Homework 9 Solutions - Practice Problems on Quantum Mechanics I | PHY 851, Assignments of Quantum Mechanics

Download Homework 9 Solutions - Practice Problems on Quantum Mechanics I | PHY 851 and more Assignments Quantum Mechanics in PDF only on Docsity! SOLUTIONS 1. In Problem 1, Homework 9, we have calculated the matrix elements of the operator x̂2 for the harmonic oscillator, eqs. (??, ??) and (??). They have selection rules ∆n = 0,±2. Using the matrix multiplication x̂4 = x̂2 · x̂2, we can write 〈n|x̂4|n〉 = ∑ n′ 〈n|x̂2|n′〉〈n′|x̂2|n〉. (1) Now the selection rules show that only three terms in this sum remain: (x4)nn = (x2)2nn + (x 2)n,n−2(x2)n−2,n + (x2)n,n+2(x2)n+2,n, (2) or, taking into account the Hermiticity of x̂2, (x4)nn = (x2)2nn + (x 2)2n,n−2 + (x 2)2n,n+2. (3) With the explicit values of the matrix elements of x̂2, we find (x4)nn = ( h̄ mω )2 [( n+ 1 2 )2 + 1 4 n(n− 1) + 1 4 (n+ 1)(n+ 2) ] = ( h̄ mω )2 3 4 (2n2 + 2n+ 1). (4) 2. a. The given function Ψ(x, 0) is the result of the displacement of the ground state wave function ψ0(x), Ψ(x, 0) = D̂(a)ψ0(x) = ψ0(x− a). (5) It is a superposition of all stationary states of the unperturbed oscillator, Ψ(x, 0) = ∞∑ n=0 Cnψn(x), (6) where the amplitudes Cn are matrix elements of the shift operator, Cn = ∫ dxψn(x)D̂(a)ψ0(x) ≡ 〈n|D̂(a)|0〉. (7) There are various ways to evaluate these amplitudes. We can use the explicit form of the harmonic oscillator eigenfunctions ψn(x) = 1√ 2nn! (mω πh̄ )1/4 e−(mω/2h̄)x 2 Hn (√ mω h̄ x ) . (8) With the dimensionless variables ξ = √ mω h̄ x, b = √ mω h̄ a, (9) 1 we write the amplitudes (7) as Cn = 1√ π2nn! e−b 2/2 ∫ dξ Hn(ξ)e−ξ 2+bξ. (10) Since the Hermite polynomials are Hn(ξ) = (−)neξ 2 dn dξn e−ξ 2 , (11) we integrate by parts n times in (10): Cn = 1√ π2nn! e−b 2/2 ∫ dξ ebξ(−)n d n dξn e−ξ 2 = 1√ π2nn! e−b 2/2bn ∫ dξ e−ξ 2+bξ. (12) The last Gaussian integral is easily evaluated by forming the complete square in the exponent which leads to the final result Cn = 1√ 2nn! bne−b 2/4. (13) The probability of finding the nth stationary state is given by Pn = C2n = 1 2nn! b2ne−b 2/2. (14) The probabilities are normalized correctly: ∞∑ n=0 Pn = e−b 2/2 ∑ n ( b2 2 )n 1 n! = 1. (15) The average degree of excitation of the original oscillator is the mean number of excited quanta 〈n〉 = ∑ n nPn = b2 2 , (16) and we came to the Poisson distribution Pn = 〈n〉ne−〈n〉 n! . (17) b. For a < (h̄/mω)1/2, b < 1, and 〈n〉 < 1, the probabilities Pn rapidly fall off as n increases. For b = 1, we have P0 = 0.6065, P1 = 0.3033, P2 = 0.0758, P3 = 0.0126. 2 distribution for the stationary states |n〉 of the oscillator centered at an unperturbed position x = 0 is Poissonian, eq. (17), with the mean value of excited quanta 〈n〉 = mωa 2 2h̄ = e2E2 2mh̄ω3 . (33) This quantity has a simple meaning: the system now has energy higher than in the presence of the field by the amount of the shift ∆E, eq. (32). This amount is transformed in the average excitation, ∆E = h̄ω〈n〉. (34) c. With no field, there is no dipole moment (stationary states have certain parity). The induced dipole moment for the ground