Quantum mechanics. Chapter The quantum mechanical formalism

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1 Chapter 5 Quantum mechanics 5.1 The quantum mechanical formalism The realisation, by Heisenberg, that the position and momentum of a quantum mechanical particle cannot be measured simultaneously renders phase space as introduced in Chapter 1 irrelevant for quantum mechanics and inevitably implies that the notion of state for a quantum mechanical system is profoundly different from that of classical mechanics. We shall not in these notes try to derive the new mathematical formalism needed for the description of quantum phenomena from more basic principles, but rather be content with the fact that so far it has been found satisfactory and consistent with observed phenomena ever since it was proposed by Dirac and von Neumann at the beginning of the 1930 ies. The formalism is conveniently described in the form of the following three postulates. We use in this chapter the standard physics notation for inner products, assumed to be linear in the second variable. Postulate 1. The state space for a quantum mechanical system is a Hilbert space H over the complex numbers, where two non-zero vectors ϕ and ψ represent the same state if ϕ = cψ for some scalar c. By normalisation one can also say that states are represented by unit vectors in H, where two unit vectors represent the same state if they deviate by a phase factor e iα. The vector space structure of H reflects the so-called sueperposition principle: if ϕ and ψ represent two different states of the system, then any non-trivial linear combination c 1 ϕ + c 2 ψ is also a possible state of the system, called a superposition of the two states. This principle, responsible for observed interference phenomena, has no analogue in classical mechanics. Postulate 2. Measurable physical quantities associated with the system are represented by self-adjoint operators on H, which are called the observables of the system. If the observable A is diagonalisable with eigenvalues λ 1, λ 2,..., then these are the possible values that the corresponding physical quantity can assume, and the orthogonal projections onto the eigenspaces E λi (A), called spectral projections, are likewise observables. 1

2 The average value of (the physical quantity represented by) A, when repeatedly measured for the system prepared in the normalised state ψ, equals ψ A ψ. A similar interpretation holds for more general observables. Before listing the third and final postulate, let us note the following consequences of Postulates 1 and 2. Suppose the observable A is diagonalisable such that A = λ i P i, i=1 where the P i are the spectral projections. Given a normalised state ψ, we then have ψ = P i ψ, (5.1) i=1 and by Pythagoras P i ψ 2 = 1. (5.2) i=1 The possible outcomes of a measurement of the physical quantity corresponding to P i are according to Posulate 2 the eigenvalues of P i which are 0 and 1. In particular, the average value equals the probability of getting the result 1, which hence equals ψ P i ψ = P i ψ 2, where we have used Pi = P i = Pi 2 (see Exercise 4.15). Hence we see that the individual terms in (5.2) have the interpretation of probabilities adding up to 1. Indeed, it is part of the interpretation of the formalism that if a measurement of the observable A yields the result λ i, then the system is in a state in the corresponding eigenspace P i H = E λi (A) right after the performance of the measurement, and the probability of this outcome is P i ψ 2. In particular, if the spectral projections are one-dimensional, corresponding to normalised eigenvectors e i, we have P i = e i e i and P i ψ 2 = e i ψ 2. Thus, given that the system is in the state ψ before the measurement, the probability that it is in state e i after the measurement is ψ e i 2. Note also that the average value of A in this case is given by ψ A ψ = λ i e i ψ 2,. i=1 Postulate 3. The state ψ(t) as a function of time t satisfies the Schrödinger equation i dψ(t) dt = Hψ(t), (5.3) 2

