RS ch5 - era- ted. lue? :' the the .tion wed and Molecular...

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Unformatted text preview: era- ted. lue? :' the the .tion wed and Molecular Vibrations and Time—Independent Perturbation Theory 5.1 Vibrations are an important form of molecular energy. The usual picture for vi- brations is a harmonic model; this chapter discusses the energies and wavefunc- tions of harmonic vibrations, both for single vibrations and for the normal coordinates of a molecule. Time-independent perturbation theory is a standard and important method for improving wavefunctions and energies and is easily in- troduced in the context of molecular vibrations. ' DIATOMIC MOLECULE VIBRATIONS Let us begin with a problem we examined in the previous chapter, namely, the Hamiltonian of a diatomic molecule; here we study the vibrational motions that we previously ignored. This problem leads us rather directly to a consideration of the eigenvalues and eigenfunctions of the harmonic oscillator, which is one of the most important problems in quantum mechanics. Then we study the gener- alization of this treatment to Vibrations of polyatomic molecules, a topic that is of consequence in the interpretation of infrared and Raman spectroscopy. The description of molecular vibrations provides an excellent example of the appli- cation of time—independent perturbation theory, so the last section of the chap- ter deals with this important approximation method for solving the Schrodinger equation. 71 72 Molecular Vibrations and Time—Independent Perturbation Theory Chapter 5 Recall from Eq. (4.6) that the Hamiltonian of a vibrating and rotating di- atomic molecule is 2 H = -E— V2 + V(r), (5.1) 2M where r is the internuclear distance and V( r) is the potential function. Here we consider r to be variable (rather than fixed, as in Chapter 4) and take V(r) to be a typical diatomic molecule potential, which usually looks like that depicted in Fig— ure 5.1. If we confine ourselves to states with small displacements from equilib— rium, we can expand the potential around its minimum at r = re. Since the first derivative vanishes at that point, we get 1 n 2 V(r) = V(re) + 5 V (re)(r -— r6.) + ‘- where V" is the second derivative of V, also known as the Hessian or force con— stant. In the following discussion, we consider only terms up through quadratic, which means that we replace the exact potential by a parabola and the oscillator motions are that of a harmonic oscillator. The Schrodinger equation for the relative motion [analogous to Eq. (4.10), but for the case of three dimensions] is HX = WX, (5.3) with the kinetic energy operator taken from Eq. (4.26) and using W as the total energy. Substituting Eq. (4.26) into Eq. (5.1) and the result into Eq. (5.3), we get —h-2 1 a2 32 1 2} ———— + +V +—k -- =W, 5.4 { 2p. r arz r 2,u.r2 (re) 2 (r re) X X ( ) where we have used k = V”(re) and 32 for the angular-momentum operator, fol— lowing standard notation. The angular eigenfunctions of this equation are the spherical harmonics, just as with the rigid rotor. Since the angular and the radial motion can be separated in Eq. (5.4), we can write X as the product X(r9 0: = YJMJ(0a ‘, (5.2) Figure 5.1 Typical diatomic r molecule potential. Sectio and ti We ni much brati( then Now and : Fina In tl hov as i in ( tha 5.2 RAIS OSCI Th eql lov pter 5 1g di- (5.1) 'e we to be 1 Fig- lilib- first {5.2) con- atic, ator 10), 5.3) otal get 5.4) fol- the iial Section 5.2 Raising and Lowering Operators for the Harmonic Oscillator 73 and the Schrodinger equation reduces to 71-21 (12 J(J + 1m2 1 2} ————— +—~——+V +—k — =W. 5.6 { 2” r drz r ZMrz (re) 2 (r re) 99 g ( ) We now assume that r = re in the centrifugal term, as this term usually varies much more slowly with r than does the potential-energy term. If we define the vi- brational energy by