state in the presence of the field is given by the expectation value of the dipole operator d̂ = ex̂, 〈d̂〉 = e ∫ ∞ −∞ dxx|ψ0(x; E)|2 = e ∫ ∞ −∞ dxx|ψ0(x− a; 0)|2 = ea = e2 mω2 E . (35) Thus, the static polarizability of the harmonic oscillator is equal to α = e2 mω2 . (36) 4. a. To find the corresponding integral operator we start with the definition. For the coordinate operator x̂ the action on a wave function is merely multiplication, x̂ψ(x) = xψ(x). (37) On the other hand, this should be expressed in terms of the kernelX(x, x′), x̂ψ(x) = ∫ dx′X(x, x′)ψ(x′). (38) The comparison of these two expressions that should be valid for an arbi- trary ψ(x) leads to the local kernel X(x, x′) = xδ(x− x′). (39) For the momentum operator and the corresponding kernel P (x, x′) we have p̂ψ(x) = −ih̄dψ dx = ∫ dx′ P (x, x′)ψ(x′). (40) ¿From here we see that the kernel is the derivative of the δ-function, P (x, x′) = ih̄ d dx′ δ(x− x′). (41) 5 Indeed, with this kernel, the last expression in (40) becomes (with the aid of the integration by parts) ih̄ ∫ dx′ [ d dx′ δ(x− x′) ] ψ(x′) = ih̄ ∫ dx′ { d dx′ [δ(x− x′)ψ(x′)]− δ(x− x′)dψ(x ′) dx′ } , (42) and the integrated term vanishes for any finite x when taken on the limits x′ → ±∞, whereas the last term provides the needed result after the inte- gration with δ(x−x′). The derivative with respect to x′ can be substituted by (with the opposite sign) the derivative with respect to x, then P (x, x′) = −ih̄ d dx δ(x− x′). (43) For the inversion we have P̂ψ(x) = ψ(−x) = ∫ dx′ P(x, x′)ψ(x′), (44) which shows that P(x, x′) = δ(x+ x′). (45) In the case of the displacement operator, D̂(a)ψ(x) = ψ(x− a) = ∫ dx′D(a;x, x′)ψ(x′), (46) that shows immediately that D(a;x, x′) = δ(x− x′ − a). (47) Finally, for the scaling transformation, M̂(α) = √ αψ(αx) = ∫ dx′M(α;x, x′)ψ(x′), (48) M(α;x, x′) = √ αδ(x′ − αx). (49) b. An operator F̂ commuting with x̂ has to be a function of x̂ only, F̂ = f(x̂). (50) The corresponding kernel is F (x, x′) = f(x)δ(x− x′). (51) 6 c. An operator Ĝ commuting with p̂ has to be a function of p̂ only, Ĝ = g(p̂). (52) The kernel of this operator contains a corresponding function g of the differentiation operator acting onto the δ-function, compare (41), G(x, x′) = g(ih̄d/dx′)δ(x− x′). (53) d. Only an operator Ĉ of multiplication by a constant c commutes both with x̂ and p̂. The corresponding kernel is C(x, x′) = cδ(x− x′). (54) e. For the factorized kernel F̂ψ(x) = f(x) ∫ dx′ g(x′)ψ(x′). (55) An arbitrary matrix element F12 of this operator in the coordinate repre- sentation has a form F12 = ∫ dxψ∗1(x)F̂ψ2(x) = ∫ dx dx′ ψ∗1(x)f(x)g(x ′)ψ2(x′). (56) For a Hermitian operator, F̂ = F̂ †, F12 = F ∗21 = ∫ dx dx′ ψ2(x)f∗(x)g∗(x′)ψ1(x′). (57) Interchanging here the variables x↔ x′ and comparing with (56), we find the condition of hermiticity f(x)g(x′) = f∗(x′)g∗(x) ; g(x) = f∗(x). (58) Thus, for a Hermitian factorized operator, the kernel should be unitary- symmetric, F̂ = F̂ † ; F (x, x′) = f(x)f∗(x′). (59) The eigenvalue problem for this operator, F̂ψ(x) = λψ(x), (60) takes the form λψ(x) = f(x) ∫ dx′ f∗(x′)ψ(x′). (61) 7