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It may be a stupid question, but why particularly for probability density expression $k~|\Psi|^2 = k~\Psi^{*}\Psi$, it's assumed that $k=1$? As it is now, then in a complex plane probability density is just a rectangular area for a complex vector. But why it has to be rectangular specifically? Why can't be $k=\pi$, so that re-defined probability density $\pi |\Psi|^2$ would mean a bounding circle area of complex vector:

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Or any other complex plane area scaling value $k$? What would be implications of that to quantum mechanics?

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    $\begingroup$ I mean, you want your probability density function to be normalized so that the $\int_{\textrm{all space}}p(x)dx=1$. So, provided that $\Psi$ is normalized in such a way that the integral of its squared-modulus is 1, this must be the choice. It's really about our definitions of probability functions, not about quantum mechanics. $\endgroup$
    – march
    Feb 10, 2022 at 18:43

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It's a normalization convention for $\Psi$ - indeed, the only sensible one. If the probability density is $k|\Psi|^2$, just absorb a $\sqrt{k}$ factor into $\Psi$. This density should not be interpreted as an area. Indeed, the real reason we square has nothing to do with $2$-dimensional geometry.

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As it is now, then in a complex plane probability density is just a rectangular area for a complex vector.

I don't think it's useful to visualize $|\psi|^2$ as the area of the rectangle whose side lengths are $\mathcal{Re}[\psi]$ and $\mathcal{Im}[\psi]$.


In the standard formulation of quantum mechanics, the states$^\dagger$ of a system are represented as elements of a Hilbert space $\mathscr H$, and observable quantities are represented as self-adjoint linear operators on $\mathscr H$. The expected value of an observable $\hat A$ in the state $\psi$ is given by $$\mathbb E_\psi[\hat A]:= \frac{\langle \psi,\hat A\psi\rangle}{\Vert \psi \Vert^2}= \frac{\langle\psi,\hat A\psi\rangle}{\langle \psi,\psi\rangle}$$

To simplify calculations, it is convenient (but not necessary) to choose $\psi$ to be normalized, i.e. $\Vert \psi\Vert^2 = 1$. If we make this choice, the expected value of the position operator $\big(\hat X\psi\big)(x) = x \psi(x)$ is given by

$$\mathbb E_\psi[\hat X] = \langle \psi, \hat X \psi\rangle = \int \mathrm dx \ \psi^*(x) \cdot \big(x \psi(x)\big) = \int \mathrm dx \ x |\psi(x)|^2$$

Comparing with the expected value of a random variable from standard probability theory, we recognize $|\psi(x)|^2$ as the probability density corresponding to the position variable.

Finally, note that if we had not normalized $\psi$, so $\Vert \psi \Vert^2 = C^2 \neq 1$, then we would find the probability density to be given by $|\psi(x)|^2/C^2$. As a result, the fact that the probability density is given by $|\psi(x)|^2$ with no additional numerical factors is merely a result of our convenient choice of normalization.


$^\dagger$In actuality, this is true only for so-called pure states. There is a more general notion of state in which they are allowed to be mixed, but that is beyond the scope of this explanation.

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