# Energy of a particle in one dimensional potential well

Consider a particle of mass $$m$$ in one dimension inside a potential well ($$-a) with infinitely high walls and the WF $$\psi(x)=A(x^2-a^2)$$ for $$|x|.

Show that the expected value of its energy is $$(10-\pi^2)/\pi^2$$ above the ground state energy.

My question: It obviously suffices to show that $$\frac{\langle E\rangle}{E_0}=\frac{10}{\pi^2}$$ Since $$\psi(x)$$ must be normalized, we have $$A=\frac{1}{4}\sqrt\frac{15}{a^5}$$ and it is easy to see that $$\langle E\rangle=\int_{-a}^a\psi(x)\cdot\frac{p^2}{2m}\psi^\dagger(x)dx=\frac{5\hbar^2}{4ma^2}$$

This means that $$E_0=\frac{\hbar^2\pi^2}{8ma^2}$$. But how do I get the ground state energy without reverse engineering? My initial thought was just $$E\psi=\frac{p^2}{2m}\psi=-\frac{\hbar^2}{2m}\partial_x^2\psi(x)=-\frac{\hbar^2 A}{m}=-\frac{1}{4}\sqrt\frac{15}{a^5}\cdot\frac{\hbar^2}{m}$$ which obviously doesn't work. Where is my error in this thought and how do I correct it?

$$E_n = \frac{n^2\pi^2 \hbar^2}{8ma^2} \tag{1}$$
$$\langle E\rangle = \frac{5\hbar^2}{4m a^2} \tag{2}$$
Take $$n = 1$$ and find
$$\langle E \rangle - E_1 = \frac{\hbar^2}{8ma^2}(\pi^2 - 10)$$