bluepilotg-2_07
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- Homework Statement
- Given the wavefunction, determine the expectation value of the particle's position
- Relevant Equations
- ##\Psi=\frac{1}{\sqrt{\pi}(b)^{1/4}}\exp(\frac{ip_0x}{\hbar}-\frac{(x-x_0)^2}{2b^2})##
##|\Psi|^2=\frac{1}{\sqrt{\pi b^2}}\exp(\frac{-(x-x_0)^2}{b^2}).##
##\braket{x}=\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dx\,x\exp(-\frac{(x-x_0)^2}{b^2}).##
##y=x-x_0 \quad x=y+x_0 \quad dy=dx.##
The boundaries remain infinite, I believe.
##\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dy(y+x_0)\exp(\frac{-y^2}{b^2}).##
##\frac{2}{\sqrt{\pi b^2}}\int_0^{\infty}dy\,y\exp(\frac{-y^2}{b^2})+\frac{2x_0}{\sqrt{\pi b^2}}\int_0^{\infty}dy\,\exp(-\frac{y^2}{b^2}).##
I then resolved the two integrals separately using the relations for Gaussian integrals in the back cover of Introduction to Quantum Mechanics by Griffiths.
##\frac{b}{\sqrt{\pi}}+x_0.##
This is not correct. The result should be ##x_0##. I have retraced my steps multiple times and have redone the calculation using a different substitution but get the same result. Would appreciate someone pointing out the problem.
##\braket{x}=\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dx\,x\exp(-\frac{(x-x_0)^2}{b^2}).##
##y=x-x_0 \quad x=y+x_0 \quad dy=dx.##
The boundaries remain infinite, I believe.
##\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dy(y+x_0)\exp(\frac{-y^2}{b^2}).##
##\frac{2}{\sqrt{\pi b^2}}\int_0^{\infty}dy\,y\exp(\frac{-y^2}{b^2})+\frac{2x_0}{\sqrt{\pi b^2}}\int_0^{\infty}dy\,\exp(-\frac{y^2}{b^2}).##
I then resolved the two integrals separately using the relations for Gaussian integrals in the back cover of Introduction to Quantum Mechanics by Griffiths.
##\frac{b}{\sqrt{\pi}}+x_0.##
This is not correct. The result should be ##x_0##. I have retraced my steps multiple times and have redone the calculation using a different substitution but get the same result. Would appreciate someone pointing out the problem.