I've always wondered why we bother with wavefunctions instead of dealing with simply $|{\Psi}|^2$ which is obviously more physically intuitive. I've been told it's because the wavefunction is more "descriptive", as it encodes probability and phase of wavefunctions. Can someone elaborate on this? I don't see how complex exponentials can describe phases - it's not apparent to me.
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Hint: what is $|e^{i\phi}|$? Therefore, what does the set of all $e^{i\phi}$ describe? – probably_someone Apr 28 '18 at 17:06
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@probably_someone It should be $1$, right? $|e^{i \phi}| = e^{i\phi}e^{-i\phi} = 1$. So, the set of all $e^{i \phi}$ would be just $1$? – sangstar Apr 28 '18 at 17:21
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possible duplicate: Form of Schrödinger equation for the probability density. – AccidentalFourierTransform Apr 28 '18 at 17:29
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@sangstar Not quite. It should be the set of all complex numbers $z$ with $|z|=1$; in other words, the unit circle. How do the points in the unit circle relate to phases? – probably_someone Apr 28 '18 at 17:32
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In general, $|\psi_1 + \psi_2|^2$ is not the same as $|\psi_1|^2 + |\psi_2|^2$. If they’re negatives of each other, for example, the first is zero and the second is not. So the relative sign, and more generally the relative phase, is important to the physical result. That’s why just the magnitude isn’t enough.
Bob Jacobsen
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Other people gave reasonable explanations why $|\Psi|^2$ is not enough, so let me explain why it actually can be enough :-), for example, for a particle in electromagnetic field. As Schrödinger noted (Nature (London) 169, 538 (1952)), you can always make the wavefunction real using a gauge transformation (https://quantummechanics.ucsd.edu/ph130a/130_notes/node296.html). Is it worth losing the freedom of choice of the gauge? I guess this depends on the specific problem.
akhmeteli
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