Basic Postulates of Quantum Mechanics

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1 Chapter 3 Basic Postulates of Quantum Mechanics 3. Basic Postulates of Quantum Mechanics We have introduced two concepts: (i the state of a particle or a quantum mechanical system is described by a wave function or state function ψ(r, t which satisfies Schrödinger 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 the state function and the procedure for doing this is given by following set of postulates: ( To every observable there corresponds an operator Â. ( The possible result of a measurement of an observable is one of the eigenvalues a n of  given by the equation Âu n = a n u n (3. where Âis an operatorand a n is eigenvaluecorrespondingto eigenfunction u n. (3 A measurement of  on a system in an eigenstate certainly leads to the result a n, the eigenvalue. (4 The average value of a large number of measurements of an observable on a system described by an arbitrary state ψ is given by  ā ψ ā = ψ Âψdr (3. provided that ψ ψdr =, and there exist suitable boundary conditions. Thus for example, the average value of momentum in x-direction for a state 33

2 34 Quantum Mechanics ψ(x,t is given by p = ( ψ (x i ψ(xdx. (3.3 x Applying to the special case of  = x, we get the average value of position x = ψ (x xψ(xdx = xψ ψdx = x ψ(x dx. (3.4 This is consistent with our interpretation of ψ(x, that ψ(x determines the probability density P(x of the particle in space, since ψ(x 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 (φ Âψ or φ  ψ to mean φ Âψdr. Furthermore, the integration may not always be over space and it may be necessary to imply integration over other continuous or discontinuous variables. 3. 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 (C n are numbers ψ = n C n ψ n (3.5 then we have Âψ = n C n Âψ n (3.6 (b They obey the laws of association and distribution. Thus if Â, B and Ĉ are three operators, we have Â( BĈ = ( BĈ, (3.7 Â( B +Ĉ =  B +ÂĈ. (3.8 (c An observable corresponds to a hermitian operator. We define the hermitian conjugate or adjoint  of an operator  by the equation (ψ Âφ = (φ  ψ = ( ψ φ (3.9

3 Basic Postulates of Quantum Mechanics 35 i.e. ψ Âφdr = ( ψ φdr. An operator is said to hermitian if  = Â, i.e. Theorems: (ψ Âφ = (φ Âψ = (Âψ φ. (3.0 (i The eigenvalues of a hermitian operator are real. (ii The eigenfunctions of a hermitian operator corresponding to different eigenvalues are orthogonal: We have eigenvalue equation and then But from Eq (3.9: Hence from Eqs (3. and (3.3: Therefore if Âu n = a n u n (3. u m Âu n = a n u m u n = a n u n u m. (3. u m Âu n = u m  u n = u n Âu m (a m a n u n u m = 0. = a m u n u m. (3.3 i m = n, a n = a n, since u n u n 0 (3.4 ii m n, u n u m = u nu m d r = 0. (3.5 If the eigenfunctions are normalized, u n u n = u nu n d r =. (3.6 Hence Eqs (3.5 and (3.6 can be written in a compact form : u n u m = u nu m d r = δ mn (3.7 where { δmn = 0 m n, δ mn = m = n. (3.8

4 36 Quantum Mechanics δ mn is called the Kronecker delta. The eigenfunctions are then said to be orthonormal. (d The eigenfunctions u n of a hermitian operator (observable form a complete orthonormal set so that any arbitrary state function ψ can be expanded in terms of them, i.e. ψ = n C n u n. (3.9 This is what we mean by a complete set. The Eq. (3.9 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 u n s are called basis vectors and C n the corresponding coordinates. It follows that u mψdr = (u m ψ = n C n (u m u n = n C n δ mn = C m, (3.0 where C m is related to the probability of finding the system described by state ψ in an eigenstate u m. Thus C m = u mψdr (3. gives the probability of operator  having the eigenvalue a m, when the system is described by a state ψ. To see this we note that the averagevalue of the operator  in state ψ is given ā = ψ Âψdr = Cm C n u mâu ndr m n = Cm C na n u m u ndr m n = Cm C na n δ mn m n = C m a m, (3. m

