The unperturbed wave functions for the infinite square well are (Equation 2.31) \psi_{n}(x)=\sqrt{\frac{2}{a} }\sin\left(\frac{n\pi }{a} x\right) .
\psi ^{0}_{n}(x)=\sqrt{\frac{2}{a} }\sin\left(\frac{n\pi }{a} x\right) .
Suppose we perturb the system by simply raising the “floor” of the well a constant amount V_{0} (Figure 7.2). Find the first-order correction to the energies.
In this case H^′=V_0 , and the first-order correction to the energy of the nth state is
E^{1}_{n}=\left\langle \begin{matrix}\psi ^{0}_{n}\begin{vmatrix} V_{0} \\ \end{vmatrix} \psi ^{0}_{n} \\ \end{matrix} \right\rangle =v_ {0} \left\langle \begin{matrix} \psi ^{0}_{n}|\psi ^{0}_{n} \\ \end{matrix} \right\rangle =v_{0}
The corrected energy levels, then, are E_{n} \approx E^{0}_{n}+V_{0} ; they are simply lifted by the amount V_{0}. Of course! The only surprising thing is that in this case the first-order theory yields the exact answer.
Evidently for a constant perturbation all the higher corrections vanish. On the other hand, if the perturbation extends only half-way across the well (Figure 7.3), then
E^{1}_{n}=\frac{2V_{0}}{a}\int_{0}^{a/2} \sin^{2}\left(\frac{n\pi }{a} x \right) dx=\frac{V_ {0}}{2}
In this case every energy level is lifted by V_0/2 . That’s not the exact result, presumably, but it does seem reasonable, as a first-order approximation.
