I know this from a quantum mechanics class:
$$\langle \psi_i | \psi_j\rangle = \delta_{ij}$$
But does it also apply for:
$$\langle \psi_i |V| \psi_j\rangle = \delta_{ij}$$
where $V$ is any arbitrary potential? That is, if I try to compute the expectation value of $V$ using 2 different states $\psi_i$ and $\psi_j$ I get $0$?
The context:
I am trying to compute the eigenvalues of the particle in a box with an added potential $V$. To do this, the strategy I decided on was to compute like so:
$$\langle H_{total} \rangle = -\frac{\hbar ^2}{2m}\frac{\partial ^2}{\partial x^2} + V$$$$\langle H_{total} \rangle = \langle -\frac{\hbar ^2}{2m}\frac{\partial ^2}{\partial x^2} + V \rangle$$ $$\langle H_{total} \rangle = -\frac{\hbar ^2}{2m}\frac{\partial ^2}{\partial x^2} + (V_0 + V_{new}) \quad ; \quad V_0 \text{ is the potential for the particle in a box, which is zero, and }V_{new} \text{ is the added potential.} $$$$\langle H_{total} \rangle = \langle -\frac{\hbar ^2}{2m}\frac{\partial ^2}{\partial x^2} + (V_0 + V_{new}) \rangle \quad ; \quad V_0 \text{ is the potential for the particle in a box, which is zero, and }V_{new} \text{ is the added potential.} $$
$$\langle H_{total} \rangle = H_0 + V_{new}$$$$\langle H_{total} \rangle = \langle H_0 \rangle + \langle V_{new} \rangle$$
These are all matrices. So I'd fill the diagonal of the $H_0$ matrix using the analytic equation for the total energy and then fill the $V_{new}$ matrix, then add the two matrices together. Finally, I would compute the eigenvalues of this final matrix.
So where does my question come into all of this? Well, I know that the only values in $H_0$ are on the diagonal, but I didn't know if that also applies to the $V_{new}$ matrix as well.
