Molecular Spectra and Molecular Structure - Lec 3S

Lecture # 3 Supplement Contents 1. 2. 3. 1. Anharmonic Oscillator, Vibration-Rotation Interaction . . . . . . . . . . . . . . . . . . Energy Levels of a Vibrating Rotor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 4 7 Anharmonic Oscillator, Vibration-Rotation Interaction The Hamiltonian is k P2R + x2 2µ 2 � �� � H= harmonic oscillator First we must re-express 1 R2 J2 2µR2 � �� � +ax3 + (1.1) non−rigid rotor in terms of a quantity whose matrix elements we know such as the displacement x. x = R − Re � � x R = Re + x = Re 1 + Re � � � � �2 � x x 1 1 ≈ 2 1−2 +3 R2 Re Re Re (1.2) power series expansion (1.3) where we truncate expansion after (x/Re )2 . The Hamiltonian operator becomes H= � � P2R k J2 2x 3x2 + x2 + ax3 + 1 − + 2µ 2 2µRe2 Re Re2 (1.4) Let us choose a basis {|v� |J�} where |v� is the harmonic oscillator basis and |J� is the rigid rotor basis. The rotational matrix elements we need are � � J|J2 |J � = J(J + 1)δJJ � (1.5) �J|constant|J � � = δJJ � constant (1.6) �J|vibr.coord.|J � � = δJJ � vibr.coord. (1.7) 2 and Be (in energy units) = 1 � J. 2µRe2 (1.8) Page 2 Thus the rotational expectation values of the Eq.(1) Hamiltonian become � 2 � � � P k 2 2x 6µBe 2 3 �J|H|J� = + x + ax + Be J(J + 1) 1 − + x �2 2µ 2 Re where 1 Re2 was replaced by (1.9) 2µBe �2 . Now we need some x matrix elements. Let √ kµ k γ= hν = �ω = � γ � �1/2 v+1 xv,v+1 = 2γ 1 [(v + 1)(v + 2)]1/2 x2v,v+2 = 2γ v + 12 x2v,v = γ � �1/2 (v + 1)(v + 2)(v + 3) x3v,v+3 = 8γ 3 � �3/2 v+1 x3v,v+1 = 3 2γ (1.10) (1.11) (1.12) (1.13) (1.14) (1.15) Remember that the harmonic oscillator part of our Hamiltonian is diagonal in the harmonic oscillator basis we have chosen. � � � � 1 2 6µBe 2 � �vJ|H|v � J � � = hν v + δJJ � δvv� + ax3vv� δjj � + Be J(J + 1) δvv� δJJ � − xvv� δJJ � + x δ (1.16) � JJ 2 Re �2 vv Note that the Hamiltonian matrix is completely diagonal in J. The remaining problem is how to arrange the H matrix now that we have two indices J and v. The Hamiltonian matrix is a super-matrix consisting of a v, v � matrix of J, J � matrices. However, since there are no matrix elements off-diagonal in J, it is convenient to alter our perspective and think of a set of single v, v � matrices, one for each value of J. Thus the Hamiltonian matrix is given by � � � � 1 1 2 6µ + Be J(J + 1) + Be 2 v + J(J + 1) (1.17) �v|H|v� = hν v + 2 2 γ� � �1/2 � �3/2 v+1 2Be v + 1 J(J + 1) (1.18) �v|H|v + 1� = 3a − Re 2γ 2γ � �1/2 � �3/2 2B e v v J(J + 1) �v|H|v − 1� = 3a − (18a) Re 2γ 2γ 3Be2 µ (1.19) J(J + 1)[(v + 1)(v + 2)]1/2 �v|H|v + 2� = γ �2 3Be2 µ (19a) �v|H|v − 2� = J(J + 1)[(v − 1)v]1/2 γ �2 � �1/2 (v + 1)(v + 2)(v + 3) (1.20) �v|H|v + 3� = a 8γ 3 � �1/2 v(v − 1)(v − 2) �v|H|v − 3� = a (20a) 8γ 3 Page 3 We are now in a position to use perturbation theory to determine the contributions of the various terms in the Hamiltonian to the eigenvalues. Note that when faced with an infinite Hamiltonian matrix it will always be necessary to use the Van Vleck transformation in order to truncate the matrix. We act as if part of the matrix isn’t there, yet we know that the Van Vleck transformed energies will be a very good approximation to the eigenvalues of the infinite matrix. In this example, the