THE HERMITE POLYNOMIAL & QUANTIZATION OF THE HARMONIC OSCILLATOR

Transcription

1 THE HERMITE POLYNOMIAL & QUANTIZATION OF THE HARMONIC OSCILLATOR TIMOTHY JONES Abstact. The hamonic oscillato possesses a singula place in quantum mechanics. It is use in a wie vaiety of moels. Hee we outline the basics.. The Hamonic Oscillato: A classical oveview Hooke fist iscovee the law of spings, that thei foce is popotional to thei isplacement. In eality, thei foce is geneally a fa moe complicate function. In fact, many complicate foces can be appoximate in a fom simila to the hamonic oscillato. Consie the potential Ux escibing some physical phenomenon. Close enough to the equilibium point x=a, we can expan this function as a Taylo polynomial, x=a {}} { Ux. Ux = Ua + = 0 x a + {}}{ Ux x! x x a + Hee the fist eivative is zeo, as we take the Taylo seies aoun an equilibium. How close is close enough? This woul epen on the egee of accuacy one seeks, though geneally it is assume that 3 Ux x 3 x a 3 Ux x x a We ae at libety to escale ou potential so that Ua = 0, wheeby we have the appoximation,. Ux = Ux! x x a It is typical to also escale the x-axis so that a = 0 an, using common teminology, Ux = Ux! x x kx Hee k is a constant. The equation of motion fo such an object becomes p t = kx, typically m x t + kx = 0 Often it is moe accuate to a a amping tem, fo example, a sping une wate is bette escibe by,.3 m x t + ν x t + kx = 0 Hee ν is some constant efine by the etaing foce F = ν x t. If a hamonic iving foce is applie, Equation.3 becomes inhomogeneous,.4 m x t + ν x t + kx = F 0 cosωt x=a

2 Equation.3 is easiest to solve. Let ν/m = γ, F 0 /m A, an ω 0 = k/m. If we suppose the solution x = expt,.5 x + γx + ω0 x = 0 + γ + ω0 = 0 = γ ± γ 4ω0 We seek a funamental set of solutions to these equations. Fom the theoy of iffeential equations we know that we equie exactly two istinct solutions. The ± in the solution to the chaacteistic equation lens us two such solutions, except in the case when γ = 4ω0. In that case, ou fist solution is.6 x t = e γt/ The chaacteistic equation suggest no futhe solution. The metho of euction of oe has us assume that.7 xt = vtx t = vte γt/ Plugging Equation.7 into Equation.5, we fin that v t = 0 an so vt = c t + c xt = c te γt/ + c e γt/ The Wonskian of these two solutions is nonzeo always, an so we have a funamental set of solutions. The othe cases ae moe easily solve an we fin, xt = a e γt/ expt γ 4ω 0 / + a e γt/ exp t γ 4ω 0 / γ 4ω 0 > 0 b te γt/ + b e γt/ γ 4ω 0 = 0 c e γt/ expit γ + 4ω 0 / + c e γt/ exp it γ + 4ω 0 / γ 4ω 0 < 0 The latte can be witten, xt = e γt/ cost γ + 4ω0 / + e γt/ sint γ + 4ω0 / The iven Equation.4 is slightly moe complicate to solve. In Equation.4 one can eplace cosωt with expiωt on the conition we take the eal pat of the solution to be foun fom this axillay equation []. If we suppose a solution x t = A expiωt + φ, we fin,.8 ω A + γiωa + ω0a e iωt+φ = F 0 /me iωt Diviing out the exponent on the left han sie an equating eal an imaginay pats of the ight yiels, Aw0 ω F 0 /m = cos φ, γωa A = F 0 /m = sin φ φ = tan F 0 m ω0 ω +γ ω γω ω ω0 The full solution to the iven ampe hamonic oscillato is then,.9 Xt = xt + x t Hee xt is the solution to the homogeneous equation we foun above one of the thee vaiations epening on the paametes. O, since we enjoy seeing lage equations in thei full foms, Xt = a e γt/ e t γ 4ω 0 / + a e γt/ e t γ 4ω 0 / b te γt/ + b e γt/ c e γt/ e it γ +4ω 0 / + c e γt/ e it γ +4ω 0 / + F 0e iωt+tan γω ω ω 0 «m ω 0 ω + γ ω

