Assignment 5 with Solutions - Introduction to Quantum Mechanics 1 | PHYS 5250, Assignments of Quantum Mechanics

Download Assignment 5 with Solutions - Introduction to Quantum Mechanics 1 | PHYS 5250 and more Assignments Quantum Mechanics in PDF only on Docsity! Quantum Mechanics - I: HW 5 Solutions Leo Radzihovsky Paul Martens October 25, 2007 1 Problem 1 To find the coordinate representation of the evolution operator for a free particle one must integrate over all paths. The tricky part here is coming up with a way of counting them. This can only be done by discretized the path into N -steps, that take time . Then we integrate over the possible locations the particle can be after each interval. Before we can take the path integral we must find the action for each infinitestmal step. S = 1 2 ∫ t+∆t t mv2 = 1 2 ∫ t+∆t t m ( δx δt )2 dt = m (∆x)2 2∆t (1) Breaking the path integral into tiny bits and plugging the action above we get U0 (xN , x0; t, 0) = √ m 2πı~ ∫ ∞ −∞ dxN−1A · · · ∫ ∞ −∞ dx1Ae ım(xN−xN−1) 2 2~ e ım(xN−1−xN−2) 2 2~ · · · e ım(x1−x0) 2 2~ (2) Since these integrals can be separated into a product of integrals one can write U0 (x, x′′; t, t′′) = ∫ ∞ −∞ dx′U0 (x, x′; t, t′)U0 (x′, x′′; t′, t) (3) where, U0 (x, x′; t, t′) = √√√√√ m2πı~ (t′ − t′′)︸ ︷︷ ︸ ′ e ım(x−x′)2 2~′ (4) We can show that these evolution operators form a close group, by calculating the product of two of them and showing that it gives us a evolution operator back. U0 (x, x′′; t, t′′) = ∫ ∞ −∞ dx′ √ m 2πı~ (t− t′) √ m 2πı~ (t′ − t′′) e ım 2~ “ x2 ′ − 2xx′ ′ + x′2 ′ + x′2 ′′ − 2x′x′′ ′′ + x′′2 ′′ ” (5) = A′A′′e ım 2~ “ x2 ′ + x′′2 ′′ ” ∫ ∞ ∞ dx′e ım 2~ “ ′+′′ ′′′ ” x′2 e − ım~ “ x ′ + x′′ ′′ ” x′ (6) 1 using ∫ ∞ −∞ dxe− a 2 x 2+hx = √ 2π a e h2 2a (7) U0 (x, x′′; t, t′′) = A′A′′e ım 2~ “ x2 ′ + x′′2 ′′ ”√ 2π~′′′ ım (′ + ′′) e m2 2~2 “ x ′ + x′′ ′′ ”2 ~′′′ ım(′+′′) (8) = √ m 2πı~′ √ m 2πı~′′ √ 2π~′′′ −ım (′ + ′′) e ım 2~ “ x2 ′ + x′′2 ′′ ” e − ım′′′ 2~(′+′′) “ x2 ′2 + x ′′2 ′′2 + 2xx ′′ ′′′ ” (9) = √ m 2πı~ (′ + ′′) e ım(x−x′′)2 2~(′+′′) (10) note ′ + ′′ = t− t′ + t′ − t′′ = t− t′′ (11) so U0 (x, x′′; t, t′′) = √ m 2πı~ (t− t′′) e ım(x−x′′)2 2~(t−t′′) (12) By successively performing such Gaussian integrals U0 (x, x′; t, 0) = √ m 2πı~t e ım(x−x′′)2 2~t (13) 2 Problem 2 In three dimension the Hamiltonian for two interacting particle is H = p21 2m1 + p22 2m2 + V (~r1 − ~r2) (14) 2.1 a: Transforming to CM coordinates We define from classical mechanics the CM ~RCM = m1~r1 +m2~r2 m1 +m2 (15) and ~r = ~r1 − ~r2 (16) we can rewrite ~r1 and ~r2 ~r1 = ~RCM + m2 m1 +m2 ~r (17) ~r2 = ~RCM − m1 m1 +m2 ~r (18) 2 the angular part of the differential equation becomes.