Okay so this has been bugging me for a while now because everyone I talked to so far didn't know the answer. I'm an A level student but I got quite interested in quantum physics and so I decided to buy Griffiths introductory QM book and just try to understand the maths and how things fit together (I'm only on the first few pages, maybe 24 I think). So far the only thing I didn't understand was that why the mod squared of the wave function is the probability density. I mean I understand that it has to be real but then why don't we square root it afterwards? Is it something that just works or there is an actual reason behind the whole thing?
[Physics] mod squared of wave function probability density
quantum mechanicswavefunction
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You should try solving this on a piece of paper. I don't think that you understand correctly how the author passes from $\partial/\partial t$ to $\partial^2/\partial^2 x$. You can not use Schrodinger's equation for $|\Psi|^2$, it is only valid for $\Psi(x, t)$:
$$ i \hbar \frac{\partial}{\partial t} \Psi(x, t) = -\frac{\hbar^2}{2 m} \frac{\partial^2}{\partial x^2} \Psi(x, t). $$
For $\Psi^*$ another equation holds, which can be obtained by conjugating both sides of the Schrodinger's equation:
$$ - i \hbar \frac{\partial}{\partial t} \Psi^*(x, t) = -\frac{\hbar^2}{2 m} \frac{\partial^2}{\partial x^2} \Psi^*(x, t). $$
Note the minus sign on the l.h.s. It comes from conjugating an imaginary unit ($i^* = -i$). This minus sign is responsible for the minus sign in your answer.
Now I trust you to carefully expand both parts of your equation and see for yourself that they are indeed equal.
In fact, it is nothing but a vague and, strictly speaking, false requirement that can be found in some physically minded books (also of very good level).
There are however physical situations where regularity of the used functions implies that they must vanish at infinity. If one is solving the stationary Schroedinger equation and the potential is sufficiently regular, the eigenvectors must be accordingly regular because of known theorems (especially due to Weyl) on elliptic regularity. In some cases regularity plus the requirement that the function belongs to $L^2$ and some control of the asymptotic value of some derivatives (arising from a nice asymptotic shape of the potential) imply that the wavefunction must vanish at infinity.
On the other hand, from a physical viewpoint, it is impossible to prepare a state in space working arbitrarily far from a given position where physical instruments are localized. It is therefore reasonable to assume that physically realisable states are described by wavefunctions vanishing outside a sufficiently large spatial region. This, in turn, implies a corresponding requirement on physically accessibile observables and their realistic realisation (it does not make sense to assume to fill in the universe with detectors). However these are physical requirements that should not be confused with mathematical restrictions.
From a pure mathematical viewpoint, instead, no requirement about a rapid decay at infinity applies on vawefunctions in $L^2$ (it is easy to construct smooth $L^2(\mathbb R)$ functions with larger and larger oscillations as $x\to \infty$ and corresponding strongly unphysical Hamiltonian operators whose these monsters are eigenfunctions). Nor any strong regularity condition applies. Indeed, these functions are defined up to zero-measure sets and all operators representing observables, in order to be properly self adjoint (not simply Hermitian) are the closure of differential operators whose final domains are therefore Sobolev-like spaces: Derivatives are required to exist in a weak sense at most.
The only exceptions are probably eigenfunctions of Hamiltonian operators with sufficiently regular potentials, where both Sobolev's regularity and elliptic regularity results can be applied and outside singularities of the potential, eigenfunctions are $C^2$ (or even smooth) in a proper sense.
Restriction to the Schwartz space may have sense because most operators have self-adjointness domains including that space (which sometimes is also a core of the operators) and also because Schwartz space is dense in $L^2$.
However it turns out to be a too strong restriction also in some elementary cases. Think of an 1D Hamiltonian operator with a potential with some mild discontinuity. It does not admit eigenfunctions of Schwartz type.
Example.
I construct here a true monster without physical meaning, but permitted by mathematical requirements of QM.
Consider a function $\psi \in L^2(\mathbb R)$ costructed this way. It attains the value $0$ everywhere except for every interval $$I_n= \left(n -\frac{1}{2(n^4+1)^{2}},\:\: n+ \frac{1}{2(n^4+1)^{2}}\right) $$ thus of length $(n^4+1)^{-2}$ and centered on every $n\in \mathbb Z$. In these intervals $$\psi(x) = n^2 \quad x\in I_n\:.$$ It is clear that $\int_{\mathbb R} |\psi(x)|^2 dx$ for some constant $C>0$ satisfies $$\int_{\mathbb R} |\psi(x)|^2 dx \leq C \sum_{n=0}^{+\infty} \frac{n^4}{(n^4+1)^2} <+\infty\:.$$ It is clear that this function oscillates with larger and larger oscillation as $|x|\to +\infty$, but these oscillations are sharper and sharper.
Next, smoothly modify $\psi$ in every interval $I_n$ to produce a smooth non-negative function $0\leq \phi(x) \leq \psi(x)$ with $\phi(n)=\psi(n)$ thus preserving finiteness of the integral of $|\phi|^2$.
Finally define $$\Psi(x) := \phi(x) + \frac{1}{1+x^2}\:.$$ It easy to see that, since $\phi(x)\geq 0$, it must be $$\Psi(x) >0\tag{1}$$ and $$\Psi \in L^2(\mathbb R)\cap C^\infty(\mathbb R)$$ because $\Psi$ is sum of two $L^2$ (smooth) functions and $L^2(\mathbb R)$ is a vector space. A possible qualitative shape of $\Psi$ is represented here
Observe that $\Psi$ is an eigenfunction (a proper eigenvector since it belongs to $L^2$), of the Hamiltonian $$H = -\frac{\hbar^2}{2m}\frac{d^2}{dx^2} + U(x)\:, $$ where $$U(x) := \frac{\hbar^2}{2m}\frac{\Psi''(x)}{\Psi(x)}$$ is smooth by construction, in particular, (1) holding.
It holds $$H\Psi = E \Psi$$ with $E=0$. Obviously all this construction is physically meaningless, but it is permitted by general requirements of QM formalism.
Best Answer
The main reason is that the probability density, constructed as you indicated, leads to a CONSERVED overall probability over all space.
In most QM books it is shown that the modulus square of the wave function, integrated over all space, does not depend on time. It can then be normalized to 1. The derivation of the conservation of the probability over all space being conserved is from the Schrodinger equation. I've not used Griffith but if it's a standard book it should have that derivation. The 'trick' is to find a current j in terms of the wave function and its complex conjugate such that the divergence of j plus the partial of the probability density with time is zero. By Gauss's theorem then the derivative wrt time of the integral of the density over space is zero.
You can then use the prob. density to get the average (I.e., expectation) values of dynamical valuable, or observables, like position and momentum, and get the Newton equations for those expectation values, and appropriate conserved quantities.
You can in more advanced books get the current density j and probability density from the Lagrangian for the quantum theory (Schrodinger or later relativistic quantum field theories) In a continuity of probability equation that conserves the probability integrated over the appropriate space.
There are also hysical reasons such as the similarity with the energy density in electromagnetism. It is also the the wave variables, E and B, squared, and then the energy is conserved when integrated over space.
The square root of the modulus square has been shown to not obey a conservation equation, so that was ruled out.
You can see a little of the math and essentially the same answer, stated with some differences, at Born Interpretation of Wave Function