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[PDF] The Hydrogen Atom
18 sept 2012 · The hydrogen atom contains two particles, the proton and the electron For a system of two particles 1 and 2 with coordinates 1x1, y1, z12 and
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141039_70321803450.pdf 1 Chapter 6
The Hydrogen Atom
18ce12>ab3>4cd0`89>a>3`0d409>ac0>
Before studying the hydrogen atom, we shall consider the more general problem of a single particle moving under a central force. The results of this section will apply to any central-force problem. Examples are the hydrogen atom (Section 6.5) and the isotropic three-dimensional harmonic oscillator (Prob. 6.3). A 920FIGURE6I92 is one derived from a potential-energy function that is spherically symmetric, which means that it is a function only of the distance of the particle from the origin: V.VFri= The relation between force and potential energy is given by (5.31) as -.V ?VFxUKyUKzi.V0FrVjrxiV+FrVjryiV#FrVjrzi (6.1) The partial derivatives in (6.1) can be found by the chain rule [Eqs. (5.53)-(5.55)]. Since ? in this case is a function of only, we have F rVjrzi rUx . and FrVjrxi rUz . =Therefore,
k rV rx r yUz .dV dr k rr rx r yUz .x r K dV dr (6.2) k rVry r xUz .y r K dV drU Kk rV rz r xUy .z r K dV dr (6.3) where Eqs. (5.57) anid (5.58) have been uised. Equation (6.1) ibecomes -.V r K dV drF x0xy+xz#i.V dVFri dr K Ú r (6.4) where (5.33) for I was used. The quantity Újr in (6.4) is a unit vector in the radial direc- tion. A central forcei is radially directed.i Now we consider the quantum mechanics of a single particle subject to a central force. The Hamiltonian operator is H2.T2xV2.VF, jmi? xVFri (6.5) where ? r jrx xr jry xr jrz [Eq. (3.46)]. Since ? is spherically symmetric, we shall work in spherical coordinates. Hence we want to transform the Laplacian opera- tor to these coordinates. We already have the forms of the operators rjrxUKrjryU and rjrz in these coordinates [Eqs. (5.62)-(5.64)], and by squaring each of these operators and UREG61FG.3F459I078XXFG F30I9X4AMI8FG5IMXPXFLCnC hha then adding their squares, we get the Laplacian. This calculation is left as an exercise. The result is (Prob. 6.4) 1 3 V, 3 ,i 3 3 i > , ,i bi 3 , 3 ,= 3 b i 3 >4cd>=, ,= bi 3 >>0`8 3> = > , 3 ,U 3 (6.6)Looking back to (5.68), which gives the operator
p- 3 for the square of the magnitude of the orbital angular momentum of a single particle, we see that 1 3 V, 3 ,i 3 3 i > , ,i .b i 3 r 3 p- 3 (6.7)The Hamiltonian (6.5) ibecomes
v-V.r 3 3A z , 3 ,i 3 3 i > , ,i , b 3Ai 3 p- 3 FVir (6.8) In classical mechanics a particle subject to a central force has its angular momentum conserved (Section 5.3). In quantum mechanics we might ask whether we can have states with definite values for both the energy and the angular momentum. To have the set of eigenfunctions of v- also be eigenfunctions of p - 32 the commutator fv
- 2>p - 3 o must vanish.We have
fv - 2>p - 3 oVf - 2>p - 3 o fF - 2>p - 3 o f -2>p- 3 oV .r 3 3A z , 3 ,i 3 3 i > , ,i , b 3Ai 3 p- 3 2>p- 3 a f -2>p- 3 oV.r 3 3A , 3 ,i 3 3 i> , ,i 2>p- 3 a b 3A b i 3 p- 3 2>p- 3 a (6.9)Recall that
p - 3 involves only = and U and not ? [Eq. (5.68)]. Hence it commutes with every operator that involves only ?. [To reach this conclusion, we must use relations like (5.47) with ? and replaced by ? and =1] Thus the first commutator in (6.9) is zero. Moreover,
since any operator commutes with itself, the second commutator in (6.9) is zero. There- fore, f -2>p- 3 oVa1 Also, since p - 3 does not involve ? and ? is a function of ? only, we have fF-2>p- 3 oVa1 Therefore, fv-2>p- 3 oVa>>`9>FVFVir (6.10) v - commutes with p - 3 when the potential-ienergy function is iindependent of = and U1Now consider the operator p
- rV.Br>,x,U [Eq. (5.67)]. Since p
- r does not involve ? and since it commutes with p- 3 [Eq. (5.50)], it follows that p - r commutes with the Hamiltonian (6.8): fv - 2>p - r oVa>>`9>FVFVir (6.11) We can therefore have a set of simultaneous eigenfunctions of v-2>p- 32 and p
- r for the central-force problem. Let K denote these common eigenfunctions: v -KVmK ?
p - 3KVLVL brr
3K2>LVa2>b2>32z ?
p - rKVArK2>AV.L2>.L b2z2>L ?
where Eqs. (5.104) aind (5.105) were usedi.120 Chapter 6 | The Hydrogen Atom
Using (6.8) and (6.13), we have for the Schrödinger equation (6.12) 12 d dt k8 d K 8, d cd , x 8K 8, rcz dt, d s- dKc.F,iK1nK
12 d dt k 8 d K 8, d cd , x 8K 8, rcfFfczi2
d dt, dKc.F,iK1nK (6.15)
The eigenfunctions of
s - d are the spherical harmonics : tfF=rxUir and since s
- d does not involve r, we can multiply : tf by an arbitrary function of r and still have eigenfunc- tions of s - d and s- 1V Therefore,
K1aF,i:
tf x xF=rxUi ?
