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 magnitude 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). (4.35) ∇ ∼∇ r · ¸ The presence of this delta function prevents the energy eigenvalue equation from being satisﬁed at the origin, since the potential only has at most algebraic singularities at that point. Thus, the second “solution” for l =0is not actually a valid solution to the eigenvalue equation. It is a spurious solution arising from the singular nature of the transformation to spherical coordinates at r =0. Thus, in the end we conclude that under these conditions we obtain exactly one regular radial function

`+1 ` φ Cr Fn,l Cr r 0 (4.36) n,l ∼ ∼ → for each value ` =0, 1, 2, of the orbital angular momentum. Note that this limiting form implies the following··· “boundary condition”

lim φn,l(r)=0 (4.37) r 0 → for the eﬀective one-dimensional problem, making the wave function act as though there were an inﬁnite potential barrier at r =0. We have gone about as far as we can go without actually specifying the potential. We thus turn to what is certainly the most important application of these ideas to atomic systems, namely, the Coulomb problem associated with a particle moving in a Coulombic potential well.

4.2 Hydrogenic Atoms: The Coulomb Problem We now assume that potential of interest is of the form

1 Zq2 V (r)= (4.38) −4πε0 r associated with the Coulomb interaction between a nucleus of charge +Zq and a single electron of charge q. We ignore the spin of the contituents, and rewrite the potential as − Ze2 V (r)= (4.39) − r 2 2 by introducing e = q /4πε0. From dimensional arguments it is possible to generate estimates of the energy and length scales associated with bound states of this system. This is a useful exercise since it allows us to express the equations of interest in dimensionless form. These scales of interest arise from the fact that the mean energy

E = T + V (4.40) h i h i h i of a bound state depends upon both its kinetic energy

P 2 T = h i (4.41) h i 2m and its potential energy Ze2 V (4.42) h i ∼− a 128 Bound States of a Central Potential where a r is a typical distance of the electron from the origin, i.e., its spread a ∆r about the∼ nucleus.h i In any stationary bound state the mean position of the particle∼ must itself be stationary, which requires that P~ =0, and so, using the uncertainty principle we have the estimate h i

2 ∆P = P 2 ~ = ~ T ~ . (4.43) h i ∼ ∆r a h i ∼ 2ma2 p Thus, an order of magnitude estimate gives

2 Ze2 E ~ . (4.44) h i ∼ 2ma2 − a To estimate the ground state energy, therefore, we minimize this with respect to the parameter a.Setting∂ E /∂a =0we ﬁnd that h i 1 2 a a = ~ = 0 (4.45) Z me2 Z where ~2 a0 = 0.51angstrom (4.46) me2 ∼ turns out to be the Bohr radius. Putting this into our expression for the energy gives the estimate 2 2 Z ~ 2 2 E = 2 = Z ε0 Z 13.6eV (4.47) −2ma0 − ∼− × where ~2 ε0 = 2 13.6eV 2ma0 ∼ In addition, we can obtain an estimate for a typical velocity for the problem through the relation 2 p ~ e 6 v = = Z = Zv0 Z 2.2 10 m/s. (4.48) m ∼ ma ~ ∼ × × In atomic physics it is useful to introduce a set of units in which e = ~ = me =1, so that a0 =1,v0 =1, and E0 =1/2. We shall not do that in the present treatment, but we will use these length and energy scales to appropriately transform the problem. For example, the radial equation for this problem takes the form

~2 Ze2 l(l +1)~2 φ00 + + En,l φ =0. (4.49) 2m n,l r − 2mr2 n,l µ ¶ Introducing a dimensionless position variable ρ = r/a thiscanbewrittenintheform

~2 d2φ Ze2 1 l(l +1)~2 1 + + En,l φ =0. (4.50) 2ma2 dρ2 2a ρ − 2ma2 ρ2 µ ¶ 2 2 2 Dividing this through by E0 = Z ε0 = ~ /2ma this reduces to the form d2φ 2 l(l +1) + λ2 φ =0 (4.51) dρ2 ρ − ρ2 − n,l µ ¶ where 2 En,l λn,l = 2 0 (4.52) −Z ε0 ≥ Hydrogenic Atoms: The Coulomb Problem 129 is a positive dimensionless quantity for the bound states that we seek. From this equation we now determine the asymptotic behavior of the function φ(ρ) for large values of ρ. Indeed, for ρ >> 1 (which corresponds to r>>a) we can neglect 1 2 terms of order ρ− and ρ− in the eigenvalue equation, which then has the asymptotic form 2 d φ 2 = λ φ (ρ ) λ = λn,l. (4.53) dρ2 →∞ λρ This has two independent solutions φ Ae± , of which the diverging solution must be rejected as being non-normalizable. From∼ the well-behaved solution we introduce a new substitution, setting λρ u(ρ)=un,l(ρ)=φn,l(ρ)e (4.54) which makes u(ρ) a slowly varying function compared to the exponential. To determine the function u(ρ) we evaluate dφ λρ = e− [u0 λu] (4.55) dρ − 2 d φ λρ 2 = e− u00 2λu0 + λ u (4.56) dρ2 − in terms of which the radial equation can£ be re-expressed as¤

