Quantum Mechanics in Three Dimensions

Transcription

1 Physics 342 Lecture 20 Quantum Mechanics in Three Dimensions Lecture 20 Physics 342 Quantum Mechanics I Monay, March 24th, 2008 We begin our spherical solutions with the simplest possible case zero potential. Asie from being uncommon, this allows us to clearly see the role of the various terms in the separation. From the solution regular at the origin, we can evelop the infinite barrier cases, in which we consier the three-imensional, spherically symmetric analogue of the infinite square well from one imension The Raial Equation When we combine the potential, which epens only on r with the angular constant l l+1)), we obtain the orinary ifferential equation for Rr): 1 R r 2 R ) l l + 1) 2 m r2 r r 2 Ur) E) = ) If we efine ur) rrr), then the above simplifies: 2 2 ) u 2 m r 2 + Ur) + 2 l l + 1) 2 m r 2 u = Eu. 20.2) Compare the term in parenthesis to the classical orbital effective potential we foun last time: U eff = Ur) + p2 φ 2 m r 2, 20.3) this is, in a sense precisely the same term we ha there if only we knew how to associate p 2 φ with the numerator of the above, we coul have written this own irectly. 1 of 10

2 20.2 No Potential Suppose we have in min no explicit potential, i.e. Ur) = 0. Then we can solve the above for u but what are the relevant bounary conitions? We pretty clearly want the wavefunction to go to zero as r 0. Near the origin, we want the probability to be finite. Think of what this means for the raial wavefunction the probability will be proportional to R R r 2 r, an this is just u u r in our new notation, but then we want u 0 or, potentially, a non-zero constant) as r 0. The point is, Rr) 1/r is allowe, but no higher power. Now we can return to the ODE 20.2) written in stanar form, this is 2 u l l + 1) r 2 = r 2 2 m E ) 2 u, 20.4) an we efine k 2 2 m E. Very far from the origin, where the constant 2 term ominates, we recover a sinusoial solution, cosk r) an sink r) an we suspect that the raial wavefunction is not normalizable that comes as no surprise, since the Cartesian solution with zero potential was also not normalizable. Still, it is instructive to look at the solutions, if only in preparation for finite range. The solutions to the above ODE are spherical Bessel/Neumann functions more explicitly, the Rr) solutions are spherical Bessel functions, ur) gets multiplie by r): u l r) = α r j l k r) + β r n l k r). 2) These are relate to the Bessel functions an Bessel s equation, of course), an can be efine via: ) 1 j p x) = 1) p p ) sin x 1 n p = 1) p p cos x x x x x. 20.6) We are intereste in the asymptotic behavior here both j p an n p reuce to cosine an sine as r as they must, an this is the source of the normalization issue. But near the origin, we have: j p x 0) = 2p p! 2 p)! 2 p + 1)! xp n p x 0) = 2 p p! x p+1), 20.7) so the well-behave solution near the origin is the j p one 1. 1 Compare with your electroynamics experience there, we cannot have a single solu- 2 of 10

3 Consier, for example, the p = 0 form, j 0 k r) = sink r) k r sink r) k r k r+ok r)2 ) k r, this is finite close to k r 0, as is clear from its Taylor expansion, 1 + Ok r). While not normalizable, we can plot the wavefunction the angular portions are unity) to get a sense for the eventual ensity. Working back to Rr), we have: an this is plotte with α = 1) in Figure Rr) = 1 r ur) = α j 0k r) 20.8) Rr) kr Figure 20.1: The zeroth spherical Bessel function this gives the raial wavefunction for a free particle in spherical coorinates. Spherical Bessel Functions We quote the result above, the ifferential equation 20.4) has solutions that look like u l r) = α r j l k r) finite at the origin). But how coul we evelop these if we in t know them alreay? Well, there s always Frobenius, that woul be one way. We coul also connect these special functions to simpler ones, as we i for the associate Legenre polynomials, for example. Here we consier yet another way to get the solutions we can use an integral transform like the Fourier transform, or Laplace transform) to simplify the ODE. What we will en up with is an integral form for the spherical Bessel functions. tion to Laplace s equation that is well-behave at both the origin an infinite. Here, the requirement is less stringent, we can allow some explosive behavior on either en, since the wavefunction, unlike E an B, is not the observable or at least, experimentally comparable object. In quantum mechanics, it is R R that is important, an the requirement is that R R r 2 be integrable. 3 of 10

