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I am reading a paper by Faddeev and Kulish ("Asymptotic conditions and infrared divergences in QED" written in 1970) and I have a question regarding the nonrelativistic problem of Coulomb scattering. The authors claim that in the nonrelativistic Coulomb scattering, the potential can be expanded in powers of the $t$, where $t$ is the time which is present in the Coulomb potential through the substitution of the distance $r$ between the scattering particle and the scatterer, which is given by $$\hat{r}(t)=\frac{\hat{p}}{m}+\hat{r},$$ where $\hat{r}$ and $\hat{p}$ are time independent operators in QM. The expansion goes as follows $$V(t)=\frac{mg}{pt}+\mathcal{O}(t^{-2})$$ where $g$ is the product of the charges of the scattering particle and the scatterer. Then, they solve the Schroedinger equation for the asymptotic regions in space, in which the Hamiltonian is comprised by the kinetic term and the first term in the expansion of $V(t)$, rather than the kinetic term alone. The wave function is given by $$\psi(\vec{r},t)=\int\frac{d^3\vec{p}}{(2\pi)^3}c(\vec{p})\exp\bigg(-i\frac{p^2}{2m}-i\frac{mg}{p}\text{sign}t\ln{\frac{|t|}{t_0}}\bigg)e^{i\vec{p}\cdot{\vec{r}}}$$ which, indeed, solves the differential equation. However, upon trying to derive an expression for the mean value $<\hat{r}(t)>$, then with the method of stationary phase, one recovers $$<\hat{r}(t)>=\frac{<\hat{p}>}{m}t-gm<\frac{\hat{p}}{p^2}>\text{sign}t\ln{|t|}+\mathcal{O}(1)$$ Doesn't the last expression contradict the initial assumption that $$\hat{r}(t)=\frac{\hat{p}}{m}+\hat{r}~ ?$$

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    $\begingroup$ what's the meaning of $\hat{r}(t)=\frac{\hat{p}^2}{2m}+\hat{r}$? $\endgroup$ Apr 12, 2022 at 16:26
  • $\begingroup$ It is a typo. Sorry.. $\endgroup$
    – schris38
    Apr 13, 2022 at 6:30

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The problem seems to lie in you taking $\hat{r}(t) = \frac{\hat{p}^2}{2 m} + \hat{r}$, whereas it should be $\hat{r}(t) = \frac{\hat{p}}{m} t + \hat{r}$. The calculation checks out when the latter is chosen.

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  • $\begingroup$ Hi Karen. Excuse my typo please. I have fixed that now. What I meant by the question is the following: it seems to me that we start with a pair of initial conditions for $\hat{r}$ and for $\hat{p}$ and then when we try to calculate the expectation values for those operators, then the former does not match the initial condition, since its second term involves a logarithmically diverging time dependent term $\endgroup$
    – schris38
    Apr 13, 2022 at 6:33
  • $\begingroup$ I am not sure I fully understand your question. First of all, there is still a typo in your question: you write $\hat{\vec{r}}(t) = \frac{\hat{\vec{p}}}{2 m} + \hat{\vec{r}}$, but it should be $\hat{\vec{r}}(t) = \frac{\hat{\vec{p}}}{m} t + \hat{\vec{r}}$. $\endgroup$
    – Karen H.
    Apr 15, 2022 at 16:00
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    $\begingroup$ (2) Not realizing that $$\langle \psi_I (t) | \hat{\vec{r}} | \psi_I(t) \rangle = - g m \left\langle \frac{\hat{\vec{p}}}{\hat{\vec{p}}^2} \right\rangle \mathrm{sign}(t) \ln |t|$$ where $| \psi_I(t) \rangle$ is the interaction-picture state, namely, $$| \psi_I(t) \rangle = e^{i t H_0} | \psi(t) \rangle$$ where $| \psi(t) \rangle$ is the state in Eq. (2) of the article. $\endgroup$
    – Karen H.
    Apr 15, 2022 at 16:05
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    $\begingroup$ Exactly. That pretty much straightforwardly follows from the expression for the state [Eq. (2) in the paper]: $$\psi_I(q, t) := \langle q | \psi_I(t) \rangle = \int \frac{d^3 \vec{p}}{(2 \pi)^3} c(\vec{p}) \exp\left[ i \vec{p} \vec{r} - i \frac{m g}{| \vec{p}|} \mathrm{sign}(t) \ln \frac{t}{t_0} \right].$$ Just calculating $\langle \psi_I(t) | \vec{r} | \psi_I(t) \rangle$ using this expression gives the answer. The only seminontrivial step in calculating this is realizing that $$\int \frac{d^3 \vec{r}}{(2 \pi)^3} \vec{r} e^{\vec{r} \vec{p}} = \mathrm{grad}_{\vec{p}} \delta (\vec{p}).$$ $\endgroup$
    – Karen H.
    Apr 18, 2022 at 23:37
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    $\begingroup$ The operator $\hat{\vec{r}}$ is spread all over the space: $$\hat{\vec{r}}=\int d^3\vec{r}\,\vec{r}|\vec{r}\rangle\langle\vec{r}|.$$ So what happens is that initially the state is localized around some $\vec{r}_0$, so that the expectation value of $\hat{\vec{r}}$ is close to $\vec{r}_0$. Then, in the course of the evolution, the state spreads out in a way that the expectation value of $\hat{\vec{r}}$ becomes the weird function of $t$ that it is (speaking of the specific case considered by Faddeev and Kulish). $\endgroup$
    – Karen H.
    Apr 20, 2022 at 15:58

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