2 l(l +1) u00 2λu0 + u =0. (4.57) − ρ − ρ2 · ¸ The classical method of solving such an equation is to assume a power series solution

∞ j s ∞ k u(ρ)= cjρ = ρ bkρ (4.58) j=0 k=0 X X where in the second form we explicitly assume the ﬁrst coeﬃcient b0 is explicitly not equal to zero, and so have pulled out the leading power in the expansion. Indeed, we know that l+1 for small r the function φn,l(r) goes to zero as r . It follows that in the same limit the function u(ρ) has the leading behavior

λρ l+1 l+2 un,l(ρ) e φ Aρ + O(ρ ). (4.59) ∼ n,l ∼ Thus we write l+1 ∞ k ∞ k+l+1 u(ρ)=ρ bkρ = bkρ . (4.60) k=0 k=0 X X Calculating the derivatives of this expression term-by-term and substituting into the differential equation for u(ρ) wecancombinetermsinpowersofρ to determine a recursion relation for the coeﬃcients:

k(k + l +1)bk =2[(k + l)λ 1] bk 1 (4.61) − −

This relation allows all coefficients to be expressed in terms of a single coefficient b0. We observe that b 2kλ 2λ lim k+1 0. (4.62) k b ∼ k2 ∼ k → →∞ k Thus, the series converges for all fixed ρ. However, in general, the function u(ρ) to which it converges will itself diverge exponentially as a function of ρ as e2λρ. This leads to unacceptable solutions of the form φ eλρ. To prevent this from happening the n,l ∼ 130 Bound States of a Central Potential power series has to terminate, so that it reduces to a polynomial function. Thus, we conclude that acceptable solutions to the eigenvalue equation have the property that the power series terminates: there exists some value of k 1 for which the corresponding coefficients vanish, i.e., for which ≥

bk+n =0 n =0, 1, 2, . (4.63) ··· For this to occur, we must have

bk =2[(k + l) λ 1] bk 1 =0. (4.64) − − This leads to the condition that (k + l)λ =1, hence that the allowed values of λ satisfy the equation 1 λ = k =1, 2, (4.65) k + l ··· Turning this back into an equation for the energies, we ﬁnd that

2 En,l = λn,lE0 = λn,lZ ε0 (4.66) − − where 1 1 λn,l = = k =1, 2, ,n=(l +1), (l +2), . (4.67) k + l n ··· ··· Thus, the spectrum of energies takes the form

2 Z ε0 En,l = n =(l +1), (l +2), . (4.68) − n2 ··· Note that for ﬁxed n, several l values produce the same energy. This is an accidental degeneracy that is not required by the rotational invariance of the problem, and reﬂects other symmetries to the hydrogen atom problem that are not apparent in the present treatment. Expressed in the more standard way, the spectrum can be written

E0 En = l =0, 1, 2, ,n 1. (4.69) − n2 ··· −

Having determined the energies, and the corresponding values of λn,l, the coeﬃcients of the power series expansion are then determined by recursion. Since the series terminates at k = n l, the acceptable solutions take the form of polynomials − n l 1 n − − k+l 1 k un,l(ρ)= bkρ − = ckρ (4.70) k=0 k=l+1 X X of order n in ρ; the functions

n 1 − λn,lr/a0 k φn,l = e− ck(r/a0) (4.71) k=l X involve polynomials of order n 1 in r and the total wavefunctions take the form − n 1 k − r r/na0 m ψn,l,m = e− CkYl (θ, φ) (4.72) a0 k=l X µ ¶ and have energies E0 En = n =1, 2 . (4.73) − n2 ··· The 3-D Isotropic Oscillator 131

For each energy level En there are (2l +1)-fold degenerate angular momentum multiplets corresponding to l =0, 1, (n 1). The degeneracy of the nth level, therefore, can be calculated as ··· − n 1 − 2 gn = (2l +1)=n . (4.74) l=0 X In spectrocopic notation the bound states n, l, m for angular momentum states with l =0, 1, 2, 3, 4, 5, are indicated with a letter| s, p,i d, f, g, h, , respectively. Thus, e.g., ··· ···

ψ2p 2, 1,m (4.75) m → | i ψ3d 3, 2,m (4.76) m → | i and so on.

