Suppose we perturb the infinite cubical well (Equation 6.30) by putting a delta function “bump” at the point$\mathbf{(}\mathbf{a}\mathbf{/}\mathbf{4}\mathbf{}\mathbf{,}\mathbf{}\mathbf{a}\mathbf{/}\mathbf{2}\mathbf{}\mathbf{,}\mathbf{}\mathbf{}\mathbf{3}\mathbf{a}\mathbf{/}\mathbf{4}\mathbf{)}$$\mathbf{:}{\mathbf{H}}^{\mathbf{\text{'}}}\mathbf{=}{\mathbf{a}}^{\mathbf{3}}{\mathbf{V}}_{\mathbf{0}}\mathbf{\delta }\mathbf{(}\mathbf{x}\mathbf{-}\mathbf{a}\mathbf{/}\mathbf{4}\mathbf{)}\mathbf{\delta }\mathbf{(}\mathbf{y}\mathbf{-}\mathbf{a}\mathbf{/}\mathbf{2}\mathbf{)}\mathbf{\delta }\mathbf{(}\mathbf{z}\mathbf{-}\mathbf{3}\mathbf{a}\mathbf{/}\mathbf{4}\mathbf{)}\mathbf{.}$

Find the first-order corrections to the energy of the ground state and the (triply degenerate) first excited states.

The anticipated value of the perturbation in the unperturbed state is the first order adjustment to the energy.

Ground state is non degenerate; Eqs. 6.9⇒

…(6.9).

localid="1658148464503" $\begin{array}{rcl}{E}^1& =& {\left(\frac{2}{a}\right)}^3{a}^3{V}_0\underset{a}{\overset{0}{\int }}{\sin }^2\left(\frac{\pi }{a}x\right){\sin }^2\left(\frac{\pi }{a}y\right){\sin }^2\left(\frac{\pi }{a}z\right)\delta \left(x-\frac{a}{4}\right)\delta \left(y-\frac{a}{2}\right)\delta \left(z-\frac{3a}{4}\right)dxdydz.\\ & =& 8V_0{\sin }^2\left(\frac{\pi }{4}\right){\sin }^2\left(\frac{\pi }{2}\right){\sin }^2\left(\frac{3\pi }{4}\right)=8V_0\left(\frac{1}{2}\right)\left(1\right)\left(\frac{1}{2}\right)=2V_0\end{array}$

First excited states:

$\begin{array}{rcl}{W}_{aa}& =& 8V_0\iiint {\sin }^2\left(\frac{\pi }{a}x\right){\sin }^2\left(\frac{\pi }{a}y\right){\sin }^2\left(\frac{2\pi }{a}z\right)\delta \left(x-\frac{a}{4}\right)\delta \left(y-\frac{a}{2}\right)\delta \left(z-\frac{3a}{4}\right)dxdydz.\\ & =& 8V_0\left(\frac{1}{2}\right)\left(1\right)\left(1\right)=4V_0.\end{array}$

$\begin{array}{rcl}{W}_{bb}& =& 8V_0\iiint {\sin }^2\left(\frac{\pi }{a}x\right){\sin }^2\left(\frac{2\pi }{a}y\right){\sin }^2\left(\frac{\pi }{a}z\right)\delta \left(x-\frac{a}{4}\right)\delta \left(y-\frac{a}{2}\right)\delta \left(z-\frac{3a}{4}\right)dxdydz.\\ & =& 8V_0\left(\frac{1}{2}\right)\left(0\right)\left(\frac{1}{2}\right)=0.\end{array}$

$\begin{array}{rcl}{W}_{cc}& =& 8V_0\iiint {\sin }^2\left(\frac{2\pi }{a}x\right){\sin }^2\left(\frac{\pi }{a}y\right){\sin }^2\left(\frac{\pi }{a}z\right)\delta \left(x-\frac{a}{4}\right)\delta \left(y-\frac{a}{2}\right)\delta \left(z-\frac{3a}{4}\right)dxdydz.\\ & =& 8V_0\left(1\right)\left(1\right)\left(\frac{1}{2}\right)=4V_0.\end{array}$

$\begin{array}{rcl}{W}_{ab}& =& 8V_0{\sin }^2\left(\frac{\pi }{4}\right)\sin \left(\frac{\pi }{2}\right)\sin \left(\pi \right)\sin \left(\frac{3\pi }{2}\right)\sin \left(\frac{3\pi }{4}\right)\\ & =& 0.\\ & & \end{array}$

$\begin{array}{rcl}{W}_{bc}& =& 8V_0\sin \left(\frac{\pi }{4}\right)\sin \left(\frac{\pi }{2}\right)\sin \left(\pi \right)\sin \left(\frac{\pi }{2}\right){\sin }^2\left(\frac{3\pi }{4}\right)\\ & =& 0.\\ & & \end{array}$

$\begin{array}{rcl}W& =& 4V_0\left[\begin{array}{ccc}1& 0& -1\\ 0& 0& 0\\ -1& 0& 1\end{array}\right]=4V_{0}D;\det (D-\lambda )=\left|\begin{array}{ccc}\left(1-\lambda \right)& 0& -1\\ 0& -\lambda & 0\\ -1& 0& \left(1-\lambda \right)\end{array}\right|\\ & =& -\lambda (1-\lambda {)}^2+\lambda \\ & =& 0\\ & & \end{array}$

$\lambda =0,\text{or}(1-\lambda {)}^2=1\Rightarrow 1-\lambda =\pm 1\Rightarrow \lambda =0.$

So the first-order corrections to the energies are $0,8V_0$.