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Solution 59 We have A = B = C, which is true when an odd number of A, B and C is true Thus either just one is a knight or all three are knights D Solution 510 We have C = A = B So the question we have to pose to A is A = A = B, ie B In words, ask A whether B is a knight D Solution 511 Let Q be the question Asking the question Q will produce the response A = Q, which we require to be A So we require that A = A = Q, ie Q In words, ask A to confirm or deny any true statement (for example 0 = 0) D Solution 512 Let Q be the question.

and finally:

Asking the question Q will produce the response A = Q, which we require to be B So we require that B = A = Q In words, ask A whether they are both the same type D Solution 513 Let Q be the question Asking the question Q will produce the response A = Q which we require to be B = A So we require that B = A == A = Q, ie B = Q In words, ask A whether B is a knight D Solution 514 Let Q be the question Choose arbitrarily to pose the question to A Asking the question Q will then produce the response A = Q The proposition whose truth we want to determine is A = B = C So we require that (A = Q) = (A = B = C).

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which is again an iterative equation which can be solved with a numerical shooting technique according to the same boundary conditions as discussed in Section 8.5. Fig. 8.16 shows the results of calculations of the three lowest energy levels of a spherical GaAs quantum dot surrounded by a finite barrier composed of Ga0.8Al0.2 As, with a sharp boundary. In fact, the formalism above, as that of the circular crosssection quantum wire, is applicable for any radial potential profile V(r), e.g. it is also valid for diffused interfaces. Again, the behaviour of the energies as a function

Similarly, Prechelt [Prechelt 1995] and Finnoff et al [Finnoff et al 1993] test the assumption that a weight becomes zero during the training process This approach is based on the observation that the distribution of weight values is roughly normal Weights located in the left tail of this distribution are.

Rearranging and simplifying we get Q = B = C That is, the question is: 'are your two companions the same type ' D Solution 517 (a) (b) (c) (d) false false false p ..

of the spatial dimension, as shown in Fig. 8.16, is as expected in confined systems, namely the confinement energy decreases as the size of the system increases.

272 Solution 5.19. - true { law ->p = p = false with p := true } true = false = { false . law true = p = p with p := false }

Figure 8.18 The probability density of the three lowest energy states in the 300 A spherical quantum dot Fig. 8.17 displays the corresponding radial components of the wave functions. It can be seen that they all have a maximum at the centre of the potential and that as the principal quantum number n increases, then the number of nodes increases. The nature of the states is perhaps better illustrated by considering the charge density. Given that successive increases in the radial coordinate r lead to spherical shells of volume proportional to r2, then the charge density between the radii r and r + Sr is proportional to r2^* (r)ip(r); this is plotted in Fig. 8.18. The lowest energy state, i.e. n = 1, has a single anti-node which is close to half the radius of the potential, while the second has two maxima in the charge density, etc. 8.8 NON-ZERO ANGULAR MOMENTUM STATES The states deduced in the previous section were of the form ^(r), and therefore were analogous to the 's' orbitals akin to hydrogen atoms. However, in atoms there exist other types of state, with the next simplest being that of the 'p' orbitals. These have wave functions of the following form:

1 PT ET(DT- W) = V^yv^p, W) - -f p ) 2

(see any quantum theory book, e.g. [4], p. 190). These states are clearly not spherically symmetric, and have non-zero angular momenta.

Solution 5.20. { law -ip = p=false with p := -ip } ->p = false { law ->p = p = false with p := p and symmetry of equivalence }

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