E_W V() J(J+1)h‘2 (57) “ ’8 2w: ’ ' then the Schrodinger equation is 712 1 d2 1 2} {—i7d—r2-r-i-Ek0-re) (5.8) Now we define a new wavefunction (a = rg“, (5.9) and substitute it into the Schrbdinger equation, giving I? d2 1 2} — ~——— + — — = E . . { drz 2 k(r r8) ll! ‘10 Finally, let us define the displacement from equilibrium: x = r — re. (5.11) In terms of x, the Schrodinger equation is W d2 1 } ~ — ~— + — 2 = . . { Zfl dxz 2 kx cl E11; (5 12) This is precisely the Schrodinger equation for a simple harmonic oscillator. Note, however, that the variable x ranges from —re to +00, rather than from —00 to 00, as it would for a harmonic oscillator in one dimension, such as that considered in Chapter 1 (Section 1.3.4). This difference will be ignored in the discussion that follows. 5.2 RAISING AND LOWERING OPERATORS FOR THE HARMONIC OSCILLATOR The Schrodinger equation (5.12) can be solved as a conventional differential equation; however, it is easier to solve if one rewrites H in terms of the raising and lowering operators 1 . 2<x + 2—) (5.13a) pm) HS?) Section 74 Molecular Vibrations and Time~|ndependent Perturbation Theory Chapter 5 i This In: and ever, a l ‘ from E uw 1p V = (ii->20 ‘ m), (5-13b) comple T( where w = (k/p.)1/2 is the classical angular frequency [Eq. (1.8)] and (p = —ih'd/dx) ative ei is the momentum operator. These equations may be inverted to give have th — H + ‘ isfy x — , / 2W (b + b ) (5.14a) Since ti _ _. #wa _' + is conw p — z —2 (b b (5.14b) value A If these expressions are now substituted into the Hamiltonian, one obtains and H = %Hw(bb+ + ab). (5.15) which] Note that we have to be careful about the ordering of b and b”, as they do not commute. Their commutator is + “w SO the ‘ = — + __. _.. __._ ' [b,b] 227 [x W,x W] (516) z = E [A x] = Using this commutator, we find that Thése cle-in-l 1 1\' _ + fl H — fiw<b b + 2). (5.17) “an 3 This form of the Hamiltonian is such that the eigenfunctions of H will also be l eigenfunctions of b+b. Thus, these eigenfunctions will satisfy the equation b+b|A) = MA), (5.18) where we define )1 to be the eigenvalue associated with b+b and we label the eigenfunctions M) with this eigenvalue. A To solve for the eigenvalues and eigenfunctions, we use the commutator g, given by Eq. (5.16) to show that g b+b(b+|,\)) = b+(1 + b+b)|)t> = (A + 1)b+l/\). (5.19) 3 > This equation shows that H“ l A) is aneig‘enfunction of b+b with an eigenvalue of )t + 1. This means that b+ acts on M) to raise the value of the eigenvalue by one. Similarly, if we consider the quantity bl A), we find that b+bb|A> = (bb+ — 1)b_)).) = (A »— Dbl/t). (5.20) Iter 5 13b) MG 14a) 14b) 5.15) b not 5.16) 5.17) obe 5.18) l the :ator 5.19) 1e of none. 5.20) Section 5.2 Raising and Lowering Operators for the Harmonic Oscillator 75 This means that b operates on IA) to lower the eigenvalue by one. There is, how- ever, a lower limit to the value of A. The limit exists because (A I b+b I A), which, from Eq. (5.18), is just A, is inherently nonnegative. This is because (A I b+ is the complex conjugate of b I A), so (A I 5% I A) is an absolute square. To avoid applying b so many times to I A) that it produces a state with a neg- ative eigenvalue, we terminate the series by choosing one of the eigenvalues A to have the value zero. The eigenfunction corresponding to A = 0 would then sat- isfy (0 | b+b I 0) = 0, which implies that bIO) = 0. (5-21) Since the eigenvalues above A = 0 can differ from that by at most an integer, it is convenient to define an integer quantum number n that is identical to the eigen- value A.Thus, we have b+b I n) = n I n), (5.22) which means that HIn) = fiw<b+b + In) = fiw<n + In), (5.23) so the energies are 1 En = fiw<n + a), n = 0, 1, (5.24) These energies are equally spaced (see Fig. 5.2), in distinct contrast to the parti~ cle-in-box spectrum of Fig. 3.2 or the rotational spectrum