How Do You Normalize a Wave Function and Verify Its Momentum Space?

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Homework Help Overview

The discussion revolves around normalizing a wave function and verifying its momentum space representation. The original poster presents a wave packet defined by a specific mathematical form and outlines tasks related to normalization and Fourier transforms in quantum mechanics.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • The original poster attempts to normalize the wave function and calculate its momentum space representation using Fourier transforms. Some participants question the convergence of the integrals involved in the calculations. Others express uncertainty about the results obtained from the integrals and suggest simplifications.

Discussion Status

Contextual Notes

Participants are navigating potential issues with convergence and the implications of their results, indicating a need for clarity on the definitions and assumptions used in the problem setup.

Tales Roberto
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Homework Statement



Consider the wave packet \psi\left(x\right)=\Psi\left(x,t=0\right) given by \psi=Ce^{\frac{ip_{0}x}{h}-\frac{\left|x\right|}{2\Delta x} where C is a normalization constant:

(a) Normalize \psi\left(x\right) to unity

(b) Obtain the corresponding momentum space wave function \phi\left(p_{x}\right) and verify that it is normalized to unity according to: \int^{\infty}_{-\infty}\left|\phi\left(p_{x}\right)\right|^{2} dp_{x}=1

(c) Suggest a reasonable definition of the width \Delta p_{x} and show that \Delta x \Delta p_{x} \geq h

The Attempt at a Solution



(a) is easy to solve and we find that C=\frac{1}{\sqrt{2\Delta x}} assuming that C is real. This way \psi=\frac{1}{\sqrt{2\Delta x}}e^{\frac{ip_{0}x}{h}-\frac{\left|x\right|}{2\Delta x}

I attempt to use Fourier Transform to calculate (b):

\phi\left(p_{x}\right)=\left(2\Pi h \right)^{-\frac{1}{2}} \int e^{\frac{-ip_{x}x}{h}} \psi dx

\phi\left(p_{x}\right)=\left(4\Pi h \right\Delta x)^{-\frac{1}{2}} \int e^{\frac{-i\left(p_{x}-p_{0}\right)x}{h}} e^{\frac{-\left|x\right|}{2\Delta x}} dx


\phi\left(p_{x}\right)=\left(4\Pi h \right\Delta x)^{-\frac{1}{2}} \left[\int_{0}^{\infty} e^{-\left(\frac{ip}{h}} + \frac{1}{2\Delta x}\right)x} dx + \int_{-\infty}^{0} e^{-\left(\frac{ip}{h} - \frac{1}{2\Delta x}\right)x} dx\right]

where p=p_{x} - p_{0}. To simplify let's write:

\beta_{1}=\left(\frac{ip}{h}} + \frac{1}{2\Delta x}\right) \beta_{2}=\left(\frac{ip}{h} - \frac{1}{2\Delta x}\right)

Then:

\phi\left(p_{x}\right)=\left(4\Pi h \right\Delta x)^{-\frac{1}{2}} \left[\int_{0}^{\infty} e^{-\beta_{1}x} dx + \int_{-\infty}^{0} e^{-\beta_{2}x} dx\right]

This integral does not converge since arguments are complex. My "feeling" is that my solution is completely wrong, please help!
 
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are you sure those integrals don't converge?
 
I calculated the integral, however i anchieved a "weird" result. I think it was wrong, i expected an exp term. Here my result:

\phi\left(p_{x}\right)=-\left(4\Pi h \Delta x\right)^{-\frac{1}{2}}\left[\frac{1}{\frac{ip}{h}+\frac{1}{2\Delta x}}+\frac{1}{\frac{ip}{h}- \frac{1}{2\Delta x}} \right]

I have nothing to do with this, may wrong!
 
i haven't tried it in detail, but it looks reasonable to me... the exponentials disappear when you take the infinite limit

try simplifying, by putting it all over the same denominator, it at least makes sense in that it goes to zero for large |p|

now calculate the magnitude square (worth plotting) to get the momentum probability density function and check if its normalised...
 

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