# Does the orthogonality of states apply when there is a potential?

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:

$$\langle H_{total} \rangle = \langle -\frac{\hbar ^2}{2m}\frac{\partial ^2}{\partial x^2} + V \rangle$$ $$\langle H_{total} \rangle = \langle -\frac{\hbar ^2}{2m}\frac{\partial ^2}{\partial x^2} + (V_0 + V_{new}) \rangle \quad ; \quad V_0 \text{ is the potential for the particle in a box, which is zero, and }V_{new} \text{ is the added potential.}$$

$$\langle H_{total} \rangle = \langle H_0 \rangle + \langle V_{new} \rangle$$

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.

• No, this isn't true. As the simplest possible example, if $V(x)$ is a constant $V_0$, then $\langle \psi_i | V | \psi_j \rangle = V_0 \delta_{ij}$ instead. – knzhou May 12 '17 at 2:20
• @knzhou ok well in that case, the end result is still somewhat what I was thinking - you get $0$ when $i\neq j$. – loltospoon May 12 '17 at 2:21
• It depends on V, but in general, the answer is no. – NickD May 12 '17 at 2:23
• @loltospoon More generally, the first equation you have says that an integral is equal to zero (for $i \neq j$), while the second says that the same integral plus an arbitrary weighting factor $V(x)$ is also zero; that can't always hold. – knzhou May 12 '17 at 2:25
• @knzhou updated the question to add the context. Is this method even valid? – loltospoon May 12 '17 at 2:40

No. If you have the regular particle in a one-dimensional box of length $L$, the free Hamiltonian has eigenvalues $|\psi_n\rangle$ with $\langle x | \psi_n \rangle=\sqrt\frac{2}{L} \sin\left(\frac{n \pi x}{L}\right)$ . If, say, $V=V_0 \frac{x}{L}$, then you can do the integral and find $\langle \psi_1|V|\psi_2\rangle=-\frac{16}{9\pi} V_0$.
Some of your equations aren't correct, by the way. The hamiltonian you're interested in is this: $$H_{total} = -\frac{\hbar ^2}{2m}\frac{\partial ^2}{\partial x^2} + V$$ In which case the expectation value $\langle H\rangle=\langle \psi|H|\psi\rangle$ is actually: $$\langle H_{total}\rangle = \int_{0}^L\left(-\psi^*(x)\frac{\hbar ^2}{2m}\psi''(x) + V(x)\psi^*(x) \psi(x)\right)dx$$ where $\psi(x)$ is some wavefunction.
Furthermore, $H_{total}$ itself isn't a matrix, it's a linear operator which acts on an infinite dimensional space! If you discretize space into $N$ points, your Hamiltonian can be approximated numerically by an $N\times N$ matrix, but as you have encountered, this matrix is not trivial to diagonalize!