Is Your Wave Packet Momentum Wave Function Correctly Normalized?

[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.

 

[vela]

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?

 

[vela]

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.

 

[vela]

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

Yup.

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.

It would affect the normalization constant but would cancel out of the other integrals, leaving your answers for the expectation values the same.

 

[mnoir]

Alright. Thank you for clearing things up for me.