Last time: Normalization of eigenfunctions which belong to continuously variable eigenvalues. often needed - alternate method via JWKB next lecture

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1 1. identities. ψδk, ψδp, ψδe, ψbox 3. trick using box normalization 4. Last time: Normalization of eigenfunctions which belong to continuously variable eigenvalues. # states # particles δθ δx dn de L 1/ L for box normalization often needed - alternate method via JWKB next lecture V(x) = αx linear potential solve in momentum representation, φ(p), and take F.T. to ψ(x) Airy functions. Semi-classical (JWKB) approx. for ψ(x) 1 / * px ( ) = [( E Vx ( )) m] 1 / i x * ψ( x) = p( x) exp ± px ( ) c h envelope variable * visualize y(x) as plane wave with x-dependent wave vector * useful for evaluating stationary phase integrals (localization, causality) **** splicing across classical (E > V) forbidden (E < V) ] Next lecture WKB Quantization Condition x + (E) p( x)d x = h x _ (E) ( n + 1 ) n= 0,1,

2 6 - Linear Potential. V(x) = αx H p = + α m x coordinate representation momentum representation x x p p h p x i h i x p note [ x, p ]= ih in both representations - prove this? h d H x H p m m i d = + α = + hα dp nd order 1st order - easier! Solve in momentum representation (a sometimes useful trick) Schr. Eq. dφ(p) dp = i ( hα E p m)φ(p) solution φ(p) = Ne ap+bp3 gives p times φ (p) gives constant times φ (p) plug into Schr. Eq. and identify, term-by-term, to get a = ie hα i b = 6hαm ( ) φ(p) = Nexp i hα Ep p3 /6m easy? φ *( p) φ( p ) = 1! N = 1!

3 f we insist on working in the ψ(x) picture, we must perform a Fourier Transform ipx / h ψ( x) = N e φ( p) dp i ψ( x) = N exp p( αx E)+ p / m dp hα odd function of p: O(p) ψ(x) = N cos (αx E)p + p3 /6m hα dp Surprise! This is a named (Airy) and tabulated integral 6-3 Now p is an observable, so it must be real. Thus φ(p) is defined for all (real) p and is oscillatory in p for all p. NEVER exponentially increasing or decreasing! T S STRANGE THAT φ(p) does not distinguish between classically allowed and forbidden regions. S THS REALLY STRANGE? f we allow p to be imaginary in order to deal with classically forbidden regions, φ(p) becomes an increasing or decreasing exponential. iθ e = cos { θ+ isin { θ sin O( p) dp = 0 even odd 1 / 3 0 * numerical tables for x near turning point i.e., x E/α analytic asymptotic functions for x far from turning point ( ) Ai( z) =π cos s 3 + sz ds since O(p) is odd wrt p p. useful for deriving f(q.n.) and for matching across boundaries. * zeroes of Airy functions [Ai(z i )=0] and of derivatives of Airy functions [Ai (z i )=0] are tabulated. (Useful for matching across center of potentials with definite even or odd symmetry.) [Two kinds of Airy functions, Ai and Bi.]

4 6-4 Ai(z) =π 1/ 0 cos s3 3 + sz ds s p(mha) 1/3 (if α >0) for our specific problem (αx E) z [ mα h ] 1/3 α Turning point E V(x) = αx α >0 x + (E) At turning point E = V( x )= αx x ( E) = E / α x Problems with linear potentials: boundary conditions V(x) αx +αx αx When there is symmetry (or 1/ symmetry) we need to know where the zeroes of d 1 4 ψ / 34 for even functions and are located. tables of zeroes of Ai(z) and A i(z) "z n " " z n " See Handout. ψ(x) { for odd functions or barrier When there is no symmetry, must match Ai (or, more precisely, a linear combination of Ai and Bi) and Ai across boundaries, but we do not have to actually look at the Airy function itself near the joining point.