5 Basic Postulates of Quantum Mechanics 37 The weighing factor C m above gives the probability of finding the Eigenvalue a m. (e The operators do not necessarily obey a commutative law, i.e. two operatorsâand B neednotgiveâ B = BÂ. If B equals BÂ,theoperators are said to commute, i.e. [Â, B] =  B B = 0 (3.3 [Â, B] is called the commutator of two operators. (f If two observables commute, then it is possible to find a set of functions which are simultaneously eigenfunctions of  and B. If they do not commute, i.e. [Â, B] 0, this cannot be done except for a state ψ which has [Â, B]ψ = 0. We now show that if u n is a simultaneous eigenfunction of  and B corresponding to eigenvalues a n and b n, then Now Further [Â, B]u n = 0. (3.4 Âu n = a n u n, Bu n = b n u n.  Bu n = Âb nu n = b n Âu n (3.5a (3.5b = b n a n u n, (3.6a Therefore i.e. BÂu n = Ba n u n = a n Bun = a n b n u n. (3.6b ( B BÂu n = (b n a n a n b n u n = 0, [Â, B]u n = 0. (3.7 The implication of this result is as follows: Since u n form a complete set so that an arbitrary function ψ can be expanded in terms of them, it follows that [Â, B]ψ = 0.

6 38 Quantum Mechanics Since ψ isarbitrary,[â, B] = 0. Thus ifasetofsimultaneouseigenfunctions of two observables  and B exist, then  and B commute. If[Â, B] = 0 and [Â,Ĉ] = 0but [ B,Ĉ] 0, then itis notpossible to find functions which are simultaneously eigenfunctions of Â, B and Ĉ. It is only possible to find eigenfunctions for  and B or for  and Ĉ. Corresponding to this, we can only make simultaneous measurements of the observables corresponding to the pair of operators  and B or to the pair  and Ĉ, it is not possible to measure B,Ĉ together. 3.3 Continuous Spectrum and Dirac Delta Functions (a Continuous Spectrum So far we have considered the case when the eigenvalues of an operator  are discrete. In case eigenvalues of an operator take on continuous values, the sum in the completeness relation (3.9 takes the form of an integral ψ(x = C(au a (xda, (3.8 where the label a corresponds to continuous set of eigenvalues and replaces the discrete label n in Eq. (3.9. For simplicity we first consider ψ to be a function of a single variable x only; the generalisation to three dimensions is straightforward. Now using (3.8 u a (xψ(xdx (u a ψ = dac(a(u a u a, (3.9 where a lies in the domain of integration of a. Now the orthogonality condition (3.8 becomes u a (xu a(xdx (u a u a = 0 when a a. (3.30 For a = a, it does not have to vanish. In fact it must be infinitely large at a = a because if it is finite at a = a, then the integral on right-hand side of Eq. (3.9 vanishes but ψ(x does not vanish in general. But this infinity must be such that u a (xψ(xdx = C(a da(u a u a, (3.3

7 Basic Postulates of Quantum Mechanics 39 so that in analogy with Eq. (3.0, we have C(a = u a (xψ(xdx (3.3 with da(u a u a =. (3.33 Thus (u a u a has the peculiar property: it is zero everywhere except at a = a [see Fig. 3.] and at a = a it is infinitely large such that its integral is. δ(a a a g a g Fig. 3. The Dirac delta function. Such a function is called Dirac δ-function and is written as δ(a a = 0 for a a = for a = a (3.34a such that daδ(a a =. (3.34b Thus we write (u a u a = u a (xu a(xdx = δ(a a, (3.35

8 40 Quantum Mechanics corresponding to orthonormality relation (3.7 for discrete set. Substituting Eq. (3.38 into Eq. (3.9, we have u a (xψ(xdx = dac(aδ(a a. Comparing it with Eq. (3.35, we have dac(aδ(a a = C(a. (3.36 This is the fundamental property of the δ-function which we require. The generalisation to three dimensions is obvious x r, a a, a a. Thus Eqs. (3.8, (3.3, (3.34, (3.35 and (3.36 respectively become ψ(r = C(au a (rda (3.37 C(a = u a (rψ(rdr (3.38 { δ(a a = 0 for a a, = for a = a. daδ(a a =, (3.39a (3.39b (u a u a = δ(a a, (3.40 C(aδ(a a da = C(a. (3.4 Here δ(a a = δ(a x a x δ(a y a y δ(a z a z. (3.4 (b Closure Relations We have ψ(r = C n u n (r (for discrete set n = C(au a (rda (for continuous set. (3.43a (3.43b Then C n = u n (r ψ(r dr, C(a = u a(r ψ(r dr. (3.44a (3.44b