of Eq. (4.42). Note that (n I b+b I n) = n. Also, since (11 I b+ is the complex conjugate of b I n), we have bIn) = W In — 1). (5.253) Figure 5.2 Energy levels and eigenfunctions of harmonic oscillator for n = O, 1, 2. (For clarity, successive wavefunctions have been shifted up so that the local zero is at the x eigenvalue.) V(x) and 11/(x) 76 Molecular Vibrations and Time—Independent Perturbation Theory Chapter 5 By a similar reasoning with (n j bbJr l n) and using the commutator given by Eq. (5.16), we find that b+ln> = Vn + 1|n + 1). (5.25b) To determine the eigenfunctions l n), we write the explicit operator form of Eq. (5.21), using Eq. (5.13a), but omitting the multiplicative constant: 72' d + —— = . . (x “w (1x) JO) 0 (5 26) This equation may be rearranged to yield 0110) Mw .. I 0) — h x ax. (5.27) After integrating and exponentiating, we get 10) = Ne‘sz/z’l = Nerf/2, (5.28) where a = ,uw/h'. The normalization constant is determined from NZ/ (“"2 dx = 1, (5.29) which gives (1 N — 4 g. (5.30) By the repeated application of b+ to {0), one can use Eq. (5.25b) to show that (13+)” In) — ("01/2 10). (5.31) Substituting Eq. (5.28) into this formula then gives N mt: _ I") = We /2H,,(Vax), (5.32) where H,z is a Hermite polynomial, some specific examples of which are as follows: 1, n = 0 2z, n = H,,(z) = 4Z2 - 2, n = . (5.33) 8z3 — 122, n = 16z4 — 48z2 + 12, n = 4 structure of the eigenfunctions is very much analogous to that for the one- dimensional box wavefunctions in Fig. 3.2, but in contrast to the box Sect Wth ity f( are t only tion, spor solut EXI ian An: Sin< NOV whit b) of 5) 7) 3) ’) EXERCISE 5.1 Usmg the definitions of b and H, prove directly that the Hamilton— Section 5.2 Raising and Lowering Operators for the Harmonic Oscillator 77 Figure 5.3 Probability densities In,l/,1(x)|2 associated with the n = 0—2 states of the harmonic oscillator, using a plotting format that is X analogous to Fig. 5.2. V(x) and t//(x)2 wavefunctions, the harmonic oscillator wavefunctions have a nonzero probabil- ity for x where the potential energy is greater than the total energy. Such regions are forbidden in classical mechanics, but can exist in quantum mechanics, although only where the wavefunction decays rapidly (usually exponentially). This mo— tion, called tunnelling, is an important example of a quantum effect and is re- sponsible for such processes as proton transfer between water molecules in solution and electronic conductivity in scanning tunneling microscopy. ian operators of Eqs. (5.12) and (5.15) are identical. Answer First we use the expressions for b and b+ in Eqs. (5.13) to show that + 2 B2 _ 2X 1 ‘1) b b 2” (x law x “w 2 . pm) 2 p z = —— + + —— — . 2,1. (x Msz W (xp pm) Since [x, p] = iii [from Eq. (5.16) or Eq. (2.36)], we have 2 a) “2+1? 5% x 2/.L7ia) 1 + : -—-—- bb 2. Now we substitute into Eq. (5.17) to get 1 W P2 + +— = — 2+ (M 2);... 21—42%). My) Ma“ which is what we wanted to show. 78 Molecular Vibrations and Time-Independent Perturbation Theory Chapter 5 seem Here EXERCISE 5.2 The parity operator P IS defined such. that The I PW) = [mm—x), amip where p = :t 1 (even or odd parity, respectively). Prove that P and the Hamiltonian in . mom Eq. (5.12) commute. Find the parity eigenvalue 1) for the first five eigenfunctions of the lor se harmonic oscillator. Answer The commutator is [P, H M = P( H 1,0) — H (Pitt). But the parity of a prod- uct is the product of the parities (That is, an even function times an odd function is odd, an odd times an odd is even, and an even times an even is even.) So PW!!!) = (PHXPW) = HP!!! because H is even (both x2 and (—ihd/d)c)2 are unchanged for x ——> —x). Therefore, [P,H]gl/=HP¢—-Hpt,lz=0, or [P,H]=O. I th II The oscillator eigenfunctions are given by Eq. (5.32). With the Hermite polynomials - . , mini] of Eq. (3.33), we see that . we w we, lie, Ila, $6, etc, are even (iparlty = 1) the r and . CI $1, (1/3, I115, (#7, etc, are odd (parity = —1). mg i_.