5 6-5 E This is not as bad as it seems because we are αx usually far from turning point at internal joining point and can use analytic asymptotic x expressions for Ai(z). linear potentials of different slope. For α > 0 there are cases (classical and nonclassical) (i) z << 0, E > V(x) classically allowed region Ai(z) π -1/ ( z { ) 1/4 sin positive 3 ( z { ) 3/ +π/ { 4 x is in here phase shift asymptotic form for z << 0. * oscillatory, but wave vector, k, varies with x -1/4 * Ai vanishes as x ± because of (-z) factor * Bi is needed for case where Airy function must vanish as x + in classical region (ii) z>> 0, E < V( x) forbidden region 3 / -1/ 1/ 4 ( / 3) z Ai(z) ( π / ) 131 z e4 34 positive decreasing well exponential behaved * not oscillatory, monotonic * Ai vanishes as x + * Bi vanishes as x ± in forbidden region asymptotic form for z >> 0. Cartoon need numerical tables to define ψ(x) in this region The two asymptotic forms of Ai(z) are not normalized, but their amplitudes (& phases) are matched. Links B.C. at x + to B. C. at x.

6 NonLecture OTHER CASE: α <0 z - ( α x+ E) α 1/3 m α h E 6-6 α < 0 for this case, need Bi(z) instead of Ai(z) x 1 / 14 / / Bi( zzz ) ( π ) exp 3 ( forbidden region, z<< 0.) 3 1 / 14 / / Bi( zz ) π cos 3 z+ π (allowed region, z>> 0.) 3 4 What is so great about V(x) =αx? ψ(x) is ugly need lookup tables, complicated solutions! Ai(z) turns out to be key to generalization of quantization of all (well behaved) V(x)! There are semi-classical JWKB ψ(x) s These blow up near turning points (i.e. on both sides). The Ai(z) s permit matching of JWKB ψ(x) s across the large gap where ψ JWKB is invalid, ill-defined. (JEFFREYS) WENTZEL KRAMERS BRLLOUN JWKB provides a way to get ψ n (x) and E n without solving differential equations or performing a FT. But actually, the differential equations are easy to solve numerically. The reason we care about JWKB is that it provides a basis for: * physical interpretation (semi-classical) * RKR inversion from E vj V J (R). * semi-classical quantization. * the link to classical mechanics is essential to wavepacket pictures.

7 6-7 (generalize on e ikx for free particle by letting k =p(x)/h depend explicitly on x (why does this not violate [x,p]=ih?) ψ JWKB = p(x) 1/ exp h p( x)d x k(x) and p(x) are classical mechanical functions of x, not QM operators. c classical envelope p(x) = [ m(e V(x) ] 1/ phase factor: adjustable to satisfy boundary conditions p(x) is pure real (classically allowed) or pure imaginary (classically forbidden). p(x) is not Q.M. momentum. t is a classically motivated function of x which has the form of the classical mechanical momentum and has the property that the λ= h varies with x in a reasonable way. p 1 / * px ( ) is probability amplitude envelope because 1 probability v so amplitude 1 v i x ipx/ h * exp px ( ) c is generalization of e to non - constant V(x). h * * node spacing λ(x) = h p(x) gives easily identifiable stationary phase region for many wiggly integrands. (Both ψ s have same λat x s.p. ) Long Nonlecture derivation/motivation.

8 6-8 Try ψ(x) = N(x)exp p( x)d x h c plug into Schr. Eq. and get a new differential equation that N(x) must satisfy d ψ + m h ( E V(x) )ψ=0 d ψ + 1 h p(x) ψ=0 ** * derived in box below 0 = ip( x) ip ± ± ( x) N N N exp px ( ) h h h c This is a new Schr. Eq. for N(x). Now make an approximation, to be tested later, that N is negligible everywhere. This is based on the hope that a slowly varying V(x) will lead to a slowly varying N(x). * dψ = N ± i h p(x) exp p( x)d x h c d ψ = N ± i h Np± i h N p ± ip N ± i h h Np exp p( x)d x h c = N ± i Np± i p h h N p h N exp p( x)d x h c 0= d ψ + p h N exp p( x)d x h c 0= N ± ip(x) N ± i p h h N exp p( x)d x h c

9 so, if we neglect N, we get p N+ pn=0 if p 0, then p 1/ p 1/ N + 1 p 1/ pn =0 dnp ( 1/ ) ( ) dnp1/ = Np 1/ + 1 p 1/ pn =0 N(x)p 1/ (x) = constant N(x) = cp(x) 1/ 6-9 OK, now we have a form for N(x) which we can use to tell us what conditions must be satisfied for N (x) to be negligible everywhere. dp 1/ N = cp 1/ = 1 p 3/ dp dp = dv p 1 m dp 1/ = p 5/ m d p 1/ = m dv dv 5 p 7/ m p dv p(x) = [ m( E V(x) )] 1/ + p 5/ m d V { ignore

