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Chapter 4 BOUND STATES of a CENTRAL POTENTIAL

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Chapter 4 BOUND STATES of a CENTRAL POTENTIAL Chapter 4 BOUND STATES OF A CENTRAL POTENTIAL 4.1 General Considerations As an application of some of these ideas we briefly investigate the properties of a particle of mass m subject to a force deriving from a spherically symmetric potential V (r). Obviously, in many of the cases where this problem is of interest, the mass m referred to is the reduced mass m m m = 1 2 (4.1) m1 + m2 of an interacting two-particle system and the radial coordinate r corresponds to the mag- nitude of the relative position vector ~r = ~r2 ~r1, although these details need not concern us here. The corresponding quantum system− is governed by a Hamiltonian operator P 2 H = + V (R) (4.2) 2m which in the position representation takes the usual form 2 ~2 ∇ + V (r), (4.3) − 2m and our goal is information about the bound state solutions ψ to the energy eigenvalue equation | i (H ε) ψ =0, (4.4) − | i whereweassumethatV (r) 0 as r , so that bound state solutions are identified as those square normalizeable→ solutions→∞ ψ ψ = d3r ψ(~r) 2 =1 (4.5) h | i | | Z for which ε 0. The≤ spherical symmetry of V (r) ( and thus of H) suggest the use of spherical coordinates for which the identity 1 ∂2 1 ∂2 ∂ 1 ∂2 2 = r + +cotθ + (4.6) ∇ r ∂r2 r2 ∂θ2 ∂θ sin2 θ ∂φ2 · ¸ holds for all points except at r =0, which is a singular point of the transformation. This last expression can be written in the form 2 2 2 2 2 2 2 ~ ∂ ~ L P = ~ = r + (4.7) − ∇ − r ∂r2 r2 where ∂2 ∂ 1 ∂2 L2 = +cotθ + (4.8) − ∂θ2 ∂θ sin2 θ ∂φ2 · ¸ 124 Bound States of a Central Potential is the operator, in this representation, associated with the square of the (dimensionless) orbital angular momentum L~ = R~ K.~ Thus, as in classical mechanics, the kinetic energy × H0 separates into a radial part and a rotational part P 2 2L2 H = r + ~ , (4.9) 0 2m 2mr2 where 2 2 2 2 1 ∂ ~ 1 ∂ Pr = ~ r = r . (4.10) − r ∂r2 i r ∂r · ¸ This last form suggests that the operator 1 ∂ P = ~ r (4.11) r i r ∂r can be viewed as the radial component of the momentum. This is a legitimate inference, and indeed Pr as defined above does correspond to the Hermitian part 1 Pr = P~ Rˆ + Rˆ P~ . (4.12) 2 · · h i of the operator P~ Rˆ. Some care must be taken, however, because (it can be shown that) · Pr is only Hermitian on the space of wave functions ψ(~r) such that limr 0 rψ(~r)=0, 1 → while the eigenfunctions of Pr all diverge at r =0as r− . Thus, Pr is an example of a Hermitian operator that is not an observable. None of this is too important for the task at hand, however. Indeed, with the aforementioned formulae we can write the energy eigenvalue equation of interest in the form 2 1 ∂2 2L2 ~ r + ~ + V (r) ε ψ(~r)=0 (4.13) −2m r ∂r2 2mr2 − ½ ¾ which is valid at all points except r =0. 2 We now make the observation that L and Lz = i∂/∂φ (or any other component − of L~ ) commute with the each other and with any function of r or Pr.Itisobvious, therefore, that they commute with H.Thus,because 2 2 H, L = L ,Lz =[H, Lz]=0 (4.14) we know that there exists a basis£ of¤ eigenstates£ ¤ common to the three operators H, L2, and Lz. These states, which we will denote by n, l, m are characterized by their associated eigenvalues {| i} H n, l, m = εn,l n, l, m (4.15) | i | i L2 n, l, m = l(l +1)n, l, m (4.16) | i | i Lz n, l, m = m n, l, m (4.17) | i | i and we denote by ψ (~r)= ~r n, l, m (4.18) n,l,m h | i the wavefunctions in the position representation associated with these states. Note that the eigenvalues of H are labeled by the principle quantum number n and by the total angular momentum quantum number `, but not by the azimuthal quantum number m, thus reflecting the necessary rotational degeneracy associated with the scalar operator H. General Considerations 125 2 It is also clear from our earlier discussions that the eigenfunctions of L and Lz can be written in the general form m ψn,l,m(~r)=Fn,l(r)Yl (θ, φ) (4.19) involving the spherical harmonics, with ` =0, 1, 2, , and m = `, , +`. Substitution of this assumed form into the energy eigenvalue equation··· results− in