3 where H is a self-adjoint operator (generally unbounded) on H, called the Hamiltonian of the system, and the derivative of the vector valued function ψ is defined as in Exercise It is important to note the significance of the self-adjointness of H: By the linearity properties of the inner product we get for two solutions ϕ(t) and ψ(t) of the Schrödinger equation d ϕ(t) ψ(t) dt = dϕ(t) dt ψ(t) + ϕ(t) dψ(t) dt = 1 1 Hϕ(t) ψ(t) + ϕ(t) i i Hψ(t) = i ( Hϕ(t) ψ(t) ϕ(t) Hψ(t) ) = 0. That is, the overlap ϕ(t) ψ(t) between the states ϕ and ψ is constant in time. In particular, the total probability ψ(t) 2 = 1 is constant, as it should be. This shows that self-adjointness, or at least symmetry, of the Hamiltonian is required for consistency of the formalism. Note also that for this to hold, H need not be time independent, although in our examples below it will be. If H is time independent one can show that the unique solution to (5.3) with initial value ψ 0 at time t = 0 is given by ψ(t) = e i Ht ψ 0, where the time evolution operator e i Ht is unitary. See Exercise 4.12 for the case when H is bounded. If H is unbounded one can use the general spectral theorem to define e i Ht. Another aspect worth noting is an inherent ambiguity in the choice of state space and observables for a given system induced by unitary transformations: Suppose a given state space H and associated observables for the system in question have been determined, and let U : H H be a unitary operator from H onto a Hilbert space H. Define, for any observable A on H, the operator A on H by A = UAU. Then A is self-adjoint (see Exercise 1). Since U preserves the inner product by Theorem 4.21, we obtain a second equivalent description, or representation, of the same system by letting H be the state space, such that Uψ H represents the same state as ψ H, and by letting A represent the same physical quantity as A. Of course, the explicit form of the observables depends on the choice of representation. A common method for obtaining a representation for a given system is by applying a quantisation procedure to the corresponding classical system (if it exists). A standard representation obtained in this way is the Schrödinger representation, where the state space is chosen to be the space of square integrable functions, also called wave functions, on the configuration space of the classical system, the observable representing a coordinate is the corresponding multiplication operator and the observable corresponding to a (generalised) momentum is the corresponding derivative operator multiplied by i. This procedure is applied in the examples below. Finally, we mention that the preceeding postulates are usually supplemented by a certain technical condition called irreducibility, which ensures that sufficiently many observables are available. One way of formulating this is to require that the state space H cannot 3

4 be split into a non-trivial direct sum H = H 1 H 2, where H 1 and H 2 are invariant under all observables. This condition may be regarded as the replacement of the property possessed by classical systems that any state can be characterised by the values assumed on the state by a certain collection of observables, such as the coordiantes and the associated (generalised) momenta. We shall, however, not make essential use of this condition below. 5.2 The free particle The free particle on a circle. We consider first the case of a particle constrained to move on a circle. More precisely we assume the configuration space is S 1, the unit circle in 2 centered at the origin. Since functions on S 1 can be identified with functions on ] π, π], when parametrising the circle by the angular variable θ, such that (x, y) = (cos θ, sin θ), we take the state space in the Schrödinger representation to be H = L 2 (] π, π]) = L 2 ([ π, π]). with the inner product normalised as in section 3.6. The two coordinate variables are represented by the multiplication operators M cos amd M sin and the momentum is represented by the differential operator D defined in Example 4.6 d). The two multiplication operators are self-adjoint by Example 4.6 a) and D is self-adjoint by Exercise The Hamiltonian is defined by analogy with the classical Hamiltonian as H = 2 2m D 2, where m is the mass of the particle. Since e n (with notation as in section 3.6) is an eigenvector for D with eigenvalue n it follows that e n is likewise an eigenvector for H with eigenvalue E n = 2 n 2 2m. Thus H is an unbounded diagonalisable operator given by and with domain H = 2 2m n 2 e n e n n Z D(H) = {ψ L 2 ([ π, π]) n Z n 4 e n ψ 2 < }. 4

5 The possible values of the (kinetic) energy of the particle are E n, n = 0, 1, 2, 3..., and for each positive energy value there correspond two momentum eigenstates of opposite momenta ± n. The time evolution operator e i Ht has the same eigenvectors e n as H with corresponding eigenvalues e i Ent (see Exercise 4.10 for the case where H is bounded). This means that e i Ht is diagonalisable and given by e i Ht = n Z e i 2m n2t e n e n, which also shows explicitly that it is unitary. Indeed, it can be shown that the solution to the Schrödinger equation i dψ(t) dt with initial condition ψ(0) = ψ 0 is given by ψ(t) = n Z = 2m D 2 ψ(t) (5.4) e i 2m n2t e n ψ 0 e n, (5.5) provided ψ 0 D(H). Note that while ψ(t) is well defined by (5.5) for all ψ 0 H, the Schrödinger equation (5.4) only makes sense if ψ(t) D(H) for all t. Hence the above statement requires showing that if ψ 0 D(H), then ψ(t) defined by (5.5) belongs to D(H) for all t and provides a solution to (5.4) with initial condition ψ(0) = ψ 0. We refrain from giving the detailed argument here (see, however, Exercise 4.12 for the case where H is bounded). The states e n, n Z, are characterised by having well defined momenta n. However, the position operators M cos and M sin have no eigenvalues, and the same holds for M θ. Consequently, states in which the particle has a well defined position do not exist. However, for any normalised state ψ represented by a wave function that vanishes (almost everywhere) outside an interval I = [θ 1, θ 2 ] we have ψ M θ ψ I. Since this holds for any subinterval I of [ π, π] we interprete the (closed) subspace L 2 (I) 1 of functions in H vanishing outside I as consisting of states for which the particle is localised in the interval I. Obviously, the function in L 2 (I) closest to a given function ψ H is the restriction of ψ to I, i.e. the orthogonal projection P I onto L 2 (I) is given by P I ψ = ψ I. In particular, the probability that the particle in a state ψ is located in the interval I is P I ψ 2 = ψ(θ) 2 dθ. which means that ψ(θ) 2 is the probability density for finding the particle at (cos θ, sin θ). I 1 We here identify a function f defined on I with the function on [ π, π] that equals f on I and vanishes outside I. 5