Using (6.16) in (6.15), we then divide both sides by : tf to get an ordinary differential equation for the uniknown function aF,iY 12 d dtk a cd ,a rcfFfczi2
d dt, d ac.F,ia1naF,i (6.17) We have shown that, for any one-particle problem with a spherically symmetric potential-energy function .F,ir the stationary-state wave functions are K1aF,i: tfF=rxUir
where the radial factor aF,i satisfies (6.17). By using a specific form for .F,i in (6.17), we can solve it for a particular problem. zxCm Up to this point, we have solved only one-particle quantum-mechanical problems. The hydrogen atom is a two-particle system, and as a preliminary to dealing with the H atom, we first consider a simpler case, that of two noninteracting particles. Suppose that a system is composed of the noninteracting particles 1 and 2. Let d z and d d symbolize the coordinates F z rx0 z rx1 z i and F d rx0 d rx1 d i of particles 1 and 2. Because the particles exert no forces on each other, the classical-mechanical energy of the system is the sum of the energies of the two particles: n1n z cn d 1o z c. z co d c. d r and the classical Hamiltonian is the sum of Hamiltonians for each particle: c1c z cc d VTherefore, the Hamiltonian operator is
c -1 c - z cc - d where c - z involves only the coordinates d z and the momentum operators 8- z that corre- spond to d zV The Schrödinger eqiuation for the systiem is
Fc - z cc - d iKFd z rxd d i1nKFd z rxd d i (6.18) We try a solution ofi (6.18) by separationi of variables, settinig K F d z rxd d i1H z F d z i H d F d d i (6.19)We have
c - z H z Fd z iH d Fd d icc - d H z Fd z iH d Fd d i1nH z Fd z iH d Fd d i (6.20)Since c
- z involves only the coordinate and momentum operators of particle 1, we have c - z RH z Fd z iH d Fd d iv1H d Fd d ic - z H z Fd z ir since, as far as c - z is concerned, H d is a con- stant. Using this eqiuation and a similar eiquation for c - d r we find that (6.20) ibecomes H d Fd d ic - z H z Fd z icH z Fd z ic - d H d Fd d i1nH z Fd z iH d Fd d i (6.21) URlGsFgD807M3GMaG01XFG6cM459I078XFG5IMPXXFLG0MG6cMG.3F459I0X78XFG5IMPXFLTGCnC hph r1 z 3 z .2 z , 3 z .2 z , jr1 d 3 d .2 d , 3 d .2 d , =o (6.22) Now, by the same arguments used in connection with Eq. (3.65), we conclude that each term on the left in (6.22) must be a constant. Using o z and o d to denote these constants, we have r1 z 3 z .2 z , 3 z .2 z , =o z rxx r1 d 3 d .2 d , 3 d .2 d , =o d o=o z jo d (6.23) Thus, when the system is composed of two noninteracting particles, we can reduce the two-particle problem ito two separate one-iparticle problems by isolving r1 z 3 z .2 z ,=o z 3 z .2 z ,rxxr1 d 3 d .2 d ,=o d 3 d .2 d , (6.24) which are separate Schirödinger equations fior each particle.Generalizing this result to
: noninteracting particles, we have r 1 =r1 z jr1 dx jrjxr1 4 ?.2 z rx2 d rVrx2 4 ,=3 z . 2 z , 3 d . 2 d ,r3 4 . 2 4 , o=o z jo dx jrjxo 4 r1 f 3 f =o f 3 f rxf=zrxdrVrx4 *06...6 7 6*S6:*::0...:6...0 Ú6 6:076 6 6 6*S6 6:...6:0 6 *S6...6...06...:6
6 ...6S:*:6 6 60*6*S6 ...6S:*: 6S*06...6...06
6 ...6S:*:6 3 f6*S6...066 6S*:6
76 *:6...6
0:06...*:6S*06...06
6 :6 6... *: ...:6r1 f V These results also apply to a single particle whose Hamiltonian is the sum of separate terms for each coordinate: r 1=r1 c .c1rx)1 c ,jr1 e .e1rx)1 e ,jr1 n .n1rx)1 n , In this case, we concilude that the wave fiunctions and energieis are ? . crxerxn,=5.c,3.e,7.n,rxo=o c jo e jo n r1 c5.c,=o
c5.c,rxr1
e3.e,=o
e3.e,rxr1
n7.n,=o
n 7.n, Examples include the particle in a three-dimensional box (Section 3.5), the three-dimensional free particle (Prob. 3.42), and the three-dimensional harmonic oscillator (Prob. 4.20). zxm 1 2 2 The hydrogen atom contains two particles, the proton and the electron. For a system of two particles 1 and 2 with coordinates . c z rxe z rxn z , and .c d rxe d rxn d , r the potential energy of interaction between the particles is usually a function of only the relative coordinates c d Fc z rxe d Fe z r and n d Fn z of the particles. In this case the two-particle problem can be simplified to two separate one-particle problems, as we now prove. Consider the classical-mechanical treatment of two interacting particles of masses l z and l d V We specify their positions by the radius vectors z and d drawn from the origin122 Chapter 6 | The Hydrogen Atom
of a Cartesian coordinate system (Fig. 6.1). Particles 1 and 2 have coordinates 1x 1 , y 1 , z 1 2 and 1 x 2 , y 2 , z 2 2 . We draw the vector r=r 2 -r 1 from particle 1 to 2 and denote the com- ponents of r by x, y, and z: x=x 2 -x 1 , y=y 2 -y 1 , z=z 2 -z 1 (6.28)The coordinates
x , y , and z are called the relative or internal coordinate4s . We now draw the vector R from the origin to the system"s center of mass, point C, and denote the coordinates of C byX, Y, and Z:
R=iX+jY+kZ (6.29)
The definition of thie center of mass ofi this two-particle syistem gives X= m 1 x 1 +m 2 x 2 m 1 +m 2 , Y= m 1 y 1 +m 2 y 2 m 1 +m 2 , Z= m 1 z 1 +m 2 z 2 m 1 +m 2 (6.30) These three equatioins are equivalent toi the vector equationi R= m 1 r 1 +m 2 r 2 m 1 +m 2 (6.31)We also have
r=r 2 -r 1 (6.32) We regard (6.31) and (6.32) as simultaneous equations in the two unknowns r 1 and r 2 and solve for them to get r 1 =R- m 2 m 1 +m 2 r, r 2 =R+ m 1 m 1 +m 2 r (6.33) Equations (6.31) and (6.32) represent a transformation of coordinates from x 1 , y 1 , z 1 , x 2 , y 2 , z 2 to X, Y, Z, x, y, z. Consider what happens to the Hamiltonian under this transformation. Let an overhead dot indicate differentiation with respect to time. The ve- locity of particle 1 is [Eq. (5.34)] v 1 =dr 1 >dt=r# 1 . The kinetic energy is the sum of the kinetic energies of the two particles: T = 1 2 m 1 ?r # 1 ? 2 + 1 2 m 2 ?r # 2 ? 2 (6.34) Introducing the timei derivatives of Eqs.i (6.33) into (6.34),i we have T=1 2m 1 aR # -m 2 m 1 +m 2 r#b?aR # -m 2 m 1 +m 2 r#b +1 2m 2 aR # +m 1 m 1 +m 2 r#b?aR # +m 1 m 1 +m 2 r#b z x ym 2 m 1 C R r 1 r 2 rFIGURE 6.1 A two-particle
system with center of mass at C. URlGsFgD807M3GMaG01XFG6cM459I078XFG5IMPXXFLG0MG6cMG.3F459I0X78XFG5IMPXFLTGCnC hpn where 1d1 d .dcFcd [Eq. (5.24)] has been used. Using the distributive law for the dot products, we find, aifter simplifying, . z d- z x d 013 z 1 d x z d x z d z x d 1b z 1 d (6.35) Let be the total mass oif the system: z x d (6.36)We define the
m of the two-particle isystem as m z d z x d Then . z d 13 z 1 d x z d m1b z 1 d (6.38) The first term in (6.38) is the kinetic energy due to translational motion of the whole system of mass is motion in which each par- ticle undergoes the same displacement. The quantity z d xx13 z 1 d would be the kinetic energy of a hypothetical particle of mass located at the center of mass. The second term in (6.38) is the kinetic energy of internal (relative) motion of the two particles. This internal motion is of two types. The distance ? between the two particles can change (vibration), and the direction of the vector can change (rotation). Note that1bz1.1b+121b1+V
Corresponding to the original coordinates K
z rx z rx z rxK d rx d rx d r we have six linear momenta: K z . z Kz z rxxRrxxxx d . d z d (6.39) Comparing Eqs. (6.34) and (6.38), we define the six linear momenta for the new coordinates , , , ?, , as x zrxx zrxx z x K mKzrxxxx m zrxx mzWe define two new moimentum vectors as
0 2 z x> z xa z xREqx0 m 2mK z x>m z xam zIntroducing these miomenta into (6.38), iwe have
.10 1 d dx 10 m x1 d dm (6.40) Now consider the potential energy. We make the restriction that ? is a function of the relative coordinates ?, , and of the two particles: =.=-Krx rx0 (6.41) An example of (6.41) is two charged particles interacting according to Coulomb"s law [see Eq. (3.53)]. Wiith this restriction ion ? the Hamiltonian funcition is . d dx 6 d m dmx= -Krx rx0... (6.42)124 Chapter 6 | The Hydrogen Atom
Now suppose we had a system composed of a particle of mass B subject to no forces and a particle of mass a subject to the potential-energy function V.xijyijz,F Further sup- pose that there was no interaction between these particles. If (b, -, 9) are the coordinates of the particle of mass B, and . xijyijz, are the coordinates of the particle of mass ai what is the Hamiltonian of this hypothetical system? Clearly, it is identical with (6.42). The Hamiltonian (6.42) can be viewed as the sum of the Hamiltonians p RM RM and p R a RaKV.xijyijz, of two hypothetical noninteracting particles with massesB and
aF Therefore, the results of Section 6.2 show that the system"s quantum- mechanical energy is the sum of energies of the two hypothetical particles [Eq. (6.23)]: E-E M KE a F From Eqs. (6.24) and (6.42), the translational energy E M is found by solving the Schrödinger equation .pV RM RM,> M -E M > MF This is the Schrödinger equa-
tion for a free particle of mass Bn so its possible eigenvalues are all nonnegative num- bers: E M k [Eq. (2.31)]. From (6.24) and (6.42), the energy E a is found by solving theSchrödinger equation
