Download Secular Equation - Advanced Quantum Chemistry and Spectroscopy - Lecture Slides and more Slides Chemistry in PDF only on Docsity! ⎟⎟ ⎟ ⎟ ⎟ ⎠ ⎞ ⎜⎜ ⎜ ⎜ ⎜ ⎝ ⎛ = ⎟⎟ ⎟ ⎟ ⎟ ⎠ ⎞ ⎜⎜ ⎜ ⎜ ⎜ ⎝ ⎛ ⎟⎟ ⎟ ⎟ ⎟ ⎠ ⎞ ⎜⎜ ⎜ ⎜ ⎜ ⎝ ⎛ ∴ NNNNNNN N N c c c m c c c MMMM MMMM MMMM MM K MMMMM K K 2 1 2 1 321 2232221 1131211 (3) written in long form: ( ) ( ) ( ) ( ) 0: 0:3 0:2 0:1 332211 3333232131 2323222121 1313212111 =−++++= =++−++= =+++−+= =++++−= NNNNNN NN NN NN cmMcMcMcMNk cMcmMcMcMk cMcMcmMcMk cMcMcMcmMk L M L L L docsity.com This can be written as: ( ) 0~~~ =− cImM I is an N x N unit matrix where Iij = δij 0 1000 0010 0001 2 1 321 2232221 1131211 = ⎟⎟ ⎟ ⎟ ⎟ ⎠ ⎞ ⎜⎜ ⎜ ⎜ ⎜ ⎝ ⎛ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ ⎟⎟ ⎟ ⎟ ⎟ ⎠ ⎞ ⎜⎜ ⎜ ⎜ ⎜ ⎝ ⎛ − ⎟⎟ ⎟ ⎟ ⎟ ⎠ ⎞ ⎜⎜ ⎜ ⎜ ⎜ ⎝ ⎛ ⇒ NNNNNN N N c c c m MMMM MMMM MMMM M K MKMMM K K K MMMMM K K Note: if m is known, can solve (3) for c1 , c2 ,…,cN . Get a non-trivial solution only if the determinant of the coefficients of the unknown {ci } is zero (4) In matrix form: 0~~ =− ImM ( ) ( ) 0 321 2232221 1131211 = − − NNNNN N N MMMM MMmMM MMMmM K MMMMM K K This is called a SECULAR EQUATION for the eigenvalues m; that is, it yields the m’s. docsity.com Procedure Given: 0~~ =− ImM = secular equation that yields mi 1.) Take one solution for m, say m1 and substitute into the set of equations given by: ( ) )1(0~~~ =− cImM Solve for cn1 , n = 1,2,…,N Since the solutions to (1) can only be solved within a constant; that is, for cn1 /c11 , the expansion coefficients can be completely specified by requiring Ψ1 =Σn=1,..,N cni Ωn to be normalized; that is, <Φi |Φi > = 1 = Σn=1,..,N |cn1 |2. Repeat steps above for next mj , solving for the {cnj }, and requiring Φj to be normal. The procedure is repeated for j = 1,2,…,N. docsity.com 2.4: Time-independent degenerate perturbation theory A. Getting the solution in principle: Procedure is effectively similar to non- degenerate case but needs modification for 2 reasons. 1.) Ek(0) – Eq(0) can not be zero when we want to calculate ak in the spectral expansion 2.) If there are 2 or more states with the same energy we don’t know which state will arise in the expansion Recall: .... .... )2(2)1()0( )2(2)1()0( +++= +++= qqqq qqqq EEEE ψλλψψψ λλ docsity.com As λ → 0, Eq →Eq(0) and Ψq → Ψq(0) But which state if there more than one with the same Eq(0)? In general therefore )0( 0 lim j j jq c ψψλ ∑=→ The sum is over the degenerate states, where the degeneracy is labeled g. To solve this problem, we need two pieces of information: a) superposition principle b) orthonormalization procedure docsity.com Thus, for the example above: 2121 22111 )0( 2 )0( 1 2121 2 2 2 1 121 * 2212 * 1222 * 211 * 11 22112211 )0( 2 )0( 2 1 )0( 1 | |0|but |2 |||| |1|ii) i) ψψ ψψψφφ ψψ ψψψψψψψψ ψψψψφφ ψφ bb bb bbbb bbbbbbbb bbbb += +== ++= +++= ++== = This is 2 equations in 2 unknowns (b1 and b2 ) since <Ψi |Ψj > can be calculated 0|;0|;1| useSimilarly,iii) )0( 3 )0( 2 )0( 3 )0( 1 )0( 3 )0( 3 === φφφφφφ to generate 3 equations in 3 unknowns to solve for (c1 , c2 , c3 ). These will be our starting wave functions! For state q, we will call these states {Φq,j(0)} docsity.com Now we can 1.) generate a complete orthonormal set of wave functions by linear combinations of the form )0( , 1 )0( , jq g j jjq cψφ ∑ = = for all degenerate eigenvalues; that is, generate g Φ’s with g sets of coefficients. 2. Use this new set of wave functions in a perturbation problem. For the following be careful of the meaning of super- and subscripts Will use: j for the degenerate level, j = 1,2,…,g q for the level of interest k for all levels, k=1,2,…,00, degenerate or not. docsity.com
