Solutions for Homework 4 on Quantum Mechanics I | PHY 851, Assignments of Quantum Mechanics

Download Solutions for Homework 4 on Quantum Mechanics I | PHY 851 and more Assignments Quantum Mechanics in PDF only on Docsity! SOLUTIONS for Homework #4 1. It is convenient to use the vector form writing all vectors V in terms of their Cartesian components Vi. The basic commutator is that between the components of the position vector x̂i and those of the momentum vector p̂i, [p̂i, x̂j ] = −ih̄δij . (1) The components of the orbital momentum operator L̂ are L̂i = (r̂× p̂)i = ijkx̂j p̂k, (2) where ijk is the fully antisymmetric tensor which has the nonzero compo- nents only if all three indices are different, and these nonzero components are equal to 1 for the right-hand order of indices (123, 231, and 312), and -1 for the left-hand order (213, 321, and 132); as always in such cases, the summation over twice repeated Cartesian indices (in our case j and k) is implied and not indicated explicitly. Note that in this expression the order of coordinate and momentum operator does not matter because the vector product contains only the products of different components of coordinate and momentum, and the latter commute, eq. (1). From eq. (1) it follows for arbitrary functions f(r̂) or g(p̂) that [p̂i, f(r̂)] = −ih̄ ∂f ∂x̂i , [x̂i, g(p̂)] = ih̄ ∂g ∂p̂i . (3) Take, for example a scalar function of momentum g(p̂2). Since the mo- mentum components commute among themselves, [L̂i, g(p̂2)] = ijk[x̂j p̂k, g(p̂2)] = ijk[x̂j , g(p̂2)]p̂k. (4) The gradient of any scalar function has a radial direction in corresponding space: ∂ ∂pj g(p2) = ∂g ∂p pj p . (5) Therefore we come to [L̂i, g(p̂2)] = ijkp̂j p̂k × function(p̂2) = 0, (6) since the result would be the i-th component of the vector product of the momentum vector by itself, and such a product is equal to zero (this fol- lows formally from the antisymmetry of ijk which is contracted with the symmetric tensor pjpk). The same conclusion holds for a scalar function of coordinates. The physical reason for the disappearance of the commu- tators of the orbital momentum with scalars is in the fact that, as we will 1 see later, the orbital momentum is the generator of spatial rotations, and scalar functions do not change (are invariant) under rotations. Now we immediately see that the operator of kinetic energy K̂ = p̂2 2m (7) is a scalar function of the momentum vector and therefore commutes with the orbital momentum. For the operator U(r̂) we obtain in a similar way [L̂i, U(r̂)] = ijkx̂j [p̂k, U(r̂] = −ih̄ijkx̂j∇kU(r̂) = −ih̄ ( r̂×∇Û ) i . (8) Again, for a potential with central symmetry, U = U(r), its gradient is directed radially, and the vector product in (8) vanishes. Since the orbital momentum operator its time independent, the Ehrenfest equation of motion for its expectation value is d dt 〈L̂〉 = 1 ih̄ 〈[L̂, Ĥ]〉. (9) The commutator with kinetic energy disappears, and the result is d dt 〈L̂〉 = −〈(r̂×∇Û)〉 = 〈(r̂× F̂)〉, (10) where F̂ = −∇Û is the force operator. In free motion, or in a central field U = U(r), the orbital momentum is conserved. 2. The Wigner distribution is usually introduced in a following way. Consider the single-particle density matrix ρ(r1, r2) ≡ ψ(r1)ψ∗(r2). (11) The usual probability density is given by the diagonal (with respect to the coordinates r1 and r2) part of (11), r1 = r2 = r, ρ(r, r) = |ψ(r)|2. (12) It is convenient to introduce the center-of-mass and relative coordinates of the pair (r1, r2), R = r1 + r2 2 , r = r1 − r2, (13) r1 = R + r 2 , r2 = R− r 2 . (14) 2 a Hermitian conjugate operator F̂ †, 〈n|F̂ †|m〉 = 〈m|F̂ |n〉∗, (31) which means F̂ † = (F̂T )∗; (32) an inverse operator F̂−1, F̂ F̂−1 = F̂−1F̂ = 1̂ (unit operator). (33) First three operators are real, P̂∗ = P̂, D̂(a)∗ = D̂(a), M̂∗(α) = M̂(α); (34) the operator k̂ = p̂/h̄ is imaginary, k̂∗ = i d dx = −k̂. (35) For real operators, transpose and Hermitian conjugation coincide. Since 〈n|P̂|m〉 = ∫ dxψ∗n(x)Pψm(x) = ∫ dxψ∗n(x)ψm(−x) =∫ dxψ∗n(−x)ψm(x) = ∫ dx (P̂ψn(x))∗ψm(x), (36) the inversion operator is Hermitian, P̂† = P̂T = P̂. (37) For the displacement operator we find 〈n|D̂(a)|m〉 = ∫ dxψ∗n(x)ψm(x− a) =∫ dxψ∗n(x+ a)ψm(x) = ∫ dx (D̂(−a)ψn(x))∗ψm(x), (38) which means D̂†(a) = D̂T (a) = D̂(−a). (39) For the scaling operator, 〈n|M̂(α)|m〉 = ∫ dxψ∗n(x) √ αψm(αx). (40) Changing variables y = αx and renaming back y → x, we obtain 〈n|M̂(α)|m〉 = 1√ α ∫ dxψn(x/α)ψm(x) = ∫ dx (M̂(α−1)ψn(x))∗ψm(x). (41) 5 Thus, M̂†(α) = M̂T (α) = M̂(1/α). (42) The first three operators cannot have eigenfunctions (not identically equal to zero) which would correspond to the eigenvalue 0; therefore for these operators the inverse operator is well defined, P̂−1 = P̂, D̂(a)−1 = D̂(−a), M̂(α)−1 = M̂(1/α). (43) In all cases we assume that the wave functions are square integrable,∫ dx|ψ(x)|2 < ∞. This is especially important for the operator k̂. Since it is imaginary, k̂† = −k̂T . But it is Hermitian on a class of square inte- grable functions (we have demonstrated this for the momentum operator p̂ = h̄k̂), k̂† = −k̂T = k̂. (44) The inverse operator for the operator k̂ does not exist since, for any ψ(x), the primitive function i ∫ dxψ(x) which should give back ψ(x) after the action by −i(d/dx) is defined only up to a constant. The reason is that the constant is an eigenfunction of the operator k̂ corresponding to the eigenvalue 0 which precludes the unique definition of the inverse operator. 4. The general solution of the Schrödinger equation for free motion with the initial condition Ψ(x, t = 0) = ψ(x) is given by the independent propaga- tion of plane waves with momentum p and energy (p) = p2/2m, Ψ(x, t) = ∫ dp 2πh̄ e(i/h̄)[px−(p 2/2m)t]φ(p), (45) where φ(p) is the probability amplitude to have momentum p in the initial wave function ψ(x), i.e. Fourier expansion of ψ(x), φ(p) = ∫ dx′ e−(i/h̄)px ′ ψ(x′). (46) Therefore Ψ(x, t) = ∫ dx′ dp 2πh̄ e(i/h̄)[p(x−x ′)−(p2/2m)t]ψ(x′), (47) or, since the initial point can be arbitrarily taken at t = t′ instead of t = 0 as in eq. (47), Ψ(x, t) = ∫ dx′ dp 2πh̄ e(i/h̄)[p(x−x ′)−(p2/2m)(t−t′)]Ψ(x′, t′). (48) This allows us to determine the Green function G(x, t;x′, t′) = ∫ dp 2πh̄ e(i/h̄)[p(x−x ′)−(p2/2m)(t−t′)]. (49) 6 This is a particular case of the general result valid for any time-independent Hamiltonian Ĥ with the spectrum {En} and corresponding stationary eigenfunctions ψn(x), Ĥψn = Enψn. (50) The Green function for the general situation is G(x, t;x′, t′) = ∑ n ψn(x)ψ∗n(x ′)e−(i/h̄)En(t−t ′). (51) In our case the sum over the spectrum becomes the integral ∫ dp/(2πh̄), the eigenfunctions are ψp(x) = exp[(i/h̄)px], and energies En → (p). The standard calculation of the Gaussian integral in (49) gives G(x, t;x′, t′) = √ m 2iπh̄(t− t′) exp [ im(x− x′)2 2h̄(t− t′) ] , (52) with the obvious generalization for the three-dimensional case, G(r, t; r′, t′) = [ m 2iπh̄(t− t′) ]3/2 exp [ im(r− r′)2 2h̄(t− t′) ] . (53) The effective propagation corresponds to the region where the exponent is not very big (to avoid wild oscillations and compensation), |x − x′| ∼ [h̄(t−t′)/m]1/2, in accordance with the previous estimates of the quantum spreading. 5. The Hamiltonian of the problem is Ĥ = p̂2 2m − fx̂, f = eE . (54) a. The Ehrenfest equations of motion can be derived by the straightfor- ward calculations of the commutators, or just by the analogy to classical mechanics, d〈x̂〉 dt = 〈p̂〉 m , (55) d〈p̂〉 dt = f. (56) The time integration leads to 〈p̂〉 = 〈p̂〉0 + ft, (57) 〈x̂〉 = 〈x̂〉0 + 〈p̂〉0 m t+ f 2m t2. (58) A semiclassical meaning of these expressions is evident. 7