is not defined.

    It is a question of agreeing to give a value to this integral. The prescription

    is

    eia(xx)da = lim

    0

    0

    eia(xx)+ada +

    0

    eia(xx)ada

    = lim0 1i(x x) + 1i(x x) = lim

    0

    22 + (x x)2

    .

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    42 Quantum Mechanics

    If x = x, (x x)2 is a fixed number, no matter how small it may be. Nowwhen 0, 2 can be neglected in comparison with (x x)2 so that thelimit = 0. If x = x

    lim0

    22 + (x x)2

    = lim

    02

    = .

    The behaviour is that of a -function, but we have to verify that its integral

    is 1. Thus we calculate

    eia(xx)da

    dx = lim

    0

    2

    2 + (x x)2 dx

    = lim0

    2

    2

    + 2

    d, = (x

    x)

    = lim0

    2tan1

    = 2 lim0

    2

    2

    = 2 .

    Thus

    1

    2

    eia(xx)da = (x x) . (3.47a)

    Its generalisation to three dimension is

    1

    (2)3

    eia(rr

    )da = (r r) . (3.47b)

    (d) Properties of -function

    f(x)(x b) = f(b)(x b) ,x(x) = 0 ,

    (x) = (x) ,(bx) =

    1

    |b|(x) ,

    (x2 b2) = 12b

    (x b) + (x + b) , b > 0

    (a x)dx(x b) = (a b) .

    These equations have meaning only in the sense of integration; for example,

    the first one means f(x)(x b)dx = f(b) .

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    Basic Postulates of Quantum Mechanics 43

    (bx) = 1|b|(x) means

    (bx)dx =1

    | b |.

    (e) Fourier Transform

    The completeness relation, in terms of eigenfunctions 12

    eiax becomes

    f(x) =12

    C(a)eiaxda , (3.48)

    1

    2

    f(x)eia

    xdx =1

    (2)

    daC(a)

    eix(aa

    )dx

    =

    daC(a)(a a)

    = C(a) .

    Therefore

    C(a) =12

    f(x)eiaxdx . (3.49)

    C(a) and f(x) are called the Fourier transforms of each other.The generalisation to three dimensions is

    f(r) =1

    (2)3/2

    C(a)eiarda , (3.50a)

    C(a) =1

    (2)3/2

    f(r)eiardr . (3.50b)

    (f) Momentum Eigenfunctions (An Example of Continuous

    Spectrum of Eigenvalues

    The momentum operator in Schrodinger representation isp = i x

    .

    Eigenvalue equation is

    pup(x) = pup(x) ,i

    xup(x) = pup(x) . (3.51)

    A solution of this equation is

    up(x) = B exp i

    px

    . (3.52)

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    44 Quantum Mechanics

    Here eigenvalue p is a continuous variable and takes on any value. Above

    we have taken the momentum in x direction. For three dimensions

    p = iand

    pup(r) = pup(r) ,up(r) = B exp

    i

    p r

    . (3.53)

    Now

    (up |up) = up(r)up(r)dr= |B|2

    e(i/)(pp

    )rdr

    = |B|23(2)3(p p) .

    Thus if we select

    B =

    1

    (2)3/2 ,

    then

    (up |up) = (p p) . (3.54)

    Thus normalised momentum eigenfunctions are

    up(r) =1

    (2)3/2

    e(i/)pr . (3.55)

    Since these eigenfunctions form a complete set we can write

    (r) =1

    (2)3/2

    C(p)e(i/)prdp , (3.56a)

    where

    C(p) =1

    (2

    )3/2 e

    (i/)pr(r)dr (3.56b)

    (r) and C(p) are the Fourier transform of each other.

    Now average value of the momentum operator is given by

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    Basic Postulates of Quantum Mechanics 45

    p = p =

    (r)

    p(r)dr

    =

    1

    (2)3

    dpdpC(p)C(p)e(i/)p

    r

    (i)e(i/)prdr=

    1

    (2)3

    dpdpC(p)C(p)

    pe(i/)(pp)rdr=

    dpdpC(p)C(p)p(p p)

    = |C(p)|2pdp . (3.57)Thus |C(p)|2 is the probability of momentum operator p having eigen-

    value p when the system is in state (r). In other words, in a measurement

    of the momentum of a particle, the probability of finding the result p is

    |C(p)|2. Thus C(p) may be regarded as the wave function, in momen-tum space just as (r) is the wave function in r space. C(p) is sometimes

    written as (p).

    3.4 Uncertainty Principle and Non-Commutativity of

    Observables

    In quantum mechanics, two observables A and B do not necessarily com-mute and obey a commutative law, i.e. in general

    [

    A,

    B] = 0 . (3.58)

    This statement is essentially equivalent to the uncertainty principle whichexpresses the limitations on our knowledge imposed by mutual disturbances

    of observations. Eq. (3.58) implies that it is not possible to find simultane-

    ous eigenfunctions of A and B i.e. we cannot have an exact knowledge ofthe result of measurement of A and B simultaneously.

