Non-complex wave function I am taking an introductory course on QM. I came across the following question:

A particle is described by the wave function $$ \psi(x) = Ae^{-ax^2}  $$
  where $A$ and $a$ are positive, real constants. If the value of $a$ is increased what effect does this have on the particle's uncertainty in position and particle's uncertainty in momentum.

I thought that the wave function must be a complex and include imaginary variables, but this wave function is not. Also I don't understand how I can relate $a$ to uncertainty.
 A: Wavefunctions are in general complex, but there is nothing preventing a specific wavefunction from being real. In fact, there are certain cases for which you can show that there is always a real wavefunction that describes the system (it's called a Schmidt decomposition, and applies when your system is made up of an even number of subsystems).
A: You are looking at a solution of the time-independent Schrodinger equation as your $\psi(x)$ does not have any time dependence,  and the basic solutions of the time-independent equation can often be real.  Linear combinations of these basic solutions can be complex.
The solutions to the time-dependent Schrodinger equation are always linear combinations of the form
$$
\Psi(x,t)=\sum_n c_n e^{-iE_nt/\hbar} \psi_n(x)
$$
and will be complex even if the time-independent functions $\psi_n(x)$ are real.
To relate $a$ to the uncertainty relation you would need to compute $\Delta x^2$ and $\Delta p^2$ using your $\psi(x)$ (which you will have to normalize) and to find how $a$ enters into the product 
$\Delta x\Delta p$. 
To give you a hint I'm including the plot of $\psi(x)^2$ for $a=1$ (black), $a=2$ (blue) and $a=1/2$ (red).

A: You are right, the "real" wave function for a particle is a complex function $\Psi(\vec{x}, t)$, which follows the time-dependent Schrödinger equation:
$$i\hbar\frac{\partial\Psi(\vec{x}, t)}{\partial t} = \hat{H}\Psi(\vec{x}, t)$$
Where $\hat{H}$ is the hamiltonian operator which gives the total energy of the 
system, consisting of the kinetic energy and the potential energy, which is given by the potential energy function $V(\vec{x}, t)$:
$$\hat{H}\Psi(\vec{x}, t) := -\frac{\hbar^2}{2m}\nabla^2\Psi(\vec{x}, t) + V(\vec{x}, t)\Psi(\vec{x}, t)$$
If the potential energy function is truly a function of time as well as position, this equation becomes very difficult to solve. Luckily, however, the potential energy is almost always only dependent on the position and therefore becomes a simpler function $V(\vec{x})$. In that case, the Schrödinger equation can be solved by assuming the wave function $\Psi(\vec{x}, t)$ to be a simple product of a function $\psi(\vec{x})$ solely dependent on the position and a function $f(t)$ that only depends on the time:
$$\Psi(\vec{x}, t) = \psi(\vec{x})f(t)$$
Then we are able to seperate the position and time dependent parts of the equation and from that find that $f(t)$ has to be equal to $e^{-iEt/\hbar}$, where $E$ is some real constant (which can be proven to be the energy of the particle). Furthermore, $\psi(\vec{x})$ must follow the so-called time-independent Schrödinger equation:
$$E\psi(\vec{x}) = \hat{H}\psi(\vec{x})$$
where $E$ is the same constant. It can be proven that $\psi(\vec{x})$ can always taken to be a real function (That is, if you do have a solution to the time-independent Schrödinger equation that isn't always realy, you can always express it as a linear combination of ones that are. It does not mean that every solution necessarily has to be real.). Then we have:
$$\Psi(\vec{x}, t) = \psi(\vec{x})e^{-iEt/\hbar}$$
Remember, that this is only under the assumption, that $\Psi(\vec{x}, t)$ CAN be written as such a product. However, all the other solutions, can then be expressed as linear combinations of these simple time-independent solutions, as Zero already pointed out.
Your question supposes a time-independent wave function that is equal to
$$\psi(x) = Ae^{-\alpha x^2}$$
(This is a one-dimensional function, so I'm leaving the arrow off) Therefore, the time-dependent wave-function you know would be
$$\Psi(x, t) = Ae^{-\alpha x^2}e^{-iEt/\hbar}$$
To find $E$, you can apply the Schrödinger equation:
$$EAe^{-\alpha x^2} = \hat{H}(Ae^{-\alpha x^2})$$
(However, you would need to know the potential energy function you used to derive that wave function)
