Assuming the eigenvalue of position operator $\hat x$ equal to $k$, can I not write:
$$\begin{align} \langle\psi_n|x|\psi_m\rangle &= \langle x\psi_n|\psi_m\rangle \\ &=\langle k\psi_n|\psi_m\rangle \\ &=k\langle\psi_n|\psi_m\rangle \\ &=k\delta_{nm} \end{align}$$
But I know that $\langle x \rangle =0$ in case of even potentials (I don't know how that happens) and what I have written above is wrong, at least in case of even potentials.
Taking the example of infinite 1D square well, the states are : $$ \psi_{n} \left(x\right)=A\sin\left(\frac{nx\pi}{L}\right)dx $$ then, $$ \langle\psi_n|x|\psi_m\rangle =A\int_{-\infty}^{\infty}\sin\left(\frac{mx\pi}{L}\right)x\sin\left(\frac{nx\pi}{L}\right)dx $$ If m=n=1, $$
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