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Type your note text. When you finish, click the note s Close button ( ). Outlook adds the note to the Notes list. To view the note again or to make changes, you can doubleclick the note to reopen it.

Fig. 14.6 Domain walls in the two-dimen-

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Consider an arbitrary configuration in a two-dimensional square lattice A domain wall is a continuous line drawn between up (+) spins and down (- ) spins The length b of the wall is the number of lattice spacings it traverses To make the definition unique, we draw a domain wall in a particular sense, such that the - spins always lie to the right of the wall, and the + spins to the left Where there is still an ambiguity, the domain wall will bend to the right An example is shown in Fig 146 By this definition, no two domain walls ever cross Some are close curves, while others begin and end on the boundary of the lattice In the set of all possible domain walls, two walls of the same shape but differ in location are counted as distinct.

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Also distinct are two walls of identical shape and location but of different sense (In the latter case, both cannot occur in the same configuration, of course) Now we impose the boundary condition that all spins on the boundary are + This simulates an external field, whose influence should become arbitrarily weak in the limit of an infinite lattice The boundary condition destroys the up-down symmetry of the model To show that spontaneous magnetization exists we shall show that, at a sufficiently low but finite temperature, the average fraction of down spins is less than ~ With our boundary condition, every domain wall is a closed curve Consider the set of all closed domain walls We classify them according to length b, and within a class of given length we give each a number i Thus any domain wall is uniquely characterized by a label (b, i).

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Let us first state some properties of the domain walls A wall of length b is a closed polygon made up of b vertical or horizontal segments of unit length If these segments were rearranged so that the closed figure resembles a square as much as possible, it will fit inside a square of side b/4 Therefore the area enclosed by a domain wall of length b is less than b 2/16 Let m (b) be the number of domain walls of length b To find a bound for m (b), imagine constructing a continuous line on a large square lattice of N sites, by successively laying down b sticks of unit length, either vertically or horizontally The number of distinct ways to do this is greater than m( b) because.

FIGURE 6.27 A Percentage Comparison is the default for the Comparison function, since it s typically the most useful way to compare various securities, irrespective of their price.

we did not require the line be closed. The first stick can go in N positions, and subsequent ones have three choices corresponding to the three possible directions: up, down, or horizontal. (For large N we can ignore constraints imposed by the boundary condition.) Therefore (14.25) Now, because the spins on the boundary are all +, every - spin is enclosed by at least one domain wall. In a particular configuration let

X(b, i) = { 0

(if the domain wall (b, i) occurs in that configuration) (otherwise)

m(b)

(14.26)

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