____________l [————________._____.__.___ EXERCiSE 5.3 Evaluate (0 i bb+be+bb+ | 0), (O | bbebib+ i O), (0 l b+b+b+beb i O), h and (0 | l7beb+b+lfL i 0), where H is the harmonic oscillator Hamiltonian. w a Answer (0 | bb+be+bb+| 0) = (1 lH! 1) = 3/2h'w; ‘ (OlbbelbeO) = (3 IHIZ) = O; is tht cons (Oib+b+b+beb|O> = 0, because biO) = 0; Eq. ( (olbbeb+b+b+Jo> = 6<3iH|3>hm = 6(3 + 32m = 21 fiw. i__._.._ ____ _________________________i 5.3 POLYATOMIC MOLECULE VIBRATIONS To S The generalization of the theory of the previous section to molecules with three - (mafi or more atoms is complex in general because of lack of separability of vibrational and rotational motion. Here we give a relatively simple treatment that circum- vents the problem, though not without some sacrifice in sophistication and rigor. The final Hamiltonian is Eq. (5.44), and the solutions of the Schrodinger equa- whe] tion are given by Eqs. (5.48) and (5.49); the manipulations leading to these results 3N 3 involve some matrix algebra that beginning students might want to skip. for e We begin with the Hamiltonian of a molecule containing N atoms moving mine in three dimensions [Eq. (2.9b)]: 752 82 H: 2 —— 2+V. (5.34) Section 5.3 Polyatomic Molecule Vibrations 79 Here, the label i identifies the atoms and a = x, y, z the Cartesian components. The potential V is assumed to describe bound motions, so we suppose that it has a minimum at an equilibrium geometry that is labelled by the coordinates xfa. For motions of the atoms near equilibrium, the potential can be expanded in a Tay- lor series that includes only a few terms. Thus, we write 8V V({xia}) = V({xfa}) + ax. e (xia _ xi) 1 62V 8 e + E g axiaax/B Ximxffxm .- xiaxx/B x113) + VA. In this expression, the first derivatives will all be zero as we expand about the minimum of the potential. VA includes the third and higher derivative terms, so we will call it the anharmonic part of the potential. This part will be neglected in the rest of the section, but in the next section we shall examine its influence on eigenvalues with the use of perturbation theory. The Hamiltonian at this point is El 62 1 e e H = ia _2—mi axiza + E % Ea’j‘3(xi“ _ xiaxxffi ‘ 9‘13), (5-36) where 2 Fm, ,1; = “(1L (5.37) axiaaxj/s xii, x3; is the matrix of force constants (the Hessian matrix) and we have omitted the constant term V({xfa}), as this can be subtracted from the total energy W as in Eq. (5.7) to define a vibrational energy E. The Schrodinger equation governing vibrational motion is Hip = Eli]. (5.38) To solve this equation, we introduce normal coordinates Qk with dimension (mass)1/2 (length), using the expression -1 xia = xi; + mi 2 E Likaky (5-39) k where the Lia, k are coefficients that can be grouped into a matrix L of dimension 3N >< 3N (i.e., the number of atoms times the number of Cartesian coordinates for each atom). For reasons that will be apparent shortly, the matrix L is deter- mined by diagonalizing the mass-weighted Hessian matrix H, which is defined by Fiat, 1'3 Hm, 1/3 = (5.40) 80 Molecular Vibrations and Time~lndependent Perturbation Theory Chapter 5 If the eigenvalues of this matrix are denoted Ak, then the eigenvalues and eigen— vectors satisfy HL = L A, (5.41) where /\ is a diagonal matrix with the Ak’s as diagonal elements. From Appen- dix A [Eq.