Back to Contents

We set up the solution of this integral by multiplying and dividing it by an exponential function that will complete the square of the argument of the exponential function in the integrand. We can discover through simple algebra that

(Eq地 1)

Thus the integral becomes

(Eq地 2)

in which Note that the limits of the integration have not changed. We can discern two reasons for that fact: 1) r represents a real number integrated between real limits, so we must integrate Z only between real numbers (that is, the imaginary components drop out of the integration and we thus ignore them), and 2) dividing zero and infinity by any finite, nonzero coefficient leaves both unchanged. Invoking the result of 64thCRC663, we thus have

(Eq地 3)

Example Application:

The Gaussian Wave Packet

In the quantum theory we often represent a particle with a wave packet, a superposition of waves with different wave numbers forming a relatively compact representation of the particle痴 location. In general we have a normalizable combination of stationary state functions,

(Eq地 A-1)

We can calculate the form of the function f(k) if we have the state function at some initial instant of time (t=0). We simply carry out the Fourier transform,

(Eq地 A-2)

We want to describe a wave packet that has the same shape in both coordinate space and momentum space, so we need the envelope of the packet to conform to the Gaussian function,

(Eq地 A-3)

To determine the value of the amplitude of that function we normalize it in accordance with the requirement of Born痴 theorem,

(Eq地 A-4)

We have from 64thCRC663

(Eq地 A-5)

Thus we get our amplitude as

(Eq地 A-6)

We thus get Equation A-2 as

(Eq地 A-7)

We then substitute that formula into Equation A-1 to get our complete state function.

efefefefef

Back to Contents