o pV R a RaKV .xijyijz,t> a .xijyijz,-E a > a .xijyijz, (6.43) We have thus separated the problem of two particles interacting according to a potential-energy function V.xijyijz, that depends on only the relative coordinates c, f, e into two separate one-particle problems: (1) the translational motion of the entire system of mass B, which simply adds a nonnegative constant energy E M to the system"s energy, and (2) the relative or internal motion, which is dealt with by solving the Schrödinger equation (6.43) for a hypothetical particle of mass a whose coordinates are the relative coordinates c, f, e and that moves subject to the potential energyV.xijyijz,F
For example, for the hydrogen atom, which is composed of an electron (o) and a pro- ton (6), the atom"s total energy is E-E M KE a i where E M is the translational energy of motion through space of the entire atom of mass M-m e Km p i and where E a is found by solving (6.43) with a -m e m p .m e Km p , and 1 being the Coulomb"s law potential energy of interaction of the electron and proton; see Section 6.5. ?? ? ? Before solving the Schrödinger equation for the hydrogen atom, we will first deal with the two-particle rigid rotor. This is a two-particle system with the particles held at a fixed distance from each other by a rigid massless rod of length s. For this problem, the vector in Fig. 6.1 has the constant magnitude sR-dF Therefore (see Section 6.3), the kinetic energy of internal motion is wholly rotational energy. The energy of the rotor is entirely kinetic, andV-k (6.44)
Equation (6.44) is a special case of Eq. (6.41), and we may therefore use the results of the last section to separate off the translational motion of the system as a whole. We will con - cern ourselves only with the rotational energy. The Hamiltonian operator for the rotation is given by the terims in brackets in (6.i43) as HV-pV R a Ra-=0 R RaC R ijja-m r m R m r Km R (6.45) where m r and m R are the masses of the two particles. The coordinates of the fictitious particle of mass a are the relative cooirdinates of m r and m R [Eq. (6.28)]. Instead of the relative Cartesian coordinates c, f, e, it will prove more fruitful to use the relative spherical coordinates rij1ij2F The 8 coordinate is equal to the magnitude ofURhG61FG6cM459I078XXFGs7y7gGsM0MICnC hpP
the r vector in Fig. 6.1, and since d z and d d are constrained to remain a fixed distance apart, we have2.3V Thus the problem is equivalent to a particle of mass k constrained
to move on the surface of a sphere of radius Because the radial coordinate is constant, the wave function will be a function of F and i only. Therefore the first two terms of the Laplacian operator in (6.8) will give zero when operating on the wave function and may be omitted. Looking at things in a slightly different way, we note that the operators in (6.8) that involve U derivatives correspond to the kinetic energy of radial motion, and since there is no radial motion, the U derivatives are omitted from 4 - V Since 1.y is a special case of 1.1V2rr the results of Section 6.1 tell us that the eigenfunctions are given by (6.16) with the U factor omitted: j . dVFrxir (6.46)
where € rather than is used for the roitational angular-momenitum quantum number.i The Hamiltonian operator is given by Eq. (6.8) with the U derivatives omitted and1V2r.yV Thus
4 -.V dk3 d r ?z 0 - dUse of (6.13) gives
x4 - j.8j x Vdk3 d r ?z 0 - d d xVFrxir.8 d xVFrxir xVdk3 d r ?z V?zr, d d xVFrxir.8 d xVFrxir 8.V?zr,
d dk3 d rxx.yrxzrxd (6.47) The moment of inertia of a system of particles about some particular axis in space as defined as z `.z d ` r d` (6.48) where d ` is the mass of the th particle and r ` is the perpendicular distance from this particle to the axis. The value of depends on the choice of axis. For the two-particle rigid rotor, we choose our axis to be a line that passes through the center of mass and is perpen - dicular to the line joining d z and d d (Fig. 6.2). If we place the rotor so that the center of mass, point , lies at the origin of a Cartesian coordinate system and the line joining d z and d d lies on the K axis, then will have the coordinates (0, 0, 0), d z will have the coor - dinates V ?r z rxyrxyrr and d d will have the coordinates Vr d rxyrxyrV Using these coordinates in (6.30), we find d z r z .d d r d (6.49) 2 R 2 rRr 1 er 6.1CTOpCvd7TGig9T1FgGX73FtGaMIG89X8DX9073yXG01FG
LMLF30GMaG73FI079GMXaG9G
G0cM429I078XFGI7y7gGX
GIM0MIRGG
CG7TG01FG8F30FIGMaGXL9TTR
126 Chapter 6 | The Hydrogen Atom
The moment of inertiia of the rotor abouit the axis we have chiosen is R 1l z 1 dz jl d 1 dd (6.50) Using (6.49), we tranisform Eq. (6.50) toi (see Prob. 6.14) R 10t d = where 0 >l z l d +-l z jl d0 is the reduced mass of the system and t>1
z j1 d is the dis- tance between l z and l d V The allowed energy levels (6.47) of the two-particle rigid rotor are o1s -sjz02 d dR rxs1yrxzrxdrVThe lowest level is
o1yr so there is no zero-point rotational energy. Having zero rotational energy and therefore zero angular momentum for the rotor does not violate the uncertainty principle; recall the discussion following Eq. (5.105). Note that ƒ increases as s d jsr so the spacing betwieen adjacent rotationial levels increases asi € increases. Are the rotor energy levels (6.52) degenerate? The energy depends on € only, but the wave function (6.46) depends on € and , where l2 is the component of the rotor"s angular momentum. For each value of €, there are dsjz values of , ranging from Fs to €. Hence the levels are - dsjz0-fold degenerate. The states of a degenerate level have dif- ferent orientations of the angular-momentum vector of the rotor about a space-fixed axis.The angles
. and , in the wave function (6.46) are relative coordinates of the two point masses. If we set up a Cartesian coordinate system with the origin at the rotor"s cen- ter of mass, . and , will be as shown in . This coordinate system undergoes the same translational motion as the rotor"s center of mass but does not rotate in space.The rotational angular momentum
Ss-sjz02
d * z+d is the angular momentum of the two particles with respect to an origin at the system"s center of mass . The rotational levels of a diatomic molecule can be well approximated by the two-particle rigid-rotor energies (6.52). It is found ("?? Section 4.4) that when a diatomic molecule absorbs or emits radiation, the allowed pure-rotational transitions are given by the selection rule s1xz (6.53) In addition, a molecule must have a nonzero dipole moment in order to show a pure-rotational spectrum. A ??? is one where only the rotational z xy m 2 m 1FIGURE 6.3 Coordinate
system for the two-lparticle rigid rotor.URhG61FG6cM459I078XXFGs7y7gGsM0MICnC hpi
quantum number changes. [Vibration-rotation transitions (Section 4.3) involve changes in both vibrational and rotational quantum numbers.] The spacing between adjacent low-lying rotational levels is significantly less than that between adjacent vibrational levels, and the pure-rotational spectrum falls in the microwave (or the far-infrared) region. The frequencies of the pure-rotational spectral lines of a diatomic molecule are then (approximately) t1E J r .E J h 1 =J rU=J RU.J=J rUh r R I1R=J rUB (6.54)
B >hKr R I ijjJ1kijrijRi- (6.55) B is called the ? of the molecule. The spacings between the diatomic rotational levels (6.52) for low and moderate values of J are generally less than or of the same order of magnitude as kT at room tem- perature, so the Boltzmann distribution law (4.63) shows that many rotational levels are significantly populated at room temperature. Absorption of radiation by diatomic mol- ecules having J1k (the J1k2r transition) gives a line at the frequency 2B; absorp- tion by molecules having J1r (the J1r2R transition) gives a line at 4B; absorption byJ1R molecules gives a line at 6B; and so on. See
. Measurement of the rotational absorption frequencies allows B to be found. From B, we get the molecule"s moment of inertia I, and from I we get the bond distance d. The value of d found is an average over the 1k vibrational motion. Because of the asym- metry of the potential-energy curve in Figs. 4.6 and 13.1, d is very slightly longer than the equilibrium bond length R e in Fig. 13.1. As noted in Section 4.3, isotopic species such as r and r have virtually the same electronic energy curve U=RU and so have virtually the same equilibrium bond distance. However, the different isotopic masses produce different moments of inertia and hence different rotational absorption frequencies. Because molecules are not rigid, the rotational energy levels for diatomic molecules differ slightly from rigid-rotor levels. From (6.52) and (6.55), the two-particle rigid- rotor levels are E A1BhJ=J rUF Because of the anharmonicity of molecular vibration (Fig.
