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Solving the Schrodinger equation for a free particle gives me this output: output image What does the $\psi^{(2,0)}$ mean here?

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closed as off-topic by march, m_goldberg, Bob Hanlon, b.gates.you.know.what, Alex Trounev Aug 8 at 7:56

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  • $\begingroup$ see GeneratedParameters $\endgroup$ – kglr Aug 7 at 15:35
  • $\begingroup$ I can see that K[1] is a generated parameter here, but the formatting of $\psi$ doesn't seem to match? $\endgroup$ – Dan Goldwater Aug 7 at 15:44
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    $\begingroup$ $\psi^{(2,0)}$ is shorthand for the second derivative of $\psi$ in its first argument. $\endgroup$ – eyorble Aug 7 at 15:47
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    $\begingroup$ .. that is, D[\[Psi][x, t], {x, 2}] $\endgroup$ – kglr Aug 7 at 15:48