Notes14-20 - QUANTU M MECHANICS , < 0, in which case...

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Unformatted text preview: QUANTU M MECHANICS , < 0, in which case E lies is condition is also sufficient dimensions (see Prob. 10, Fig. 5a is sufficiently broad n corresponding to a larger re 6a, b, and 0 show a series in Fig. 5c, d, and e for suc- re) values of E; both signs 6b show the eigenfunctions 73/ levels of a particle bound by an extension of the fore- any higher discrete energy than that corresponding to (b) ..|/ ial and larger (less negative) from (a) to (b) to (c) and is ilope can both be continuous THE SCHRODINGER WAVE EQUATION 37 Thus, for a potential energy that approaches a finite constant value as a: —-) i 00, there may be a finite number of discrete energy levels, or in some cases an infinite number [if V(x) falls off slowly enough for large [ml], depending on V(r) and the mass of the particle. However, if V(z) ——> + so as a: —-> i on, an argument like that given above shows that there will always be an infinite number of discrete energy levels; apart from arbitrary multiplying constants there will be just one eigenfunction u(:z:) for each of these. CONTINUOUS ENERGY EIGENVALUES It is possible to find eigenfunctions that obey the boundary and continuity conditions for all energy eigenvalues that exceed the smaller of the two numbers V(+ 00) and V(— 00 ). If, for example, the potential energy has the form illustrated in Fig. 5a, then solutions of the wave equation can be found for all positive values of E. This is because the solutions for large III are of the form A sin alzl + B cos alzl a = + (22?), (8.6) and there is no reason Why both terms should not be kept. Thus it is always possible to adjust the phase of each of the wave functions for large [ml (which is equivalent to adjusting the ratios A /B for the solutions for large positive and negative 2:) so that they join together smoothly when continued in to a: = 0. K The classical terms periodic (or multiply periodic) and aperiodic are sometimes used to designate the particle motions associated With discrete and continuous energy eigenvalues, respectively. DISCRETE AND CONTINUOUS EIGENVALUES IN THREE DIMENSIONS We shall assume Without further discussion that all the foregoing results can be taken over in a natural way for the three—dimensional wave equa- tion (8.2). We can expect that, if V(r) ——> + on as r —> oo in all directions, there will be an infinite set of discrete energy levels extending to + 00. If V(r) is bounded as r —> oo in some direction, there may be a finite or an infinite number of discrete levels, depending on the form of V. In this case, the discrete energy levels cannot exceed the smallest value that V(oo) has in any direction. For values of E larger than this smallest V( 00 ), the energy eigenvalues cover a continuous range extending to + co . SHONE—DIMENSIONAL SQUARE WELL POTENTIAL As a simple explicit example of the calculation of discrete energy levels of a particle in quantum mechanics, we consider the one-dimensional QUANTU M MECHAN|CS (15) Fly. 7 One-dimensional square well potential with (a) perfectly rigid walls and (b) finite potential step. motion of a particle that is restrained by reflecting walls that terminate a region of c'onstant potential energy. Two simple types of potential energy are considered. Figure 7a shows a situation in which V(:c) = 0 for —a < a: < a, and V(:c) = +00 for Isl > a, corresponding to per- fectly rigid, impenetrable walls at the points a: = is. In Fig. 7b the increase in potential energy at the walls is abrupt but finite, so that V(2:) = V0 for [ml > a; because of its appearance, this is often called a square well potential. The motion of a classical particle with total energy E less than V0 is the same for both these potentials; but, as we shall see, the quantum-mechanical behavior is different. In general, an abrupt finite increase in potential energy at the boundaries of a region forces a particle toward the interior of the region. Such a potential may be thought of as a limiting case of a potential of the type shown in Fig. So, for which the force —dV/dx is always directed in toward x = 0. The force associated with a square well potential is zero except at the boundaries, so