Do even potentials produce states which are not orthonormal? Do even potentials produce states which are not orthonormal? For example, in the case of infinite square well, assuming that the states are represented by:
$$ \psi_n = \frac{\sqrt{2}}{a} \sin\left(\frac{n\pi x}{a}\right) $$
While proving their orthonormality, if I apply


*

*A general (I mean neither even nor odd) potential i. e. the limits run from $0 \to a$,the eigenstates come out to be orthonormal. 

*An even potential, i.e. the limits run from $-a \to a$, I find that the eigenstates are not orthonormal. 
I wonder how can that be so? Aren't the eigenstates supposed to be orthonormal always?
It also further results in making the expectation values of the position operator being equal to zero. 
 A: If two states, $|\alpha\rangle$ and $|\beta\rangle$, are eigenstates to different energies, i.e. $H |\alpha\rangle=E_\alpha |\alpha\rangle$ and $H |\beta\rangle=E_\beta |\beta\rangle$, they are orthogonal.
Consider 
the expression $\langle \alpha |H|\beta\rangle$: it can be simplified by acting with $H$ on $|\beta\rangle$:
$$\langle \alpha |H|\beta\rangle= \langle \alpha |\left(H|\beta\rangle\right)=E_\beta \langle \alpha |\beta\rangle,$$
or, since $H$ is Hermitian, by acting on $\langle \alpha|$:
$$\langle \alpha |H|\beta\rangle= \left(\langle \alpha |H\right)|\beta\rangle=\left(H^\dagger |\alpha\rangle\right)^\dagger|\beta\rangle=\left(H|\alpha\rangle\right)^\dagger|\beta\rangle=\left(E_\alpha |\alpha\rangle\right)^\dagger|\beta\rangle=E_\alpha \langle \alpha |\beta\rangle.$$
We have arrived to identity
$$E_\alpha \langle \alpha |\beta\rangle=E_\beta \langle \alpha |\beta\rangle,$$
which implies that either


*

*$E_\alpha \neq E_\beta$ and $\langle \alpha |\beta\rangle=0$ – the states are orthogonal,

*$E_\alpha=E_\beta$ – in this case the eigenstates might not be orthogonal. As an example, consider two orthogonal eigenstates to the same energy $|u\rangle,|v\rangle$. Then, $\frac{|u\rangle+|v\rangle}{\sqrt{2}}$ is eigenstate as well – you can easily check it – but it is not orthogonal to either $|u\rangle$ or $|v\rangle$. (Eigenspace can always be orthogonalized, though by the means of Gram-Schmidt process.)


If you prefer less abstract derivation of this fact, you can always write down the expression $\langle \alpha |H|\beta\rangle$ as an integral, i.e.
$$ \langle \alpha |H|\beta\rangle = \int_{-\infty}^{\infty} \alpha^* (x) \left(-\frac {d^2}{d\text{x}^2} + V(x)\right) \beta(x) \text{d}x$$
and try the same exact process as above:


*

*observe that $\beta(x)$ is an eigenstate,

*try to make the double derivative act on $\alpha^*(x)$ by using integration by parts (and assuming that both $\alpha$ and $\beta$ vanish at infinity),

*compare the two results.


No assumptions about the form of potential are needed – this fact does not depend on it. 