6 The free particle on a line. We next consider the case of a particle constrained to move on a line. More precisely, we assume the configuration space is and take the state space in the Schrödinger representation to be with standard inner product ϕ ψ = H = L 2 (), ϕ(x)ψ(x)dx. The coordinate x is represented by the multiplication operator M x with domain of definition D(M x ) = {f L 2 () x 2 f(x) 2 dx < }, (5.6) which is self-adjoint (see Exercise 2 for details). The momentum is represented by the differential operator d i dx suitably extended to a self-adjoint operator as will be described next. It is convenient to define d dx first on the space S() of rapidly decreasing functions. These are C -functions whose derivatives all decay faster that any power at ±. More precisely, S() = {f C () x N f (n) (x) is a bounded function of x for all n, N = 0, 1, 2,... }. Obviously, S() contains the space C0 () and therefore the closure of S() in H equals H. Moreover, it is clear that S() is invariant under d dx. The following lemma is instrumental for the use of Fourier analysis in the theory of differential equations. Lemma 5.1 The Fourier transformation maps S() bijectively onto itself and the following formulas hold for f S(): f (p) = (ip) n ˆf(p) (5.7) x n f(p) = i n ˆf (n) (p). (5.8) The first formula follows by integration by parts, and the second by differentiating both sides of the formula defining ˆf and then differentiating under the integral sign on the righthand side. See Exercise 3 for details. As a consequence of (5.7) we obtain Applying the inversion formula, this yields 1 i 1 i f (p) = p ˆf(p). d dx f = F M p Ff, f S(). Since M p is a self-adjoint operator with domain D(M p ) = {f L 2 () p 2 f(p) 2 dp < }, (5.9) 6

7 this formula can be used to extend the operator 1 d i dx to a self-adjoint operator D given by whose domain is D( D) = {f L 2 () D = F M p F (5.10) p 2 ˆf(p) 2 dp < }, see Exercise 1. Hence the observable representing the momentum will be chosen to be D. Since M p has no eigenvalues, the same holds for D and we conclude that there are no states in H in which the particle has a well defined position or a well defined momentum. Arguing as above for the particle on a circle, if the particle is the normalised state ψ, then ψ(x) 2 is the probability density for observing the particle at position x. An equivalent description of the particle is obtained by applying the Fourier transformation F to the Schrödinger representation. The Hilbert space H = FH is again L 2 (), where we denote the variable by p instead of x. The momentum observable now becomes F ( D) F = M p by (5.10). The position observable, on the other hand, becomes F M x F = D, which is obtained in the same way as (5.10) by using the expression for F given in Theorem The so obtained representation is commonly referred to as the momentum representaion, since the probability density for a particle in the normalised state ψ for having momentum p now equals ψ(p) 2 by the same arguments as previously used for the position observable in the Schrödinger representation. The Hamiltonian is again obtained from the classical Hamiltonian by simple substitution as H = 2m D 2, in the Schrödinger representation, and in the momentum representaion as H = 2 2m M p 2. Thus the Schrödinger equation in the momentum representaion takes the form i d ψ(t, p) = dt The solution to this equation is readily obtained as 2m p2 ψ(t, p). ψ(t, p) = e i 2m p2t ψ 0 (p), provided the initial value ψ(0, ) = ψ 0 belongs to D(H ). Note that the time evolution operator e i H t is again unitary and equals the multiplication operator defined by the function e i 2m p2t taking values on the unit circle in C. 7