[ −1 sin θ ∂ ∂θ ( sin θ ∂ ∂θ ) + 1 sin2 θ m2 ] Θm` (θ) = ` (`+ 1) ~2Θm` (θ) (49) − ∂ 2 ∂φ2 Φm (φ) = m2Φm (φ) (50) Φ (φ) = eımφ, where m ∈ Z (51) 3 Problem 3 Suppose we have 3 bosons in a box measured to have states n = 3, n = 3, and n = 4. We can build the wave function from the single state wave functions ψn (x) = √ 2 L sin (nπ L x ) (52) by first finding the product wave function Ψ(product)n1n2n3 (x1, x2, x3) = ( 2 L ) 3 2 sin (n1π L x1 ) sin (n2π L x2 ) sin (n3π L x3 ) (53) and then symmetrizing Ψn1n2n3 (x1, x2, x3) = √ 1 3! [ Ψ(product)n1n2n3 (x1, x2, x3) + Ψ (product) n1n2n3 (x2, x3, x1) +Ψ(product)n1n2n3 (x3, x1, x2) + Ψ (product) n1n2n3 (x3, x2, x1) +Ψ(product)n1n2n3 (x2, x1, x3) + Ψ (product) n1n2n3 (x1, x3, x2) ] . (54) These wave function don’t change at all under the exchange of the two n = 3 particle which means that this sum of six wave functions can be simplified to a sum of three wave functions Ψn1n2n3 (x1, x2, x3) = 1√ 3 ( 2 L )3/2 [ sin nπ L x1 sin nπ L x2 sin mπ L x3+ sin nπ L x2 sin nπ L x3 sin mπ L x1 + sin nπ L x3 sin nπ L x1 sin mπ L x2 ] , (55) with n = 3 and m = 4. 4 Problem 4 4.1 a: Distinguishable Each particle has a choice of three states which is 3 ∗ 3 ∗ 3 = 27 (56) total 5 4.2 b: Identical bosons There are no restriction on the number of particle allow in each states. I prefer to visualize this as an Einstein solid. Here are a few examples of configurations one particle in each state 0|0|0 (57) two particles in a and one particle in c 0 0| |0 (58) all three particles in b |0 0 0| (59) The number of independent combinations is( N| +N0 ) ! N|!N0! = (2 + 3)! 2!3! = 10 (60) 4.3 c: Identical fermions cells can hold either no particle or one particle, the only possibility is one in each state. 5 Problem 5 5.1 a: Translation operator The transformation for H is T † HT = p2 2m + V (x+ ) (61) H is only invariant if  = na, where n ∈ Z. Therefore, [Tna,H] = 0, if n ∈ Z (62) 5.2 b: Bloch states T is unitary so it can be put in the form. T = e ıa ~ p (63) Where p† = p. We use p here because for the translation this quantity turns out to be the momentum operator, but as a general statement this should just be some hermitian operator. From part ’a’ we know that eigenstates of H are also eigenstates of Tna. Because T is unitary it’s eigenvalues must be |λk| = 1 (64) 6 ie. λk = eıka (65) so Taψk (x) = eıkaψk (x) = ψk (x+ a) (66) so ψk,n (x) = eıkxuk,n (x) , where uk,n (x) = uk,n (x+ a) (67) This is know as Bloch’s Theorem. 5.3 c: S.E. for Bloch waves − ~ 2 2m ∂2 ∂x2 ( eıkxuk,n (x) ) + V (x) eıkxuk,n (x) = En (k) eıkxuk,n (x) (68) taking the derivative, where k is just a parameter − ~ 2 2m ∂2xuk,n (x)− ı ~2k m ∂xuk,n (x) + ( ~2k2 2m + V (x) ) uk,n (x) = En (k)uk,n (x) (69) A nice way to get the above result is by using the relation below[ p2, f (x) ] = p [p, f ] + [p, f ] p = −~2∂2xf − 2ı~∂xfp (70) 5.4 d: Eigenvalue properties note that ψk,n (x) = eıkxuk,n (x) (71) for outside the range 0 < k < 2πa , a.k.a the first Brillouin zone k = 2π a n+ k′, where n ∈ Z and 0 < k′ < 2π a (72) ψk,n (x) = eık ′x eı 2π a nxuk,n (x)︸ ︷︷ ︸ ≡uk′ (x) (73) It is clear from this and from Eq.67 that there is no physical distinction between states with eigenvalues k and k+s2π/a (s an integer) and thus the eigenvalue of the uk,n(x) must be periodic in k with period 2π/a, i.e., En(k) = En(k+2π/a). This should also be clear from direct examination of Eq.69 for a periodic V (x), particularly after Fourier-transforming the equation. 6 Problem 6 Compute The ground state for N-particles in a box of size L when the particles are identical, spin-polarized. 7