9 Basic Postulates of Quantum Mechanics 4 Substituting in Eq. (3.43a, we have for discrete set ( u n (r ψ(r dr u n (r Thus it follows that ψ(r = n ( = n u n (ru n (r ψ(r dr. u n (ru n (r = δ(r r. n The corresponding relation for continuous set is (3.45a u a (ru a(r da = δ(r r. (3.45b These are known as Closure Relations. These are equivalent to completeness relations, since one can write (e.g. discrete set ψ(r = ψ(r δ(r r dr = ψ(r u n (ru n (r dr n = n ( u n (r ψ(r dr u n (r = n C n u n (r. (3.46 (c A Simple Representation of δ-function Consider the function e ia(x x da. Nowe ia(x x isanoscillatingfunctionandtheaboveintegralisnotdefined. It is a question of agreeing to give a value to this integral. The prescription is ( e ia(x x 0 da = lim e ia(x x +εa da+ ε 0 ( = lim ε 0 i(x x +ε ( ε = lim ε 0 ε +(x x. 0 i(x x ε e ia(x x εa da

10 4 Quantum Mechanics If x x, (x x is a fixed number, no matter how small it may be. Now when ε 0,ε can be neglected in comparison with (x x so that the limit = 0. If x = x ( lim ε 0 ε ε +(x x = lim ε 0 ε =. The behaviour is that of a δ-function, but we have to verify that its integral is. Thus we calculate ( e ia(x x ε da dx = lim ε 0 ε +(x x dx Thus = lim ε 0 [ = lim tan η ε 0 ε] ( π = lim ε 0 ( π = π. e ia(x x da = δ(x x. π Its generalisation to three dimension is (π 3 e ia (r r da = δ(r r. (d Properties of δ-function ε ε +η dη, η = (x x (3.47a (3.47b f(xδ(x b = f(bδ(x b, xδ(x = 0, δ( x = δ(x, δ(bx = b δ(x, δ(x b = b( δ(x b+δ(x+b, b > 0 δ(a xdxδ(x b = δ(a b. These equations have meaning only in the sense of integration; for example, the first one means f(xδ(x bdx = f(b.

11 Basic Postulates of Quantum Mechanics 43 δ(bx = b δ(x means (e Fourier Transform δ(bxdx = b. The completeness relation, in terms of eigenfunctions π e iax becomes Therefore π f(x = π f(xe ia x dx = (π = dac(aδ(a a = C(a. C(a = π C(ae iax da, (3.48 dac(a e ix(a a dx f(xe iax dx. (3.49 C(a and f(x are called the Fourier transforms of each other. The generalisation to three dimensions is f(r = C(ae ia r da, (π 3/ C(a = f(re ia r dr. (π 3/ (3.50a (3.50b (f Momentum Eigenfunctions (An Example of Continuous Spectrum of Eigenvalues The momentum operator in Schrödinger representation is Eigenvalue equation is p = i x. pu p (x = pu p (x, i x u p(x = pu p (x. (3.5 A solution of this equation is ( i u p (x = Bexp px. (3.5

12 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 and p = i pu p (r = pu p (r, ( i u p (r = Bexp p r. (3.53 Now (u p u p = u p (ru p(rdr = B e (i/ (p p r dr = B 3 (π 3 δ(p p. Thus if we select B = (π 3/, then Thus normalised momentum eigenfunctions are (u p u p = δ(p p. (3.54 u p (r = (π 3/e(i/ p r. (3.55 Since these eigenfunctions form a complete set we can write ψ(r = C(pe (i/ p r dp, (π 3/ (3.56a where C(p = (π 3/ e (i/ p r ψ(rdr (3.56b ψ(r and C(p are the Fourier transform of each other. Now average value of the momentum operator is given by