8 The solution to the Schrödinger equation i d dt ψ(t) = 2m D 2 ψ(t) in the Schrödinger representation with initial value ψ 0 is obtained by applying the inverse Fourier transformation to the solution in the momentum representation with initial nalue ˆψ 0 : ψ(t) = F (e i 2m p2 t ˆψ0 ). Note that two integrations are involved on the righthand side, one for each of the two Fourier transformations. It turns out that one of the integrations can be performed explicitly, yielding the result m (x y)2 im ψ(t, x) = e 2 t ψ 0 (y) dt, 2π it valid for ψ 0 D(H). This formula can e.g. be used to analyse the spreading in time of the wavefunction due to the uncertainty of momentum as can be found in most text books on introductory quantum mechanics. 5.3 The harmonic oscillator The one-dimensional harmonic oscillator is the system consisting of one particle constrained to a line and acted on by a linear force. Thus the state space is the same as for the free particle, H = L 2 (), and the position and momentum operators are also the same. The Hamiltonian, on the other hand, is obtained by adding the potential, now regarded as a multiplication operator, which is quadratic to the free particle Hamiltonian. More precisely we define H 0 = 2 2m D kx2, (5.11) where as usual D = 1 d i dx, and k is a positive constant. Since both operators in the above sum are unbounded, appropriate attention has to be paid to the domain of H 0. Our strategy will be to take as the domain of H 0 the space S() of rapidly decreasing functions and to find an orthonormal basis of eigenvectors Ω n, n = 0, 1, 2, 3,..., in this space. Once this has been accomplished we can extend H 0 to a self-adjoint operator H in a similar way as done for the operator D in Example First, let us note that since both D and M x 2 evidently map S() into itself, the same holds for H. Second, let us introduce the two operators where A = d dx + cx and A = d + cx (5.12) dx c = mk 8 (5.13)

9 For f S() we then have that is (A Af)(x) = d2 f dx 2 + c2 x 2 f(x) cf(x), H 0 = 2 2m (A A + c) on S(). (5.14) In addition, we have the relation A A = AA 2c on S(), which is commonly written as where is the commutator of A and A. satisfies [A, A ] = 2c, (5.15) [A, A ] = AA A A From (5.14) it is seen that a state Ω 0 S() which AΩ 0 = Ω 0 + cω 0 = 0 (5.16) is an eigenstate for H 0 with eigenvalue E 0 = 2 2mc. But (5.16) is a linear first order differential equation whose solutions are proportional to e cx2 /2. Since c > 0, this function belongs to S() (see Exercise 4). Hence we set Ω 0 (x) = a e cx2 /2, (5.17) where the constant a > 0 is chosen such that Ω 0 has norm 1. Next we observe that, if Ω S() is an eigenstate for A A with eigenvalue λ then, as a consequence of (5.15), the state A Ω, provided it is non-zero, is an eigenstate of A A with eigenvalue λ + 2c: A A(A Ω) = A (A A + 2c)Ω = A (λ + 2c)Ω = (λ + 2c)A Ω. Since S() is evidently invariant under A, we conclude by induction that the states Ω n = a n (A ) n Ω 0, n = 0, 1, 2, 3,..., (5.18) all of which are non-zero as will be seen below, are eigenstates for H 0 with corresponding eigenvalues E n = 2 2m (2nc + c) = ω(n + 1 ), n = 0, 1, 2, 3,..., (5.19) 2 where ω = k m is the frequency of the classical harmonic oscillator. Here the coefficients a n > 0 are chosen such that Ω n has norm 1. We show that E n, n = 0, 1, 2,..., is the exaustive list of energy levels for H 0 (and its extension H) by establishing the following 9

