Tales Roberto
Nov4-09, 06:08 PM
1. The problem statement, all variables and given/known data
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
3. 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 lets 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!
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
3. 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 lets 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!