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By Percy Williams Bridgman

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0)t . Since T is translation-invariant, Bloch-states are eigenstates of T. )t / N √ √ We get x|q = eiqx / N , q|x = e−iqx / N , x|x = x|q q|x = q 1 N eiq(x−x ) = δx,x q Proof that |q is an eigenstate of T: x|T |q = x|T |x x = x |q √ δx ,x+1 + δx ,x−1 eiqx / N x √ √ = eiq(x+1) + eiq(x−1) / N = eiq + e−iq eiqx / N = (2 cosh q) x|q ≡ T (q) x|q Now we can compute 0|W (l)|0 = 0|T l |0 = 0|T l |q q|0 q T l (q) 0|q q|0 = = q 1 N T l (q) q This step uses that the Bloch states are also eigenstates of the product Tl .

Every second spin: ↑·↑·↑·↓·↓·↑· ↓·↑·↓·↑·↓·↑· ↓·↑·↑·↓·↑·↓· ↑·↓·↑·↓·↓·↑· This list contains only 2N/2 configurations. The relative weights of one of these configurations follows from combining the weights of all the configurations in the first list which correspond to the given configuration in the second (these differ only in the spins which were disregarded in going from the first list to the second). However, the second list, and the corresponding statistical weights, can also be regarded as a statistical ensemble in its own right, that is, a new ensemble of spins s , whose statistical weights are described by a new Hamiltonian H [s ], with new couplings, fields etc.

For the choice with p˜(˜ x, t˜) = p(x(˜ x), t(t˜)). D ˜ z = 2, D = D. 2: One example for the ballistic deposition model. Particle A is released randomly and sticks at the first site where it has an occupied nearest neighbor. Particle B can fall to the ground. 1 A simple model: Ballistic deposition (BD) The following growth models work a bit as tetris: We release a particle from a randomly chosen point above the surface. In the BD-model with the nearest neighbor sticking rule, the particle sticks to the first site along its trajectory that has an occupied nearest neighbor.

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