- #1

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The conditions for real solutions are according to Kreyszig's Functional Analysis that E, in the following equation ( HY=EY),

**is in the resolvent set of the Hamiltonian operator.**

I would like to ask for confirmation of that this is the right approach:

1. Having ##H\psi = E\psi##, one solves the equation with E treated as a constant first. One uses no initial conditions. The true form of the energy is ## E= \frac{h^2k^2}{8\pi^2m}## . The solution is here now called ##\eta(q)##

2. Following Kreyszig's book on this, I quote

"The solution ##\eta(q)## can now be used to represent any ##\psi \in L ^2(-\infty,+\infty)## as a wave packet in the form:

\begin{equation}

\psi(q) = \frac{1}{\sqrt{2\pi}} \lim_{a\rightarrow \infty} \int_{-a}^{a}\phi(k)\eta(q)dk

\end{equation}

where

\begin{equation}

\phi(k) = \frac{1}{\sqrt{2\pi}} \lim_{a\rightarrow \infty} \int_{-a}^{a} \psi(q)\eta(q)dq

\end{equation}

Should this be exercised on the solution :

\begin{equation}

\eta(q) = e^{-2i\gamma q} + 3/4 + \frac{i\hbar}{2\gamma}

\end{equation}

How would the integral and thus the wavepacket look like? Based on what is in the book, I am not sure on what to make out of ##\phi(k)## and ##\psi(q)##, basically, how does one find these two auxiliary functions?

Hope this question was clear

Thanks