10 Theorem 5.2 The set (Ω n ) n=0,1,2,... is an orthonormal basis for L 2 (). Proof. Note first that, for f, g S(), f Ag = f(x)(g (x) + xg(x))dx = lim f(x)g (x)dx + = lim { [f(x)g(x)] = ( f (x) + xf(x))g(x)dx = A f g. xf(x)g(x)dx } f (x)g(x)dx + xf(x)g(x)dx (5.20) This shows that the domain of the adjoint of the (unbounded) operator A contains S() and that A coincides with A on this space. Combining (5.14) and (5.20) it follows that f, H 0 g = H 0 f, g, f, g S(), (5.21) showing that the the domain of definition for the adjoint of H 0 contains S() and that the adjoint coincides with H 0 itself on this space. We say that H 0 is symmetric on S(). Since the eigenfunctions Ω n belong to S(), the proof of Lemma 4.11 still applies to show that Ω n is orthogonal to Ω m for n m. Hence (Ω n ) n=0,1,2,... is an orthonormal set in L 2 (). It remains to show that {Ω n n = 0, 1, 2,... } = {0}. We do this in two steps: 1) We show that span{ω n n = 0, 1, 2,... } contains the function ϕ p given by ϕ p (x) = e ixp Ω 0 (x), x, for all p. 2) We show that It then follows that {ϕ p p } = {0}. (5.22) {Ω n n = 0, 1, 2,... } = span{ω n n = 0, 1, 2,... } {ϕ p p } = {0} as desired. In order to verify 1), we note that Ω n is of the form Ω n (x) = P n (x)e cx2 /2, where P n is a polynomial of order n. Indeed, this is obviously the case for n = 0 and then it follows easily by induction for all n, since A (P n (x)e cx2 /2 ) = (2cxP n (x) P n(x))e cx2 /2, 10

11 where 2cxP n (x) is a polynomial of order n + 1 while P n(x) is of order n 1, if P n (x) is of order n. Hence we have that span{ω n n = 0, 1, 2,... } = {QΩ 0 Q polynomial}, and we need to show that for any given p and any ε > 0 there exists a polynomial Q such that ϕ p QΩ 0 2 = a 2 e ixp Q(x) 2 e cx2 dx < ε 2. (5.23) Let T n (x) be the n th Taylor polynomial for exp, T n (x) = Since the Taylor expansion for exp converges uniformly to exp on any given bounded interval [, ], there exists for each > 0 an n such that Moreover, we have and hence e ixp T n (ixp) < ε 2a T n (ipx) n k=0 n px k k=0 k! x k k!. for x [, ]. exp( xp ), e ixp T n (ixp) 2 e cx2 4e 2 xp cx2. Since the last function is integrable (verify this) we can choose > 0 such that 4e 2 xp cx2 < ε2 2a 2. x > With this choice of and n we obtain ϕ p T n (ixp)ω 0 2 = a 2 e ixp T n (ixp) 2 e cx2 dx + x > ( ) ε 2 a 2 2a dx + 4a 2 e 2 xp cx2 dx x > < ε2 2 + ε2 2 = ε2. a 2 e ixp T n (ixp) 2 e cx2 dx Thus, setting Q(x) = T n (ixp), we have established (5.23) and 1) is proven. In order to verify 2), assume that f L 2 () is orthogonal to all ϕ p, p, that is f(x)ω 0 (x)e ixp dx = 0 for all p. By definition of the Fourier transformation this means that F(fΩ 0 ) = 0. Since F is an isometry this implies fω 0 = 0 and hence f = 0, since Ω 0 is a positive function. This proves 2) and thereby the theorem. 11

12 Having found an orthonormal basis of eigenvectors for H 0, this operator can be extended to a self-adjoint operator H by the same procedure as used in Example We define H by the formula 2 Hψ = ω (n ) Ω n ψ Ω n (5.24) on the domain or in Dirac notation D(H) = {ψ H H = ω n=0 (n )2 Ω n ψ 2 < }, n=0 (n ) Ω n Ω n. n=0 The energy levels of the ocsillator are given by (5.19). The time evolution operator e i i H is unitary and given by e i i H = n=0 e iω(n+ 1 2 )t Ω n Ω n. Thus the solution to the Schrödinger equation (5.3) with initial value ψ 0 is in this case given by ψ(t) = e iω(n+ 1 2 )t Ω n ψ 0 Ω n, (5.25) provided ψ 0 D(H). n=0 emark 5.3 The eigenfunctions Ω n have independent interest and are well known from classical analysis. In case c = 1 they are called Hermite functions and the polynomials P n are, up to normalisation, called the Hermite polynomials. The first few of them are given in Exercise 6. A detailed analysis of the Hermite functions allows one to carry out the sum in (5.25) obtaining an integral formula for the solution (see eg. J.Glimm and A.Jaffe: Quantum Physics, p. 19.) As a consequence of Lemma 5.1 we see that if c = 1 then F H 0 = H 0 F, i.e. F commutes with H 0 on S(). Applying both sides to Ω n we get H 0 (FΩ n ) = ω(n )FΩ n. Since the eigenstate for H 0 corresponding to eigenvalue ω(n+ 1 2 ) is unique up to a constant multiple we get ˆΩ n = c n Ω n 2 We are strictly speaking only showing that H is an extension of H 0 restricted to span{ω n n = 0, 1, 2,... }. That H as defined by (5.24) coincides with H 0 on all of S() requires an additional argument that will not be given here. See e.g. M.eed and B.Simon: Methods of modern mathematical physics, Vol.1 for a detailed analysis of the space S(). 12

