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One of the main advantages of this formalism is that it is independent of the form of the one-particle Hamiltonians He and Hh, and indeed calculations can be performed on any system in which the standard electron and hole wave functions can be calculated [77,132,133]. Multiplying equation (6.13) on the left by w and integrating over all space, then the total exciton energy follows simply as the expectation value:

assume sxXk = (Zi:k^i<N:a[i]xXi) + a[k-l]xX ~ }

The neighborhood function is usually a function of the distance between the coordinates of the neurons as represented on the map, i.e.

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Now, whatever the form of the relative motion term Ur of equation (6.14), it will be a function of the electron-hole separation, which is quantified in terms of the three variables, x = xe Xh\, y = \ye yh\ and a = \ze Zh . The denominator D in equation (6.16) can therefore be written as:

S = sxX + a[k-l] . We have thus determined that A sxXk = (Ei:k^i<N:a[i]xXi) A SxX ~ = (Zi So the loop body is k,s := k-l ,sxX + a[k-l] and the complete algorithm is as follows: k,s := N,0 ; { Invariant: O^k^N k } A sxXk = Bound function:

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Since Ue and Ul are functions of z only, then the integrations over the x y plane will involve only the relative motion term Ur. Correspondingly, the integrations over the electron and hole coordinates, ze and Zh will yield a result which is a function of the electron and hole separation (a) only, i.e. it is possible to write:

A 5 = sxX + a[k-l]

hmn,kj(t) = h(\\Cmn - Ckj\\2,t)

Although the one-particle wave functions, Ue and Uh, in principle, extend to 00, in practice, from a computational viewpoint, it is necessary to define them between 'effective infinities' . As discussed in 3, these are defined so that:

do k > 0 k,s := k-l ,

and similarly for Uh. Using these results, the denominator D in equation (6.17) can be re-written as:

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{ s = (Zi:0^ Note that this derivation of the algorithm constitutes a formal proof of Homer's rule. Note also how the calculation of S (as opposed to guessing what it should be) avoids making a 'one-off error. With problems like this one, it is very easy for array indices to be 'one-off. For example, we might have guessed the value sxX + a[k] for S. The consequences are often noticed immediately, but not always. And, in the digital world, a small error of this nature can be disastrous! Exercise 13.13. Verify the correctness of the initialization and the termination condition. D

with the coordinates Cmn,Ckj G K2. With increasing value of ||c^ CjH2, (that is, neuron kj is further away from the winning neuron mn), hmn,kj > 0. The neighborhood can be defined as a square or hexagon. However, the smooth Gaussian kernel is mostly used: hmn,kj(t) = n ( t ) e 2-2 > (4.20)

where the Dirac S-functions have been introduced in order to ensure that contributions are included from both ze Zh = a and Zh ze = a. If the integration over Zh is performed first, then the first term only has a finite value when Zh = ze a and the second term when Zh = ze + a; hence:

The problem of evaluating XM for M ^ 0 seems a trivial one but, in fact, has been studied for thousands of years and an algorithm that is optimal for all values of M is not known. This subsection discusses a number of algorithms for this purpose. Elementary Algorithm. The simplest solution is to use the invariant

Making the substitution z = ze a in the first term, and relabelling z as ze in the second, finally gives the denominator in equation (6.16) as:

in a loop that initializes k to 0 and continually increments it until it equals M:

where rj(t) is the learning rate factor and cr(t) is the width of the kernel. Both n(t) and a(t) are monotonically decreasing functions. The learning process is iterative, continuing until a "good" enough map has been found. The quantization error is usually used as an indication of map accuracy, defined as the sum of Euclidean distances of all patterns to the codebook vector of the winning neuron, i.e.

the Hamiltonian takes a block-diagonal form:

{ (KM }

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