many-body state in second quantization The ground state of a system of N particles is represented as 
$$
\mid \Psi \rangle = \frac{1}{\sqrt{2^NN!}}\big( \hat{a}_1^{\dagger} + \hat{a}_2^{\dagger} \big)^{N} \mid 0\rangle
$$
or similarly
$$
\mid \Psi \rangle = \frac{1}{\sqrt{2^NN!}}\big(e^{-i\phi/2} \hat{a}_1^{\dagger} + e^{i\phi/2}\hat{a}_2^{\dagger} \big)^{N} \mid 0\rangle
$$
How do I get an insight into what kind of state is this ?
I understand what these kinds of states mean:
$$
\mid \Psi \rangle = \frac{1}{{(N/2)!}}(\hat{a}_1^{\dagger})^{N/2} (\hat{a}_2^{\dagger})^{N/2} \mid 0\rangle
$$
or 
$$
\mid \Psi \rangle = \frac{1}{{(N)!}}(\hat{a}_2^{\dagger})^{N} \mid 0\rangle
$$
Note: Check - ref., Eq. 9, 11, 28
 A: To avoid discussing symmetrization, I will assume you are talking about bosons.
$$
\mid \Psi \rangle = \frac{1}{{(N/2)!}}(\hat{a}_1^{\dagger})^{N/2} (\hat{a}_2^{\dagger})^{N/2} \mid 0\rangle
$$
is a state where you have $\frac{N}{2}$ type-1 particles and $\frac{N}{2}$ type-2 particles. You have in total $N$ particles, or equivalently, $N$ excitations of the vacuum.
The state
$$
\mid \Psi \rangle = \frac{1}{\sqrt{2^NN!}}\big( \hat{a}_1^{\dagger} + \hat{a}_2^{\dagger} \big)^{N} \mid 0\rangle
$$
also contains $N$ excitations of the vacuum. But this time, the excitations are not simply $\hat{a}_1^{\dagger}$ or $\hat{a}_2^{\dagger}$ ; instead, each excitation is the symmetric superposition $\frac{\hat{a}_1^{\dagger} + \hat{a}_2^{\dagger}}{\sqrt{2}}$.
To understand how these two states are different, imagine you want to measure how many 1-states ans 2-states are present in your system. If you do this measurement in the first case, you will find exactly $\frac{N}{2}$ 1-states and $\frac{N}{2}$ 2-states.
Now if you perform this measurement in the second case, each individual $\frac{\hat{a}_1^{\dagger} + \hat{a}_2^{\dagger}}{\sqrt{2}}$ state has a $1/2$ probability of being in the 1-state and $1/2$ of being in the 2-state. Assuming $N$ is large, you will probably find about half of the particles in the 1-state and the other half in the 2-state : the probability distribution of states will be peaked around this $1/2:1/2$ value with a certain width.
