Problems for Edward F. Redish |
We defined the delta function as
(a) Use the definitions to argue the following properties of the delta function:
where the set of x_{i} are the solutions (there may be more than one) of the equation
Solution
(a) We defined the delta
function as the limit of our "bead model" of the string:
the string is replace be a series of discrete mass points. The Dirac state |x> is
supposed to correspond to only one of the beads displaced. This is like the basis
vectors we used for a linear space describing a finite set of coupled oscillators.
If only the bead at point x is displaced upward, then all the other beads
must be at zero. The overlap the states where two different beads are displaced
must be taken to be zero just as we took e
Since the only time we get a non-zero value of the delta function is when the two arguments are equal, we can consider it to be a function of the difference, non-zero only when its argument is zero. This gives the result in the first paragraph.
Since the integral of the δ(x-x') over any function f(x') gives f(x), if we take f to be the constant 1, we see that the integral over a δ is going to be 1. Since it is only non-zero at one point, the only way we could get the width x the height to be 1 when the width is 0 is to take the height to be infinite.
(b) We can get this result by changing variables from x to g. Change variables to z = g(x):
dz = (dg/dx) dx
and writing x = g^{-1}(x) we get the result
The square brackets is now our new function that we are integrating with a delta function. It will give the value of the square bracketed function when z = 0. Since z = g(x), if we let x_{1} be the solution of the equation 0 =g(x_{1}), then our answer will be
This, of course, has assumed that there is only one solution to 0 = g(x). If there are many, this means that the curve of g(x) is not monotonic. It turns around. To do the integral, we have to break it up into pieces from minimum to maximum. For each of those pieces, when the argument crosses 0, we will get another contribution like the one we have above, yielding the result with the sum.
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Last revision 10. December, 2004.