Quantum Physics II (8.05) Fall 2013 Assignment 2

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1 Quntum Physics II (8.05) Fll 2013 Assignment 2 Msschusetts Institute of Technology Physics Deprtment Due Fridy September 20, 2013 September 13, :00 pm Suggested Reding Continued from lst week: 1. Griffiths section Introduction to liner lgebr, Griffith s Appendix nd Shnkr Ch. 1. Bsic foundtions of quntum mechnics: 1. Griffiths Ch.3. Griffiths does not go into s much depth with Dirc nottion s we do in lecture. Problem Set 2 1. Squre well with delt function [10 points] Consider the one-dimensionl infinite squre well 0 x. We dd delt function t the middle of the well V (x) = V 0 δ x 2, V 0 > 0, (1) with V 0 lrge vlue with units of energy. In fct, V 0 is lrge compred to the nturl energy scle of the well: V 0 (! 2 ) γ 1. (2) m 2 The dimensionless number γ is tken to be lrge. The delt function is creting brrier between the left-side nd the right-side of the well. As the delt function intensity V 0 becomes infinite we cn get singulr sitution. Clculte the ground stte energy, including corrections of order 1/γ but ignoring higher order ones. Compre with the energy of the first excited stte. Wht is hppening to the energy difference between these two levels? 2. Nodes in wvefunctions [10 points] We hve written the Schrödinger eqution in the form ψ +(E U(x))ψ = 0. 1

2 Physics 8.05, Quntum Physics II, Fll et ψ k be the energy eigenstte with energy E k nd ψ k+1 be the energy eigenstte with energy E k+1 greter thn E k. () Show tht ( ) b b ψ k+1 ψ k ψ k ψ k +1 = (E k+1 E k ) dx ψ k ψ k+1. (1) (b) et now, b with < b be two successive zeroes of ψ k (x) nd ssume, for convenience tht ψ k (x) > 0 for < x < b. By mking use of (1) show tht ψ k+1 must chnge sign in the intervl (, b). Tht is, ψ k+1 must hve t lest one zero in between ech pir of zeroes of ψ k. Hint: consider the sign of ech side of eqution (1) under the ssumption tht ψ k+1 does not chnge sign in (, b). 3. Developing the vritionl principle [10 points] () Consider normlized tril J wvefunctions ψ(x) tht re orthogonl to the ground stte wvefunction ψ 1 : dx ψ 1 (x)ψ(x) = 0. Show tht the first excited energy E 2 is bounded s: E 2 dx ψ (x)hψ(x). This result hs cler generliztion (tht you need not prove): tril wvefunctions orthogonl to the lowest n energy eigensttes give n upper bound for the energy of the (n +1)-th stte. (b) Assume we cn use rel wvefunctions nd consider the functionl J dx ψ(x)hψ(x) F(ψ) = J. dx ψ(x)ψ(x) This functionl hs remrkble property: it is sttionry t the energy eigensttes! You will do computtion tht confirms this for specil cse, while giving you insight into the nture of the criticl point. et us tke ψ(x) = ψ 2 (x)+ ǫ n ψ n (x), (3) This is the first excited stte perturbed by smll dditions of the other energy eigensttes: the ǫ s re ll tken to be smll. Evlute the functionl F for this wvefunction including terms qudrtic in the ǫ s but ignoring terms cubic or higher order. Confirm tht ll liner terms in ǫ s cncel, showing tht the functionl is indeed sttionry t ψ 2 (x). Does nyǫ dropout to qudrtic order? Discuss the nture of the criticl point (mximum, minimum, flt directions, sddle). n=1

3 Physics 8.05, Quntum Physics II, Fll One-dimensionl ttrctive potentils hve bound stte [10 points] (Bsed on Exercise of Shnkr (p.163) prt (b).) Use the vritionl principle to prove tht ny ttrctive potentil in one dimension must hve t lest one bound stte. We tke n ttrctive potentil to be one where the potentil goes to zero t plus nd minus infinity: lim x ± V (x) = 0, it is piecewise continuous, never positive, nd not equl to zero. Note tht it follows tht V (x) = V (x). To do this, consider the tril wvefunction ( α ) 1/4 αx 2 /2 ψ α (x) = e, π nd try to show tht the expecttion vlue E(α) of Ĥ on this stte! 2 d 2 E(α) = dx ψ α (x) Hψ ˆ α (x), H ˆ = V (x). 2m dx 2 cn be mde negtive for suitble choice of α. Finding the contribution of the potentil term to E(α) is chllenging. For rbitrry ttrctive V (x) it cn t be clculted explicitly, but finding bound for it suffices. A bound cn be obtined by finding point x 0 where the potentil is continuous nd tkes negtive vlue (such point must exist). Suppose V (x 0 ) = 2v 0 > 0. Since the potentil goes to zero t plus nd minus infinity, there is finite intervl [x 1,x 2 ] bout x 0 (with x 1 < x 0 < x 2, Δ x 2 x 1 ) for which V (x) v 0. Explin how the potentil term cn be bounded by replcing V (x) by potentil Ṽ tht stisfies Ṽ(x) = v 0 for x [x 1,x 2 ] nd zero elsewhere. 5. Vritionl nlysis of the potentil V (x) = αx 4 [20 points] We re considering the SE! 2 d 2 ψ + αx 4 ψ = Eψ. 2m dx 2 () Perform chnge of coordintes, setting x = βu nd determine the constnt β so tht the differentil eqution tkes the form 1d 2 ψ 4 +(u e)ψ = 0. 2 du 2 How is E given in terms of the unit-free constnt e?

4 Physics 8.05, Quntum Physics II, Fll (b) Use nlgebricmnipultor tht cnhndledifferentil equtions todetermine the vlue of the constnt e for the ground stte energy to six digits ccurcy (e 0.67). For this try integrting the eqution numericlly strting t u = 0 setting ψ(0) = 1 nd ψ (0) = 0 (why is this derivtive zero?). Only for discrete vlues of e the solution does not blow up s u becomes lrge. The lowest such vlue of e is the one you re looking for. (c) Write cndidte wvefunction for the vritionl principle nd determine n upper bound for the first excited energy. (d) Use the lgebric mnipultor to determine the next-to-lowest vlue of e (to three digits ccurry) nd compre with your vritionl estimte. 6. A property of mtrices [5 points] We cn define function of mtrix M by power series. If f(z) is function with Tylor series expnsion f(z) = f n=0 n z n, then we define f(m) n=0 f n M n. et M be the mtrix 0 i M = i 0 Show tht e imθ tkes the form e imθ = A(θ)1 + B(θ)M, (4) where 1 is the 2 2 identity mtrix nd A nd B re functions you must determine. Wht is the lgebric property of mtrix M of rbitrry size tht would led to this result?

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