that the particle is acted on by no force except a sudden impulse directed toward the origin as it passes the points a: = ia. PERFECTLY RIGID WALLS It was shown in Sec. 8 that the wave function must vanish at the points a: = i-a, when the potential energy has the form shown in Fig. 7a. From Eq. (8.5) the wave equation for Izl < a is simply (9.1) 2m * ME) (9.2) which has the general solution “($)=Asina:c+Bcos:z:c a=+( QUANTU M MECHANICS (b) it}: (a) perfectly fleeting walls that terminate a v0 simple types of potential . situation in which V(a:) = 0 :l > a, correSponding to per— nts 2: = id. In Fig. 7b the is abrupt but finite, so that earance, this is often called a iical particle with total energy otentials; but, as we shall see, rent. In general, an abrupt oundaries of a region forces a 1. Such a potential may be f the type shown in Fig. 5a, for l in toward a: = 0. The force zero except at the boundaries, ‘rce except a sudden impulse points a: = id. .ion must vanish at the points the form shown in Fig. 7a. 1 a is simply (9.1) = + (9.2) THE SCHRGDINGER WAVE EQUATION 39 Application of the boundary conditions at a: = ia gives Asinaa+BcOSaa=0 —Asinaa+BcOSaa=0 from which we obtain Asinaa=0 Bc05aa=0 Now we do not want both A and B to be zero, since this would give the physically uninteresting solution 11. = 0 everywhere. Also, we cannot make both sin ac and cos aa zero for a given value of a or E. There are then two possible classes of solutions: For the first class A = 0 and cos em = 0 and for the second class B = 0 and sin aa = 0 Thus aa = mr/ 2, where n is an odd integer for the first class and an even integer for the second class. The two classes of solutions and their energy eigenvalues are then u(a:) = B cos 13—255 n odd u(x) = A sin 7% n even 2 2 2 E = «8;; in both cases It is evident that n = 0 gives the physically uninteresting result u = 0 and that solutions for negative values of n are not linearly independent of those for positive n. The constants A and B can easily be chosen in each case so that the eigenfunctions 14(2) are normalized. There is thus an infinite sequence of discrete energy levels that cor- respond to all positive integer values of the quantum number n. There is just one eigenfunction for each level, and the number of nodes of the nth eigenfunction that are within the potential well is n — 1. These results are in agreement with the discussion of Sec. 8. It is interesting to note that the order of magnitude of the lowest or ground-state energy level is in agreement with the uncertainty relation (3.1). The position uncertainty of order a implies a momentum uncertainty at least of order fi/a, which in turn implies a minimum kinetic energy of order 712/ ma”. FINITE POTENTIAL STEP When the potential energy has the form shown in Fig. 7b, it is necessary to supplement the general solution (9.2), which is still valid for lzl < a QUANTUM MECHANICS since Eq. (9.1) is unaltered there, by a solution for M > a. The wave equation in this region is 2 —L@+Vuu=Eu 2m d1:2 which has the general solution for E < V0 (bound states) 2m(Vg — E) F u(:c) = Cr” + D65” {3 = + I: h, (9.3) The boundary conditions at 2: = i no discussed in Sec. 8 require that we set D = 0 if Eq. (9.3) is to represent the solution for :c > a, and C = 0 if the solution is for a: < ——a. We now impose on the solutions (9.2) and (9.3) the requirements that u and du/dz be continuous at :v = j; a. A sin aa + B cos aa = Ce‘“ aA cos aa — aB sin aa = —fiCe—5“ —A sin aa + B cos aa = De‘“ aA cos aa + «:3 sin aa = BDe-fi“ from which we obtain 2A sin aa = (C — D)e‘5“ 2aA cos aa = —B(C — D)e—5“ "‘ 2B cos aa = (C + D)e—5“ 2aB sin aa =’ 5(C + D)e"" (9.5) Unless A = O and C = D, Eqs. (9.4) have as their consequence Similarly, unless B = 0 and C = —D, Eqs. (9.5) give l l l l 1‘ 1-!5: a cot aa = —B (9-5) l l l l l a tan aa = 13 (9.7) Now it is impossible for Eqs. (9.6) and (9.7) to be valid at once, since / on elimination of )3 this would require that tan” out = — 1, which in turn would make a imaginary and 13 negative, contrary to Eq. (9.3). Also, we do not want A, B, C, and D all to vanish. Thus the solutions may again be divided into two classes: For the first class ....V We ....,.9 A=0 C=D and a tan em = {3 and for the second class B=0 C=—D acot