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I would like to Plot a function like $e^{ix^2}$, similar to the one seen in this video at 1:26: https://www.youtube.com/watch?v=p7bzE1E5PMY&t=86s. I think what I'm trying to do is to plot the imaginary part on one axis, and the real part on the other, both as a function of x. Does anybody know how that works? I hope it isn't to complicated, since this isn't my task, just trying to boost my understanding of quantum mechanics.

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    $\begingroup$ The video and your description don't correspond. The video plots pos. vs. vel., not real and imaginary parts. Which do you want? Possibly useful functions: ReIm[], ParametricPlot[], ParametricPlot3D[]. $\endgroup$ – Michael E2 Jun 8 at 17:00
  • $\begingroup$ @MichaelE2: The OP unwittingly linked to a later time-stamp in the video than they intended to. I've corrected the URL. $\endgroup$ – Michael Seifert Jun 8 at 17:16
  • $\begingroup$ @MichaelSeifert Ah, thanks. I have little patience with video explanations as well. So already put out by a non-self-contained Q, I gave up. $\endgroup$ – Michael E2 Jun 8 at 18:17
  • $\begingroup$ Ah, I'm sorry for that, didn't realise there was a time stamp in the hyperlink @MichaelE2. Thanks for correcting it Michael Seifert. $\endgroup$ – Jonathan Bollig Jun 10 at 20:38
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As noted in the comments, the function you want is ParametricPlot3D. You create the "graph" as a parametric curve with x as the parameter, and whose y and z coordinates are $\Re(\psi)$ and $\Im(\psi)$ respectively.

psi[x_] = Exp[I x^2];
ParametricPlot3D[{x, Re[psi[x]], Im[psi[x]]}, {x, -3, 3}, 
 AxesOrigin -> {0, 0, 0}, Ticks -> {Automatic, None, None}, Boxed -> False]

enter image description here

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  • $\begingroup$ for fun you could add `ParametricPlot3D[{x, Re[psi[x]], Im[psi[x]]} // Evaluate, {x, -3, 3}, AxesOrigin -> {0, 0, 0}, Ticks -> {Automatic, None, None}, Boxed -> False, ColorFunction -> Function[{x, y, z}, Hue[z]]]' $\endgroup$ – chris Jun 8 at 17:24
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    $\begingroup$ @chris: probably ColorFunctionScaling -> False, ColorFunction -> Function[{x, y, z}, Hue[Arg[psi[x]]/(2 \[Pi])]]] would have more physical meaning, but let's walk before we run. :-) $\endgroup$ – Michael Seifert Jun 8 at 17:29
  • $\begingroup$ Awesome, that was exactly what I was looking for. Thanks for taking your time! $\endgroup$ – Jonathan Bollig Jun 10 at 20:39

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