4.3 The 3-D Isotropic Oscillator The Hamiltonian P 2 1 H = + mω2R2 (4.77) 2m 2 3 P 2 1 = α + mω2X2 (4.78) 2m 2 α α=1 X · ¸ is separable in Cartesian coordinates, and has eigenstates

ψ (~r)= ~r n = ~r nx,ny,nz = ψ (x)ψ (y)ψ (z) n h | i h | i nx ny nz that are the products of 1D oscillator wave functions

1/4 π− q2 /2 ψ (x)= e− x H (q ) nx n nx x √2 x nx! where Hnx (q) is the Hermite polynomial, qx = βx, and β = mω/~. The spectrum of the isotropic oscillator Hamiltonian are the energies p 3 En =(n + )~ω n = nx + ny + nz 0, 1, 2, . 2 ∈ { ···} which are generally degenerate due to the many diﬀerent ways a positive integer n can be represented as the sum of three other positive integers. We note for future reference that the product form of the eigenfunctions implies that

β2r ψ(~r)=e− fn(~r) where fn(~r) is of polynomial order in r. Because H is clearly a scalar with respect to rotations (being a function of P~ P~ · and R~ R~) we know that there also exists an basis of eigenstates · ψ (~r)= ~r n, l, m = Fn,l(r)Yl,m(θ, φ) n,l,m h | i 2 common to the operators H, L , and Lz. Indeed, we know that the function φn(r)= rFn,l(r) satisiﬁes the equation

2 2 ~ `(` +1)~ 1 2 2 φ00 + + mω r εn,l φ =0. (4.79) −2m n,l 2mr2 2 − n,l · ¸ 132 Bound States of a Central Potential

As before, we simplify by multiplying through by 2m/~2 to obtain (surpressing indices) − `(` +1) 4 2 φ00 + β r ν φ =0, (4.80) − r2 − · ¸ 2 4 2 2 2 where ν =2mεn,l/~ and β = m ω /~ . As in the Coulomb problem we first determine 2 the asymptotic behavior of the solution for r>>β− , in which limit the differential equation above simplifies to 4 2 φ00 β r φ =0, (4.81) − which has two solutoins, which at large r are dominated by the behavior exp( β2r2/2). The decaying solution corresponds to the behavior already noted in the Cartesian± form, while the growing one is clearly unacceptable. We thus introduce a new function u(r) such that β2r2 φ(r)=u(r)e− in terms of which β2r2 2 φ0(r)=e− [u0 β ru] − β2r2 2 4 2 2 φ00 = e− u00 2β ru0 + β (r β )u . − − The equation for the functoin u(r) is then£ found to take the form ¤

2 2 `(` +1) u00 2β ru0 β + ν u =0. − − r2 − · ¸ We assume a power series solution consistent with the small r behavior deduced earlier, i.e.,

∞ k+`+1 u(r)= bkr b0 =0. 6 k=0 X Substitution into the differential equation for u leads, eventually, to a series of equations. The first equation of interest comes from setting the coefficients of r`+2 equal to zero and requires for its solution that b1 =0.For k>1 the resulting equations reduce to a recursion relation of the form 2 (k +2)(k +2` +3)bk+2 = (2k +2` +3)β ν bk. − From this we deduce that all coefficeints bk £with k odd are identically¤ zero. From the large k behavior of the even terms we deduce that if the series does not terminate it will converge to a function that for large r behaves as exp(β2r2), which leads again to the unacceptable solution. Thus, as in the Coulomb problem, the acceptable solutions must terminate, i.e., there exists an even integer k such that bk =0, but 0=bk+2 = bk+4 = . Thiscanonlyhappenif,forsomeevenintegerk, 6 ··· ν =2β2 (k + ` +3/2) k =0, 2, 4, ··· With the definition of ν and β this implies that

εn,l = ~ω (n +3/2) n = k + `k=0, 2, 4, ··· Clearly, for a given value of n, the allowed ` values of orbital angular momentum satisfy the relation εn,l = ~ω (n +3/2) ` = n k − Thus, more than one value of ` is generally allowed for each value of n. For n even the allowed values of ` include ` =0, 2, ,n, while for odd n the allowed values include ` = 1, 3, ,n. There is, again, an accidental··· degeneracy in this system above that required by the··· rotational invariance of the Hamiltonian. It is a straightforward exercise to show that the degeneracy of the nth level of the isotropic oscillator is (n +1)(n +2)/2.