out = —{3 QUANTU M MECHANICS ition for [x] > a. The wave (bound states) n(Vo — E)]§ ,. (9.3) ssed in Sec. 8 require that we )lution for :c > a, and C’ = 0 ) and (9.3) the requirements os aa = —fl(C’ — D)e‘fi“ (9.4) in aa = 6(0' + D)e‘fl“ (9.5) as their consequence (9.6) (9.5) give (9.7) :97) to be valid at once, since ;an2 aa = —1, which in turn .trary to Eq. (9.3). Also, we [‘hus the solutions may again ss ztanaa=fl acot aa = —fl THE SCHRODINGER WAVE EQUATION 41 EN ERGY L EVELS The energy levels are found from a numerical or graphical solution of Eqs. (9.6) and (9.7) with the definitions for a and 6 given in Eqs. (9.2) and (9.3). A simple graphical method for effecting this solution is described here, since it shows quite clearly the way in which the number of discrete levels depends on V0 and a. We put E = aa, 1; = 6a, whence Eq. (9.7) becomes E tan E = n, with £2 + 112 = 2me1? ....—_—-——i” Since E and g are restricted to ositive es, the ener levels ma be ound in this case tom the intersections in the first quadrant of the curve of E tan E plotted against E,_Vji_t_h_the circle of knowwdmw F, "—— '1 he construction is drawn in Fig". 8 for three values of V002; for each of the two smaller of these values there is one solution of Eq. (9.7) and for the largest there are two. Figure 9 is a similar construction for the solution of Eq. (9.6) in ‘which the energy levels are obtained from the intersections of the same circles with the curve of —-E cot E in the first quadrant. The smallest value of Voa,2 gives no solution, and the two larger values each give one. Thus the three increasing values of Voa2 give altogether one, two, and three energy levels, respectively. Fig. 8 Graphical solution of Eq. (9.7) for three values of Vac“; the vertical dashed lines are the first two asymptotes of 'q = Etan E. QUANTUM MECHANICS Fly. 9 Graphical solution of Eq. (9.6) for three values of Voa’; the vertical dashed line is the first asymptote of 1, -=- —s cot s. It is clear from Figs. 8 and 9 that, for a given particle mass, the energy levels depend on the parameters of the potential energy through the combination Voa”. For Voa2 between zero and #271,2/ 8m, there is just one energy level of the first class; for Van between rzhz/ 8m and four times this value, there is one energy level of each class, or two altogether. As Voa2 increases, energy levels appear successively, first of one class and then of the other. It is not difficult to see from Eq. (9.2) that, when ordered according to increasing eigenvalues, the nth eigenfunction has n — 1 nodes. PARITY It follows from the foregoing discussion that the eigenfunctions of the first class are even with respect to change in sign of :c [u( -:c) = u(:c)], whereas the eigenfunctions of the second class are odd [u(-—:c) = —u(:c)]. This division of the eigenfunctions into even and odd types is not accidental and will now be shown to be a direct consequence of the fact that the potential energy function V(z) is symmetric about a: = 0. If we change the sign of a: in the wave equation (8.5) — %d2;(f) + V(:c)u(x) = Em) (9.3) QUANTU M MECHANICS values of Van“; the vertical 5. for a given particle mass, the >f the potential energy through . zero and «aha/8m, there is just between «fizz/8m and four times :h class, or two altogether. As sively, first of one class and then In Eq. (9.2) that, when ordered : nth eigenfunction has n — 1 .at the eigenfunctions of the first gn of a: [u(—x) = u(a:)], whereas e odd [u(—x) = —u(x)]. This and odd types is not accidental )nsequence of the fact that the trio about x = 0. If we change (9.8) THE SCHRéDINGER WAVE EQUATION 43 and if V(—x) = V(x), we obtain it: d’u( — 2:) _ 2m dxz + V($)u(—x) = Eu(_x) Then 24(2) and u(—-x) are solutions of the same wave equation with the same eigenvalue E. Suppose at first that there is only one linearly independent eigenfunction that corresponds to this energy level; then u(:z:) and u(—x) can differ only by a multiplicative constant: u(—:z:) = eu(:z:) ' (9.9) Changing the sign of :1: in Eq. (9.9) gives u(:z:) = eu(—x). From these two equations it follows at once that ez=1 or e=i1 Thus all such eigenfunctions of a symmetric potential are either even or odd with respect to changes of sign of 2:. Such wave functions are said to have even or odd parity. If an eigenvalue has more than one linearly independent eigenfunc- tion, the foregoing argument breaks down, and these eigenfunctions need not have a definite parity; they need not be even or odd. However, we can easily see that linear combinations of such eigenfunctions can be found such that each has even or odd parity. Suppose that an eigenfunc— tion u(x) does not have a definite parity. It can always be written u(x) = u.