COURSE INTRODUCTION: WHAT HAPPENS TO A QUANTUM PARTICLE ON A PENDULUM π 2 SECONDS AFTER IT IS TOSSED IN?

Transcription

1 COURSE INTRODUCTION: WHAT HAPPENS TO A QUANTUM PARTICLE ON A PENDULUM π SECONDS AFTER IT IS TOSSED IN? DROR BAR-NATAN Follows a lecture give by the author i the trivial otios semiar i Harvard o April 9, 989. Abstract. This subject is the best oe-hour itroductio I kow for the mathematical techiques that appear i quatum mechaics i oe short lecture we start with a meaigful questio, visit Schrödiger s equatio, operators ad expoetiatio of operators, Fourier aalysis, path itegrals, the least actio priciple, ad Gaussia itegratio, ad at the ed we lad with a meaigful ad iterestig aswer. Cotets. The Questio. The Solutio 3. The Lemmas 3 4. The Morals 5. The Questio Let the complex valued fuctio ψ = ψt, x be a solutio of the Schrödiger equatio t = i x + x ψ with ψ t= = ψ. What is ψ t=t = π? I fact, the major part of our discussio will work just as well for the geeral Schrödiger equatio, where: t = ihψ, H = x + V x, ψ t= = ψ, arbitrary T, ψ is the wave fuctio, with ψt, x represetig the probability of fidig our particle at time t i positio x. H is the eergy, or the Hamiltoia. x is the kietic eergy. V x is the potetial eergy at x. Date: This editio: March 7, 999; First editio: April 9, 989.

2 . The Solutio The equatio t = ihψ with ψ t= = ψ formally implies ψt, x = e it H ψ x = e i T it V ψ x. By Lemma 3. with = ad settig x = x we thus get: ψt, x = e i T e i T V e i T e i T V... e i T e i T V ψ x. Now usig Lemmas 3. ad 3.3 we fid that this is: c deotes the ever-chagig uiversal fixed umerical costat c dx e i x x T / e i T N V x... Repackagig, we get ψt, x = c dx... dx exp dx e i x x T / e i T N V x i T k= xk x k i T T/ Now comes the big ovelty. keepig i mid the picture x dx e i x x T / e i T N V x ψ x. V x k ψ x. k= x x x k x x ad replacig Riema sums by itegrals, we ca write ψt, x = c T dx T W x x kt T t T Dx exp i dt ẋ t V xt ψ x, where W x x deotes the space of paths that begi at x ad ed at x, W x x = {x : [, T ] R : x = x, xt = x }, ad Dx is the formal path itegral measure. This is a good time to itroduce the actio L: T Lx := dt ẋ t V xt. With this otatio, ψt, x = c dx ψ x W x x Dxe ilx.

3 Let x c deote the path o which Lx attais its miimum value, write x = x c + x q with x q W, ad get ψt, x = c dx ψ x Dx q e ilx c+x q. W I our particular case L is quadratic i x, ad therefore Lx c + x q = Lx c + Lx q this uses the fact that x c is a extremal of L, of course. Pluggig this ito what we already have, we get: ψt, x = c dx ψ x Dx q e ilx c+ilx q = c W dx ψ x e ilx c W Dx q e ilx q. Now this is excellet ews, because the remaiig path itegral over W does ot deped o x or x, ad hece it is a costat! Allowig c to chage its value from lie to lie, we get ψt, x = c dx ψ x e ilxc. Lemma 3.4 ow shows us that x c t = x cos t + x si t. A easy explicit computatio gives Lx c = x x, ad we arrive at our fial result: ψ π, x = c dx ψ x e ix x. Notice that this is precisely the formula for the Fourier trasform of ψ! That is, the aswer to the questio i the title of this documet is the particle gets Fourier trasformed, whatever that may mea. 3. The Lemmas Lemma 3.. For ay two matrices A ad B, e A+B = lim e A/ e B/. Proof. sketch Usig Taylor expasios, we see that e A+B ad e A/ e B/ differ by terms at most proportioal to c/. Raisig to the th power, the two sides differ by at most O/, ad thus as required. Lemma 3.. e A+B = lim e A+B = lim e A/ e B/, e itv ψ x = e itv x ψ x. Lemma 3.3. e i t ψ x = c 3 dx e i x x t ψ x.

4 Proof. I fact, the left had side of this equality is just a solutio ψt, x of Schrödiger s equatio with V = : t = i xψ, ψ t= = ψ. Takig the Fourier trasform ψt, p = π e ipx ψt, xdx, we get the equatio ψ t = ip ψ, ψ t= = ψ. For a fixed p, this is a simple first order liear differetial equatio with respect to t, ad thus, tp i ψt, p = e ψ p. Takig the iverse Fourier trasform, which takes products to covolutios ad Gaussias to other Gaussias, we get what we wated to prove. Lemma 3.4. With the otatio of Sectio ad at the specific case of V x = x ad T = π, we have x c t = x cos t + x si t. Proof. If x c is a critical poit of L o W x x, the for ay x q W there should be o term i Lx c + ɛx q which is liear i ɛ. Now recall that Lx = T dt ẋ t V xt, so usig V x c + ɛx q V x c + ɛx q V x c we fid that the liear term i ɛ i Lx c + ɛx q is T dt ẋ c ẋ q V x c x q. Itegratig by parts ad usig x q = x q T =, this becomes T dt ẍ c V x c x q. For this itegral to vaish idepedetly of x q, we must have ẍ c V x c, or This is the famous F = ma of Newto s, ad we ẍ c = V x c. have just rediscovered the priciple of least actio! I our particular case this boils dow to the equatio ẍ c = x c, x c = x, x c π/ = x, whose uique solutio is displayed i the statemet of this lemma. 4

5 4. The Morals Schrödiger s equatio is related to some ifiite dimesioal path itegrals. These path itegrals ca sometime be evaluated, with iterestig ad useful results. The Fourier trasform fits withi some -parameter family of uitary operators, defied by U t = e ith for t other tha π. The same techiques lead to explicit formulas for U t for ay t. We d better work harder, to uderstad how all of this fits ito some bigger coheret picture. Istitute of Mathematics, The Hebrew Uiversity, Giv at-ram, Jerusalem 994, Israel address: drorb@math.huji.ac.il 5