4 Our goal is to solve: with k 2 2 m E 2 [ ] l l + 1) u r 2 k 2 u = ) as usual. We make the ansatz: ur) = r p e r x fx) 20.10) where p, fx) an the integration limits x can be complex here) are to be etermine by our solution. This is similar to our usual Frobenius ansatz, but with an integral rather than a sum. First we nee the erivatives: u r) = p r p 1 e r x fx) + r p x e r x fx) u r) = p p 1) r p 2 e r x fx) + 2 p r p 1 x e r x fx) + r p x 2 e r x fx). Now if we input this into the ODE, we get: 20.11) 0 = r p 2 [p p 1) l l + 1) + 2 x r p + x 2 + k 2) r 2] e r x fx) ) The analogue of the Frobenius inicial equation is the r 0 portion of the above: p p 1) l l + 1) = 0 this has solutions: p = l an p = l + 1. How shoul we choose between these? Think of the case p = l then in front of the integral, we have r l 2 which, for l > 0 will lea to ba behavior at r = 0. If we take p = l + 1 for l > 0, we will have a vanishing term in front of the integral. So take p = l + 1. We are left with 0 = r l fx) [ 2 l + 1) x + x 2 + k 2 ) r ] e r x [ = r l fx) 2 l + 1) x + x 2 + k 2 ) ] 20.13) e r x. Notice the replacement in the secon line, r acting on er x. This comes from the observation that er x = r e r x. 4 of 10

5 We still on t know what our limits of integration are, nor o we have fx). All we have one so far is choose a value for p. Keep in min that the choices we make: p, fx), an the integration path itself, all serve to limit our solution we will not obtain the most general solution to 20.9). We can use the prouct rule to rewrite our integral again: [ 0 = r l fx) 2 l + 1) x + x 2 + k 2 ) ] e r x { = r l fx) 2 l + 1) x) e r x [ + fx) x 2 + k 2) e r x] e r x [ fx) x 2 + k 2 ) ] } ) Again in the interests of specialization we coul make the above zero by setting: 0 = 2 l + 1) x fx) [ fx) x 2 + k 2 ) ] [ 0 = fx) x 2 + k 2 ) e r x] 20.15) there are other ways to get zero from the integral, but we are not currently intereste in them. These two equations are enough to set the function fx) an at the same time, etermine the integration region. Taking the ODE first from inspection, a goo guess is fx) = x 2 + k 2 ) q, then: This gives us 2 l + 1) x fx) = [ fx) x 2 + k 2)] q = l ) fx) = x 2 + k 2) l ) With fx) in han, we nee to choose integration limits for the secon equation in 20.15). Suppose we think of this as one-imensional integration for x C, we coul have in min a complicate contour integral), then: x 2 + k 2) e r x x f x=x 0 = ) 5 of 10

6 If we set x 0 = i k, x f = i k, then each enpoint vanishes separately. This suggests we take: i k ur) = r l+1 e r x x 2 + k 2) l ) i k This can be rewritten by taking z = 1 i k x, then ur) = r i k l) 1 k r) l e i k r) z 1 z 2) l z ) 1 Now, the integral form of the spherical Bessel functions is: j l x) = xl 2 l+1 l! +1 1 e i x z 1 z 2) l z, 20.21) an noting that constants on t matter since we will normalize the raial wavefunction at the en, anyway), we can write u l r) = A r j l k r) 20.22) Spherical Barrier We cannot exten the free particle range to infinity, but it is possible to cut off the spherical Bessel solutions at a particular point we imagine a perfectly confining potential like the infinite square well, but w.r.t. the raial coorinate. This imposes a bounary conition if we take { 0 r R Ur) = r > R 20.23) then the free particle solutions escribe above have ur = R) = 0, so as to match with the perfect zero outsie the infinite spherical) well. With the cut-off in place, the solutions are normalizable, we just integrate from r = [0, R], which is more manageable than an infinite omain. But, we have to choose a value for k that makes j l k R) = 0 the zeroes of the Bessel function are, like cosine an sine, finite in any given interval. The 6 of 10

7 spacing is not so clean as simple trigonometric functions. If this were sine or cosine, we woul just set k R = n π or a half-integer multiple an be one. With the spherical Bessel functions, it is possible to fin zero-crossings an also to etermine how many zero crossings are in an interval), but there is no obvious formula. For the l = 0 case, we o, in fact, know the zero crossings, since j 0 k r) = sink r) k r, an k R = n π for integer n gives zero. Then, the usual story: Our bounary conition has provie quantization of energy. Remember that we get E out of this process. The quantize k = n π R = 2 m E gives E = 2 n 2 π 2 2 m R ) for integer n. Now, this is relatively uninteresting as a three-imensional solution to Schröinger s equation: There is no angular component since we solve with l = 0, an Y 00, φ) = 1 2 is the only available term. We have π for l = 0, an n-inexe set of wavefunctions normalize): ψr,, φ) = sin n π R r) 2 π R r ) This groun state state with lowest energy) is spherically symmetric, it shares the symmetry of the potential itself an iea to which we shall return). In general, we can enote the n th zero crossing of the l th spherical Bessel function as β nl The full spatial wavefunction reas: ) βn l r ψ n l m r,, φ) = A n l j l Yl m, φ) ) R with normalization A n l. The energy, then, is given by k R = β n l E n l = 2 β 2 n l 2 m R ) an we can use this in the full time-epenent solution: Ψ n l m r,, φ, t) = ψ n l m r,, φ) e i E n l t ) Each energy E n l is share by the 2l + 1 values of m that come with the angular solution. In Figure 20.2, we see the probability ensity unnormalize) for the l = 0 an n = 1, 2, 3 states top row) an the ensities for ψ 11 1, ψ 110 an ψ of 10

