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Let us consider the quantum scattering problem on the line with the Hamiltonian $$H=-\frac{d^2}{dx^2}+ V(x),$$ where $V(x)=1$ when $x\in (0,a)$, and $V(x)=0$ otherwise.

It is easy to see that $H$ has no discrete spectrum (e.g. no bound states).

What explicitly are the Moeller operators $\Omega^{\pm}$?

A reference would be helpful.

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    $\begingroup$ The basic calculations are §III.6-7 in A. Messia's Quantum Mechanics v.1 (1961). $\endgroup$
    – Igor Khavkine
    Commented Aug 18, 2018 at 16:21
  • $\begingroup$ @IgorKhavkine The reference you mentioned does not seem to discuss Moeller operators at all. It studies scattering theory only using stationary methods, unless I missed something. My goal is to understand the dynamic approach to scattering theory, e.g. by writing down the Moeller operators in the above example which seems to be simplest non-trivial one. $\endgroup$
    – asv
    Commented Aug 18, 2018 at 18:32
  • $\begingroup$ It does seem to be possible to write down Moeller operators, but the justification I know is purely intuitive, which does not use the mathematical definition of Moeller operators. $\endgroup$
    – asv
    Commented Aug 18, 2018 at 18:36
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    $\begingroup$ Sorry, your question was ambiguous. In one spatial dimension, you can "evolve" along $x$ too. I thought you were talking about $x$-evolution, since you didn't mention time at all. $\endgroup$
    – Igor Khavkine
    Commented Aug 19, 2018 at 7:19

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Assuming that you want to compute $\Omega^+ = \lim_{(t-t') \to \infty} e^{-iH(t-t')} e^{-iH_0(t'-t)}$ with $H_0 = -\frac{\partial^2}{\partial x^2}$, this operator has the explicit integral representation:

\begin{align*} (\Omega^+ \Phi)(x) &= \lim_{(t-t') \to \infty} \int\frac{dk}{2\pi} \, e^{-ik^2 (t-t')} \int dx' \, \psi_{k}(x) \bar{\psi}_{k}(x') \int\frac{dk'}{2\pi} \, e^{-ik'^2(t'-t)} \int dx'' \, e^{ik'(x'-x'')} \Phi(x''), \end{align*}

where $\psi_k(x)$ is a complete set of generalized eigenfunctions for $H$, with $H\psi_k = k^2 \psi_k$. Leaving as an exercise all the interchanges of integral signs, the result has the form

\begin{align*} (\Omega^+\Phi)(x) &= \int dx'' \int\frac{dk}{2\pi} \, [\alpha_k \psi_{k}(x) + \beta_k\psi_{-k}(x)] e^{-ikx''} \Phi(x'') \\ & \quad + \lim_{(t-t')\to \infty} \int dx'' \, \int\frac{dk}{2\pi} \int\frac{dk'}{2\pi} e^{-i(k^2-k'^2) (t-t')} \psi_{k}(x) \, F(k,k') \, e^{-ik'x''} \Phi(x''), \end{align*}

where we have used $$ \int dx' \, \bar{\psi}_k(x') e^{ik'x'} = 2\pi [\alpha_k \delta(k-k') + \beta_k \delta(k+k')] + F(k,k') , $$ with $\alpha_k+\beta_k = 1$ related to the transmission and reflection coefficients of $V(x)$ and $F(k,k')$ is less singular than the $\delta$-functions. Since the eigenfunctions $\psi_k(x)$ are in principle known, all of this can be computed explicitly.

The remaining integral under the limit is essentially a double Fourier transform in $k^2$ and $k'^2$ evaluated at large $(t-t')$. But since $F(k,k')$ and the rest of the integrand are less singular than a $\delta$-function, then result of the integral should go to $0$ as $(t-t') \to \infty$ (on the other hand, the Fourier transform of a $\delta$-function does not decay to $0$ in at least some directions).

Thus, the final result should be $$ (\Omega^+\Phi)(x) = \int dx'' \int\frac{dk}{2\pi} \, [\alpha_k \psi_{k}(x) + \beta_k\psi_{-k}(x)] e^{-ikx''} \Phi(x'') . $$ All the heuristic intermediate steps can be justified using explicit formulas, which are in principle available for this example.

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    $\begingroup$ I'm not sure about your first formula. Normally, you'd have to use the spectral measure(s) of $H$ somewhere, but you just use Lebesgue measure $dk$ (this works for $H_0$). Also, you simply refer to "generalized eigenfunctions," but of course the solution space is two-dimensional and we need to know which solution exactly we're supposed to use. (Another issue is that both operators have multiplicity $2$; we have to hope that the change of variable $E=k^2$ takes care of this.) $\endgroup$
    – Christian Remling
    Commented Aug 19, 2018 at 15:41
  • $\begingroup$ @ChristianRemling, you're right about the spectral measure in general. But here, everything simplifies once you look at the exact solutions. The multiplicity 2 issue is resolved by including both $k = \pm \sqrt{E}$ eigenfunctions in the $k$-parametrization. And of course, $\psi_k(x)$ has to be normalized by appropriate $k$-dependent factors for the first formula to work as written. $\endgroup$
    – Igor Khavkine
    Commented Aug 19, 2018 at 16:20
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    $\begingroup$ @IgorKhavkine: The (half line) spectral measures should still have a density, so look like $f(k)\, dk$, and you could obtain $f$ as something like $\textrm{Im }y'(0,k)/y(0,k)$, with $y$ being the solution that is $e^{ikx}$ near infinity. I guess you could just give a name to this without saying what it is, and then your sketch should work as intended. $\endgroup$
    – Christian Remling
    Commented Aug 19, 2018 at 17:42
  • $\begingroup$ @IgorKhavkine: I still do not understand your final expression for $\Omega^+$. It should be a (generalized) function in two variables. What are these two variables in your notation? It seems that the right hand side depends only on one variable $x$. $\endgroup$
    – asv
    Commented Aug 20, 2018 at 15:59
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    $\begingroup$ @orbits All good points! Shees, did I miss a bunch of things when I quickly typed up the answer. I inserted the wave function $\Phi(x)$ which I always thought was there, but didn't actually write. From $\frac{1}{k'-k\pm i\epsilon} = P\frac{1}{k'-k} \mp i\pi \delta(k'-k)$ you do get a $\delta$-function. The remaining question is whether the remainder $F(k,k')$ is really sufficiently regular for the second term to go to $0$ as $t-t' \to \infty$. $\endgroup$
    – Igor Khavkine
    Commented Aug 21, 2018 at 7:03

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