\magnification=\magstep3

$$
   {\vec{F} } =  { \int {- k \vec{x}} } 
$$

$$
   {{U} } =  { {{1}\over{2}} k x^2 } 
$$


$$ \eqalign{
  {\Psi(x)}  &=  { A e^{- \alpha x} } \cr
  {\Psi(x)}  &=  { A x e^{\alpha x} } \cr
  {\Psi(x)}  &=  { A e^{- \alpha x^2} } \cr
}
$$

$$
  { {{-\hbar^2}\over{2m}} {{\partial^2 \Psi}\over{\partial x^2}} + U \Psi = E \Psi  }
$$

$$
  { {{-\hbar^2}\over{2m}} {{\partial^2 \Psi}\over{\partial x^2}} + 
       { {{1}\over{2}} k x^2 \Psi } 
       = E \Psi  }
$$

$$
  { {{\partial^2 \Psi}\over{\partial x^2}} 
       = - (  {{2m} \over {\hbar^2}} ) 
          ( E -  { {{1}\over{2}} k x^2 } ) \Psi  }
$$

$$
  { {{\partial^2 \Psi}\over{\partial x^2}} 
       = (  {{2m} \over {\hbar^2}} ) 
          ( { {{1}\over{2}} k x^2 } - E ) \Psi  }
$$

$$
  { {{\partial^2 \Psi}\over{\partial x^2}} 
       = (  {{m} \over {\hbar^2}} ) 
          { k x^2 } \Psi  
       - (  {{2m} \over {\hbar^2}} ) 
           E \Psi  }
$$

$$
  { {{\partial^2 \Psi}\over{\partial x^2}} 
       = (  {{k m} \over {\hbar^2}} ) 
           { x^2 \Psi }
       - (  {{2 m E} \over {\hbar^2}} ) 
           { \Psi }  
    }
$$

$$
  { {{\partial^2 \Psi}\over{\partial x^2}} 
       =  - 2 \alpha \Psi 
         + 4 \alpha^2 x^2 \Psi
    }
$$




\bye
