Ground state Dirac formalism In Dirac formalism, the action of the 'lowering' operator acting on a ground state energy eigenstate is given by
$$\hat{a}|0\rangle=0$$
Notation wise, it is clear what the left hand side express means.
What about the $0$ on the right hand side? Is this just matter of notation? 
I am sure $0$ is not the energy eigenvalue for a particle in an energy eigenstate corresponding to the ground state. Any particle has a positive energy eigenvalue in the ground state.
 A: The expression you wrote $\hat{a} |0 \rangle = 0$ is a great expression to force you to think carefully about what these symbols all mean. Good question!
Remember that quantum states live in a Hilbert space, which is a vector space plus a few extra properties. It often is useful to form analogies with other vector spaces that we are used to - for example, consider vectors in the XY Cartesian plane. In this plane, $\vec{x}$ and $\vec{y}$  can be picked as basis vectors. Then any vector in the whole plane can be expressed as $\vec{v} = a \vec{x} + b\vec{y}$.
Now, you see that if $a$ and $b$ are both 0, then $\vec{v} = 0$ is the 'null vector'. If you like, it is a linear combination of all basis vectors with coefficient 0 for each basis vector.
Now back to the Hilbert space: the state $|0\rangle$ is a basis vector in the Hilbert space, much like $\vec{x}$ in the XY plane. Just like in XY, general states in the Hilbert space can be written as $|\psi\rangle = \alpha|0\rangle + \beta|1\rangle + \ldots$
If you pick all coefficients to be 0, then you have just described the 'null vector' of the Hilbert space, and that is what is meant by the right hand side of the equation $\hat{a} |0\rangle = 0$. One note: the null vector is not a physical state that a system can have (you can think about the fact that it is not normalizable, for example). But it is a state in the Hilbert space nonetheless!