8 20.3. EXAMPLE OF SIMILAR PROBLEM PDE + BC) Lecture 20 trivially relate to ψ 11 1 ). Keep in min that our local ensity is, now ψn l m ψ n l m r 2 sin. Notice that as with the one-imensional square well, the number of noes in the r irection top row) is irectly relate to the energy of the state. r r Figure 20.2: Probability ensities ψ ψ r 2 sin in this case) for ψ 100, ψ 200, ψ 300 top row) an ψ 11 1, ψ 110, ψ 111 bottom row). The horizontal axis of each plot is the raial irection, the vertical irection goes from 0 to π in Example of Similar Problem PDE + BC) Before we begin our onslaught the Coulomb potential an Hyrogenic wavefunctions let s review some of the other applications of our current techniques, just to highlight the familiarity of the proceure of solving the Schröinger equation, even while its interpretation is new. First of all, we recognize that solving the PDE requires some bounary conitions in the above infinite potential, we wante regular solutions on the interior of the sphere of raius R. This is similar to solving Laplace s equation for the electrostatic potential insie an outsie a istribution. Consier a sphere, if we are insie some spherically symmetric istribution of charge, an the origin r = 0 is enclose, then solutions must go like r p. If we are solving for the potential outsie the istribution, where spatial infinity is inclue, we expect V r p. All of this is in the name of bounary conitions. The same is true for quantum mechanical problems: Our omain of interest or, in some cases, the energy scale of interest) efines the types of solution we accept physically. Take a uniformly charge spherical shell with surface charge σ. We rotate 8 of 10

9 20.3. EXAMPLE OF SIMILAR PROBLEM PDE + BC) Lecture 20 the shell with angular velocity ω = ω ẑ about the ẑ axis). Our goal is to fin the magnetic fiel, outsie the sphere, say. Well, for r R, we know that the magnetic vector potential satisfies: 2 A = 0 A out r Ain r ) r=r = µ 0 K ) First, from the physical setup, we automatically know the surface current: K = σ v = σ ω R sin ˆφ. So we know pretty quickly that A sin ˆφ. Now sin is not one of your usual Legenre polynomial solutions to Laplace s equation, so there is something strange going on here. Suppose we take the ansatz: A = Ar, ) ˆφ, then the vector Laplace equation reas: 0 = 2 A = 2 A) ˆφ + A 2 ˆφ ) The absence of Legenre polynomials comes as no surprise we are not solving Laplace s equation remember that 2 ˆφ 0, the spherical basis vectors are position epenent). One can easily if teiously) calculate 2 ˆφ = 1 r 2 sin 2 ˆφ, so the equation we are trying to solve looks like: 2 A A r 2 sin 2 = ) an this looks more like the Schröinger equation. If we use our usual separation of variables, factoring Ar, ) into Ar, ) = A r r) A ), then the above becomes r r 2 A ) r [ + A r 1 A sin sin A ) 1 ] sin 2 = ) We can make our separation ansatz as before, with constant l l + 1): r 2 A r) [ 1 A sin r sin A ) 1 sin 2 = l l + 1) A r ] = l l + 1) ) For the angular equation, we now nee to solve: sin sin A ) + l l + 1) sin 2 1) A = ) But that is very close to an equation we encountere in our evelopment of Yl m, think of [4.25]: sin sin A ) + l l + 1) sin 2 m 2 ) = ) 9 of 10

10 20.3. EXAMPLE OF SIMILAR PROBLEM PDE + BC) Lecture 20 This was solve by the associate Legenre function Pl m cos ) that makes up the epenence of the spherical harmonics. Comparing 20.34) to 20.35), we see that for our current purposes, we are intereste in m = ±1 it oesn t really matter which sign we choose, take m = 1). So the angular portion is A ) Pl 1 cos ). We can go further, though the bounary conition A out r R) A out A in r R) A in ) = µ0 σ ω R sin 20.36) suggests that we want l = 1 the associate Legenre polynomials are orer l in sin an cos ). In fact, if we look up P1 1 cos ) = sin, just what we want. With l = 1 in han, we can return to the raial equation: r 2 A r) r A r = l l + 1) = 2 A r r) = α r + β r ) This secon orer ifferential equation has actually given us both the interior an exterior cases. For the interior, we set β = 0 so that the potential oes not blow up at the origin. For the exterior, we set α = 0 to get goo behavior at r. We can finish the whole job now: A out = α r 2 P 1 1 cos ) ˆφ A in = β r P 1 1 cos ) ˆφ, 20.38) an P1 1 cos ) = sin perfect). Just apply the bounary conition 20.36) an continuity A out = A in ) 2α R 3 sin β sin = µ 0 σ ω R sin ˆφ α R 2 = β R 20.39) to get α = 1 3 µ 0 σ ω R 4 an β = 1 3 µ 0 σ ω R. The en result is A out = µ 0 σ ω R 4 sin 3 r 2 ˆφ A in = 1 3 µ 0 σ ω R r sin ˆφ ) Homework Reaing: Griffiths, pp Homework assigne in Lecture of 10