PHYS 307:  Computational Physics Lab (QM)

Winter 2008

Analytic Solution to Homework #4

1. (a) The dimensionless Schrödinger equation, in the same units as previously, is
   $$\psi^{\prime\prime} - U\psi = -\xi^2\psi,$$
   where $s=x/a$ , $^\prime\equiv d/ds$ , $U=2ma^2V/\hbar^2$ , and $\xi^2=2ma^2E/\hbar^2$ . For the infinite square well, with $U=0$ for $\vert s\vert<1$ and $U=\infty$ for $\vert s\vert\ge1$ , the equation reduces to
   $$\psi^{\prime\prime} + \xi^2\psi = 0, ~~~~\vert s\vert < 1,$$
   with $\psi(\pm1)=0$ . The even and odd solutions are, respectively, $\psi = A\cos\xi s$ and $\psi=B\sin\xi s$ , so the boundary conditions imply
   $$\xi = \left\{\begin{array}{ll} (n+{\ensuremath{\textstyle\frac12}})\pi & ~~\text{(even)}\\ n\pi & ~~\text{(odd)}\\ \end{array}\right.$$
   Clearly the lowest energy mode has $\xi=\xi_0={\ensuremath{\textstyle\frac12}}\pi$ , so the (normalized) ground-state wave function is
   $$\psi_0(s) = \cos{\ensuremath{\textstyle\frac12}}\pi s,$$
   and
   $$E_0 = \frac{\pi^2\hbar^2}{8ma^2}\,.$$
   Now suppose that
   $$V = \left\{\begin{array}{ll} \epsilon E_0 & ~~~(\vert s\vert<{\en... ...ac12}}<\vert s\vert<1)\\ \infty & ~~~(\vert s\vert\ge1)\\ \end{array}\right.$$
   To solve this system, we will write down expressions for the even solution valid for $0\le s\le{\ensuremath{\textstyle\frac12}}$ , and ${\ensuremath{\textstyle\frac12}}\le s\le1$ , apply the boundary condition $\psi(1)=0$ , and then impose continuity in $\psi$ and $\psi^\prime$ at $s={\ensuremath{\textstyle\frac12}}$ . For $0\le s\le{\ensuremath{\textstyle\frac12}}$ ,
   $$\psi^{\prime\prime} + \eta^2\psi = 0$$
   where

   $$\eta^2 = \xi^2 - \epsilon\,\xi_0^2.$$ (1)

   
   
   The even solution is
   $$\psi = A \cos\eta s.$$
   For ${\ensuremath{\textstyle\frac12}}\le s\le1$ , we have
   $$\psi =B \cos\xi s + C \sin\xi s,$$
   which is more conveniently written as
   $$\psi = D \sin \xi(1-s)$$
   in order to satisfy the boundary condition at $s=1$ . At $s={\ensuremath{\textstyle\frac12}}$ , continuity requires that
   

   $$A \cos{\ensuremath{\textstyle\frac12}}\eta = D\sin{\ensuremath{\textstyle\frac12}}\xi$$

   $$-A\eta \sin{\ensuremath{\textstyle\frac12}}\eta = D\xi\cos{\ensuremath{\textstyle\frac12}}\xi,$$

   
   so
   $$\eta\tan{\ensuremath{\textstyle\frac12}}\eta = \xi\cot{\ensuremath{\textstyle\frac12}}\xi.$$
   As with the finite-well problem, we must solve this equation numerically, in conjunction with the above definition of $\eta$ (Eq. 1). See the numerical solutions.

   (c) First-order perturbation theory gives us the following expression for the change in the ground-state energy due to the perturbation in $V$ :
   

   $$\delta E_0 = \int_{-1}^1\,\psi_0^2\,\delta V\,ds$$

   $$= \int_{-1/2}^{1/2}\,\psi_0^2\epsilon E_0\,ds$$

   $$= \epsilon E_0 \int_{-1/2}^{1/2}\,\cos^2{\ensuremath{\textstyle\frac12}}\pi s\,ds$$

   $$= {\ensuremath{\textstyle\frac12}}\epsilon E_0 \int_{-1/2}^{1/2}\,(1+\cos\pi s)\,ds$$

   $$\noalign{\vskip 6pt} = \epsilon E_0 \,\left({\ensuremath{\textstyle\frac12}} +{\textstyle\frac1\pi}\right)$$

   $$\noalign{\vskip 6pt} = 0.8183 \,\epsilon E_0.$$

   
   See the numerical solutions for a comparison between the analytic, numerical, and perturbative solutions.

2. (c) The perturbation theory result is
   

   $$\delta E_0 = \int_{-1}^1\,\psi_0^2\,\delta V\,ds$$

   $$= \epsilon E_0 \int_{-1/2}^{1/2}\,\cos^2{\ensuremath{\textstyle\frac12}}\pi s\,(1-2\vert s\vert)\,ds$$

   $$= 2\epsilon E_0 \int_0^{1/2}\,\cos^2{\ensuremath{\textstyle\frac12}}\pi s\,(1-2s)\,ds$$

   $$\noalign{\vskip 6pt} = \epsilon E_0 \,\left({\textstyle\frac14} +{\textstyle\frac2{\pi^2}}\right)$$

   $$\noalign{\vskip 6pt} = 0.4526 \,\epsilon E_0.$$

   
   

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