13 Basic Postulates of Quantum Mechanics 45 p = p = ψ (r pψ(rdr = (π 3 dp dpc (p C(pe (i/ p r ( i e (i/ p r dr = (π 3 dp dpc (p C(p pe (i/ (p p r dr = dp dpc (p C(ppδ(p p = C(p pdp. (3.57 Thus C(p is the probability of momentum operator p having eigenvalue p when the system is in state ψ(r. In other words, in a measurement of the momentum of a particle, the probability of finding the result p is C(p. Thus C(p may be regarded as the wave function, in momentum space just as ψ(r is the wave function in r space. C(p is sometimes written as φ(p. 3.4 Uncertainty Principle and Non-Commutativity of Observables In quantum mechanics, two observables  and B do not necessarily commute and obey a commutative law, i.e. in general [Â, B] 0. (3.58 This statement is essentially equivalent to the uncertainty principle which expresses the limitations on our knowledge imposed by mutual disturbances of observations. Eq. (3.58 implies that it is not possible to find simultaneous eigenfunctions of  and B i.e. we cannot have an exact knowledge of the result of measurement of  and B simultaneously. We now show explicitly that Eq. (3.58 leads to the uncertainty principle.  and B being observables are hermitian. Let ā and b denote the average values of large number of measurements of  and B respectively. Define  ā = δâ α, B b = δ B β. (3.59

14 46 Quantum Mechanics It is clear from Eq. (3.59 that α and β are hermitian operators so that we have α α = α, β β = β. The mean square deviations of the measured values, of  and B about the mean are given by ( a = ( a = α, ( b = ( B b = β. (3.60 Now ( a α = = = = ψ (xααψ(xdx (α ψ αψdx (αψ αψdx αψ dx. Then ( α β = αψ dx( βψ dx (αψ βψdx. (3.6 This follows from the Schwartz inequality ( ( f dx g dx f gdx. Hence ( a ( b (αψ βψdx = (α ψ βψdx = ψ αβψdx = ψ αβ +βα ψdx+ ψ αβ βα ψdx = P iq, (3.6

15 Basic Postulates of Quantum Mechanics 47 where P = ψ αβ +βα ψdx, Q = ψ αβ βα i ψdx. Since αβ+βα and i αβ βα 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. (3.6 gives ( a ( b P +Q Q = ψ αβ βα i ψdx = ψ iâ B B ψdx, (3.63 since from Eq. (3.59 αβ βα =  B BÂ. In particular if  = p, B = x,[ˆp,ˆx] = i ( a = ( p, ( a = ( x, ( p ( x 4 ψ ψdx = 4. The root mean square deviation is often called the uncertainty (standard deviation, i.e. p = ( p, x = ( x. (3.64 Hence p x /, (3.65 where p and x denote the uncertainties in the measured values of p and x. If [Â, B] = 0 there is no mutual disturbance, and the result of simultaneous measurements of observables  and B can be known exactly.

16 48 Quantum Mechanics Example Momentum and energy operators for a free particle are given by so that  = p i x B = H = p m m ] [ p,h ψ(x = [ i x m = 0 x (3.66 ] x which is true for an arbitrary function ψ(x. Thus [ p,h] = 0. ψ(x The energy and momentum of a free particle can be known exactly, simultaneously. 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. If  denotestheaveragevalueforalargenumberofmeasurements of an operator  for an arbitrary state function ψ, show that A is real if  is hermitian. 3. The state function for a free particle moving in x-direction is given by ψ(x = Ne ( x /δ +ip 0x/. Normalise this wave function. Find the state function φ(p in momentum space. (i Show that for the state ψ(x given above x = 0, p = p 0, x = δ, p = p 0 + δ.

17 Basic Postulates of Quantum Mechanics 49 Hence show that ( x = (x x = δ so that ( p = (p p = δ x p =. (ii Using the relations p = p φ(p dp p = p φ(p dp show that ( Useful integral p = p 0, p = p 0 + δ. + x n e αx dx = α n+ π(n! n n! 3.3 Show that for a particle of mass m, moving in a potential V(r, [H, r] = i p m, [ p,h] = [ p,v( r] = i V. Using the aboveresults and the fact that H is hermitian, show that m d r = p, dt d p = V. dt (Note that Newton s 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.

18 50 Quantum Mechanics 3.5 A particle is in a state ψ(x = ( 5πx sin, x a a a = 0 elsewhere, show that the probability for the particle to be found with momentum p is given by 00πa sin (pa/ (p a / 5π. 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 ψ(x = x, x a, ψ(x = 0, x > a, find the probability that a measurement of energy will give the result ( E n = π m 4a n. 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 a particle moving in a potential since then p,h 0 [cf. Problem 3.3]. 3.8 a If a particle is in a state ( ψ(x = 4e x πδ δ find the probability of finding it in the momentum eigenstate u p (x = π e i px. b If it is in a state u p (x = π sinnx, 0 x π show that the probability of finding it with momentum p is given by φ(p = π sin 4( ( p nπ. ( p n