Interpretation of the wave function in quantum mechanics I just started watching the coursera lectures on the basics of quantum mechanics and one of the first lectures were on deriving Schrodinger's equation and its interpretation it under Born's interpretation. What I want to ask is what the wave function \begin{equation} \psi({\bf r},t) \end{equation} return and represent. Now I know that
\begin{equation} |\psi({\bf r},t)|^2dxdydz \end{equation} is the probability of finding the quantum particle described by \begin{equation} \psi({\bf r},t) \end{equation} in the volument element \begin{equation} dV = dxdydz \end{equation} at time T. But im not sure what the wave function $\psi$ returns.
Could someone please explain in laymans term the return type of the $\psi$ function and what the lone $\psi$ function represents.
 A: If this were computer science, we might say $\psi$ takes a $d$-tuple of reals ($r$) and another real ($t$) and returns a complex number with the attached unit of $L^{-d/2}$ in $d$ dimensions (with $L$ being the unit of length).1
If you want any more of an interpretation, well then you've already given it: $\psi(r,t)$ is the thing such that $\int_R\ \lvert \psi(r,t) \rvert^2\ \mathrm{d}V$ is the probability of the particle being observed in the region $R$ at time $t$. You can loosely think of it as a "square root" of a probability distribution.
The reason the "square root" interpretation is not quite right, and probably the reason you aren't satisfied with the $\int_R\ \lvert \psi(r,t) \rvert^2\ \mathrm{d}V$ definition, is that any particular instance of $\psi(r,t)$ carries extraneous information beyond what is needed to fully specify the physics. In particular, if we have $\psi_1$ describing a situation, then the wavefunction defined by $\psi_2(r,t) = \mathrm{e}^{i\phi} \psi_1(r,t)$ gives identical physics for any real phase $\phi$.
So the return value of the wavefunction itself is not a physical observable -- one always takes a square magnitude or does some other such thing that projects many mathematically distinct functions onto the same physical state. Even once you've taken the square magnitude, $\lvert \psi(r,t) \rvert^2$ arguably isn't directly observable, as all we can measure is $\int_R\ \lvert \psi(r,t) \rvert^2\ \mathrm{d}V$ (though admittedly for arbitrary regions $R$).

1You can check that $-d/2$ is necessarily the exponent. We need some unit such that squaring it and multiplying by the $d$-dimensional volume becomes a probability (i.e. is unitless). That is, we are solving $X^2 L^d = 1$, from which we conclude $X = L^{-d/2}$.
A: The wavefunction returns a complex number whose modulus-squared is a probability density and whose phase is related to the probability current, i.e., where probability is flowing to.
If you write in the form
$$\Psi({\mathbf x},t) = \sqrt{\rho({\mathbf x},t)}\exp\left(i\frac{S({\mathbf x},t)}{\hbar}\right)\text{,}$$
where $\rho(\mathbf{x},t)\geq0$ and $S(\mathbf{x},t)$ is real, then $\rho(\mathbf{x},t)$ is the probability density of measuring the position of the particle at $\mathbf{x}$ at time $t$.
Plugging this form into the Schrödinger equation with Hamiltonian $\hat{H} = -\frac{\hbar^2}{2m}\nabla^2 + V(\mathbf{x})$ gives:
$$\begin{eqnarray*}
\left[\frac{1}{2m}\left|\nabla S\right|^2 + V\right] + \frac{\partial S}{\partial t} &=& \frac{\hbar^2}{2m}\rho^{-1/2}\nabla^2\sqrt{\rho}\text{,}\\
\frac{\partial\rho}{\partial t} + \nabla\cdot\left(\frac{1}{m}\rho\nabla S\right) &=& 0\text{.}
\end{eqnarray*}$$
The first equation is not important for our immediate purposes, but the second one is the continuity equation for the probability density $\rho$, forcing probability to be conserved. In other words, if we define
$$\mathbf{J} = \frac{\rho}{m}\nabla S\text{,}$$
then where $\rho$ is the probability density, $\mathbf{J}$ is the probability current representing where the probability is flowing.
A: I will answer in layman terms as from your age in your profile I would not expect a very strong background in the necessary mathematics.
When we have a function f(x) it returns a value at the point x, a real number.  If it is the potential, V(r)=1/r we are able to calculate the potential and solve simple problems or enter the potential in complicated equations and solve complicated problems. This is a real function.
One can extend functions using complex numbers. Complex number functions are in reality two functions:
f1(x,y,z,t) + f2(x,y,z,t)*i
Here i is the square root of the real number (-1). Using complex numbers simplifies the form of equations generally and allows for easier manipulation of theoretical quantities.
The Schrodinger equation is an equation in complex numbers and thus
\begin{equation} \psi(r,t) \end{equation}
is a solution of the equation in our everyday space and time with two functions attached, it returns a complex value with two real numbers, the second one attached to i .
Our measurements in real life return real numbers, so as it is Psi has no use. That is why it is squared with its complex conjugate , to give a real number which will represent the probability in the volume element. This last is a postulate, the Born rule, and it has been found to work and has not been falsified experimentally, and thus we accept the quantum mechanical framework of nature.
