Relation between matrix elements of momentum and position operators

[Fantini]

Hello. I'm having trouble understanding what is required in the following problem:

Find the relation between the matrix elements of the operators $\widehat{p}$ and $\widehat{x}$ in the base of eigenvectors of the Hamiltonian for one particle, that is, $$\widehat{H} = \frac{1}{2M} \widehat{p}^2 + V(\widehat{x}).$$

I don't understand what kind of relation he is asking for. The eigenvectors for the Hamiltonian satisfy the equation $\widehat{H} \varphi_i = E_i \varphi_i$, but I don't know how to use that. The answer is $$\frac{ \langle i \, | \, p \, | j \rangle}{\langle i \, | \, x \, | j \rangle} = \frac{iM(E_i - E_j)}{\hbar},$$ but it doesn't enlighten me.

How do I find an arbitrary matrix element of an operator?

 

[Ackbach]

Just thinking out loud here. First off, we can tidy up the notation so that everything matches: $\hat{H} \, |i\rangle =E_i \, |i\rangle$, and the same for $j$. Let's remember some of the canonical commutation relations, in case they prove useful:
\begin{align*}
[x,p]&=i\hbar \\
[x,x]&=0 \\
[p,p]&=0 \\
[AB,C]&=A[B,C]+[A,C]B.
\end{align*}
The last allows us to compute
\begin{align*}
[x,H]&=\left[x,\frac{p^2}{2M}+V(x)\right] \\
&=\frac{1}{2M} \, [x,p^2]+\underbrace{[x,V(x)]}_{=0} \\
&=-\frac{1}{2M} \, [p^2,x] \\
&=-\frac{1}{2M} \, \left( p[p,x]+[p,x]p \right) \\
&=\frac{i \hbar}{M} \, p.
\end{align*}
The reason I computed this commutator is because the answer has $E_i$ and $E_j$ in it, leading me to think that we're going to have to calculate something with an $H$ in it. If you calculate $[x,H] \, |j\rangle$ two different ways, you get
$$[x,H] \, |j\rangle = \frac{i\hbar}{M} \, p \, |j\rangle = (E_j-H) \, x \, |j\rangle.$$
Here, I have done
\begin{align*}
[x,H] \, |j\rangle&=(xH-Hx) \, |j\rangle \\
&=xH \, |j\rangle - Hx \, |j\rangle \\
&=x E_j \, |j\rangle-Hx \, |j\rangle \\
&=E_j \, x \, |j\rangle -Hx \, |j\rangle \\
&=(E_j-H) \, x \, |j\rangle.
\end{align*}

It seems to me that these computations might help. Does this give you any ideas?