matrix will be treated entirely by first and second order perturbation theory and no diagonalization will be carried out. For our choice of basis, the zero-order Hamiltonian is diagonal by definition. � � � � (0) H(0) = hν v + 12 + Be J(J + 1) δJJ � δvv� = EvJ The perturbation terms are what is left over � � 2Be 6µBe2 (1) 3 2 H = ax − J(J + 1)x + J(J + 1)x δJJ � Re �2 (1.21) (1.22) The first order corrections to the energy are given by µ but γ� 2 = Thus � � � � 1 6µ J(J + 1) v, J|H(1) |v, J = Be2 2 v + 2 γ� (1.23) � B2 � E (1) = 6 hνe v + 21 J(J + 1) (1.24) 1 hν . This is a harmonic vibrational correction to the rotational constant. Compare with the leading term in Dunham’s expression for Y11 ∼ −αe . Note that this term causes B(v) to increase as v increases. This is puzzling because everyone knows that B(v) decreases for real diatomics. The second order corrections are E (2) = � �� � � v, J|H(1) |v � , J v � , J|H(1) |v, J (0) (0) E(v,J) − E(v� ,J) v� = � v (1.25) Note that for each of the three allowed off-diagonal matrix elements of equations 18-20, there will be two nonzero terms in the summation over v � . From the �v|H|v ± 3� matrix elements we get � � a2 v(v − 1)(v − 2) − (v + 1)(v + 2)(v + 3) (2) E±3 = 3hν 8γ 3 2 a (2) E±3 = − 3 [(3v 2 + 3v + 2)] 8γ hν (0) (1.26) (1.27) (0) The 3hν in the denominator of (26) comes from Ev,J − Ev±3,J . From the �v|H|v ± 2� matrix elements we get (2) 9Be4 µ2 J 2 (J + 1)2 [(v − 1)v − (v + 1)(v + 2)] 2hν γ 2 �4 � � 18Be4 J 2 (J + 1)2 v + 21 9Be4 µ2 J 2 (J + 1)2 [2v + 1] = − =− γ 2 �4 · hν (hν)3 E±2 = (2) E±2 (1.28) Page 4 This is a harmonic vibrational correction term to a centrifugal distortion constant. This will not agree with Dunham’s result because we only kept terms in the expansion of B in powers of Rxe through the second power. This second order correction to the energy is actually a 2 × 2 = 4th power correction in � �4 x also contribute. Re x Re . Terms in the B expansion through From the �v|H|v ± 1� matrix elements we obtain (2) 9a2 (v 3 − (v + 1)3 ) 4Be2 J 2 (J + 1)2 [v − (v + 1)] 12aBe J(J + 1)[v 2 − (v + 1)2 ] − + (2γ)3 hν 2γRe2 hν Re (2γ)2 hν � � � 2 � � � Be J(J + 1) v + 12 9 a 2Be2 J 2 (J + 1)2 a 2 +6 =− (3v + 3v + 1) − hνγ 2 Re 8 hνγ 3 γRe2 hν E±1 = (2) E±1 (1.29) The first term of (29) should be added to a similar term which occurs in equation (27). The sum will be the first term in equation (30) below. The second term may be simplified using µ 1 = hν γ�2 giving − 4Be3 J 2 (J+1)2 (hν)2 and 1 Be 2µ = Re2 �2 which should be compared with Dunham’s Y02 . This term is the harmonic oscillator contribution to the centrifugal distortion constant. In summary � 2 � � � �� � � � 6B J(J+1) v+ 1 �2 18Be4 J 2 (J+1)2 (v+ 12 ) ( 2 ) 4Be3 J 2 (J+1)2 e a 15 7 a + hνγ v + 12 + 60 − − E (2) = − hνγ 3 2 4 Re (hν)2 (hν)3 (1.30) � �2 The first term is an anharmonic contribution to ωe xe v + 12 and to the zero point energy. In fact, this is why ωe xe is called the anharmonicity constant. The second term is an anharmonic correction to the rotational energy. Note that this correction is of the same � � sign as the harmonic correction we obtained in first-order in equation (24). B = Be − αe v + 12 . However the sign of a is negative for realistic potential curves. If |a| is large enough, the negative anharmonic contribution to αe will be larger than the positive harmonic contribution and B(v) will decrease as v increases. The third term is the harmonic contribution to centrifugal distortion. EJ = BJ(J + 1) − DJ 2 (J + 1)2 . � � The fourth term is a harmonic vibrational correction to the centrifugal distortion constant D = De − βe v + 12 . 