17s_engylvlstrct

17s_engylvlstrct - MIT OpenCourseWare http://ocw.mit.edu...

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Unformatted text preview: 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 . Lecture #17 Supplement Contents A. Energy Level Structure of 2 Π and 2 Σ States: Case (a) and Case (b) Limits . . . . . . 1 B. Matrix Elements for 2 Π and 2 Σ States Including Π ∼ Σ Perturbation . . . . . . . . . 4 C. Parity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 A. Energy Level Structure of 2 Π and 2 Σ States: Case (a) and Case (b) Limits The 2 Π Hamiltonian matrix was shown to be 2 Π ± 1 / 2 A + B ( x 2 − 2) B ( x 2 − 1) 1 / 2 3 / 2 2 Π ± B ( x 2 − 1) 1 / 2 − 1 / 2 A + Bx 2 1 / 2 x ≡ J + 2 1 Case ( a ) limit | A − 2 B | Bx Use second-order perturbation theory to express the e ff ect of the o ff-diagonal matrix element on the energy. The second order correction is valid when | A − 2 B | Bx . ( Δ E V ). E 3 / 2 = 1 A + B ( x 2 − 2) + B 2 ( x 2 − 1) 2 A E 1 / 2 = − 2 1 A + Bx 2 − B A 2 ( x 2 − 1) Notice that if we define three new e ff ective parameters the energy level expressions take on particularly 1 5.80 Lecture #17 Supplement Page 2 simple forms. A e ff = A − 2 B B 2 B 3 / 2 e ff = B + A B 2 B 1 / 2 e ff = B − A Thus E 3 / 2 = 1 A e ff + B 3 / 2 e ff ( x 2 − 1) 2 E 1 / 2 = − 1 A e ff + B 1 / 2 e ff ( x 2 − 1) 2 Thus we have two stacks of energy levels for which the extrapolated x = 1 levels are symmetrically split on either side of E Π and whose e ff ective rotational constants (just like for 1 Σ molecule) can be averaged to give the true B Π rotational constant. A second limiting case is obtained when A = 0. This is the case ( b ) limit. If A = 0, we must diagonalize the 2 Π Hamiltonian matrix. We solved the problem for two levels separated by 2 d , with an o ff-diagonal matrix element V connecting them, and for an average energy E o . In the present problem E o = E Π − B + Bx 2 d = − B V = B ( x 2 − 1) 1 / 2 V 2 1 / 2 E ± = E o ± d 1 + d ⎡ 2 ⎤ 1 / 2 E ± = E Π + B ( x 2 − 1) ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ 1 − B ( x 2 − 1) 1 / 2 ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ − B = E Π + B ( x 2 − 1) Bx Observe the di ff erence in energy between an E + level for x → x + 1 and E − level for x → x . (Note the ± does not refer to parity in this problem.) E + ( x + 1) − E − ( x ) = B ( x 2 + 2 x + 1 − x − 1 − 1 − x 2 − x + 1) = 0 Thus pairs of levels corresponding to “ Ω = 3 / 2”, J + 1 and “ Ω = 1 / 2”, J are exactly degenerate. If A is not quite zero, then these pairs will be slightly split and the splitting is called the spin-splitting. At high 5.80 Lecture #17 Supplement Page 3 enough J , a 2 Π molecule will always develop this case ( b ) energy level pattern. The spacing between these degenerate pairs of levels is E + ( J + 1) − E + ( J ) = B ( x 2 + 2 x + 1 − 1 − x − 1 − x 2 + 1 + x ) = 2 Bx Note that the spacing between successive rotational levels of a...
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This note was uploaded on 11/28/2011 for the course CHEM 5.74 taught by Professor Robertfield during the Spring '04 term at MIT.

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17s_engylvlstrct - MIT OpenCourseWare http://ocw.mit.edu...

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