(A.14)], we note that L is an orthogonal matrix (L = L4). This means that the inverse of Eq. (5.39) is Qk = mf/ZLia.k(-xia F— xia’a)‘ Let us now express the Hamiltonian in Eq. (5.36) in terms of normal coor- dinates. This requires a straightforward application of the chain rule to the ki- netic—energy operator, using Eq. (5.42), and direct substitution of Eq. (5.39) into the potential energy. The result is 71—2 62 H z "T ; gLiaikLm’k'SQkanr 1 + ‘2‘ 2 2 Hims 2 2 Lia,ijB,k’Qka’a (5-43) £01 11; k k’ where we have replaced F by H, using Eq. (5.40). This expression can be simpli- fied by moving the sums over k and k’ to the outside and invoking orthogonali— ty of the L matrix in the kinetic energy and Eq. (5.4-1) in the potential energy. In both cases the result yields zero if k at k’, so the double sum reduces to the single- sum result H= 2{—h—2fi—+:—LAQ2} (544) k 2 act 2 k k ' ' A comparison of this equation with Eq. (512) indicates that Eq. (5.44) is a sum of harmonic oscillator Hamiltonians in which the reduced mass ,u is unity and the force constant k is the same as M. This means that the harmonic oscillator fre- quency for mode k is wk = (Ak)1/2. Equation (5.44) demonstrates that the role of the normal-coordinate transformation given by Eq. (5.39) is to convert the Hamiltonian of Eq. (5.36), wherein all degrees of freedom are coupled together, to the uncoupled sum in Eq. (5.44). Although the orthogonal matrix L that accomplishes this transfor- mation has dimensions 3N >< 3N, there are only 3N — 6 vibrational coordi- nates for most polyatomic molecules (3N —- 5 for linear molecules). The reason is that six of the frequencies that result from diagonalization of the H matrix are zero. These frequencies correspond to the three coordinates that cause the center of mass of the molecule to translate and three other coordi- Sect? nate mod Cha} sepa can 1 only in th strict notet tatio quir< (5.4‘ anal and and 1 The and t wher eigen when )ter 5 igen— 5.41) upen— ieans 5.42) :oor— le ki- ) into 543) mph- )nali— gy. In ngle- 5.44) 1 sum )7 and )r fre- inate 5.36), m in isfor— iordi- . The the H i that Sordi- tation are quires imposing the constraints of conservation of angular momentum on Eq. Section 5.3 Polyatomic Molecule Vibrations 81 nates that correspond to overall rotation of the molecule. For all six of these modes, the force constant M is zero, and the only contribution to the energy in Eq. (5.44) comes from the kinetic-energy terms. We have already noted (in Chapter 4) that translational motions of the center of mass are completely separable from internal motions, so this contribution to the energy expression can be ignored. The corresponding separation of rotation and vibration works only in the limit of infinitesimal vibrational motions. We will assume this limit in the development that follows, which means that the sum in Eq. (5.44) is re- stricted to the 3N — 6 terms for which Ak is nonzero. However, it should be noted that because all coordinates have zero-point motions, vibration and ro- always coupled to some extent. To treat this problem properly re- (5.44), which is beyond the scope of the text. Let us now introduce raising and lowering operators into Eq. (5.44). By analogy with Eqs. (5.13), these are defined by to 1/2 iP bk = (Qk + (5.45a) and +— 4 “r bk — (2),.) Qt wk , (5.45b) and the commutation relation is [15,, 1);] = 5W. (5.46) The Hamiltonian thus reduces to 3N—6 1 H = 2 fiwk<bek + a), (5.47) k=1 and the energy eigenvalues are 3N—6 1 E = 2 hwk<nk + '2'), k=1 where nk is the quantum number associated with the kth vibrational mode. The eigenstates can be expressed as products of functions taken from Eq. (5.32) as 3N—6 ak 1/4 e—aka2/2 1n1n2n3 "av—6) = H F—‘anlval—k Qk), (5-49) k=1 7" 2""nkl)1/2 where ak = (ck/7i. 82 Molecular Vibrations and Time-Independent Perturbation Theory Chapter 5 EXERCISE 5.4 The water molecule has three nonnal modes, whose frequencies are (01 = 3,833 cm’l, wz = 1,649 CHI—1, and (03 = 3,943 elm-1. What is the energy of the (112) state? What is the energy difference between (112) and (100)? Answer Use E = 2 (n, + Qwa to give s,modes E(100) = (3/2)a)1 + (1/2)w2 + (1/2)w3 = 8546 cm—1, E(112) = 18081 cm—l, E(112) — E(100) = 9,535 cm-I. 5.4 TIME-INDEPENDENT PERTURBATION THEORY The oscillator problems that we have considered in this chapter can often