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  • I'm trying to solve the Schrödinger's equations using the method of Laplace transforms, but I can't figure out the inverse transform once I've reached this point.
  • I'm a high school student, so I'm self-taught and I haven't much experience with the theory behind the Inverse Laplace transform.
  • I'm sorry for the bad typing of my formulas but this is my first post.
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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $\ds{\bbox[5px,#ffd]{}}$ Basically, you want to compute $\ds{\bbox[5px,#ffd]{\left.\int_{0^{+}\ -\ \infty\ic}^{0^{+}\ +\ \infty\ic}\! \expo{\alpha x\!\root{s}}\!\!\expo{ts} {\dd s \over 2\pi\ic}\,\right\vert_{\ t\ >\ 0}}}$ where $\ds{\alpha \equiv \pars{\ic - 1}\root{m \over h} \mbox{is a}\ constant}$.


\begin{align} &\bbox[5px,#ffd]{\left.\int_{0^{+}\ -\ \infty\ic}^{0^{+}\ +\ \infty\ic}\! \expo{\alpha x\!\root{s}}\!\!\expo{ts} {\dd s \over 2\pi\ic}\,\right\vert_{\ t\ >\ 0}}\\[5mm] = &\ -\int_{-\infty}^{0}\expo{\alpha x\root{-s}\ic}\expo{ts} \,{\dd s \over 2\pi\ic} - \int_{0}^{-\infty}\expo{\alpha x\root{-s}\pars{-\ic}}\expo{ts} \,{\dd s \over 2\pi\ic} \\[5mm] = &\ -\int_{0}^{\infty}\expo{\alpha x\root{s}\ic}\expo{-ts} \,{\dd s \over 2\pi\ic} + \int_{0}^{\infty}\expo{\alpha x\root{s}\pars{-\ic}}\expo{-ts} \,{\dd s \over 2\pi\ic} \\[5mm] = &\ -\,{1 \over \pi}\int_{0}^{\infty}\sin\pars{\alpha x\root{s}} \expo{-ts}\,\dd s \\[5mm] = &\ \stackrel{s\ \mapsto\ s^{2}}{=}\,\,\, -\,{2 \over \pi}\int_{0}^{\infty}s\sin\pars{\alpha xs} \expo{-ts^{2}}\,\dd s \\[5mm] = &\ \bbx{-\,{\alpha \over 2\root{\pi}} \,{x\exp\pars{-\alpha^{2}x^{2}/\bracks{4t}} \over t^{3/2}}\,,\qquad \Re\pars{x} > 0} \\ & \end{align}
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  • $\begingroup$ Thank you so much for your precious help. Just to be sure, is a=(i-1)sqrt(m/h)? $\endgroup$ Commented Oct 6, 2020 at 12:15
  • $\begingroup$ I'm also worried about what happens to the t variable: at the beginning I have a function of position and time, f(x,t), then I use the Laplace transform to obtain a new function H(x,s). Shouldn't I expect to obtain another function of x and t after applying the inverse transform? $\endgroup$ Commented Oct 6, 2020 at 13:38
  • $\begingroup$ @GabrielePrivitera Yes, $\alpha$ is the one you wrote in your comment. I didn’t see any time $t$ in your question. Thanks. $\endgroup$ Commented Oct 6, 2020 at 21:01
  • $\begingroup$ thank you for your answer. As far as the time-dependency is concerned, I'll explain "my" method to solve the problem from which I derived the function stated at the beginning of my question. To solve the Schrodinger's equation, I apply the Laplace transform to both sides of my equation. So I start from a function f(x,t) and I obtain a new differential equation involving G(x,s). I solve the new equation and find the expression of G. Than, through the inverse Laplace transform, we should obtain another function F(x,t), which is time dependent $\endgroup$ Commented Oct 7, 2020 at 12:19
  • $\begingroup$ @GabrielePrivitera I just fixed it. I guess I misread your question. Thanks. $\endgroup$ Commented Oct 7, 2020 at 18:16

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