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.

  • $\begingroup$ Are you sure $H|\vec{k}\rangle=E(\vec{k})|\vec{k}\rangle$? Note that $|\vec{k}\rangle$ is not an eigenstate of $V$. $\endgroup$
    – Andrew
    Commented Apr 27, 2021 at 2:29
  • $\begingroup$ No. But $H_0|\vec{k}> = E(\vec{k})|\vec{k}>$. "Free state" and in-state are engenstates of $H_0$ and $H$ respectively. $\endgroup$ Commented Apr 27, 2021 at 4:15
  • $\begingroup$ Ah sorry, I misunderstood your notation. I understand the issue now and have added an answer. $\endgroup$
    – Andrew
    Commented Apr 27, 2021 at 10:14

1 Answer 1


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).

  • $\begingroup$ Thanks for the answer. I actually thought I ruled out any singularity of $<\vec{k}|\phi^+_{\vec{p}}>$ at the same energy from the normalization condition $\int \frac{d^3\vec{k}}{(2\pi)^3} |<\vec{k}|\phi^+_{\vec{p}}>|^2$. Too naive, the integration measure easily cures the singularity and this kind of singularity is common in scattering process. $\endgroup$ Commented Apr 27, 2021 at 12:16
  • $\begingroup$ The lesson for me is, be careful with states like $|\vec{k}>$ that are not properly normalized. $\endgroup$ Commented Apr 27, 2021 at 12:19

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