3 Figue.. Examples of thee possible solution types fo the homogeneous ampe hamonic equation.. Elementay quantization of the Hamonic Oscillato in one an thee imensions in cooinate epesentation The one imensional case can be extene easily to the thee imensional case; one might not be satisfie with such a solution. We will also emonstate the solution in spheical cooinates. With a hamonic oscillato potential, Schöinge s equation in one imension becomes,. ψ m x + mω x = Eψ, η = mω E x, K = ω ψ η = η Kψ A solution to this equation can be foun with the taitional appoach... Asymptotic behavio an the Hemite Polynomial. η K : ψ as η = Ae η / ψη = Afηe η / ψ η = fη η η fη + η fη e η / η fη. η η fη + K fη = 0 η We intouce the ansatz, base on the mile tem an the oveall stuctue of Equation., that fη will have a eivative-cyclic solution such as the Legene an Laguee functions with the kenel exp η. Inee the fist two eivatives suggest a coect iection,.3 η e η = ηe η, η e η = e η + 4η e η In accoance with the Laguee polynomial solution specifically the Roiguez fomulation, we woul ty the following solution:.4 fη = n n η αe η n e η We sample the n = case, f η = α +4η, f η = α8η, f η = 8α 8 η8η+k 4η = 0 3

4 This equation is satisfie when K = 4 o geneally K = n +. This is the quantization conition. The eae might not be satisfie with ou methoology, an so we wish to e-emonstate this conition with anothe appoach. Hee we suppose a seies solution to fη. fη = a j η j j + j + a j+ ja j + K a j η j = 0.5 j=0 j=0 a j+ = j + K j + j + a j As j the seies will appoach zeo at a ate of /j which is a conition fo ivegence. We thus nee the seies to teminate fo some j = n K = n + so that a n+ = 0. Fom ou efinition of K, we have foun the enegy levels to be.6 E = n + ω.. Nomalization of wave function. The solution we emonstate is calle a Hemite polynomial,.7 Hη = n n η e η n e η Popeties of this function can be foun with epeate activation of the eivatives, H n η = n n η e η n ηe η = n n η e e η η.8 = n n 3 η e η n 3 A continuation of this activation woul thus fin, η n η e η η e η η e η η η e η.9 H n η = n H n + ηh n H n+ = nh n + ηh n This implies that,.0 η H nη = η n e η n η n + n e e η η = ηh n η H n+ η = nh n η The nomalization equation is, via integation by pats,. ψn η ψ nηx = A n mω e η ηn n+ e η ηn+ H n ηη = Use of Equation.0 an epeate integation by pats will iminish the stan-alone eivative an the Hemite polynomial until we have,. A mω n n! e η = A mω n n! π = Thus we have.3 ψ n x = Whee a most geneal solution is mω /4 e η / n n η e π n n! η n e η Ψ = C n ψ n n=0 4