13 for some constant c n. This formula shows that the Fourier transformation is a diagonalisable operator with the Hermite functions as eigenfunctions and corresponding eigenvalues equal to c n. One can show (see Exercise 5) that c n = ( i) n, that is the Fourier transformation has four different eigenvalues ±1, ±i. Exercises Exercise 1 Let A be a (possibly unbounded) self-adjoint operator on H 1 and let U B(H 1, H 2 ) be unitary, where H 1 and H 2 are Hilbert spaces. Show that the operator UAU with domain of definition D(UAU ) = {y H 2 U y D(A)} is a selfadjoint operator on H 2. Show also that A and UAU have the same eigenvalues and that E λ (UAU ) = UE λ (A) for all λ. Exercise 2 Let M x denote the multiplication operator on L 2 () defined by the identity funtion id : x x with domain D(M x ) = {f L 2 () xf(x) 2 dx < }. Show that M x is self-adjoint. Hint. That D(M x ) D(Mx) and that M x coincides with M x on D(M x ) follows by using the Cauchy-Schwarz inequality for the integral xf(x)g(x)dx. To obtain D(M x) D(M x ) one can assume f D(Mx) and then insert for g in the integral just mentioned the function that equals xf(x) on the interval [, ] and vanishes elsewhere, and which belongs to D(M x ) (why?). One can then deduce that xf(x) 2 dx is bounded by a constant for all > 0 and consequently that f D(M x ). Exercise 3 a) Prove eq. (5.7) for n = 1 by using partial integration (see e.g. the proof of eq. (5.20)). Use induction to show (5.7) for arbitrary n N. b) Show eq. (5.8) for n = 1 by writing ˆf(p) ˆf(p 0 ) + i xf(p 0 ) p p 0 = ( e ixp e ixp ) 0 + ixe ixp 0 f(x)dx p p 0 e ix(p p0) 1 + ix p p 0 f(x) dx and split the integral into a contribution I 1 from [, ] and a contribution I 2 from \ [, ]. Then estimate I 1 by using the inequality e ix(p p0) 1 + ix p p 0 p p 0 e, x, p p 0 1, 13

14 which can be obtained by Taylor expanding the exponential function, and estimate I 2 by using e ix(p p0) 1 + ix p p 0 3 x, x, which can be obtained from the mean value theorem applied to cos and sin. By choosing > 0 large and p close enough to p 0 one then shows that the first expression above can be made arbitrarily small in a similar way as in the estimates used for proving 1) in the proof of Theorem 5.2. Use induction to deduce (5.8) for arbitrary n N. c) Use a) and b) to conclude that F maps S() into itself. Exercise 4 a) Verify that Ω 0 belongs to S(), by using appropriate asymptotic properties of exp. b) Show that QΩ 0 belongs to S() for any polymomial Q. Exercise 5 and deduce that a) Show by using Lemma 5.1 that where c n is defined as in emark 5.3. F A = i A F c n = ( i) n c 0, b) Use that F is isometric to argue that c n = 1 and then use the form of Ω 0 to conclude that c 0 = 1 e x2 /2 dx = 1, 2π and hence that c n = ( i) n, n = 0, 1, 2,.... Exercise 6 is odd. a) Show that Ω n is an even function if n is even and an odd function if n b) The Hermite polynomial H n is defined as follows: Let Ω n (x) = P n (x)e x2 /2 be the n th (normalised) eigenfunction for the harmonic oscillator with c = 1. Then H n (x) = b n P n (x) = 2 n x n +..., that is the normalisation factor b n is chosen such that the highest order term in H n (x) has coefficient 2 n. Show that H 0 (x) = 1 H 1 (x) = 2x H 2 (x) = 4x 2 2 H 3 (x) = 8x 3 12x. 14