Problems for Intermediate Methods in Theoretical Physics Edward F. Redish

In lecture, we introduced the Dirac delta function in order to be able to write

.

We defined the delta function as

.

(a) Use the definitions to argue the following properties of the delta function:

• The delta function is a generalization of the Kroenecker delta, δ_ij = 0 for i ≠ j; that is, δ(x-x') = 0 if x ≠ x'.
• The delta function is only a function of the difference between x and x'.
• In some (difficult to determine) sense, δ(x-x') is infinite for x = x'.

(b) For some applications we will have to use the delta function of a function of x rather than just of x itself. (For example when we want to integrate over momentum states -- p -- but want to conserve energy -- E = p²/2m.) Show that:

.

where the set of x_i are the solutions (there may be more than one) of the equation

g(x) = 0.

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 ei·e = 0 if i ≠ j; that is, the vector with a one in the i-th place (and zeros everywhere else) dotted into the vector with a one in the j-th place (and zeros everywhere else) is 0 if i is not equal to j.

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 x₁ be the solution of the equation 0 =g(x₁), 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.

RETURNS

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Last revision 10. December, 2004.