5 Hee C n ae coefficients which can only be etemine when the specific physical situation is given. 3. The Spheical Hamonic Oscillato Next we consie the solution fo the thee imensional hamonic oscillato in spheical cooinates. It is obvious that ou solution in Catesian cooinates is simply, 3. E = ω n x + n y + n z + 3 Less simple, but moe eifying is the case in spheical cooinates. With the convesions, 3. we have, 3.3 x y z = x = sin θ cos φ y = sin θ sin φ z = cos θ sin θ cos φ cos θ cos φ sin θ sin φ sin θ sin φ cos θ sin φ sin θ cos φ cos θ sin θ 0 We invet the matix, sin θ cos φ cos θ cos φ sin θ sin φ sin θ sin φ cos θ sin φ sin θ cos φ cos θ sin θ sin θ Thus, 3.5 θ φ = sin θ cos φ sin φ sin θ cos θ sin φ cos φ 0 cos θ cos φ cos θ sin φ sin θ sin θ cos φ sin φ sin θ cos θ cos θ cos φ cos θ sin φ sin θ sin φ sin θ cos φ sin θ 0 θ φ x y z Of couse, x = x + θ x θ + φ x φ So we also have, x 3.6 y = sin θ cos φ cos θ cos φ sin φ sin θ sin φ sin θ cos φ cos θ sin φ sin θ cos θ z sin θ 0 θ φ Hee we begin an asie on the angula momentum opeatos. This will be useful because it will bing us half the solution. We have, 3.7 L x L y = 0 z y z 0 x L z y x 0 We convet this, 0 cos θ sin θ sin φ cos θ 0 sin θ cos φ sin θ cos φ cos θ cos φ sin φ cos sin φ sin θ sin φ θ cos θ sin φ sin θ sin θ sin φ sin θ cos φ 0 cos θ sin θ 0 5 x y z θ φ

6 We let u = cos φ sin φ, u = u φ, A = 0 0 We quickly note that Au = u an u an u ae othogonal, so we have: 3.8 Lxy A cos θ sin θau sin θu cos θ u = L z sin θu T u sin θ 0 cos θ sin θ 3.9 Lxy = L z 0 Au Au cot θ Popely, none of the tems involves. Thowing in the pope i Lx 3.0 = sin φ cos φ L y i cos φ θ cot θ sin φ φ, θ φ θ φ we fin, L z = i With some algeba we can show that 3. L = L x + L y + L z = θ + cot θ θ + sin θ φ We know that L Y θ, φ = ll + Y θ, φ. We also have that the Laplacian in spheical cooinates is, = + + θ + cot θ θ + sin θ φ = + L 3.3 Thus, in thee imensions an spheical cooinates, the Schöinge equation is, 3.4 m + ψ + L m ψ + V ψ = Eψ By sepaation of vaiables, the aial tem an the angula tem can be ivoce. The solution to the angula equation ae hyogeometics. The eae is efee to the supplement on the basic hyogen atom fo a etaile an self-containe eivation of these solutions. Ou esulting aial equation is, with the Hamonic potential specifie, R + m E m ω ll + m R = 0 We can quickly solve this equation by applying the SAP metho Simplify, Asymptote, Powe Seies. We set the stage by fist ewiting the above equation in a fom which will late simplify ou pocess, R + m ω me ll + φ R = 0 Wite = m ω, then ou fist asymptote is towas infinity whee the tems ominate; ou secon is towas zeo whee / ominates an we fin that 3.7 R R = R e / 6

7 The next tem is, 3.8 R = R + R ll + R To solve such equations, we suppose thee exist a function z which will bing this equation into a moe conventionally solvable fom. We will etemine what z is by the consequences of this eman. With We now solve 3.9 R z R R = R z = R z + R z + z z R + R z z ll + R = 0 This equation is solvable if the coefficients ae popotional, o even bette, equal: z ll = = R ll + + R ll + R = 0 This equation has solutions easily foun as, 3. R = A exp l + l z + B exp z ll + ll + But we can solve fo z an fin that z = ll + ln, wheeby, R = A l+ + B l We ismiss A since we ae consieing the infinitesimally nea zeo asymptote. Next we suppose ou complete solution is a powe seies via, 3. R = l e / k a k k = k a k k+l e / Ou fist eivative is, 3.3 R R = k = k = k a k k + l k+l e / k+l+ e / a k k+l e / k + l a k k+l e / k + l Ou secon eivative is, a k k + lk + l k+l k + l + k+l + k+l+ e / k = k a k k+l e / k + l + k + lk + l Recalling that, R + 7 ll + me = 0

