# Wave packet momentum wave function

knawd

## Homework Statement

A wave packet is described by the momentum-space wave function A(p)=C when 0<p<p0, and A(p)=0 for all other values of p. Here C is a constant.

i) Normalize this wave function by solving for C in terms of p0.

ii) Calculate the expectation values <p> and <p2>. From these compute the standard deviation in terms of p0.

## Homework Equations

For normalization: h(integral from 0 to p0)A*(p)dp=1, h being Planck's constant.
<p>=h(integral from 0 to p0)pA*(p)dp
<p2>=h(integral from 0 to p0)p2A*(p)dp
(I am not entirely sure if this equation for <p2> is correct, it may be <p2>=h(integral from 0 to p0)p2A*(p)A(p)dp)
standard deviation=<p2> - <p>2

## The Attempt at a Solution

i) When I normalized the wave function from 0 to p0, I got C=1/sqrt(h*p0).

ii) This is the part I'm struggling with. Typically <p> would be equal to zero when integrated from -infinity to +infinity, but this is from 0 to p0. Doing this I am getting <p>=p0.
For <p2> I am integrating from 0 to p0 and getting <p2>=(1/3)p02.
One of these values (or both) can't be correct because when I try to calculate the standard deviation=<p2> - <p>2 = (1/3)p02 - p02= -(2/3)p02. I'm pretty sure standard deviations can't be negative. So the part of the question I really can't figure out is what I'm doing wrong when I try to find the expectation values.

## Answers and Replies

Staff Emeritus
Homework Helper
Your integrals are all wrong. For normalization, you want

$$\int_0^{p_0} A^*(p)A(p)\,dp = 1$$

Similarly, the expectation values are

$$\langle p \rangle = \int_0^{p_0} A^*(p)pA(p)\,dp$$

and

$$\langle p^2 \rangle = \int_0^{p_0} A^*(p)p^2A(p)\,dp$$

knawd
Okay. So using those integrals I'm getting:

C=1/sqrt(hp0)

<p>=p0/2

<p2>=p02/3

stdev=(1/12)p02

Also, am I right in using h in front of all of the integrals for this problem?

Staff Emeritus
Homework Helper
Okay. So using those integrals I'm getting:

C=1/sqrt(hp0)

<p>=p0/2

<p2>=p02/3

stdev=(1/12)p02
The last quantity is the variance, not the standard deviation.

Also, am I right in using h in front of all of the integrals for this problem?
No. Why do you keep putting it there?

mnoir
So the standard deviation is sqrt(<p2>-<p>2), which equals p0/sqrt(12)?

My textbook says to put 2*pi*hbar (=h) in front of the integrals, I wasn't sure if that had something to do with me getting strange answers though.

Staff Emeritus