I am working out of Sakurai and he says the integral with respect to

[Berko]

I am working out of Sakurai and he says the integral with respect to p` of exp(ip`(x`-x``)/h-bar) is 2-Pi-h-bar times the dirac delta function of (x`-x``). (Sorry, I am not sure how to make actual integrals).

I was wondering why this is true? For example, if x` = x``, isn't the integral infinity? And, if not, why is it well defined given the oscillations involved?

Thank you for your time.

P.S. This is page 54, equation 1.7.31 out of Sakurai's Modern Quantum Mechanics, second edition.

 

[dextercioby]

The explanation you look for can be found in any book on Fourier analysis and distribution theory.

 

[tiny-tim]

Hi Berko!

Technically, the dirac delta function is not a function, it's a distribution (or "generalised function").

It is not intended to be on its own, it is always inside another integral.

ie, a distribution A(x) is only intended to be inside an ∫A(x)f(x) dx …

its "value" for any particular value of x, eg A(0), is not really relevant.

(see http://en.wikipedia.org/wiki/Distribution_(mathematics)" )

 

[homology]

One way to think of it might also be: what is the Fourier transform of $$\delta(x-x')$$?

 

[The_Duck]

A way to think about this is to do

$$\int_{-A}^A e^{ikx} dk = [\frac{1}{ix}e^{ikx}]_{k=-A}^{k=A} = \frac{1}{ix}(e^{iAx} - e^{-iAx}) = 2 \frac{\sin{Ax}}{x}$$

Now let A go to infinity. You get a function that is highly oscillatory every except right at x=0. You can do the integral of the function over all x; the result is 2 pi. So if you integrate this function against another function f(x), the oscillations will make everything cancel out except right in the region x=0, and the result will be 2*pi*f(0). So this suggests that you identify

$$\lim_{A \to \infty} 2 \frac{\sin{Ax}}{x} = 2 \pi \delta (x)$$

 

[homology]

@The_Duck: sometimes I wish there were "LIKE" buttons on here, i'd click it. Nice derivation.