4.6), the average internuclear distance increases with increasing vibrational quantum
number i so as increases, the moment of inertia I increases and the rotational con- stant B decreases. To allow for the dependence of B on i one replaces B in E A by B FThe mean rotational constant
B for vibrational level is B 1B e .a e = rKRUi where B e is calculated using the equilibrium internuclear separation R e at the bottom z 2 R 2 1 z 2 R 2 z 2 R 2 z 2 R 2 xxxyer 6.1CTO-C6cM429I078XFGI7y7g4IM0MIG9PTMI20X7M3G
0I93T707M3TR
128 Chapter 6 | The Hydrogen Atom
of the potential-energy curve in Fig. 4.6, and the d248l.2r ‡8r.l.2r a)rv632 ga)r ( i.l . c e is a positive constant (different for different molecules) that is much smaller than B e . Also, as the rotational energy increases, there is a very slight increase in av- erage internuclear distance (a phenomenon called )o .82tvgl3as2i.r8.2r ). This adds the term FhDJ 2 =Jj1U 2 to E rot , where the centrifugal-distortion constant ˆ is an extremely small positive constant, different for different molecules. For example, for 12 C 16 O, B 0 = 57636 MHz, c e -540 MHz, and D-0.18 MHz. As noted in Section 4.3, for lighter diatomic molecules, nearly all the molecules are in the ground r-0 vibrational level at room temperature, and the observed rotational constant is B 0 . For more discussion of nuclear motion in diatomic molecules, see Section 13.2. For the rotational energies of polyatomic molecules, see mr7 oial saT)Fl73r7, chaps. 2-4.EXAMPLE
The lowest-frequency pure-rotational absiorption line of 12 C 32S occurs at 48991.0 MiHz.
Find the bond distanice in
12 C 32S. The lowest-frequency rotational absorptiion is the J-0R1 line. Equations (1.4i), (6.52), and (6.51) giive h3-E upper FE lower -1 =2U0 2 2ad 2 F0 =1U0 2 2ad 2 which gives d-=hK44 2 3aU 1K2 . Table A.3 in the Apipendix gives a- m 1 m 2 m 1 jm 2 -
12=31.97207U
=12j31.97207U 16.02214v10
23g-1.44885v10 F23 g
The SI unit of massi is the kilogram, andi
d-1 24h 3 0R1 a 1K2 -1 24
6.62607v10
F34 J s =48991.0v10 6 s F1U=1.44885v10
F26 kgU 1K2 -1.5377v10 F10 m-1.5377 Å The J-1 to J-2 pure-rotational tranisition of 12 C 16O occurs at
230.538
GHz. =1 GHz-10 9 Hz.U Find the bond distiance in this molecule.i ( ‰ i7o8Š 1.1309 v10 F10 m.U ? ??? ? The hydrogen atom consists of a proton and an electron. If o symbolizes the charge on the proton =e-j1.6v10 F19CU, then the electron"s charge is Fe.
A few scientists have speculated that the proton and electron charges might not be exactly equal in magnitude. Experiments show that the magnitudes of the electron and proton charges are equal to within one part in 10 21. See G. Bressi et al., 'FfiuaOEodua‰, x , 052101 (2011) (available online at arxiv.org/abs/1102.2766). We shall assume the electron and proton to be point masses whose interaction is given by Coulomb"s law. In discussing atoms and molecules, we shall usually be considering isolated systems, ignoring interatomic and intermolecular interactions.
URoG61FGergIMyF3GvX0MLCnC hpa
Instead of treating just the hydrogen atom, we consider a slightly more general prob- lem: the hydrogenlike atom, which consists of one electron and a nucleus of charge 8. For v.zr we have the hydrogen atom; for .dr the Gs K ion; for .nr the Si dK ion; and so on. The hydrogenlike atom is the most important system in quantum chemistry. An exact solution of the Schrödinger equation for atoms with more than one electron cannot be obtained because of the interelectronic repulsions. If, as a crude first approximation, we ignore these repulsions, then the electrons can be treated independently. (See Section 6.2.) The atomic wave function will be approximated by a product of one-electron functions, which will be hydrogenlike wave functions. A one-electron wave function (whether or not it is hydrogenlike) is called an orbital. (More precisely, an orbital is a one-electron spatial wave function, where the word dd means that the wave function depends on the electron"s three spatial coordinates >, `, and a or 2, - r and 0. We shall see in Chapter10 that the existence of electron spin adds a fourth coordinate to a one-electron wave func
- tion, giving what is called a spin-orbital.) An orbital for an electron in an atom is called an atomic orbital. We shall use atomic orbitals to construct approximate wave functions for atoms with many electrons (Chapter 11). Orbitals are also used to construct approximate wave functions for molecules. For the hydrogenlike atom, let (>, `, a) be the coordinates of the electron relative to the nucleus, and let ,.rjKxkKzrV The Coulomb"s law force on the electron in the hydro- genlike atom is [see Eq. (1.37)] V.= d J*... y i d x, i (6.56) where ,?i is a unit vector in ithe r direction. The minuis sign indicates an atitractive force. The possibility of small deviations from Coulomb"s law has been considered. Exper- iments have shown that if the Coulomb"s-law force is written as being proportional to i =dK r then ?zy =zM V A deviation from Coulomb"s law can be shown to imply a nonzero photon rest mass. No evidence exists for a nonzero photon rest mass, and data indicate that any such mass must be less than zy =Fz xaA A. S. Goldhaber and M. M. Nieto,OE8c4c'
` b 82, 939 (2010) (arxiv.org/abs/0809.1003); G. Spavieri et al., ƒ2c ' ` c€ˆ, 61, 531 (2011) (www.epj.org/_pdf/HP_EPJD_classical_and_quantum_ approaches.pdf). The force in (6.56) is central, and comparison with Eq. (6.4) gives RF?i??Ri. d ?J*... y i d V Integration gives F. d J*... yx x z i d xRi.= d J*... y i (6.57) where the integration constant has been taken as 0 to makeF.y at infinite separation
between the charges. For any two charges z and d separated by distance i zd r Eq. (6.57) becomes F. z d J*... y i zd (6.58) Since the potential energy of this two-particle system depends only on the relative coordinates of the particles, we can apply the results of Section 6.3 to reduce the prob- lem to two one-particle problems. The translational motion of the atom as a whole simply adds some constant to the total energy, and we shall not concern ourselves130 Chapter 6 | The Hydrogen Atom
with it. To deal with the internal motion of the system, we introduce a fictitious par- ticle of mass e1 m e m N m e m N (6.59) where m e and m N are the electronic and nuclear masses. The particle of reduced mass e moves subject to the potential-energy function (6.57), and its coordinates F rUK1UK8i are the spherical coordinates iof one particle relatiive to the other ( : ). The Hamiltonian for the internal motion is [Eq. (6.43)] HV1.2 e? .Ze rl r (6.60) Since ? is a function of the coordinate only, we have a one-particle central-force prob- lem, and we may apply the results of Section 6.1. Using Eqs. (6.16) and (6.17), we have for the wave functioin cFrUK1UK8i1RFriY
mlKF1UK8iUKl1 UKUKUzUKK2m2tl (6.61)
