Part 7, 8, 9, Outline, Schedule

PHYS 424 Notes

Part 8

  1. Chapter 2, cont.
    1. Delta Functions
      1. Definition: d(x) = 0, x¹0, d(x) = ¥, x=0, with òd(x)dx = 1
      2. òf(x)d(x)dx = f(0) for any f(x)
      3. òf(x)d '(x)dx = f '(0) for any f(x)
      4. [Rigorously, d(x) is given by òf(x)d(x)dx = limn->¥ Ö(n/p) ò f(x)e -nx2 dx, with the limit defined outside the integration.]
      5. ò dk eik(x-x ') = 2pd(x-x')                                                                         (8-1)
      6. Hence f(x)=(2p) -1/2 ò F(k)eikx dk <=> F(k)=(2p) -1/2 ò f(x)e -ikx dx
    2. Free Particle
      1. Schroedinger Equation and solution

        [p2/(2m)]y = Ey = [hbar2k2/2m)]y

        Solved by yk(x) = Ak e ikx, where complex exponentials make more sense than trig functions for states that should have definite momentum. The parameter k may be either positive or negative, and both k and -k yield solutions for the same E, so there are indeed two solutions as there should be for a second-order equation. The complete Y is

        Yk(x,t) = e ikx -i hbar k2 t / (2m)           (8-2)

        and since pop|k> = hbar k |k>, from popYk = -i hbar dYk/dx = hbar k Yk   , hbar k is reasonably regarded as the value of the momentum in |k>. Therefore cos(kx) and sin(kx) are linear combinations of the states that have definite momenta (of hbar k and - hbar k).

        However, there is a problem: <k'|k> = ò dx ei(k-k')x = 2 pd(k-k') Although the states of different moment are orthogonal, the normalization integral is 2pd(0) which cannot be adjusted to be 1! There is no normalized state which corresponds to a free particle of definite momentum. This fact is not a mathematical anomaly, it is a simple consequence of the Uncertainty Principle.

      2. Wave packets

        The actual physical states can be written, however, as sums over the definite-momentum states in the usual fashion. Since there are now a continuous distribution of states, instead of discrete numbers of states, the sum will have to become an integral. Hence a general wave function will be

        Y(x,t) = (2p)-1/2 ò -¥ ¥ dk f(k) exp{i[kx - hbar k2 t / (2m)]}           (8-3)

        We can use (8-1) to get the coefficients f(k) from the values of Y(x,t) at any set time, say t=0:

        òdx e -ik'x Y(x,0) = (2p)-1/2 ò dx dk f(k) ei(k-k')x
        = (2p)1/2 ò dk f(k) d(k-k')

        f(k') = (2p)-1/2 òdx e -ik'x Y(x,0)             (8-4)

        The wave function (8-3) is called a "wave packet," and the most useful cases are where f(k) involves a narrow range of values of k.

Last Revised 05/03/02

Part 7, 8, 9, Outline, Schedule