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By Paul C. W. Davies, David S. Betts (auth.)
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Extra resources for Quantum Mechanics
Sample text
16) is identical to the Schrodinger equation for u(x). As the system is non-degenerate there can only exist one stationary state for each value of E, so u(x) and u( - x) must represent the same solution. 17) However, reversing the sign of x, u( - x) = eu(x). 18) we conclude u(x) = e2u(x) implying e = ± 1 and, as claimed, u( - x) = ± u(x). 19) The reason is simple. As u(x) must be either an even or odd function of x, 1u 12 is even. Hence xl u 12 is odd, and the contribution to the integral for x > 0 exactly cancels that from x < o.
8). 21) where C and D are constants. To save labour one can make use of the fact that, as the potential is symmetric in x, solutions will have definite even or odd parity. We then need only solve in the half-space x> 0 and deduce the solution for x < 0 by reflection. For x> 0, e(Jx is unacceptable because flu 12 dx will blow up as x --+ 00 implying that ucannot be normalized. Therefore D = O. 8). 23) respectively. 25). No solutions exist in terms of elementary functions, and numerical or graphical techniques are necessary.
Continuity in u and du/dx at x = ± a and the normalization condition then fixes all the four constants A, B, C and D, but there is now no restriction on 0( or K or, hence, on E. In the region E> Yo, then, the energy levels form a continuum and the wave functions do not decline at large Ixl; the particle is thus unbound and has an appreciable probability of being found far from the well. 5 BOUNCING BALL Consider the problem of a ball which bounces vertically on a rigid horizontal plane. 26) where x is the vertical height above the plane and 9 is the acceleration due to gravity (Fig.