where Y ml is a spherical harmoniic, and the radial funiction RFri satisfies .2 e kR? rR? r lFl i2
er R.Ze rl rR1ERFri (6.62)
To save time in writiing, we define the coinstant as a>rl 2 ee (6.63) and (6.62) becomes R? rR? rl E ae Z ar. lFl i rAR1 (6.64)
r We could now try a power-series solution of (6.64), but we would get a three-term rather than a two-term recursion relation. We therefore seek a substitution that will lead to a two- term recursion relation. It turns out that the proper substitution can be found by examining the behavior of the solution for large values of . For large , (6.64) becomes R? rl E aeR1 UKKrK
(6.65) C R C n l rZsFIGURE 6.5 Relative
spherical coordinatles.URoG61FGergIMyF3GvX0MLCnC hnh
which may be solved uising the auxiliary eqiuation (2.7). The soilutions are yLh3x1.)vA
e > t 2 n>tU4 (6.66)
Suppose that ƒ is positive. The quantity under the square-root sign in (6.66) is nega- tive, and the factor multiplying 0 is imaginary:1U2cr
x?tkU>8cc1e (6.67)
where (6.63) was used. The symbol rin (6.67) indicates that we are giving the behavior of 1U2 for large values of 0; this is called the ... 7 * behavior of the function. Note the resemblance of (6.67) to Eq. (2.30), the free-particle wave function. Equation (6.67) does not give the complete radial factor in the wave function for posi- tive energies. Further study (... 6...:6...0Ú pages 21-24) shows that the radial function for 1e remains finite for all values of 0, no matter what the value of ƒ Thus, just as for the free particle, all nonnegative energies of the hydrogen atom are allowed. Physically, these eigenfunctions correspond to states in which the electron is not bound to the nucleus; that is, the atom is ionized. (A classical-mechanical anal- ogy is a comet moving in a hyperbolic orbit about the sun. The comet is not bound and makes but one visit to the solar system.) Since we get continuous rather than discrete allowed values for 1e8 the positive-energy eigenfunctions are called yC Cm mCCy C The angular part of a continuum wave function is a spherical harmonic. Like the free-particle wave functions, the continuum eigenfunctions are not normaliz- able in the usual sense.We now consider the
*: states of the hydrogen atom, with oe1 (For a C m j2ec. cK2x1) In this case, the quantity in parentheses in (6.66) is positive.
Since we want the wave functions to remain finite as 0 goes to infinity, we prefer the minus sign in (6.66), and in order to get a two-term recursion relation, we make the substitution 1U2? .U 1U2 (6.68) a.)vA e t b n>t (6.69) where in (6.68) stands for the base of natural logarithms, and not the proton charge. Use of the substitution (6.68) will guarantee nothing about the behavior of the wave function for large 0 The differential equation we obtain from this substitution will still have two linearly independent solutions. We can make any substitution we please in a differential equation; in fact, we could make the substitution 1U2? U 1U2 and still wind up with the correct eigenfunctions and eigenvalues. The relation between € and would naturally be 1U2? .tU 1U21Proceeding with (6.68), we evaluate
and 18 substitute into (6.64), multiply by U t U8 and use (6.69) to get the following differential equation for 1U26
U t 1 1tU.tU t2 31t
.n .t2U.1 n24 ?e (6.70) We could now substitute a power series of the form ? R ?e U (6.71) for . If we did we would find that, in general, the first few coefficients in (6.71) are zero. If is the first nonzeiro coefficient, (6.71i) can be written as ? R ? U 8cc Ce (6.72)
132 Chapter 6 | The Hydrogen Atom
Letting 0>9Vy, and then defining
0 as 0 >B 0xy , we have 71F 010 B 0xy 8 0xy 18 y F 010 0 8 0 , 0 -0 (6.73) (Although the various substitutions we are making might seem arbitrary, they are standard procedure in solving differential equations by power series.) The integer s is evaluated by substitution into thie differential equatiion. Equation (6.73) iis
7?8?18
y v?8? (6.74) v ?8?1 F 010 0 8 0 , 0 -0 (6.75)Evaluating
7K and 7? from (6.74) and subistituting into (6.70)i, we get
8 2 vUx?2yx2?8V2-8 2 vKxy 2 xyx?2ib V1V2-V2-y?8V(?(x1?v10
(6.76)To find
s, we look at (6.76) for810. From (6.75), we have
v?0?1 0 , vK?0?1 1 , vU?0?12 2 (6.77)Using (6.77) in (6.76i), we find for
8100 ?y 2 xyV( 2
V(?10 (6.78)
Since
0 is not zero, the terms in parentheses must vanish: y 2 xyV( 2V(10. This is a
quadratic equation in ithe unknown s, with the roots y1(, y1V(V1 (6.79) These roots correspond to the two linearly independent solutions of the differential equa- tion. Let us examine them from the standpoint of proper behavior of the wave function.From Eqs. (6.68), (6.74), and (6.75), we have
. ?8?1d V-8 8 y F 010 0 8 0 (6.80)Since d
V-811V-8xz, the function .?8? behaves for small r as
0 8 y . For the root y1(, .?8? behaves properly at the origin. However, for y1V(V1, .?8? is proportional to 1 8 (x1 (6.81) for small r. Since (10, 1, 2,z, the root y1V(V1 makes the radial factor in the wave function infinite at the origin. Many texts take this as sufficient reason for rejecting this root. However, this is not a good argument, since for the relativistic hydrogen atom, the (10 eigen- functions are infinite at810. Let us therefore look at (6.81) from the standpoint of quadratic
integrability, since we certainly require the bound-state eigenfunctions to be normalizable. The normalization integral [Eq. (5.80)] for the radial functions that behave like (6.81) looks like i0 2.2 2 8 2 t8 i0 1 8 2( t8 (6.82) for small r. The behavior of thie integral at the lowier limit of integratiion is 1 8 2(V1 810(6.83)
URoG61FGergIMyF3GvX0MLCnC hnn
For (1zrxdrxnr.r (6.83) is infinite, and the normalization integral is infinite. Hence we must reject the root y1F(Fz for (rzV However, for (1yr (6.83) is finite, and there is no trouble with quadratic integrability. Thus there is a quadratically integrable solution to the radial equatioin that behaves as 8 Fz for small U Further study of this solution shows that it corresponds to an energy value that the experimental hydrogen-atom spectrum shows does not exist. Thus the 8 Fz solution must be rejected, but there is some dispute over the reason for doing so. One view is that the z+8 solution satisfies the Schrödinger equation everywhere in space except at the origin andhence must be rejected [ˆU page 156; B. H. Armstrong and E. A. Power, ‰ €'
lVC 262 (1963)]. A second view is that the z+8 solution must be rejected because the Hamiltonian operator is not Hermitian with respect to it (U U Section 10.5). (In Chapter 7 we shall define Hermitian operators and show that quantum-mechanical operators are required to be Hermitian.) Further discussion is given in A. A. Khelashvili and T. P. Nadareishvili, ‰ €' ., wZC 668 (2011) (see arxiv.org/abs/1102.1185) and inY. C.