2. Energy Levels of a Vibrating Rotor: Dunham’s Expression for E(v, J ) Derived from E(r) Following is an excerpt from Microwave Spectroscopy by C. Townes and A. Schawlow, pages 9–11, which describes the results of Dunham’s inversion of the potential energy, V (r), expressed as a power series in the dimensionless displacement coordinate ξ, into a power series in the rotational and vibrational quantum numbers, J(J + 1) and (v + 1/2). The Rydberg-Klein-Rees procedure is exactly the reverse of this, converting E(v, J) into V (r). Dunham’s Solution for Energy Levels Dunham1 has calculated the energy levels of a vibrating rotor, by a Wentzel-Kramers-Brillouin method, for any potential which can be expanded as a series of powers of (r − re ) in the neighborhood of the potential minimum. This treatment shows that the energy levels can be written in the form Page 5 EvJ = � � Y�j �,j 1 v+ 2 �� J j (J + 1)j (2.1) where � and j are summation indices, v and J are, respectively, vibrational and rotational quantum numbers, and Y�j are coefficients which depend on molecular constants. The effective potential function of the vibrating rotor may be written in the form U = a0 ξ 2 (i + a1 ξ + a2 ξ 2 + . . . ) + Be J(J + 1)(1 − 2ξ + 3ξ 2 − 4ξ 3 + . . . ) (2.2) where ξ = (r − re )/re , Be = h/8π 2 µre2 . The term involving Be J(J + 1) allows for the influence of the rotation on the effective potential. Examination of the Harmonic Oscillator part of the potential energy, U, will give a value for a0 . U = 1/2k(r − re )2 = 2π 2 µωe2 (r − re )2 = 2π 2 µωe2 ξ 2 re2 if Be ≡ h 8π 2 µre2 then U = ∴ a0 = h hωe2 2 ξ 4Be ωe2 4Be Dunham1 shows that the first 15 Yli ’s are Y00 = Be /8(3a2 − 7a21 /4) Y10 Y20 ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ 2 2 2 ⎪ = ωe [1 + (Be /4ωe )(25a4 − 95a1 a3 /2 − 67a2 /4 ⎪ ⎪ ⎪ ⎪ ⎪ 2 4 ⎪ + 459a1 a2 /8 − 1155a1 /64)] ⎪ ⎪ ⎪ ⎪ ⎪ 2 2 2 = (Be /2)[3(a2 − 5a1 /4) + (Be /2ωe )(245a6 − 1365a1 a5 /2 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 2 2 3 ⎪ − 885a2 a4 /2 − 1085a3 /4 + 8535a1 a4 /8 + 1707a2 /8 ⎪ ⎪ ⎪ ⎪ ⎪ 3 2 2 ⎬ + 7335a1 a2 a3 /4 − 23, 865a1 a3 /16 − 62, 013a1 a2 /32 + 239, 985a41 a2 /128 − 209, 055a61 /512)] Y30 Y40 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 2 2 2 ⎪ = (Be /2ωe )(10a4 − 35a1 a3 − 17a2 /2 + 225a1 a2 /4 ⎪ ⎪ ⎪ ⎪ ⎪ 4 ⎪ − 705a1 /32) ⎪ ⎪ ⎪ ⎪ ⎪ 3 2 2 ⎪ = (5Be /ωe )(7a6 /2 − 63a1 a5 /4 − 33a2 a4 /4 − 63a3 /8 ⎪ ⎪ ⎪ ⎪ ⎪ 2 3 3 ⎪ + 543a1 a4 /16 + 75a2 /16 + 483a1 a2 a3 /8 − 1953a1 a3 /32⎪ ⎪ ⎪ ⎪ ⎭ 2 2 4 6 − 4989a1 a2 /64 + 23, 265a1 a2 /256 − 23, 151a1 /1024) (2.3) Page 6 Y01 = Be {1 + (Be2 /2ωe2 )[15 + 14a1 − 9a2 + 15a3 − 23a1 a2 Y11 ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ + 21(a21 + a13 )/2]} ⎪ ⎪ ⎪ ⎪ ⎪ 2 2 2 = (Be /ωe ){6(1 + a1 ) + (Be /ωe )[175 + 285a1 − 335a2 /2 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 2 ⎪ + 190a3 − 225a4 /2 + 175a5 + 2295a1 /8 − 