(x) + we) where u.(x) = &[u(x) + u(—:c)] is even, uo(x) = 11;[u(:c) — u(—x)] is odd, and u. and u, are linearly independent. Then if the wave equation (9.8) is symmetric, we can write it as h? dzu. h: dzuo _ _ in dzz + (V — E)u. — fin dxz + (V — E)uo _ 0 (9.10) 0’1 Changing the Slign of a: in Eq. (9.10), we obtain hz dine h’ d’ua .. _ in dx2 + (V — E)“, 5,7, dz. - (V - E)u., — 0 (9.11) Addition and subtraction of Eqs. (9.10) and (9.11) show that u. and u, are separately solutions of the wave equation with .the same eigenvalue E. A SIMPLIFIED SOLUTION Knowledge that the solutions possess a definite parity sometimes simplifies the determination of the energy levels, since then we need only find the solution for positive 2:. Even solutions have zero slope and odd solutions have zero value at x = 0. If, for example, we wish to find the even solu- 44 QUANTUM MECHANICS tions, Eqs. (9.2) and (9.3) can be replaced at once by u(x)=BcOSaz 0<x<a u(2:) = 06—5” 2: > 0. Instead of making both u and du/dx continuous at a: = a, it is enough to make the ratio (1 /u) (du/ dz) continuous at a: = (1, since the normalizing constants B and C are eliminated thereby. This gives Eq. (9.7) at once. Similarly, the odd solutions are u(2:)=Asina2: 0<x<a u(x)=Cc—’ x>a Then continuity of (1 / u) (du/ dz) at :1: = a immediately gives Eq. (9.6). PROBLEMS 1. Use the arguments of Sec. 6 to set up a differential equation for \b that involves a second time derivative of ‘p, in the case of a free particle. Discuss any solutions that this equation has that are not shared by the free-particle Schrodinger equation. 2. Show that the free-particle one-dimensional Schrodinger wave equation (6.8) is invariant with respect to galilean transformations. Do this by showing that, when the transformation x’ = x —— at, t' =t is applied, the transformed wave function ¢’(:z:’,t’) = f (1:,t)‘p(:z:,t) satisfies Eq. (6.8) with respect to the primed variables, where f involves only 2, t, h, m, and 9. Find the form of f, and show that the traveling wave solution ¢(:c,t) = Ae“""“’" transforms as expected. 3. How must a wave packet \6 fall off for large r in order that the volume integral of P and the surface integral of S. in Eq. (7.4) converge? 4. Show directly that (1),) is real for a wave packet. 5. Show that for a three—dimensional wave packet is» = its.) + <pzx>> 6. Calculate the energy levels and plot the eigenfunctions for the three bound states in the potential of Fig. 7b when Voag = (BM/m. Compare with the first three states for the potential of Fig. 7a. 7. Discuss the relation between the energy levels for the potential of Fig. 7b‘ and those for the potential V(:z:) = + an, :2: < 0; 17(2) = 0, 0 < z < a; V(.1:) = V0, 1: > a. 8. Show that if the potential energy V(r) is changed everywhere by a constant, the time-independent wave functions are unchanged. What is the effect on the energy eigenvalues? 2. SI-La'rt‘callg Samma‘fin'c Ipo'l’ebnll‘a/f X = $5,139 Co: 45 My; v sine sl'n 45, Z: rcarél -1 [.L 9., (72 2 mg 9; H T2 9)" 9' 77’ Sir) 9 9‘9 9‘9 f// ( Ix) H ‘U m \3 § 9. ,A ,. I “I ,, fir,ng ., __ ffiafiEfl’g‘IMQ—Zdfia Wigowi‘ 5‘" S i£.:_?\€ "ti—Z L ,, 4:1 * I (wk, , 0A0— ME a S! 0d OJ 0— J . "Mitiganuhe ,_ Q, fl?“ 7‘ If w}; f ‘ fa, L , 7 _ 7 f? 7’”. @375) h :"MMQE,if” ‘9)..‘fiwffi * Anzi/ar ' M’bmznhm' ' V Qg/uqhan .' N, ;;z 5; (w 5;) f j e = o :5— 5 fat. . RD) f * 7‘ (’Y) /* é) , . ,, - ¥ -;,. 2. A 2 x _ . ., ‘krz _ 7 - +-— _ 2— ++; [w ¥ _ a. J .2 — E x ,, . ffiv “w, H , _. 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This note was uploaded on 10/21/2009 for the course PHYS PHYS 5307 taught by Professor Asokray during the Fall '08 term at UT Arlington.

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Notes14-20 - QUANTU M MECHANICS , &amp;lt; 0, in which case...

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