Cantelaube, arxiv.org/abs/1203.0551. Taking the first root in (6.79), we have for the radial factor (6.80) . -801d F-8 8 ( v-80 (6.84) With y1(r Eq. (6.76) becomes 8v? j-d(jdFd-80v?j-dib FzFd-Fd-(0v1y (6.85)
From (6.75), we havei
v -801 ? 01y 0 8 0 (6.86) v?1 ? 01y 0 0 8 0Fz 1 ? 01z 0 0 8 0Fz 1 ? 91y-9jz0 9jz 8 9 1 ? 01y -0jz0 0jz 8 0 v?1 ? 01y
0-0Fz0
0 8 0Fd 1 ? 01z0-0Fz0
0 8 0Fd 1 ? 91y-9jz09 9jz 8 9Fz v?1 ? 01y -0jz00 0jz 8 0Fz (6.87) Substituting these iexpressions in (6.85i) and combining sumsi, we get ? 01y
60-0jz0
0jz jd-(jz0-0jz0 0jz j# di bFd-Fd-(Fd-0 Ú 0 ...8 0 1ySetting the coefficieint of
8 0 equal to zero, we giet the recursion reilation 0jz 1 d-jd-(jd-0Fdib Fz0-0jz0jd-(jz0-0jz0
0 (6.88) We now must examine the behavior of the infinite series (6.86) for large U. The result of the same procedure used to examine the harmonic-oscillator power series in (4.42) suggests that for large U the infinite series (6.86) behaves like d d-8V (See Prob. 6.20.) For
large U the radial function (6.84) behaves like . -80xd F-8 8 ( d d-8 18 ( d - 8 (6.89)Therefore,
.-80 will become infinite as U goes to infinity and will not be quadrati- cally integrable. The only way to avoid this "infinity catastrophe" (as in the harmonic- oscillator case) is to have the series terminate after a finite number of terms, in which case the d F-8 factor will ensure that the wave function goes to zero as U goes to infinity. Let the134 Chapter 6 | The Hydrogen Atom
last term in the series be 9 8 91 Then, to have
9xb 2> 9x32. all vanish, the fraction multi-
plying 0 in the recursion relation (6.88) must vanish when 0F91 We have 3- =9x(xbUF3ib Vb2>>9Fa2>b2>32. (6.90)
† and are integers, and wie now define a new initeger : by4j9x(xb2>>4Fb2>32>
2. (6.91)
From (6.91) the quanitum number
must satisfy (m4Vb (6.92)Hence
ranges from 0 to 4Vb1 E nergy L evelsUse of (6.91) in (6.90) gives
-4Fib Vb (6.93)Substituting -
j=Voa a oKbd 3 U bK3 [Eq. (6.69)] into (i6.93) and solving foirƒÚ
we get oFVi 3 4 3 d 3 oa a bFVi 3 zd a 3 a 4 3 u 3 (6.94) where b joa a i 3 Kzd 3 [Eq. (6.63)]. These are the bound-state energy levels of the hydrogenlike atom, and they are discrete. shows the potential-energy curve [Eq. (6.57)] and some of the allowed energy levels for the hydrogen atom = iFbU1 The crosshatching indicateis that all positive einergies are allowed. It turns out that all changes in : are allowed in light absorption and emission. The wavenumbers [Eq. (4.64)] of H-atom spectral lines are then f jb nFpBFo 3 Vo b uBFd 3 oa a buB b 4 3 b Vb 4 3 3 j. b 4 3 b Vb 4 3 3 (6.95) where .Fba 1>4
Vb is the OE7 06*: ...: for hydrogen.Degeneracy
Are the hydrogen-atom energy levels degenerate? For the bound states, the energy (6.94) depends only on :. However, the wave function (6.61) depends on all three quantum numbers :, , and , whose allowed values are [Eqs. (6.91), (6.92), (5.104), and (5.105)] z x ym 2 mRym2mr
C ym 2 m ym2mFIGURE 6.6 Energy levels
of the hydrogen atlom. URUG61FGSMD3g4W090FXGergIMyF34v0MLGf9uFXGAD3807M3TCnC hnP41zrxdrxnr. L
(1yrxzrxdr.rx4=z L l1=(rx=(Kzr.rxyr.rx(=zrx( L i Hydrogen-atom states with different values of or , but the same value of :, have the same energy. The energy levels are degenerate, except for41zr where and must
both be 0. For a given value of :, we can have : different values of For each of these values of , we can have d(Kz values of The degree of degeneracy of an H-atom bound-state level is found to equal 4 d (spin considerations being omitted); see Prob. 6.16. For the continuum levels, it turns out that for a given energy there is no restriction on the maximum value of ; hence these levelsi are infinity-fold deigenerate. The radial equation for the hydrogen atom can also be solved by the use of ladder operators (also known asS...*0#...*:
); see Z. W. Salsburg, ‰ 6€6'7 Ú
36 (1965).VrVz??? ? ?
€? s4H C. fx(f3Cefw)5y
Using (6.93), we have for the recursion relation (6.88) 0Kz 1 di 4b0K(Kz=4
?0Kzg?0Kd(Kd 0 (6.99) The discussion preceding Eq. (6.91) shows that the highest power of 0 in the polynomial v?8 1 0 x 0 8 0 [Eq. (6.86)] is 914=(=zV Hence use of -1i 4b [Eq. (6.93)] in .?8 1d =-8 8 ( v?8 [Eq. (6.84)] gives ithe radial factor in tihe hydrogen-atom as . 4( ?8 18 ( d =i8 4b R 4=(=z 01y 0 8 0 (6.100) where b >J y 2 d d d [Eq. (6.63)]. The complete hydrogenlike bound-state wave func- tions are [Eq. (6.61i)] 4(l 1. 4( ?8 6 l( ?2rx1 1. 4( ?8 x (l ?2 z Cd d fl1 (6.101) where the first few itheta functions are igiven in Table 5.1.How many nodes does
.?8 have? The radial function is zero at 81Cr at 81y for (1yr and at values of 0 that make v?8 vanish. v?8 is a polynomial of degree4=(=zr and it can be shown that the roots of v?8 1y are all real and positive. Thus,
aside from the origin and infinity, there are4=(=z nodes in .?8 V The nodes of the
spherical harmonics are discussed in Prob. 6.41. y5v|xEB)f)HCSfAHCev|w)(5|Cf|xC1|Hydg
For the ground state of the hydrogenlike atom, we have41zrx(1yr and l1yV The
radial factor (6.100) iis . zy ?8 1 y d =i8 b (6.102)The constant
y is determined by niormalization [Eq. (5.8i0)]: x y x d r C y d =di8 b 8 d xt81z136 Chapter 6 | The Hydrogen Atom
Using the Appendix iintegral (A.8), we finid
. baV8r.3y
i b z x3 d Fi8xb (6.103)Multiplying by 6
aa .bxVr bx32 we have as the grouind-state wave functioin
K baa .b bx3 y i b z x3 d Fi8xb (6.104) The hydrogen-atom energies and wave functions involve the reduced mass, given by (6.59) as . l d l ) l d jl ) . l d bjl d xl ) . l d bja1aaa b .a1 l d (6.105) where l ) is the proton mass and l d xl ) was found from Table A.1. The reduced mass is very close to the electron mass. Because of this, some texts use the electron mass instead of the reduced mass in the H atom Schrödinger equation. This corresponds to assuming that the proton mass is infinite compared with the electron mass in (6.105) and that all the internal motion is motion of the electron. The error introduced by using the electron mass for the reduced mass is about 1 part in 2000 for the hydrogen atom. For heavier atoms, the error introduced by assuming an infinitely heavy nucleus is even less than this. Also, for many-electron atoms, the form of the correction for nuclear motion is quite complicated. For these reasons we shall assume in the future an infinitely heavy nucleus and simply use the electron mass ini writing the Schrödiniger equation for atoims. If we replace the reduced mass of the hydrogen atom by the electron mass, the quan - tity ... defined by (6.63) becomes b a . a , 3 l d d 3 .a1 3b >€ (6.106) where the subscript zero indicates use of the electron mass instead of the reduced mass. b a is called the Meöono vs.dC since it was the radius of the circle in which the electron moved in the ground state of the hydrogen atom in the Bohr theory. Of course, since the ground-state wave function (6.104) is nonzero for all finite values of 0, there is some probability of finding the electron at any distance from the nucleus. The electron is certainly not confined to a circle. A convenient unit for electronic energies is the zazxfoe,ueaf (eV), defined as the kinetic energy acquired by an electron accelerated through a potential difference of 1 volt (V). Po- tential difference is defined as energy per unit charge. Since d.b1a3b vbaFb
> and b>ˆ> .b>2 we have b>ˆ.b1a3b vbaFb
> (6.107)EXAMPLE
Calculate the ground-istate energy of the hydrogen atom using iSI units and convert the result to electronvolts. The H atom ground-state energy is given by (6.94) with4.b and i.b as
o .F d xu 3 3a1 Use of (6.105) for gives
o.Fa1V1ba
vba F b > rVb1a3b vbaFb
> r V13a avba F >>0r 3V1 b vba
Fb3 > 3 x‰? 3 r 3 > i 3 4 3 o.FV31b vba Fb >rVi 3 x4 3 ‡dVb>ˆrxVb1a3b vbaFb
>rH
URUG61FGSMD3g4W090FXGergIMyF34v0MLGf9uFXGAD3807M3TCnC hni o1V.rF jˆ,.i
R 4 R ,1VrF jˆ (6.108)
a number worth remembering. The minimum energy needed to ionizei a ground-state hydrogen atom is 13.i598 eV. Find the 41R energy of "B Rx in eV; do the minimum amoiunt of calcula - tion needed. ( ‰: 0 V kFk eV.)1lFobm1
Find naf for the hydrogen-atom groundi state.
