Question

Calculate the first-order energy corrections for all levels of an infinite square well potential
(between x = 0 and x = L) with a perturbation H' = αx(L-x)

Answer 1

Answer on Question #47580 – Physics – Quantum Mechanics

Question.

Calculate the first-order energy corrections for all levels of an infinite square well potential (between $x = 0$ and $x = L$) with a perturbation $H' = \alpha x(L - x)$.

$$H' = \alpha x (L - x)$$ $$E_n^{(1)} = ?$$

Solution.

Pesturbation theory helps to find an approximate solution of a problem. From the pesturbation theory we know that the first-order energy corrections is the following:

$$E_n^{(1)} = H_{nn}' = \int_{-\infty}^{+\infty} \bar{\psi}(x) H' \psi(x) dx$$

$\psi(x)$ is the wave function;

$H'$ is the pesturbation.

In the case of an infinite square well potential the wave functions is equal to:

$$\psi(x) = \sqrt{\frac{2}{L}} \sin \left(\frac{\pi n x}{L}\right); n \in \mathbb{Z}$$

So, we can calculate the first-order energy corrections:

$$\begin{aligned} E_n^{(1)} = H_{nn}' &= \int_{-\infty}^{+\infty} \bar{\psi}(x) H' \psi(x) dx = \frac{2\alpha}{L} \int_{0}^{L} \sin^2 \left(\frac{\pi n x}{L}\right) x (L - x) dx = \\ &= \frac{2\alpha}{L} \int_{0}^{L} \frac{1}{2} \left(1 - \cos \left(\frac{2\pi n x}{L}\right)\right) (x L - x^2) dx = \\ &= \frac{\alpha}{L} \int_{0}^{L} x L - x^2 - x L \cos \left(\frac{2\pi n x}{L}\right) + x^2 \cos \left(\frac{2\pi n x}{L}\right) dx \end{aligned}$$

Thus, we received the sum of four integrals. Let calculate each of them.

$$\int_{0}^{L} x L dx = \frac{L^3}{2};$$ $$\int_{0}^{L} x^2 dx = \frac{L^3}{3};$$ $$\int_{0}^{L} x \cos \left(\frac{2\pi n x}{L}\right) dx = 0$$ $$\int_{0}^{L} x^{2} \cos \left(\frac{2\pi n x}{L}\right) dx = \frac{L^{3}}{(2\pi n)^{2}}$$

Finally,

$$\begin{aligned} E_{n}^{(1)} &= \frac{\alpha}{L} \int_{0}^{L} x L - x^{2} - x L \cos \left(\frac{2\pi n x}{L}\right) + x^{2} \cos \left(\frac{2\pi n x}{L}\right) dx = \frac{\alpha}{L} \left(\frac{L^{3}}{2} - \frac{L^{3}}{3} + \frac{L^{3}}{(2\pi n)^{2}}\right) \\ &= \alpha L^{2} \left(\frac{1}{6} + \frac{1}{(2\pi n)^{2}}\right) \end{aligned}$$

Answer.

$$E_{n}^{(1)} = \alpha L^{2} \left(\frac{1}{6} + \frac{1}{(2\pi n)^{2}}\right)$$

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