# Why particularly probability density is defined as $|\Psi|^2=\Psi \Psi^{*}$?

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:

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

• 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. Feb 10, 2022 at 18:43

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.

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.