8 ou net equation thus equies that a k k+l e / me k + l + 3 k + lk + l + ll + = 0 O moe simply, 3.5 k+l e / me k + l + 3 kk + l + a k a k = 0 We seek to match the coefficients of, since they must vanish inepenently, wheeby, 3.6 k+l e / me k + l + 3 k + k + l + 3 a k = 0 a k+ This gives us the ecusion elation, 3.7 a k+ = me + k + l + 3 k + k + l + 3 Requiing this seies to teminate to pevent non-physical behavio is ou quantization conition, wheeby we must have, 3.8 k f me k f + l + 3 = 0 = E = ω k f + l + 3 This ecusion elationship an eigenvalue fomula thus efine a thee imensional hamonic oscillato. a k 4. Algebaic solution The time inepenent Schöinge equation fo this fist oe potential appoximation is: 4. h m i x imωx ψ m x + mω x ψ = Eψ + mωx ψ = Eψ m i x h i x + imωx ω ψ = Eψ a a + ω ψ = Eψ Hee we have: 4. a ± = h m i x ± imωx It is easy to show that a a + a + a = ω, an as well, we have: 4.3 a + a + ω 8 ψ = Eψ

9 We now intouce the following chain of logic: a + a + ω a + ψ = a + a a + + ωa + ψ = a + a a + ω ψ + ωψ = E + ωa + ψ a + a + ω a + ψ = E + ωa + ψ As well, a a + ω 4.4 a ψ = E ωa ψ It is now obvious to see why a + is calle the aising opeato an a is calle the loweing opeato. We o of couse equie a minimum loweing, a goun level. This equiement puts upon us that ψ o a ψ o = 0 = ψ o x = A o e mω x. We finally conclue that E o = ω/, an 4.5 ψ n x = A n a + n e mω x, E n = n + /ω 4.. Fock states an a moe matue evelopment of the algebaic solution. A moe canonical appoach to the pevious section is as follows. We wite the Hamiltonian as 4.6 H = m P + mω Q This can be witten imensionlessly with the intouction of substitute opeatos, mω 4.7 q = Q 4.8 p = mω P 4.9 H = ω p + q In paallel with the pevious section, we consie the opeatos, 4.0 a = q + ip a = q ip H = ω aa + a a Fom basic quantum mechanics we know that [Q, P ] = i, thus [q, p] = i, an so [a, a ] =, wheeby 4.3 H = ωaa + a a = ωaa = ωa a + The Hamiltonian thus shaes its eigenvalue spectum with N = a a, an 4.4 [N, a] = a aa aa a = [a, a]a = a 4.5 [N, a ] = a aa a a a = a [a, a ] = a Thus we fin, N n = n n = Na n = an n = n a n 9 Na b n = n ba b n

10 Thus a b n is an eigenvecto of N with eigenvalues n b. We equie these eigenvalues be positive fo the following eason. An opeato is equie to have a eal expectation value since the expectation value is what we woul measue, i.e. we equie that fo an opeato R, ψ R ψ = ψ R ψ This implies R is Hemitian, an given R ψ = ψ then we have, ψ R ψ = ψ ψ = ψ R ψ = ψ ψ Since the eigenvalues ae eal, then the squae nom of these eigenvectos follow the fom: 4.8 n a a n = n N n = n n n 0 an so we conclue that thee must be a n b = 0 cutoff As well, Na n = a N + n = n + a n n aa n = n + n n We compae equation with the fact that N n + = n + n + an conclue that 4. a n = n + = C n = n aa n = n + a n = n + n + A simila agument yiels a n = n n. It follows that 4. n = a n 0 n! These ae the so-calle Fock states. Refeences [] W. E. Boyce, R. C. DiPima, Elementay Diffeential Equations an Bounay Value Poblems, John Wiley & Sons, Inc., 996 [] Wikipeia contibutos, Hamonic oscillato, Wikipeia, The Fee Encyclopeia, oscillato&oli= accesse Mach,