be rep- resented with the Hamiltonian H : H0 + AV, (5.50) where H0 is a zeroth-order Hamiltonian for which solutions to the Schrodinger equation may be obtained easily and V is a time-independent perturbation that is small compared to H0 in some sense. For oscillator problems, we are often able to choose H0 as the harmonic oscillator Hamiltonian and V as the anharmonic part of the potential. The parameter )k eventually will be set to unity, but we will use it to keep track of the order of the perturbation. This means that terms which contain M“ are of kth order in the perturbation expansion. Let us denote the eigenfunctions of H0 by (1)9, and the corresponding eigen- values as E S. This implies that Ht]th = Em. (5.51) In the perturbation theory approximation, we assume that the exact wave- function 1,0,, may be expanded in powers of A as w. = #1 + wt“ + W53) + (5.52) The corresponding energy expansion is E, = E}, + ABS) + 295$?) + (5.53) Substituting these expansions into the full Schrodinger equation (H 111,, = Ent/Jn) and equating like powers of )1 gives Harte, = 13%;; (zeroth order), (5.54a) V¢g + Hod)? = ESP¢2 + E9, 2” (first order), (5.54b) . . f . mfg—1) + Hots)” = 2E5,“ fr“ (jth order, j 2 1). (5.54c) k=0 , The zeroth-order equation is satisfied by assumption. To solve Eq. (5.54b), we expand (1),?) in terms of the complete set of zeroth-order states, that is, Sec wh4 wor to l by 1 and Ont am] Section 5.4 Time-Independent Perturbation Theory rare :1) = Ecgzikbg (555) 112) k” 0 , . where an is a coefficient. It follows from Eq. (5.55) that «fink/>2 )) = 0; in other words, the perturbation causes changes to the wavefunction that are orthogonal to the zeroth-order states. Substitution of Eq. (5.55) into Eq. (5.54b), followed by multiplication by (1)9, and integration, then leads to E2” = <¢21V1¢2>, (5.56) and it follows that 0 o (1) _ <¢lel¢n> an — —-————Eg _ E2 (k ¢ n). (5.57) One can develop expressions for higher order terms in a similar manner. For ex- 'eP' ample (2) _ (459.1 V l ¢2><¢2 l V I (159.) 50) E, — 2 m. 5.58 m ES—E‘k’ ( ) ger The only time that Eqs. (5.57) and (5.58) run into serious trouble is when the hat state n is degenerate with one of the states k, as that makes the denominators in bk“ those equations vanish, and as a result, the perturbation theory corrections to n,” the energy diverge. In that case we have to use an alternative approach, degen- Ylu erate perturbation theory, in which one first finds the exact eigenvalues associat- [Ch ed with a Hamiltonian matrix contructed from the degenerate states. In this calculation, the terms in the perturbation that couple the degenerate states are “1' included in the Hamiltonian, so that after diagonalization, the degeneracy is lift- ed. Perturbation theory is then applied to the diagonalized states, including those 51) terms in the perturbation that were not included in the first step. (e EXERCISE 5.5 Prove that, for a system with only two states, if V has no diagonal ma- ! trix elements, the average energy is the same in the zeroth order as in the second order. ‘2) Answer Since V has no diagonal matrix elements, (a | V I a) = (B l V | 3) = 0, where the two states are labeled a and B. Using Eq. (5.58), we find that 3) E512) : E30) + <a|V[013>(3|:/|01) in) E a — E fl and (2) _ (0) (filVlCXXUlVlB) 3) E5 —E,3 +———————Eg_Eg . 