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Fig. 1.14 illustrates the effect that such diffraction would have on the 'free-electron' curves. At wavevectors which satisfy von Laue's condition, the energy bands are disturbed and an energy gap opens. Such an improvement on the parabolic dispersion curves of earlier, is known as the nearly free electron model. The space between the lowest wavevector solutions to von Laue's condition is called the first Brillouin zone. Note that the reciprocal lattice vectors in any particular direction span the Brillouin zone. As mentioned above a face-centred cubic lattice has a body-centred cubic reciprocal lattice, and thus the Brillouin zone is therefore a three-dimensional solid, which happens to be a 'truncated octahedron' (see, for example reference [1], p. 89). High-symmetry points around the Brillouin zone are often labelled for ease of reference, with the most important of these, for this work, being the k = 0 point, referred to as T', and the < 001 > zone edges, which are called the 'X' points.

the players in advance of the game). What property should the initial position satisfy for one of the players to always have a winning strategy D Exercise 12.3 (The Grid Game). In this game, players A and B are provided with a grid of any size (see Figure 12.4). On each turn, player A draws a solid horizontal line or a solid vertical line between two adjacent points in the grid. Player B draws a dashed horizontal line or a dashed vertical line between two adjacent points. One player cannot play in a place that the other has already played. Player A wins by completing a closed curve of solid lines; player B wins by preventing A from completing a closed curve. Who should win and what is the winning strategy D

B.1.2

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2.1 THE INFINITE WELL The infinitely deep one-dimensional potential well is the simplest confinement potential to treat in quantum mechanics. Virtually every introductory level text on quantum mechanics considers this system, but nonetheless it is worth visiting again as some of the standard assumptions often glossed over, do have important consequences for one-dimensional confinement potentials in general. The time-independent Schrodinger equation summarises the wave mechanics analogy to Hamilton's formulation of classical mechanics [22], for time-independent potentials. In essence this states that the kinetic and potential energy components sum to the total energy; in wave mechanics, these quantities are the eigenvalues of linear operators, i.e.

Exercise 12.4. Consider an urn filled with a number of balls each of which is either black or white. There are also enough balls outside the urn to play the following game. We want to reduce the number of balls in the urn to one by repeating the following process as often as necessary. Take any two balls out of the urn. If both have the same colour, throw them away, but put another black ball into the urn; if they have different colours, then return the white one to the urn and throw the black one away. Each execution of the above process reduces the number of balls in the urn by one; when only one ball is left the game is over. What, if anything, can be said about the colour of the final ball in the urn in relation to the original number of black balls and white balls Hint: denote the original number of black balls and white balls by bo and WQ, respectively. Let b and w be the number of black balls and white balls after an arbitrary number of executions. Then, on the next execution the values of b and w are changed according to the following rules: b := b-2 ; b := b + l if the balls are both black, w := w-2 ; b := b+l if the balls are both white, b := b-l if the balls are different colours. Can you see what is invariant about b and/or w D

where the eigenfunction w describes the state of the system. Again in analogy with classical mechanics the kinetic energy operator for a particle of constant mass is given

4

Exercise 12.5. (a) Suppose a daisy has 16 petals arranged symmetrically around its centre. There are two players. A move involves removing one petal or two adjacent petals. The winner is the one who removes the last petal. Who should win and what is the winning strategy (b) Generalize your solution to part (a) to the case that there are initially n petals and a move consists of removing between 1 and m adjacent petals (where ra is fixed in advance of the game). (c) Two players are seated at a rectangular table which is initially bare. They each have an unlimited supply of circular coins of varying diameter. The players take it in turns to place a coin on the table, such that it does not overlap any coin already on the table. The winner is the one who puts the last coin on the table. Who should win and what is the winning strategy (Harder.} What, if anything, do you assume about the coins in order to justify your answer n

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