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The inverse operator is $L^{-1} = 1/2 \cdot (x-1)^{-1} d^2/dx^2 - 1/2 \cdot (x-1)^{-2} d/dx$. Its eigenfunctions are derivatives of Airy functions, $Ai^{\prime } ((2\lambda )^{1/3} (x-1))$, $Bi^{\prim …

The exponentiated operator shifts $f$ by one unit in $x$, i.e., $$ \frac{1}{e^{d/dx} (f(x))} = \frac{1}{f(x+1)} $$ It's not quite clear what is desired by "getting rid of the denominator" - the result …

No, $U(t,s)$ is, in general, not self-adjoint. If we, for ease of formulation, denote the parameters $t,s$ as "times", then $U(t,s)$ is given by the time-ordered exponential $$ U(t,s) = T\exp \left( \ …

Paraphrasing L.I.Schiff, "Quantum Mechanics", the $S$-matrix $S=\langle \beta | \alpha^{+} \rangle $ is the amplitude of the final asymptotic state $\beta $ contained in what became of an initial asym …

You haven't really defined $(d/dy)^{-1} $, but let's pick, for instance, $(d/dy)^{-1} f(y) = \int_{0}^{y} dy' f(y')$. Then, your two expressions for $H(y)$ in general are not equal. A simple counterex …

You can of course write $$ k(x,t) = \exp \left[ e^{-t^2 /2} p\left( -\frac{d}{dt} \right) e^{t^2 /2} \right] \delta (x-t) $$ Not clear whether that affords you any sort of simplification you may be po …