According to my knowledge (which is feeble) we will also get the same result if used direct values. For example, if the probability of something happening is 4% or 0.04, we should make an arrow of 0.04 length, but we make one of 0.02, why?
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1$\begingroup$ Exact duplicate of : physics.stackexchange.com/q/57595 $\endgroup$– user108787Commented Sep 17, 2016 at 22:19
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$\begingroup$ Just to be clear, Feynman uses real value amplitudes in his book examples, and I think he avoids complex numbers, but in actual life, complex valued amplitudes are the norm, so to get a value that corresponds with real life, you need a real, non complex, number. Squaring does that. $\endgroup$– user108787Commented Sep 17, 2016 at 22:34
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2$\begingroup$ Partly because, mathematically, wavefunctions are vectors in a $L^2$ Hilbert space, which is complex-valued. Squaring the amplitude, rather $\Psi^{*} \Psi = |\Psi|^2$ is one way to ensure that you get real-valued probabilities, which is also related to the fact that according to Sturm-Liouville theory (of which the Schrodinger equation is of such a form), the S-L operator yields real eigenvalues, and so on... $\endgroup$– Dr. Ikjyot Singh KohliCommented Sep 17, 2016 at 23:05
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1$\begingroup$ @IkjyotSinghKohli thanks very much for that, I always wanted to connect this point to math in a more formal way, because there is often some subtle aspect I fall over on. $\endgroup$– user108787Commented Sep 17, 2016 at 23:19
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1$\begingroup$ @CountTo10 NP. Yes, I very much agree. $\endgroup$– Dr. Ikjyot Singh KohliCommented Sep 18, 2016 at 3:13
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