    We now show explicitly that Eq. (3.58) leads to the uncertainty prin-

    ciple. A and B being observables are hermitian. Let a and b denote theaverage values of large number of measurements of

    A and

    B respectively.

    Define A a = A ,B b = B . (3.59)

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    Basic Postulates of Quantum Mechanics 47

    where

    P = +

    2dx ,

    Q =

    i

    2dx .

    Since +2

    and i2

    are hermitian operators and the average value of

    a large number of measurements of a hermitian operator is real, therefore

    P and Q are real numbers.

    Thus Eq. (3.62) gives

    (a)2(b)2

    P2 + Q2

    Q2

    = i 2

    dx2= iAB B A

    2dx

    2 , (3.63)since from Eq. (3.59) = AB B A.

    In particular if A =

    p, B =x, [p, x] = i

    (a)2 = (p)2 ,

    (a)2 = (x)2 ,

    (p)2(x)2 2

    4|

    dx|2 = 2

    4.

    The root mean square deviation is often called the uncertainty (standard

    deviation), i.e.

    p =

    (p)2 , x =

    (x)2 . (3.64)

    Hencepx /2 , (3.65)

    where p and x denote the uncertainties in the measured values of p

    and x. If

    [ A, B] = 0there is no mutual disturbance, and the result of simultaneous measure-

    ments of observables A and B can be known exactly.

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    48 Quantum Mechanics

    Example

    Momentum and energy operators for a free particle are given by

    A = p i xB = H = p2

    2m

    2

    2m

    2

    x2(3.66)

    so that

    p, H

    (x) =

    i

    x

    2

    2m

    2

    x2 (x)

    = 0

    which is true for an arbitrary function (x). Thus

    [p, H] = 0 .The energy and momentum of a free particle can be known exactly, simul-

    taneously. In other words, it is possible to find a wave function which is a

    simultaneous eigenfunction of both momentum and energy.

    3.5 Problems

    3.1 IfA denotes the average value for a large number of measurementsof an operator A for an arbitrary state function , show that Ais real if A is hermitian.

    3.2 The state function for a free particle moving in x-direction is given

    by

    (x) = N e(x2/22+ip0x/) .

    Normalise this wave function. Find the state function (p) in mo-

    mentum space.

    (i) Show that for the state (x) given above

    x = 0 ,p = p0 ,

    x2 = 2

    2 ,

    p2 = p20 +1

    2

    2

    2.

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    Basic Postulates of Quantum Mechanics 49

    Hence show that

    (x)2 = (x x)2

    = 2

    2

    (p)2 = (p p)2=

    1

    2

    2

    2

    so that

    xp =1

    2.

    (ii) Using the relations

    p =

    p|(p)|2dp

    p2 =

    p2|(p)|2dpshow that

    p = p0 ,

    p2

    = p20 +

    1

    2

    2

    2 .Useful integral

    +

    x2nex2

    dx =1

    n+1

    2

    (2n)!

    22nn!

    3.3 Show that for a particle of mass m, moving in a potential V(r),

    [H,r] = ipm

    ,

    [

    p, H] = [

    p, V(

    r)] = iV .

    Using the above results and the fact that H is hermitian, show that

    md

    dtr = p ,

    d

    dtp = V .

    (Note that Newtons law is valid for expectation values.)

    3.4 Using the result

    [H, x] = i p

    m,

    for a particle of mass m moving in x-direction in a potential V(x),

    show that the average value of its momentum in a stationary state

    with discrete energy is zero.

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    50 Quantum Mechanics

    3.5 A particle is in a state

    (x) =1

    asin

    5x

    a , |x| a= 0 elsewhere ,show that the probability for the particle to be found with momen-

    tum p is given by

    100a

    2

    sin2(pa/)

    (p2a2/2 252) .

    3.6 A particle of mass m is confined by an infinite square well potential

    V(x) = 0,

    |x

    | a; V(x) =

    ,

    |x

    |> a. If the particle is in the state

    (x) = x , |x| a ,(x) = 0 , |x| > a ,

    find the probability that a measurement of energy will give the

    result

    En =2

    2m

    24a2

    n2 .

    3.7 For a free particle, find a wave funcion which is a simultaneous

    eigenfunction of both momentum and energy. This is not so for aparticle moving in a potential since then |p, H| = 0 [cf. Problem3.3].

    3.8 a) If a particle is in a state

    (x) = 2

    2

    14

    ex2

    2

    find the probability of finding it in the momentum eigenstate

    up(x) = 12e ipx.b) If it is in a state

    up(x) =

    2

    sin nx, 0 x

    show that the probability of finding it with momentum p is given

    by

    |(p)|2 = 12

    sin4(p n)2pn2

    2 .