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<x\right>=A^2\int_{-\infty}^\infty x e^{-\frac{2am x^2}{\hbar}}\,dx=0

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

\displaystyle \left<p\right>=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}

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5 thoughts on “Griffiths 1.9

  1. Richard

    At the risk of sounding stupid–my calculus is kinda rusty–in 1.9(a) above, why can you just drop the e^(it) portion when you integrate?

    Reply
    1. arnulfo Post author

      e^(it) does not affect the magnitud.
      The term in the integration is the wave funtion times its conjugate. That cancels out the e^(it) factor

      Reply

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