3) Adding these and dividing by two gives 3) E53) + E? = Eff” + 52;” 2 2 This means that the average energy is the same in the zeroth, first, and second orders 84 Molecular Vibrations and Time—Independent Perturbation Theory Chapter 5 5.5 EXAMPLES A simple example of first order perturbation theory is a harmonic oscillator in its ground state, subject to a quartic perturbation V = ~yx4, with 'y a constant. In this case, E5,” = Willem). (5.59) If we express x in terms of the raising and lowering operators, then, for the ground state, we have h 2 (1) = + 4 E0 y<2mu> (01(1) + b ) 10>, (5.60a) 01‘ 1 71‘ 2 E5) = 3y<——> . (5.60b) 2px» _——H__ EXERCISE 5.6 Derive Eq. (5.60b) from Eq. (5.60a). Answer To go from the first equation to the second in this derivation, we have used several consequences of Eqs. (5.25). First, note that b 10) = 0 and (0 I b+ = 0, so any term with a leading b+ or a trailing b must vanish. Second, the only nonzero products of b and 13* must have the same number of raising and lowering operators, as the num- ber of raising operations must match the number of lowering operations in order to con- vert IO) into (0 |. As a result, the only nonzero products coming from (b + b+)4 are bb“bb+ and bbb+b+. We can then apply Eqs. (5.25) directly to evaluate these two terms, yielding l and 2, respectively, for a sum of 3, which is the coefficient in Eq. (56%). Alternatively, one can use the commutator relation of Eq. (5.16) to express bb+ = b+b + 1 for both of the terms bb+bl7+ and .bbb'*b+. This is done twice for the bb‘ibb+ term,yielding (b+b + 1)2. Then each of the b+b -‘- 1 terms that result gives unity, as the leading H” and trailing b give a vanishing result when applied to (0! and [0) re- spectively. The commutator can also be applied to the bbb‘LbJr term, starting with the bb+ in the middle to give bb+ + bb+bb+ and then using the commutator again to reduce each of these terms to unity. The three factors of unity again sum to 3. Note that a cubic potential would not contribute in the first order in this example because of symmetry, but would show up in the second order. Another example is the interaction of a diatomic molecule with a static elec- tric field E (the Stark effect). Here, we take the diatomic vibrations to be de- scribed by a harmonic oscillator and assume that the dipole moment u of the molecule interacts with the field E via a perturbation V = u - E. (5.61) Now we expand p. in a Taylor series about the diatomic equilibrium position: [.L=p.e+,.r,;'x+~'. (5.62) The constant term Me contributes an energy )ue - E in first order and zero in higher orders. This contribution to the interaction energy is called the first-order Stark effect, and the magnitude of the energy shift resulting from it varies lin- Sugge early r high S} linear becaus Evaluz portan This is energy It write d where plete t] This fo the eqi COHSta] The ov This is SUGGESTED Althoug ed in ev ters 1 ar textboo. P. W. Atl York, Suggested Readings 85 early with E. The first-order Stark-effect term will vanish for molecules with high symmetry. For these molecules the most important effect comes from the linear term. In the first order, this term is EL“ = (n in; - Exln) = o, (5.63) because the perturbation has odd symmetry. The second-order term is (2)_ (njxikxk’xln) ,. 2 En — I; Ew(n __ (“e ' Evaluation of the coordinate matrix element [using Eq. (5.25)] leads to two im- portant intermediate states, k = n :t 1. The final result is (2) : ‘(l‘é ' E)2 E, 2sz . (5.65) This is an example of the second-order Stark effect, and we see that the shift in energy here is proportional to the square of the applied field. It turns out that it is also possible to solve this problem exactly. To do so, we write down the complete Hamiltonian of the perturbed oscillator, namely, fi2 612 l r “ 5% + + (u; - E>x, (566) where we have (for simplicity) ignored the constant term [Le - E. If we now com— plete the square with the two potential terms, we get H=_£2d_2+l array Mug-EV + — — 5. 