20_580ln_fa08

20_580ln_fa08 - MIT OpenCourseWare http/ocw.mit.edu 5.80...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
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 .
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
5.80 Lecture #20 Fall, 2008 Page 1 of 6 pages Lecture #20: Transformations between Basis Sets: 3-j, 6-j, and Wigner-Eckart Theorem Last time: effects of “Remote Perturbers”. What terms must we add to the effective H so that we can represent all usual behaviors with minimum number of parameters. Today: A taste of spherical tensor algebra. Suppose we want to evaluate matrix elements of H SO for atoms. We have a choice between two basis sets: coupled | L S J M J uncoupled | L M L S M S If we take the simplified form of H SO H SO = ζ (nLS) L S = ζ ( nLS ) L z S z + 1 2 ( L + S + L S + ) , we can set up a matrix for H SO in either the coupled or uncoupled basis. One basis is more convenient than the other but all necessary matrix elements are explicitly evaluable because all of the quantum numbers we need to evaluate the matrix elements appear explicitly in either basis set. But H SO , J 2 = 0 H SO , J z = 0 H SO , L z 0 H SO , S z 0 . This means that H SO is fully diagonal in | L S J M J but massively off-diagonal in | L M L S M S . We see that H SO is diagonal in | LSJM J here: J = L + S J 2 = L 2 + S 2 + 2 L S L S = 2 1 ( J 2 L 2 S 2 ) LSJM J H SO LSJM J = 1 2 ζ ( nLS ) [ J(J + 1) L(L + 1) S(S + 1) ] . There are cases when it is not possible to evaluate matrix elements in the “wrong” basis set because the necessary quantum numbers do not appear explicitly in the basis set. A famous example is the Zeeman Hamiltonian that cannot be expressed in the coupled basis set, which is best for H SO .
Background image of page 2
5.80 Lecture #20 Fall, 2008 Page 2 of 6 pages H Zeeman = µ Bohr [ L Z + 2 S Z ] B Z (B Z is magnetic field) LSJM J L Z L SJ = ? M
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

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.

Page1 / 7

20_580ln_fa08 - MIT OpenCourseWare http/ocw.mit.edu 5.80...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online