10 6-10 N = c 5 4 m p 9/ dv But we have made several assumptions about N * N << ip h N =+ icm h p 3/ dv * N << i p h N = icm h p 3/ dv * N all of this is satisfied if 5 mh 4 i << p h N = c h p+3/ dv p 3 <<1 s this the JWKB validity condition? Spirit of JWKB: if initial JWKB approximation is not sufficiently accurate, iterate: p(x) ψ 0 (x) ψ 0 (x) p 1 (x) p 1 (x) ψ 1 (x) (ordinary JWKB) (first order JWKB) d ψ0 p1 eg.. + ψ p ( x) 0 = 0 1 = h h ψ0 ( x) 1 / i x ψ1( x) = p1( x) exp ± p1 ( x ) h c d ψ 0 1 / see ** Eq. on p. 6-8 iterative improvement of accuracy p 1 (x) is not smaller than p 0 (x). t has more of the correct wiggles in it. END OF NONLECTURE

11 6-11 Resume Lecture 1 / i x ψ( x) p( x) exp ± px ( ) { h c envelope V provided that d is negligible adjustable phase shift. AND hm dv << 1 px ( ) 1 3 same as (x) dp < or d λ λ << p required for N ( x) to be negligible Next need to work out connection of ψ JWKB (x) across region where JWKB approx. breaks down (at turning points!). dλ at turning point because p(x) 0 BUT ALL S NOT LOST near enough to a turning point all potentials V(x) look like V(x)=αx! Now our job is to show that asymptotic ARY and JWKB are identical for a small region not too close and not too far on both sides of each turning point. THS PERMTS ACCURATE SPLCNG OF ψ(x) ACROSS TURNNG PONT REGON!

12 V(x) E 6-1 V(x) = αx α > 0 approx. linear V(x) a-ary a-ary JWKB a JWKB x Region E > V(x) classical ψ a ARY z = (αx E) α ~ π 1/ ( z) 1/4 sin 3 ( z)3/ +π 4 mα h 1/3 at turning point E = V(a) =αa so z = (x a) mα h 1/3 αx-e = (x a) α << 0 when x << a Region / splice using a-airy. Region E < V(x) forbidden region, z >> 0 ψ a ARY ~ π 1/ z 1/4 e /3z3/ Now consider ψ JWKB for a linear potential and show that it is identical to a-airy! ψ JWKB ~c ± p(x) 1/ exp p( x)d x h a both c + and c additive terms could be present p(x) [ m( E V(x) )] 1/

13 6-13 x < a classical, p is real, ψ JWKM oscillates x > a forbidden, p is imaginary, ψ JWKB is exponential pretend V(x) looks linear near x = a p(x) = [ mα( a x) ] 1/ (l-jwkb) linear x a x a p( x)d x = (mα) 1/ a x = (mα) 1/ 3 ( ) 1/ d x a x ( )3/ a = (mα) 1/ 3 ( a x)3/ x Region ψ 1 / θ θ l JWKB( x) ~ p( x) [ Ae i + Be i ] 1 / = px ( ) C sin( θ+ φ) Define the JWKB phase factor, θ(x): 1 / 1 x α θ = = h m px ( ) a h 3 Now compare θ(x) to z(x) but, earlier, z = (x a) mα h 1/3 p = ( mαh) 1/3 ( z) 1/ ( a x) p 1/ = ( mαh) 1/6 ( z) 1/4 3 / θ = 3 ( z)3/ for exponential factor for pre-exponential factor Thus, putting all of the pieces together

14 6-14 ψ l JWKB = (mαh) 1/6 ( z) 1/4 C sin ( z)3/ =ψ a ARY p 1/ f C = (mαh) 1/6 π 1/ φ = π4 θ φ ψ l JWKB exactly splices onto ψ a ARY with a ϖ/4 phase factor (shifted from what the argument of sine would have been if one had started the phase integral at x = a Similar result in Region ψ JWKB ~Ae f(x) + Be +f(x) at x + f(x) B =0 ψ l JWKB = A(mα) 1/4 (x a) 1/4 exp mα 1/ h (x a)3/ 3 which is equal to ψ a ARY if A = (mαh) +1/6 π 1/ Final step: ψ JWKB ψ a ARY, ψ JWKB ψ a ARY require A = C/ perfect match on opposite sides of turning point. Ai(z) valid in region where ψ JWKB is invalid.