Complex Nature of the Wavefunction I am doing this problem and I realized the wavefunction is real. But we also already showed that the wavefunction needs to be complex. Why is it that the wavefunction given here is real? I first taught $N$ could be complex, but it cannot since when you compute it to normalize $\Psi (\mathbf{x})$, it is a function of $a_o$ and $r$ which are both reals.

 A: It is complex, just with the imaginary component being zero.
A: An eigenfunction like the hydrogen ground state referred to in the question separates into time and space parts:
$$ \Psi(r,\theta,\phi,t) = \psi(r,\theta,\phi) ~ e^{-iEt/\hbar} $$
where $E$ is the energy of the state. However the exponential part isn't a physical observable so we tend to ignore it and just write:
$$ \Psi(r,\theta,\phi,t) = \psi(r,\theta,\phi) $$
In the case of the hydrogen ground state $\psi(r,\theta,\phi)$ is real, but this is only part of the wavefunction and the whole wavefunction is complex.
A: If you define the spatial probability density $\rho(x,t) = \psi^*(x,t)\psi(x,t)$ and compute its derivative with respect to time, you obtain after some algebra
$$\frac{\partial \rho}{\partial t} = -\frac{\partial}{\partial x} \left[\frac{i\hbar}{2m} \left(\frac{\partial \psi^*}{\partial x}\psi - \psi^* \frac{\partial \psi}{\partial x}\right)\right]\equiv -\frac{\partial}{\partial x} J$$
where $J(x,t)$ is called the probability current.  It obeys a continuity equation $\frac{\partial \rho}{\partial t}+\frac{\partial J}{\partial x} = 0$, which means that if $\rho$ changes in some region, then there must be some flux of $J$ through the region's boundary.  You can think of it as being the probability-oriented counterpart to the familiar charge/current continuity equation for electromagnetism.
If you write $\psi(x,t) = \alpha(x,t) e^{i\phi(x,t)}$ where both $\alpha(x,t)$ and $\phi(x,t)$ are real (we can always do this) and plug the result into the expression for $J$, we find that
$$J(x,t) = \frac{\hbar \alpha}{m} \frac{\partial \phi}{\partial x}$$
The interpretation is that spatial variations in the wavefunction's phase correspond to a non-zero probability current.

You can always multiply any element of the Hilbert space by a global phase factor, so you can take a purely real wavefunction and make it complex by multiplying it by $i$. However, in order to have a fully real $\psi$ in the first place, it must be that the phase is spatially constant - meaning that the probability current must vanish.
In 1D, this is always true of energy eigenfunctions$^\dagger$ (disregarding the trivial $e^{-iEt/\hbar}$ factor which arises from time-evolution).  One can see this by noting that since $\rho$ is constant in time for such states, $\frac{\partial J}{\partial x}=0$; from there, since an energy eigenfunction must be normalizable, it follows that $A\rightarrow 0$ as $x\rightarrow \pm \infty$ so $J=0$ everywhere.
In higher dimensions, $\nabla \cdot \mathbf J = 0$ does not imply that $\mathbf J=0$.  In particular, when a state has angular momentum about some axis, there will be a non-vanishing probability current corresponding to an angular "flow".  However, when angular momentum is not present (as is the case for $m=0$ orbitals, then the probability current vanishes, as does the spatial variation of the phase.
$^\dagger$As a side note, this is not true of "generalized" (i.e. non-normalizable) eigenfunctions, like those of the free particle on a line.
A: Hydrogen ground state identified by quantum numbers $(n=1,\ell =0,m=0)$ has no dependancy on phase, so ground state has phase factor $e^{i \phi} = 1$. However for other Hydrogen energy levels situation can be different. For example Hydrogen state identified with quantum numbers $(2,1,\pm 1)$ does depend on phase and as such state's wavefunction has full complex form :
$$ \psi _{2,1,\pm 1}=\mp {\frac {1}{8{\sqrt {\pi }}a_{0}^{3/2}}}{\frac {r}{a_{0}}}e^{-r/2a_{0}}\sin \theta ~e^{\pm i\phi } $$
