Griffiths 1.9

A particle of mass $m$ is in the state $\Psi(x,t)=A e^{-a\left( \frac{mx^2}{\hbar}+i t\right)}$, where $A$ and $a$ are positive real constants.

1. Find A
2. Find the potential energy function $V(x)$
3. Calculate the expected values of $x, x^2, p, p^2$
4. Find $\sigma_x$ and $\sigma_p$. Is their product consistent with the uncertainty principle?

$\displaystyle \int_{-\infty}^\infty A^2 e^{-2a\left( \frac{mx^2}{\hbar}\right)}\,dx=2A^2\sqrt{\frac{\hbar}{2am}}\int_{0}^\infty \sqrt{\frac{2am}{\hbar}} e^{-2a\left( \frac{mx^2}{\hbar}\right)}\,dx=2A^2\sqrt{\frac{\hbar\pi}{2am}}=1$

$\displaystyle \displaystyle A^2=\sqrt{\frac{2am}{\pi \hbar}}$

$\displaystyle i \hbar \frac{\partial\Psi}{\partial t}= \hat{H}\Psi = -\frac{\hbar^2}{2m}\frac{\partial^2\Psi}{\partial x^2}+\hat{V}\Psi$

$\displaystyle i \hbar \frac{\partial\Psi}{\partial t}= \hbar a\Psi$

$\displaystyle -\frac{\hbar^2}{2m}\frac{\partial^2\Psi}{\partial x^2}= \hbar a(\frac{\partial}{\partial x} (x \Psi))$

$\displaystyle -\frac{\hbar^2}{2m}\frac{\partial^2\Psi}{\partial x^2}= a( \hbar-2x^2a m)\Psi$

$\displaystyle \hbar a\Psi = a( \hbar-2x^2a m)\Psi+V\Psi$

$\displaystyle V(x)=2a^2 mx^2$

$\displaystyle \left=A^2\int_{-\infty}^\infty x e^{-\frac{2am x^2}{\hbar}}\,dx=0$

$\displaystyle \left=A^2\int_{-\infty}^\infty x^2 e^{-\frac{2am x^2}{\hbar}}\,dx=-\frac{\hbar }{m}A^2\int_0^\infty (-\frac{2m x^2}{\hbar}) e^{-\frac{2am x^2}{\hbar}}\,dx$

$\displaystyle \left=-\frac{\hbar }{m}A^2\frac{\partial}{\partial a}\sqrt{\frac{\pi \hbar}{2ma}}=\frac{\hbar }{2ma}A^2\sqrt{\frac{\pi \hbar}{2ma}}$

$\displaystyle \left=\frac{A^2\hbar}{2i}\int_{-\infty}^\infty (-\frac{4ma x}{\hbar}) e^{-\frac{2am x^2}{\hbar}}\,dx=0$

$\displaystyle \left< p^2\right>=-A^2\hbar^2\int_{-\infty}^\infty e^{-\frac{am x^2}{\hbar}}\frac{\partial}{\partial x}\left[ (-\frac{2ma x}{\hbar}) e^{-\frac{am x^2}{\hbar}}\right]\,dx$

$\displaystyle \left< p^2\right>=2A^2\hbar^2\int_0^\infty {\left( -\frac{2ma x}{\hbar}\right)}^2 e^{-\frac{2am x^2}{\hbar}}\,dx$

$\displaystyle \left< p^2\right>=2A^2\hbar^2\sqrt{\frac{2ma }{\hbar}}\int_0^\infty y^2 e^{-y^2}\,dy$

$\displaystyle \left< p^2\right>=-A^2\hbar^2\sqrt{\frac{2ma }{\hbar}}\int_0^\infty y (-2y) e^{-y^2}\,dy$

$\displaystyle \left=A^2\hbar^2\sqrt{\frac{2ma }{\hbar}}\int_0^\infty e^{-y^2}\,dy$

$\displaystyle \left=A^2\hbar^2\sqrt{\frac{2ma }{\hbar}}\sqrt{\pi\int_0^\infty e^{-\rho^2}2\rho\,d\rho}=A^2\hbar^2\sqrt{\frac{2\pi ma }{\hbar}}$

$\displaystyle \sigma_p^2 \sigma_x^2=A^2\hbar^2\sqrt{\frac{2\pi ma }{\hbar}}\frac{\hbar }{2ma}A^2\sqrt{\frac{\pi \hbar}{2ma}}=A^4\hbar^2\pi\frac{\hbar }{2ma}=\frac{\hbar^2}{4}$