MIT OpenCourseWare http://ocw.mit.edu 5.80 Small-Molecule Spectroscopy and Dynamics Fall 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.5.80 Lecture #9 Fall, 2008 Page 1 Lecture #9: The Born-Oppenheimer Approach to Transitions-Selection Rules -Relative Intensities First of 3 lectures illustrating simplest patterns in 3 main types of transitions - mostly for diatomicmolecules permanent µ pure rotation (microwave) ∆v ≡ 0 change in µ rotation-vibration (IR) ∆v ≈ ±1 electronic symmetries rotation-vibration-electronic (UV – VIS) ∆v = any How does the Born-Oppenheimer Approximation help us to predict what to expect in the spectrum? Begin reading Hougen monograph http://physics.nist.gov/Pubs/Mono115/contents.htmlChapter 6 of Bernath KEY TOPICS * Electric dipole transitions: e ∑ rα→ Mif (R) α vector electrons in body integrate over r * DIRECTION COSINES LAB XYZ → body xyz integrate over θ, φ, χ * Selection Rules symmetry and propensity * Hönl-London rotational linestrength factors sum over MJ Last time I was concerned with how to go H → H °+ H ′→ EevJ exact simplified missingexact stuff This was mostly formal. As spectroscopists we care much less about how to compute spectra ab initio than how to extract information from real spectra. The reason the Born-Oppenheimer approximation is so important is that it enables us to simplify our interpretation of spectra.5.80 Lecture #9 Fall, 2008 Page 2 It is very helpful to think of Eevr = Ti + Gi(v) + Fi,v(J) Φi(r;R) |ΩJM〉χi,v(R) Vi,J(R) and that all electronic properties vary slowly with R, and all observable quantities normally varysmoothly with v,J. All non-smooth variations should be explained by resonances in an energy denominator of aperturbation expansion. Expect to find patterns in spectra that can be represented as power series in (v + 1/2) and J(J + 1). BRUTE FORCE EivJ = Tei + ∑Y,m (v +1 / 2 ) [J(J +1) ]m ,m The Y,m are “molecular constants”. They are of no special importance except as intermediate step inEevr → Vi,J(R). For the present, we must concentrate on how to go from spectrum → Y,m. To do this we need to know what will appear in the spectrum:* selection rules * relative intensity patternselectric dipole transitions i vi ΩiJiMiPif ∝εL ·oscillating electric field in LAB µb body (dipole antenna)body-fixed coordinates of e– with respect to center of mass. fΩfJfMf vf µ = erα α ∑ 2zˆz)x 5.80 Lecture #9 Fall, 2008 Page 3 In the spirit of Born-Oppenheimer we get rid of all electronic coordinates by integrating over r. Only the electronic wavefunctions and ∑rα depend on r. α electrons  e i ∑rα f ≡ vector in Mif (R) bodyα r frame transition dipolemoment function xxˆ+ yyˆ + zzˆ(3 integrals) 3 Mif components next we integrate over θ,φ: the orientation of body z with respect to LAB XYZ (for polyatomics wewould need 3 Euler angles). z φ z θ DIRECTION COSINES εL·=( X + εYY + εZZ)· M xˆ+ M yˆ + Mµb εX  ( x y ·ˆ( )y X x ≡ cos X,xˆLb a 3 × 3 matrix LAB body ⎛ X ˆx Xyˆ Xˆz  Zˆz ⎞ ⎟⎟⎟ ⎜⎜⎜ α θ, φ)=( ⎝ ⎠ It requires 3 Euler angles to define XYZ with respect to xyz, but θ, φ are only 2 needed for a diatomicmolecule. 5.80 Lecture #9 Fall, 2008 Page 4 In order to specify r in both LAB and body, need one more angle. Phase choice — conventionally used in ab initio calculations. Unexpected result below. [Why do we care? Electronic coordinates. Nuclei are by definition on the z axis.] This is the transformation that relates LAB to body (fixed choice of x = π/2). Does not need to be Hermitian. Needs only to be unitary α–1 = α†. Check! Note that, when θ = φ = 0 (z along Z), we can see unexpected effect of arbitrary phase choice. 0 −1 0 ⎛ ⎞ OK, now we are ready to do the θ,φ integration. |〈θφ|ΩJM〉|2 is probability ofOnly factor in ε·Mif integral that depends on θ,φ is |ΩJM〉 finding z pointing in θφ direction with respect to XYZ. 〈ΩiJiMi|α⎝⎜⎜⎜xyzXYZ Lb|ΩfJfMf〉θ,φ ⎟ ⎟⎟⎠ i.e. Y↔x(0,0 1 0 0 0 )= X↔–y (extra rotation about Z by π/2) 0 1 direction cosine matrix elements MJ Selection rules BRANCH ∆J = 0, ±1 TYPE OF BAND ∆Ω = 0 for Mz,if ≠ 0 “parallel” or ∆Ω = ±1 for Mx,if or My,if ≠ 0 “⊥” POLARIZATION ∆MJ = 0 for εz ≠ 0 “π”-polarized, ∆M = ±1 for εx or εy ≠ 0 “σ” 2 5.80 Lecture #9 Fall, 2008 Page 5 So we have ΩfJfMfPif ∝∑ εS vi Mb,if (R) vf ΩiJiMi αSb S,b  R θ,φ sum inside of | |2 overall intensity polarization dependencerotational selection rules sub-band selection rule OK. Now let’s look at specific cases. Pure Rotation Spectrum i ≡ f vi ≡ vf 1Λ (only Ω = Λ) if we restrict consideration to singlet states, Ωi = Ωf ∆Ω = 0 ↔ Mz,ii ≠ 0 Mx,ii = My,ii = 0 αSb ME’s are only not so simpleproduct of 2possibility for polyatomics only one component oftransition moment is non-zero factors: bodyMoreover, if light is linearly polarized, we can choose Z as polarization axis, thenand LAB only εZ ≠ 0 ↔ ∆M = 0 So we have simplified it to Pif ∝ ε2 vi Mz,ii (R) vi ΩiJiMi αZz ΩiJfMi θφ R∝Intensity Next we consider selection rules for 2 factors in this equation. µ = er is odd with respect to i Body-fixed inversion: i [not LAB inversion: I] which defines parity in atoms i Φi(r;R) ≠ ± Φi(r;R) (i on total wavefunction) not a guaranteed symmetryexcept for homonuclear molecule g,u symmetry ** no pure rotation spectrum for homonuclear Mii(R) = 0** yes pure rotation spectrum for heteronuclear Mii(R) ≠ 0 25.80 Lecture #9 Fall, 2008 Page 6 reflection thru planecontaining internuclear axisAll diatomic molecules have σv(xz) and σv(yz) symmetry elements. This means that Mx,ii = My,ii = 0 for all diatomic molecules.  So, for diatomic molecule we have only one non-zero component of M(R) (unless homonuclear). Expand in powerThe Dipole Moment Function (Permanent) series about RedM (or some other) + convenient point) Mz,ii (R) = Mz,ii (Re dR R =Re (R − Re) + 1 d2M (R − Re )2 2 dR2 Q Q2 Now we can take vibrational matrix elements. = Mz,ii (Re) +