Equations (3.89) fori
naf and (6.7) for V R ? give j naf1 j?ya2?jt1V2 R R j?yV R ?jt jV R ?1 r R ? r8 R xR 8 j r? r8 Vr 8 R 2 R 2 R ?1 r R ? r8 R xR 8 j r? r8 since 2 R ?1(.(xr,2 R ? and (1k for an state. From (6.104) with i1ri we have ? 1 Vr R b V R d V8 b i so r? r81V Vr R b V R d V8 b and r R ? r8 R 1 Vr R b V R d V8 b FUsing t
18 R jpBmj?jt8jt?jt? [Eq. (5.78)], we haive j naf1V2 R R j r b R k k , k 0 r bd VR8 b VR 8d VR8 b 18 R jpBmj?jt8jt?jt? j1V2 R R b R k t? k pBmj?jt? , k 0 8 R bd VR8 b VR8d VR8 b1jt812
R R b R 1d R k b where Appendix integral A.8 and b 1 k 2 R d R were used. From (6.i94), d R k b is minus the groundi-state H-atom energy, and (6.108) gives naf1rF jˆF (See
also Sec. 14.4.) Find naf for the hydrogen-atom R) k state using (6.113)i. ( ‰: 0 d R R k b1.rF jˆ‡ 1
FkjˆF,
Let us examine a significant property of the ground-state wave function (6.104). We have 8 1.c R xe R xn R , r R F For points on the + axis, where e1k and n1ki we have 8 1.c R , r R1RcRi and
? rkk .cijkijk,1 Vr R .i b, R dViRcR b
(6.109) ,���slY.hc shows how (6.109) varies along the + axis. Although ? rkk is continuous at the origin, the slope of the tangent to the curve is positive at the left of the origin but negative 100( x, 0, 0) x er 6.1CTOiC DT2G73G01FG
1rgIMyF3490MLGyIMD3gX4
T090FGc9uFGaD3807M3XR
138 Chapter 6 | The Hydrogen Atom
at its right. Thus 8>x8 is discontinuous at the origin. We say that the wave function has a at the origin. The cusp is present because the potential energy . =17O S x4d # , becomes infinite at the origin. Recall the discontinuous slope of the particle-in-a-box wave functions at the wallsi of the box. We denoted the hydrogen-atom bound-state wave functions by three subscripts that give the values of p, , and In an alternative notation, the value of is indicated by a letter:Letter R
†V 01234567V ??€ The letters r r r R are of spectroscopic origin, standing for sharp, principal, diffuse, and fundamental. After these we go alphabetically, except that " is omitted. Preceding the code letter for r we write the value of p. Thus the ground-state wave function >Ú##
is called > Úe or, more simply, 1 . Wave Functions for ? z22 For3=S0 we have the states >
S## 0+>SÚ1Ú
0+>SÚ#
0 and >
SÚÚ
- We denote > S## as > Se or simply as 2 . To distinguish the three 2 functions, we use a subscript giving the value and denote them as S8 Ú 0+S8 #0 and S8
1Ú - The radial factor in the wave function depends on p and , but not on , as can be seen from (6.100). Each of the three 2 wave functions thus has the same radial factor. The 2 and 2 radial factors may be found in the usual way from (6.100) and (6.99), followed by normalization. The results are given in ?.Note that the exponential factor in the
3=S radial functions is not the same as in the a
Úe function. The complete wave function is found by multiplying the radial factor by the ap- propriate spherical hiarmonic. Using (6.101i), Table 6.1, and Taible 5.1, we have +Se=Ú 4ÚxS
y 7 Sg z xS yÚ17, Sg zO17,xSg
(6.111) +S8 1Ú =Ú 4ÚxS
y 7 g z xS ,O17,xSg
+:7+1+O 1i2 (6.112) TABLE 6.1 Radial Factors in the Hydrogenlike-Atom WaveFunctions
a Úe =Sy 7 g z xS +O 17,xg a Se =Ú RS y 7 g z xS yÚ17, Sg zO17,xSg
a S8 =Ú SR y 7 g z xS + + ,O17,xSg
a e =S R y 7 g z xS yÚ1S7, gcS7 S , S S g S zO 17,x g a 8 = S R y 7 g z xS y 7, g17 S , S g S zO 17,x g a 2 = ÚR # y 7 g z xS + + , S O 17,x g 1 920F9IGURE6.343XF957E8GXFR6A3GM9P4LFX9nURe3rGRgCyC hap T181 c Ve . eF8 i 2 8> j nF8 ab =2aF8>1tra1,The HHyd
T181 e Ve l. eF8 i 2 > j nF8 ab =2aF8>1a f1,1b
3The HHrd
ognAt T e H T AmBfB T Bsut T si T fat T ,swugAmltc T wgcmgA T igpfswB T m, T fat T abcwsvt,Am.t T -gEt T ix,p( fms,B T -0+#ÚS*6...:TvwgSaBTBsutTsiTfatTwgcmgATix,pfms,B ToatTa 4TigpfswTug.tBTfatTwgcmgATix,p(
fms,B T ltws T gf T aVcoTt6ptSfTiswTVTBfgftB ter C 6.16T CO1n-P1ic-1lF
C ocFb-1lF oat TSwsngnmAmfb
T si T im,cm,v T fat T tAtpfws, T m, T fat T wtvms, T si T BSgpt T -atwt T mfB T psswcm,gftB T Amt T m, T fat T wg,vtB T r T fs T aUcao1,TfsT,Uc,oTg,cT TfsT Uc TmBT5)P Th3 :1d[ T ?0? 8 c1Vkd 04 raRv 8 ?` 84r,o1 (? 8 1a 8
1a f1,1ca1c,1c T he HH3d
qt T ,s- T gB.- T qagf T mB T fat TSwsngnmAmfb
T si T fat T tAtpfws, T agEm,v T mfB T wgcmgA T psswcm,gft TT ntf-tt, T r T g,c T aUcaT-mfaT,sTwtBfwmpfms,Ts,TfatTEgAxtBTsiT,Tg,cT 6TqtTgwtTgB.m,vTiswTfatTSwsngnmAmfbT si T im,cm,v T fat T tAtpfws, T m, T g T fam, TBSatwmpgA
T BatAA T pt,ftwtc T gf T fat T swmvm,] T si TT m,,tw T wgcmxB T r T g,c T sxftw T wgcmxB T aUca.TqtTuxBfTfaxBTgccTxSTfatTm,im,mftBmugATSwsngnmAmfmtBThe HH3dTiswTgAAT ? ? ? ? ? ? z ? x y z ? x y z ? x y m?m ? C C m yzyzyz om|H yCdgACv84l0g9Gs930F984Er4D9s4e3G89
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rVaR9 c4LF9sURe3rGRg1920FX9g4MF9 ge4DF9rg9UgFE9rR94DD9XX84l0g199ThR9gGMF93Fd3gi930XFgF9sURe6
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Rr r r
r r R140 Chapter 6 | The Hydrogen Atom
possible values of 1 and 82 keeping r fixed. This amounts to integrating (6.115) over 1 and81 Hence the probability of finding the electron between r and ,j2, is
a 3f =,U 3 , 3 >2,> 0 3r a0 r a V: tf =12>8UV 30`8>1>21>28?a
3f =,U 3 , 3 >2, (6.116) since the spherical hiarmonics are normalizeid: 0 3r a0 r a V: tf =12>8UV 30`8>1>21>28?b ??