the··· following ordinary differential equation ~2 1 d2 `(` +1)~2 rFn,l + + V (r) εn,l Fn,l =0 (4.20) −2m r dr2 2mr2 − µ ¶ for the radial functions Fn,l(r). This equation is independent of the eigenvalue m of Lz, confirming our labeling of the eigenfunctions based upon the expected rotational degeneracy of the system. Thus, we can anticipate that each energy eigenvalue εn,l will be 2` +1 fold degenerate due to the rotational invariance of H. Since the differential equation does, generally, depend upon the quantum number `, we can expect a different series of eigenvalues εn,l for each value of l. Additional accidental degeneracies may arise, however, (and in fact do arise when V (r)=kr2, which corresponds to the 3D harmonic oscillator, and when V (r)= k/r, which corresponds to the Coulomb potential.) As a further simplification− it is customary to introduce a new function φn,l(r)=rFn,l(r) (4.21) which obeys the equation ~2 `(` +1)~2 φ00 + + V (r) εn,l φ =0. (4.22) −2m n,l 2mr2 − n,l · ¸ This latter equation is of the precise form ~2 φ00 +[Veff (r) εn,l] φ =0. −2m n,l − n,l obeyed by a particle moving in a 1-dimensional effective potential `(` +1) 2 V (r)= ~ + V (r) (4.23) eff 2mr2 consisting of the so-called “centrigugal barrier” `(`+1)~2/2mr2 in addition to the central potential experienced by the particle. Note that the wavefunction is defined only for r>0. Moreover, the normalization condition 2 2 r dr dΩ ψn,l,m =1 (4.24) Z ¯ ¯ leads, along with the standard normalization¯ condition¯ for the spherical harmonics dΩ Y m 2 =1 (4.25) | ` | Z to a normalization condition ∞ 2 2 ∞ 2 dr r Fn,l(r) = dr φ (r) =1 (4.26) | | n,l Z0 Z0 ¯ ¯ ¯ ¯ 126 Bound States of a Central Potential for the functions φn,l which make them correspond to a particle whose wave function is restricted to the positive real axis. As we will see, the boundary condition obeyed by these functions for most cases of interest corresponds to one in which φn,l 0 as r 0, as though there were an infinite barrier at the origin confining to the particle→ to the→ region r>0. To see this, we consider the general properties of the solution to Eq.(4.22) for small values of r. First, we rewrite the equation in the form `(` +1) φ00 + v(r)+kn,l φ =0 (4.27) n,l − r2 n,l · ¸ where 2mV (r) 2 2m εn,l v(r)= kn,l = | | (4.28) ~2 ~2 and we are assuming the εn,l < 0, as is appropriate to bound state solutions. We further assume the following: 1. V (r) is bounded except near r =0. 2. V (r) M/r as r 0 for some positive constant M. In other words, the magnitude of| V diverges| ≤ at the→ origin no more quickly than the Coulomb potential. 3. With these constraints, we now assume a solution near the origin which is of the form of a power law in r, i.e., we assume that φ (r) Crs+1 r 0 (4.29) n,l ∼ → for some positive constant s. This assumed form corresponds to the actual radial s wave function having the behavior Fn,l(r) Cr as r 0. ∼ → If we now substitute these limiting forms into the energy eigenvalue equation we obtain the following algebraic relation s 1 s 1 s+1 2 s+1 s(s +1)r − `(` +1)r − v(r)r + k r =0. (4.30) − − − n,l Now as r 0, the last two terms are at most of order rs and become negligible compared → s 1 to terms of order r − . Thus, for this form to be consistent at small r we must have s(s +1)=`(` +1), which has two solutions s = `s= (` +1). (4.31) − In principle, this gives us two linearly independent solutions to the second order differential equation. The solution with s = ` gives a solution for which, at small r, `+1 ` φ (r) Cr Fn,l(r) Cr (4.32) n,l ∼ ∼ while the solution with s = (` +1)gives a solution which for small r has the limiting behavior − C C φ (r) Fn,l(r) . (4.33) n,l ∼ r` ∼ r`+1 Of these, the solutions of the second type can be rejected on physical grounds. First, if `>0, then the solution of the form φ(r) C/r` is not normalizeable ∼ a dr C 2 ε 0. (4.34) | | r2` →∞ → Zε Hydrogenic Atoms: The Coulomb Problem 127 In the l =0case the situation is a little different. In fact, if l =0, the kinetic energy operator acting on any function proportional to 1/r gives a delta function, i.e., for r near zero, 2 2 1 Fn,0(r) =4πδ(~r).
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