Notes4 - FINITE SQUARE WELL As a last example consider the finite square well potential “V0 for —-a<x<a V“ I l o for Ix> a[1145

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Unformatted text preview: FINITE SQUARE WELL As a last example, consider the finite square well potential “V0, for —-a<x<a, V“) I l o, for Ix] > a, [1145] where V0 is a (positive) constant (Figure 2.17). Like the delta-function well, this potential admits both bound states (with E < 0) and'scatteritig states (with'E > 0). We'll-look first at the bound states. .. '— In the region 2: < -—a the potential is zerois'o the Schrodingerequation reads hZ d211, (12¢. 2 _— = E 9 = 1 2m (1’):2 w 0: dxz K V; where . «/—2 E x a + [2.146] I is real and positive. The general solution is tux) = A exp(-—-ch) + B exp(tcx), but the first temt blows up (as x —->' —oo), so the physically admissible solutiOn (as beforef-see Equation 2.119) is ' em = Be”, tori-é —a. [2.147] ' ' ' FIGURE 2.17: The finite square well (Equation 2.145). in the region —a < x < a, V(x) 2 —V0, and the Schrodinger equation reads a? (12110 (131,1; "354:2 ‘VW=E‘1[” 0" cm I‘m" where t/ 2m (E + V0) h Although E is negative, fo-tr bound states, it must be great-er than —V(j, by the old theorem E > Vmin (Problem 2.2); so I is also real and positive. The general 39 - . l . [2.148] Ill solution is tflx) = C'sinax) + D cos(Lr.), for — a < x a, ‘ {2.149] where C and D are arbitrary constants. Finally, in the region x > a the potential is again zero; the general solution is tflx) = F exp(-icx) + GeXpch)‘, but the second term blows up (as x —> 00), so we are left with $00 '= Few”, for x > a. [2.l50] The next step is to impose boundary conditions: 15' and dW/dfit‘ continuous at —a and +a. But we can save a little time by noting that this otential is an even function, so we can assume with no loss of generality that the solutions are either even or odd (Problem 2.1(c)). The advantage of this is that we need only impose the boundary conditions on one side (say, at +a); the other side is then automatic, since tfl—x) 2 “£1,000. I’ll work outgthe e‘vensolutions; you get to do the odd ones in Problem 2.29. The cosine is even (and the sine is odd), so I’m looking for solutions of the form Fe‘“, for x > a, 1.00:) = DcosClx), for 0 < x < a. [2.151] 1/x(—-x), for x < 0. The continuity of in), at x = a, says Fe‘” = D ocean), [2.152] and the continuity of d tit/(ix, says HKFe_m = —ID sinfla). [2.1531 Dividing Equation 2.153 by Equation 2.152, we that I K = ItanCla). [2.154]; \/ (zo/zF—t 1t Sit/2 2n ' 511/2 2 FIGURE 2.18: Graphical solutiou to Equation 2.156, for zo = 8 (even states). This is a formula for the allowed energies, since K and l areboth functions of E. To solve for E, we first adept some nicer notation: Let 2: E Ia, and 20 a ZmVo. $2.155] According to Equations 2.146 and 2.148, 051+ 13) = 2mV0/h2, so m = '23 — zz, and Equation 2.154 reads tanz = ,/ (zo/z)2 +1. [2.156] This is a transcendental equation for z (and hence for E) as a function of Zn (which is a measure of the “size” of the well). It can be solved numerically, using a computer, or graphically, by plotting tanz and ,/ (mg/2:)2 — 1 on the same grid, and looking for points of intersectibn (see Figure 2.18). Two limiting cases are of special interest: 1. Wide, deep well. If :0 is very large, the intersections occur just slightly below 3,, = HIT/2, with 11 odd; it follows that 1 7 1 n-rr-h- -—-—-—. .1 7 2m(2ct)2 E2 5] EH+VOE But E + V0 is the energy above the bottom of the well, and on the right side we have precisely the infinite square well energies, for a well of wifiih: 2n (see Equation 2.27)—uor rather, half of them, since this n is odd. (The other ones, of course, come from the odd wave functions, as you’ll discover in Ptobleritt 21.290810 the finite sq-guare well goes over to the infinite square well, as V0 -—> do: 'liiloWe-v'er. for any finite V0 there are only a finite number of bound states. 2. Shallow, narrow well. As 2:0 decreases, there are fewer ante state-s, until finally (for m < H / 2, where the lowest odd state disappeaer ewe remains. It is interesting to note, however, that there is always one bottttldt state, no matter how “weak” the well becomes. _;_;';;______;_;__;:M;;z;_;;:;__ W g5; "£53m ,_ 343/ _____£pfaefl&m..___, : / X/Z‘WW‘” - ( ) E <2: E, __—:._,.,.___ __ .___.~ _—_.___—__M__,—__——___—.WW _ 591% .: . E .2 m1 «9W? 46;sz <59 em“ cam /‘ .5? .9 Mt .- .- mefim Vfldéfl : A a. fi‘ x .. __._.4 gm'mmi ‘z/gx * I 14—- a}; 3‘14, _ . 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This note was uploaded on 12/10/2011 for the course PHYS 4221 taught by Professor Cflo during the Spring '11 term at CUHK.

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Notes4 - FINITE SQUARE WELL As a last example consider the finite square well potential “V0 for —-a<x<a V“ I l o for Ix> a[1145

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