Standard tableaux

In a tableau corresponding to a partition of n, there are, of course, n different arrangements of the way the first n integers may be entered. Among these there is a subset that Young called standard tableaux. These are those for which the numbers in any row increase to the right and downward in any column. Thus, we have for 3,2  

The operators Vi and Hi corresponding to T, are, of course, different and, in fact, have no permutations in common other than the identity. The first important result is that 

This is so because there is some pair of numbers appearing in the same row of 7) that must appear in the same column of J), if it is later. To see this suppose that the entries in the tableaux are (THi and (Tj)ki, where k and I designate the row and column in the shape. Let the first difference occur at row m and column n. Thus, Tj )m Ti)mn, but Ti)mn must appear somewhere in Tj. Because of the way standard tableaux are ordered it must be Tj)m n, where m m and n n. Now, also by hypothesis, Tj)mn = (Ti)mn, since this is in the region where the two are the same. Therefore, there is a pair of numbers in the same row of 7) that appear in the same column of Tj. Calling these numbers p and q, we have 

One should not conclude, however, that T yA/i =HVj = OifT) Ty. Although true in some cases, we see that it does not hold true for the first and last of the tableaux above. No pair in a row of the last is in a column of the first. In fact, the nonstandard tableau 

This would also be true for the operators written in the other order. 

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A standard tableau is defined as one in which the numbers increase when one reads from left to right in each row and from top to bottom in each column. It can be shown Mi) that the dimension of a representation is equal to the number of standard tableaux associated with the corresponding diagram. The reader is also referred to the literature Mi) for methods of calculating characters of the representations from the diagrams. 

We stated above that there is an inequivalent irreducible representation of S associated with each partition of n, and we use the symbol fx to represent the number of standard tableaux corresponding to the partition, X. Using induction on n. Young proved the theorem... 

We will choose arbitrarily to work with the first of the standard tableaux of a given partition. With this we can form the two Hermitian algebra elements... 

The general result states that the number of linearly independent functions from the set UTT, S(y) i = 1,..., / is the number of standard tableaux with repeated elements that can be constructed from the labels in the H product. As a general principle, this is not so easy to prove as some of the demonstrations of linear independence we have given above. The interested reader might, however, examine the case of two-column tableaux with which we are concerned. Examining the nature of the tt, for this class of tableau, it is easy to deduce the result using ffVff. This is all that is needed, of course. The number of linearly independent functions cannot depend upon the representation. 

Of these two schemes, it appears that the standard tableaux functions have properties that allow more efficient evaluation. This is directly related to the occurrence of the J f on the outside of OAfVAf. Tableau functions have the most antisymmetry possible remaining after the spin eigenfunction is formed. The HLSP functions have the least. Thus the standard tableaux functions are closer to single determinants, with their many properties that provide for efficient manipulation. Our discussion of evaluation methods will therefore be focused on them. Since the two sets are equivalent, methods for writing the HLSP functions in terms of the others allow us to compare results for weights (see Section 1.1) of bonding patterns where this... 

We note finally that if P = for a particular product function the standard tableaux function and HLSP function are the same. 

Transformations between standard tableaux and HLSP functions... 

Since the standard tableaux functions and the HLSP functions span the same vector space, a linear transformation between them is possible. Specifically, it would appear that the task is to determine the a,y s in... 

The perceptive reader may already have observed that the functions we use can take many forms. Consider the non-Hermitian idempotent f/g)VJf. Using the permutations interconverting standard tableaux, one finds that (f/g) PMitj S i = 1,...,/ is a set of linearly independent functions (if S has no double occupancy). Defining a linear variation function in terms of these. 

After this digression we now return to the problem of determining the HLSP functions in terms of the standard tableaux functions. We solve Eq. (5.118) by... 

The tableaux in the last paragraph are, of course, not unique. In any row either orbital could be written first, and any order of rows is possible. Thus, there are 2 X 3 = 48 different possible arrangements for each. We have made them unique by setting a[Pg.91]

The element is a projector for the first component of the irreducible representation basis. Using standard tableaux functions we can select a function of a given symmetry and a given spin state with... 

The S5munetry standard tableaux functions are not always so intuitive as those in the first case we looked at. Consider, e.g., the configuration Islp lp lpfisalsb, for which there are two standard tableaux and no other members in the constellation. 

To obtain the symmetry functions in terms of HLSP functions we can transform the standard tableaux functions using the methods of Chapter 5. The transformation matrix is given in Eq. (5.128) ... 

There are two linear combinations of the standard tableaux functions that comprise a pair of E S5munetry. The E projectors are... 

In Chapter 15 we give an extensive treatment of the it system of benzene, but here we outline briefly some ofthe symmetry considerations. We consider the configuration P P2P3 Pa Pi Pe, where Pi stands for a C2 p orbital at the / C atom, numbered sequentially and counterclockwise around the ring. The set of five standard tableaux is... 

The transformation from standard tableaux functions to HLSP functions is independent of the spatial symmetry and so we need the 4-matrix in Eq. (6.24) again. This time the results are... 

The calculations in this illustration were not done with a minimal basis set, since, if such were used, they would not show the correct behavior, even qualitatively. This happens because we must represent both F and F in the same wave function. Clearly one set of AOs cannot represent both states of F. Li does not present such a difficult problem, since, to a first approximation, it has either one orbital or none. The calculations of Fig. 8.4 were done with wave functions of 1886 standard tableaux functions. These support 1020 s mimetry functions. We will discuss the arrangement of bases more fully in Chapter 9. 

This is the number of linearly independent standard tableaux or Rumer functions that the entire basis supports. 

It is instmctive to examine the symmetry of the standard tableaux function of highest EGSO population given in Table 10.2. The effects of the two symmetry... 

Our ability to represent the wave flinetion for allyl as one standard tableaux function should not be considered too important. If we had ordered our 2/7 orbitals differently with respect to particle labels, there are cases where the A2 function would require using both standard tableaux functions. 

Since there are only four terms, we give the whole wave function for the smallest calculation. In terms of standard tableaux functions one obtains... 

The reader will recall that a given configuration has different standard tableaux functions and HLSP functions if and only if it supports more than one standard tableaux function (or HLSP function). 

The normalization and overlap integrals of the two standard tableaux functions may be written as a matrix... 

The SCVB method can also be used to study the tt system of the allyl radical. As we have seen already, only one of the two standard tableaux ffinctions is required because of the symmetry of the molecule. We show the results in Table 10.4, where we see that one arrives at 85% of the correlation energy from the largest MCVB calculation in Table 10.2. There is no entry in Table 10.4 for the EGSO weight, since it would be 1, of course. 

The basis described was used to generate one Is occupied and four virtual RHF orbitals. Using these afull calculation yields 250 standard tableaux functions, which may be combined into 125 functions of symmetry. The results for energy, bond distance, and vibrational frequency are shown in Table 10.5. We see that the agreement for is within 0.1 eV, for Re is within 0.01 A, and for cOe is within 20 cm . Even at the equilibrium nuclear separation, the wave function is dominated... 

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