459a1 a2 ⎪ ⎪ ⎪ ⎪ ⎪ 2 + 1425a1 a3 /4 − 795a1 a4 /2 + 1005a2 /8 − 715a2 a3 /2 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 3 2 2 ⎪ + 1155a1 /4 − 9639a1 a2 /16 + 5145a1 a3 /8 ⎪ ⎪ ⎪ ⎪ ⎪ 2 3 ⎬ + 4677a1 a2 /8 − 14, 259a1 a2 /16 + 31, 185(a41 + a15 )/128]} ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 3 2 ⎪ Y21 = (6Be /ωe )[5 + 10a1 − 3a2 + 5a3 − 13a1 a2 ⎪ ⎪ ⎪ ⎪ ⎪ 2 3 ⎪ + 15(a1 + a1 )/2] ⎪ ⎪ ⎪ ⎪ ⎪ 4 3 Y31 = (20Be /ωe )[7 + 21a1 − 17a2 /2 + 14a3 − 9a4 /2 + 7a5 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 2 2 + 225a1 /8 − 45a1 a2 + 105a1 a3 /4 − 51a1 a4 /2 + 51a2 /8⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 2 3 2 ⎪ − 45a2 a3 /2 + 141a1 /4 − 945a1 a2 /16 + 435a1 a3 /8 ⎪ ⎪ ⎪ ⎪ ⎭ 2 3 4 5 + 411a1 a2 /8 − 1509a1 a2 /16 + 3807(a1 + a1 )/128] ⎫ Y02 = −(4Be3 /ωe2 ){1 + (Be2 /2ωe2 )[163 + 199a1 − 119a2 + 90a3 ⎪ ⎪ ⎪ ⎪ ⎪ − 45a4 − 207a1 a2 + 205a1 a3 /2 − 333a21 a2 /2 + 693a21 /4 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 2 3 4 ⎪ + 46a2 + 126(a1 + a1 /2)]} ⎪ ⎪ ⎪ � � ⎬ � � � 19 4 3 2 + 9a1 + 9a1 /2 − 4a2 Y12 = − 12Be ωe ⎪ 2 ⎪ ⎪ ⎪ 5 4 ⎪ ⎪ Y22 = −(24Be /ωe )[65 + 125a1 − 61a2 + 30a3 − 15a4 ⎪ ⎪ ⎪ ⎪ 2 2 2 ⎪ + 495a1 /4 − 117a1 a2 + 26a2 + 95a1 a3 /2 − 207a1 a2 /2 ⎪ ⎪ ⎪ ⎪ ⎭ 3 4 + 90(a1 + a1 /2)] (2.4) (2.5) Y03 = 16Be5 (3 + a1 )/ωe4 Y13 Y04 ⎫ ⎪ ⎪ ⎬ = (12Be6 /ωe5 )(233 + 279a1 + 189a12 + 63a31 − 88a1 a2 − 120a2 + 80a3 /3) ⎪ ⎪ ⎭ = (64Be7 /ωe6 )(13 + 9a1 − a2 + 9a21 /4) (2.6) � It should be noted that Be is generally much smaller than ωe . For most molecules the ratio Be2 ωe2 is of the order of 10−6 , although for light molecules such as H2 it approaches more nearly to 10−5 . In such cases more terms are required in the expressions for the various coefficients. If Be /ωe is small, the Y ’s can be related to the ordinary band spectrum constants as follows: Y10 ≈ ωe Y20 ≈ −ωe xe Y30 ≈ ωe ye Y01 ≈ Be Y11 ≈ −αe Y21 ≈ γe Y02 ≈ −De Y12 ≈ −βe Y40 ≈ ωe ze Y03 ≈ He (2.7) Page 7 where these symbols refer to the coefficients in the Bohr theory expansion for the molecular energy levels: � FvJ = ωe 1 v+ 2 � � �2 � �3 � �4 1 1 1 − ω e xe v + + ωe ye v + + ωe ze v + 2 2 2 +Bv J(J + 1) − De J 2 (J + 1)2 + He J 3 (J + 1)3 + . . . (2.8) � � � �2 where Bv = Be − αe v + 12 + γe v + 12 . . . (cf.2 , p. 92, pp. 107-108). Sandeman3 has extended Dunham’s treatment to include other terms of the same order of magnitude which involve higher powers of the vibrational quantum number. For the special case of the Morse potential function, Dunham shows that all the Y�0 ’s except Y10 and Y20 vanish and all but the first terms in the expressions for Y10 and Y20 are zero. Because of the simplicity of the expressions obtained with the Morse function, and because it does give a quite good fit to the actual potential in the region of r = re , the Morse function has been widely used. Several important relationships between constants have been derived for Morse potential functions. These are often useful for estimating otherwise unknown parameters. The Kratzer relation De = 4Be3 ωe2 . The Pekeris relation αe = � �1/2 6 �� − Be2 ω3 xe Be3 ωe The constant Y00 is exactly zero for Harmonic and Morse oscillators, but an approximate value for general oscillators is � �2 Be − ωe xe αe ωe αe ωe 1 Y00 = + + . 12Be Be 4 12Be