Consider a particle of mass $m$ in a harmonic oscillator potential $V(x)=\frac{1}{2}m\omega^2x^2$. I am given that $\psi(x,t)$ is an eigenfunction of the ladder operator $a=\frac{1}{\sqrt{2m\hbar\omega}}(m\omega x+ip)$ with eigenvalue $\alpha(t)=\alpha_0e^{-i\omega t}$. Here $\alpha_0$ is a real constant and $x$ and $p$ are the position and momentum operators.
I have shown that the uncertainties are $\Delta x=\sqrt{\frac{\hbar}{2m\omega}}$ and $\Delta p=\sqrt{\frac{\hbar m\omega}{2}}$, and hence the uncertainty principle is saturated for this wavefunction $\psi(x,t)$, that is $\Delta x \Delta p=\frac{\hbar}{2}$. I am asked: what is the physical interpretation of this wavefunction? I thought that $\psi(x,t)$ was the ground state wavefunction since this saturates the uncertainty principle, but surely it cannot be since $\psi$ would have to obey $a\psi=0$ if this were the case. Or does it just imply that $\psi(x,t)$ is a Gaussian wavefunction? (Is it then true that only Gaussian wavefunctions have such minimum uncertainty?)
Your state is not the ground state but rather a so-called "coherent state". In general coherent states are eigenstates of the annihilation operator and are considered to be "maximally classical" states. For example, your state represents a ground state that has been shifted from the origin of the oscillator (so that it is no more an eigenstates of the Hamiltonian, i.e. it is not a stationary state). This state oscillates back and forth and the average of the position operator in this state follows the classical motion of a particle in an harmonic trap. During the oscillation it maintains a Gaussian shape.
Regarding the uncertainty principle, the more concentrated $\psi$ is, the more spread out its Fourier transform is: it is not possible to arbitrarily concentrate both. This is exactly the uncertainty principle for the $x$ and $p$ operators. It is well known that the functions that minimise uncertainty are Gaussians.