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,
$$ <x>=A\int_{-\infty}^{\infty}x\sin^{2}\left(\frac{x\pi}{L}\right)dx $$
if we apply an even potential then the equation gets reduced to$$ <x>=\frac{1}{L}\int_{-L}^{L}x\sin^{2}\left(\frac{x\pi}{L}\right)dx=0 $$ while in case of a potential(neither even nor odd), the equation leads to $$<x>=\frac{2}{L}\int_{0}^{L}x\sin^{2}\left(\frac{x\pi}{L}\right)dx=L/2 ~? $$
Here $n=m=1$ but $ <x>=0 $ for even potentials, which is confusing me! It should be $k$ right?
Best Answer
The $|\psi_n\rangle$ you are using are eigenstates of the Hamiltonian, $\hat{H}$, not eigenstates of the position operator $\hat{X}$. There's no sense in which you may write $\langle\psi_m|\hat{X}|\psi_n\rangle=(\text{some constant constant})*\langle\psi_m|\psi_n\rangle$, because it is NOT true that $\hat{X}|\psi_m\rangle=k|\psi_m\rangle$.