Non-relativistic scattering amplitude For the nonrelativistic scattering in a potential $V$, the scattering amplitude $\vec{p}$ to $\vec{k}$ is proportional to $$<\vec{k}|V|\phi^{+}_\vec{p}>,$$ with $$E(\vec{k})=\frac{\vec{k}^2}{2m}=E(\vec{p})$$ (conservation of energy), where $|\phi^{+}_\vec{p}>$ is the in-state $\Omega_+|\vec{p}>$ which satisfies the Lippmann Schwinger equation $$|\phi^{+}_\vec{p}>=|\vec{p}>+\frac{1}{E(\vec{p})-H_0+i\epsilon}V|\phi^{+}_\vec{p}>$$ and is an eigenstate of the full Hamiltonian $H=H_0+V$ with eigenvalue $E(\vec{p})$, $H|\phi^{+}_\vec{p}>=E(\vec{p})|\phi^{+}_\vec{p}>$. See Sakurai eqn 7.1.34 or any quantum textbook.
Here comes the question, I somehow get zero from the matrix element $<\vec{k}|V|\phi^{+}_\vec{p}>=<\vec{k}|H-H_0|\phi^{+}_\vec{p}>=(E(\vec{p})-E(\vec{k}))<\vec{k}|\phi^{+}_\vec{p}>=0$, where I let $H$ act to the right and $H_0$ to the left.
 A: I agree with these steps:
\begin{equation}
\langle \vec{k} | V | \phi_\vec{p}^+ \rangle = \langle \vec{k} | H-H_0 | \phi_\vec{p}^+ \rangle = \left(E(\vec{p}) - E(\vec{k})\right) \langle \vec{k} | \phi_\vec{p}^+ \rangle
\end{equation}
However, before you conclude the right hand side is zero, we should check that $\langle \vec{k} | \phi_\vec{p}^+ \rangle$ is really finite. We might be worried that it's not, because the denominator in the definition of $|\phi_\vec{p}^+\rangle$ looks like it will go to zero when $E(\vec{p})=E(\vec{k})$.
So let's expand that part out carefully.
\begin{eqnarray}
\langle \vec{k} | \phi_\vec{p}^+ \rangle &=& \langle \vec{k} | \vec{p} \rangle + \langle \vec{k} | \frac{1}{E(\vec{p})-H_0+i\epsilon} V | \phi_\vec{p}^+ \rangle \\
&=& \delta^{(3)}(\vec{k}-\vec{p}) + \frac{1}{E(\vec{p})-E(\vec{k})+i\epsilon} \langle \vec{k} | V | \phi_\vec{p}^+ \rangle
\end{eqnarray}
where to get to the second line, $H_0$ acts on the left.
As a result...
\begin{eqnarray}
\langle \vec{k} | V | \phi_\vec{p}^+ \rangle &=& \left(E(\vec{p}) - E(\vec{k})\right) \langle \vec{k} | \phi_\vec{p}^+ \rangle \\
&=& \left(E(\vec{p}) - E(\vec{k})\right) \left[ \delta^{(3)}(\vec{k}-\vec{p}) + \frac{1}{E(\vec{p})-E(\vec{k})+i\epsilon} \langle \vec{k} | V | \phi_\vec{p}^+ \rangle \right]  \\
&=& \langle \vec{k} | V | \phi_\vec{p}^+ \rangle
\end{eqnarray}
This result is good news in the sense that it is a true statement and self-consistent, but bad news in the sense that you haven't learned anything new about the system. But, there is definitely a lesson to learn here, since that "1/energy difference" singularity contains a lot of physics and you need to pay careful attention to it as you continue your studies.
Aside: You don't need to worry about the delta function $\delta^{(3)}(\vec{k}-\vec{p})$ (or even the one hiding in $(E(\vec{k})-E(\vec{p})+i\epsilon)^{-1}$ via the identity $(x+i\epsilon)^{-1}=\mathcal{P}[1/x]-i\pi \delta(x)$) since if you integrated $(E(\vec{k})-E(\vec{p})) \delta^{(3)}(\vec{k}-\vec{p})$ against a test function, the result would be zero. A way to see this is that $\int dx f(x) x \delta(x) = 0$ (assuming $f(x)$ is less singular than $1/x$ at $x=0$, which should be true for reasonably behaved test functions).