5.67 2M dx2 2 WU x /.Lw2 2 sz ( ) This form of the Hamiltonian indicates that the potential is still harmonic, with the equilibrium position centered on x = —(u; - E)/p.w2. In addition, there is a constant term on the right that represents an overall shift to each eigenvalue. The overall energy expression is thus ;~E2 E, = m6(n + — (5.68) This is exactly the same as En = E 20) + E22), with E 22) given by Eq. (5.65). SUGGESTED READINGS Although the harmonic oscillator and time-independent perturbation theory are treat- ed in every quantum chemistry and quantum mechanics text (cited at the end of Chap- ters 1 and 2, respectively), polyatomic vibrations are less commonly treated. Among the textbooks that deal with this subject are the following: P. W. Atkins and R. S. Friedman, Molecular Quantum Mechanics, 3rd ed. (Oxford, New York, 1997). 86 Molecular Vibrations and 'l'Ime-lndependent Perturbation Theory Chapter 5 E. B.Wilson, Jr., J. C. Decius, and P. C. Cross, Molecular Vibra tions (Dover, New York, 1955). S. Califano, Vibrational States (Wiley, New York, 1976). PROBLEMS 5.1 Sketch the n = 6 wavefunction of the harmon nodes are there? 5.2 The fourth Hermite polynomial is ic oscillator and its square. How many H4(z) = 16z4 — 4822 + 12. Find the most probable position for the n = 4 harmonic oscillator. Is it closer to the origin (x = O) or to the turning point (where the kinetic energy vanishes)? Compare your answer with the n = 0 state, and comment on the relation of the two results to the Bohr correspondence principle of Section 3.1.5 (H int: In what location does the classical oscillator spend the most time?) 5.3 Which of the following are zero for a harmonic (0 l bbb+b+ l 0); (0 l b+bb+b 10); (Olb+bb+10). 5.4 Evaluate the following for a harmonic oscillator: <2[(b + b+)2|0>. 5.5 Consider the H20 molecule in its ground electronic state. (a) Write down the Hamiltonian governing the molecule ’5 vibrational motions, tak- ing the potential to be harmonic. Express your answer using raising and lower- ing operators. (b) Give expressions for the ground—state ener wavefunction in normal coordinates. (c) The first neglected terms in the Hamiltonian are the cubic anharmonicities. Consider the cubic term V = CmeQZ, where the Qi’s are the normal coordi- nates. Express V in terms of raising and lowering operators. ((1) Evaluate the first—order perturbation theory expression for the contribution of the cubic term in (c) to the total energy of the ground state. (e) With what excited states does the ground state mix if V is included in the sec- ond order? Specify the quantum numbers of the states. oscillator (l 0) = ground state)? gy and wavefunction. Express the 5.6 Use first-order perturbation theor y to determine the ground-state energy of the quartic oscillator 2 H=L+yx4. 2.“ Use a harmonic oscillator to define the ze roth-order Hamiltonian. In the first order, what choice of harmonic frequenc y gives the lowest zeroth-plus first-order energy? What is the ground-state energy for this frequency? Prt 1" Problems 87 5.7 Consider a two-dimensional rigid rotor that is perturbed by interacting with an ap- plied external field. The Hamiltonian is H = J 2/21 + ,uE cos qfi, where ,u is the di- pole moment and E the applied electric field. (a) If E = 0, what are energy levels and wavefunctions of the rotor? (b) How do these energy levels Change if the perturbation is treated as a first— order problem? (c) Now consider the ground state of the rotor. What is the second—order expression for the ground-state energy? Give explicit expressions for all sums and inte— grals, and evaluate at least one integral. Equation (5.57) permits a calculation of the first-order perturbed wavefunction, and Eq. (5.58) allows a calculation of the second-order energy. As an example, consider the four cubic anharmonicity terms for the two-di- mensional harmonic oscillator. (For example, these terms arise in the description of H20 if the bend is ignored.) The anharmonicities are Val : aixi, Vaz = 02x; Vas “axixz, and _ 2 V124 — a4x1x2, where a1. a2, a3, a4 are constants. Write the first-order perturbed wavefunction arising from the ground state [0,0) of the two-dimensional oscillator. ...
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