as can be seen from (5.72) and (5.80). The function a 3 =,U, 3 2 which determines the probabil- ity of finding the electron at a distance r from the nucleus, is called the ; see . For the 1Z ground state of H, the probability density VcV 3 is from Eq. (6.104) equal to O F3,Kg times a constant, and so Vc be V 3 is a maximum at ,?a (see Fig. 6.14). However, the radial distribution function a be =,U 3 , 3 is zero at the origin and is a maximum at ,?g (Fig. 6.9). These two facts are not contradictory. The probability density VcV 3 is proportional to the probability of finding the electron in an infinitesimal box of volume2 >20>212 and this probability is a maximum at the nucleus. The radial distribution func-
tion is proportional to the probability of finding the electron in a thin spherical shell of inner and outer radii r and ,j2,2 and this probability is a maximum at ,?g1 Since c be depends only on r, the 1Z probability density is essentially constant in the thin spherical shell. If we imagine the thin shell divided up into a huge number of infinitesimal boxes each of volume2 >20>212 we can sum up the probabilities Vc
be V 3 >2 >20>21 of being in z 2 x 2 1 y m C x x 1 y m C x x 1 y m C x x x xxx z zz zx 1 zx 1 z x 1FIGURE 6.9 Plots of the
radial distribution lfunction r =U 2 2 for the hydrogen l atom. URUG61FGSMD3g4W090FXGergIMyF34v0MLGf9uFXGAD3807M3TCnC h-h each tiny box in the thin shell to get the probability of finding the electron in the thin shell as being 1> ze 1 d . L'sXXV The volume .
L'sXX of the thin shell is J n4F,K2,i
n = J n 4, n -J4, d x2, where terms in F2,i d and F2,i n are negligible compared with the dr term. Therefore the probability of being iin the thin shell is 1> ze 1 d . L'sXX -a dze F: yy i dx J4, d x2,-a dze RFJ4i =zjd v d J4, d x2,-a dze , d x2, in agreement with (6.116). The 1 s radial distribution function is zero at ,-y because the volume J4, d x2, of the thin spherical shell becomes zero as r goes to zero. As r increases from zero, the probability density 1> ze 1 d decreases and the volume J4, d x2, of the thin shell increases. Theiir product 1> ze 1 d J4, d x2, is a maximum at ,-gV1lFobm1
Find the probability ithat the electron ini the ground-state Hi atom is less than a idistance a from the nucleus. We want the probability tihat the radial coordiniate lies between 0 anid a . This is found by taking thei in?nitesimal probabiliity (6.116) of beingi between r and ,K2, and summing it over the range from 0i to a . This sum of in?nitesiimal quantities is thei de?nite integral x ? g y a d3f , d x2,-J g n ? g y O =d,jg , d x2,-J g n O =d,jg k=, d g d=d,g d J=dg n p rB g y x-JRO =d F =FjJi=F=zjJiv-yVndn where a zy was taken from Table 6.1 and the Appendix integral A.7 was used. Find the probability ithat the electron ini a d8 z H atom is less thani a distance a from the nucleus. iUse a table of integrals or the websitei integrals.wolfram.com. (Answer:
0.00366.) . Hf3CIgxy5dH|3(,HCev|w)(5|R
The factor
O it2 makes the spherical harmonics complex, except when t-yV Instead of working with complex wave functions such as (6.112) and (6.114), chemists often use real hydrogenlike wave functions formed by taking linear combinations of the complex func- tions. The justification for this procedure is given by the theorem of Section 3.6: Any lin- ear combination of eigenfunctions of a degenerate energy level is an eigenfunction of the Hamiltonian with the same eigenvalue. Since the energy of the hydrogen atom does not depend on m, the d8 z and d8 =z states belong to a degenerate energy level. Any linear com - bination of them is ian eigenfunction of ithe Hamiltonian with tihe same energy eigeinvalue. One way to combine these two functions to obtain a real function is d8 +z zd Fd8 =z Kd8 z i-z Jzd4 k 7 g r Fjd x,O =7,jdg xLiEx1xHTLx2 (6.118) where we used (6.112), (6.114), and O i2 -HTLx2 ixLiEx2V The zjzd factor normal- izes d8 Y ? 1d8 1 d x2`-z d R?1d8 =z 1 d x2`K ? 1d8 z 1 d x2`K ?F d8 =z i nd8 z x2`K ?F d8 z i nd8 =z x2`r - z dFzKzKyKyi-z
142 Chapter 6 | The Hydrogen Atom
Here we used the fact that t
n and t Vn are normalized and are orthogonal to each other, since , tv e = V`i U+ `i c3iV , tv e t`i c3iVeThe designation
t > for (6.118) becomesi clearer if we note tihat (5.51) gives t > Vn s?tv 5Kt >V 2Kt
(6.119)A second way of combining the functions is
t a xn `?t =t n Vt Vn UVn s?tv 5Kt c2c focF focicV 2Kt
(6.120) t a Vn s?tv 5Kt aV 2Kt
(6.121)The function
t e is real and is often denoted by t e Vt b Vn ?v t 5Kt bV 2Kt
(6.122) where capital stands for the number of protons in the nucleus, and small is the coordinate of the electron. The functions t > 8ct a8 and t
b are mutually orthogo- nal (Prob. 6.42). Note that t b is zero in the plane, positive above this plane, and negative below it.The functions
t Vn and t n are eigenfunctions of 0z t with the eigenvalue: tr t 1 The reasoning of Section 3.6 shows that the linear combinations (6.118) and (6.120) are also eigenfunctions of 0z t with eigenvalue tr t1 However, t
Vn and t n are eigenfunctions of 0 z b with eigenvalues:Vr andxr1 Therefore, t > and t a are not eigenfunc- tions of 0 z b 1 We can extend this procedure to construct real wave functions for higher states. Since ranges from V9 tox98 for each complex function containing the factorV`2d2i
there is a function with the same value of and but having the factor x`2d2i1 Addition and
subtraction of these functions gives two real functions, one with the factor cra c=2d2iU8 the other with the factor c focc=2d2iU1 lists these real wave functions for the hydrogenlike atom. The subscripts on these functions come from similar considerations as for the t > 8ct a8 and t
b functions. For example, the 43 >a function is proportional to (Prob. 6.37). The real hydrogenlike functions are derived from the complex functions by replacing `di K=tvU nKt with v VnKt c foc=2d2iU or v VnKt cra c=2d2iU for dCeH for dVe the i factor is n K=tvU nKt for both real and complex functions. In dealing with molecules, the real hydrogenlike orbitals are more useful than the complex ones. For example, we shall see in Section 15.5 that the real atomic or- bitals t > 8ct a8 and t
b of the oxygen atom have the proper symmetry to be used in constructing a wave function for the 3 t d molecule, whereas the complex 2 orbitals do not.URbGergIMyF3X7OEFG.IXP709XTCnC h-n
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