quantum mechanics - Approximate expression for the ground state of hopping Hamiltonian

In second quantization, the Hamiltonian describing the hopping process between two neighboring sites is given ($N$ - number of particles and $M$ - number of sites) by:

$$\hat{\mathcal H} = J\sum\limits_{\langle i,j \rangle}\hat{a}^{\dagger}_{i}\hat{a}_{j}$$

It can be diagonalized using Fourier series

$$\hat{a}_{i} = \frac{1}{\sqrt{M}}\sum\limits_{\mathbf{k}} \hat{b}_{\mathbf{k}}e^{-i\mathbf{k}\cdot\mathbf{R}_i}$$

The ground state is given by

$$|\mathrm{GS}\rangle = \frac{1}{\sqrt{N!}}\left( \hat{b}^{\dagger}_{\mathbf{k} = \mathbf{0}}\right)^N | 0 \rangle \approx C\ e^{\sqrt{N}\hat{b}^{\dagger}_{\mathbf{k} = \mathbf{0}}}| 0 \rangle$$

How is the approximation justified ($C$ ensures normalization)?

I calculated a few quantities using both states. First of all both states can be represented using site creation operators $\hat{a}_i$ in the following way: $$|\rm GS_{1}\rangle = \frac{1}{\sqrt{N!}}\left[\frac{1}{\sqrt{M}}\sum\limits_{i=1}^{M}\hat{a}_{i}^{\dagger} \right]^{N} |0\rangle$$ $$|\rm GS_2 \rangle = \prod\limits_{i=1}^{M}e^{\sqrt{\frac{N}{M}}(\hat{a}^{\dagger}_i - \hat{a}_i)}|0\rangle$$ so the first state is $\rm SU(M)$ coherent state while the second one is the product of Glauber coherent states. A few quantities of interest:

1. 
   

   Energy - expectation value of the Hamiltonian: $$\langle \hat{\mathcal H} \rangle_{1} = 2JN$$ $$\langle \hat{\mathcal H} \rangle_{2} = 2JN$$

   
   
   


2. 
   

   Particle density and fluctuations: $$\langle \hat{n}_{i} \rangle_{1} = \frac{N}{M},\ \ \ \langle \Delta \hat{n}_{i}^{2} \rangle_{1} = \frac{N}{M}\left(1 - \frac{1}{M} \right)$$ $$\langle \hat{n}_{i} \rangle_{2} = \frac{N}{M},\ \ \ \langle \Delta \hat{n}_{i}^{2} \rangle_{2} = \frac{N}{M}$$

   
   


3. 
   

   Total number of particles and fluctuations: $$\langle \hat{N} \rangle_{1} = N,\ \ \ \langle \Delta \hat{N}^{2} \rangle_{1} = 0$$ $$\langle \hat{N} \rangle_{2} = N,\ \ \ \langle \Delta \hat{N}^{2} \rangle_{2} = N$$

   
   

As @NorberSchuch mentioned in his answer below particle density fluctuations are practically the same and coincide in the thermodynamic limit.

Answer

The basic idea is that the term with the highest weight in the exponential series is exactly the desired term. Further, all of the weight is in terms with closeby particle number, and small fluctuations in this number do not matter in many cases.

(Note: I will omit the $k=0$ subscript and just write $b^\dagger$ in the following for convenience.)

First, we have that
$$ e^{\sqrt{N}b^\dagger} |0\rangle = \sum \frac{\sqrt{N}^k}{k!} (b^\dagger)^k |0\rangle\ . $$ Let us now study which term in the series has the highest weight. To this end, we have to make sure that the corresponding states have the same normalization. This is, if we define a normalized state $$ |k\rangle = \frac{(b^\dagger)^k}{\sqrt{k!}}|0\rangle\ , $$ each of which appears in the series with a probability $$ |c_k|^2 = \frac{N^k}{k!}\ . $$ This is an (unnormalized) Poisson distribution, whose mean is exactly at $k=N$, as desired.

Note that this does in fact not mean that the two states are close in any distance measure (in fact, they aren't). However, it tells us that most of the weight in the sum is in states with $|k-N|\le c\sqrt{N}$, as the variance of the Poisson distribution is $N$. Since in many applications, we in fact care about the number of particles per site (which is an intensive quantity and well-defined as the system size goes to infinity), and any such fluctuation in $k$ will vanish as $N\rightarrow\infty$, both states will have the same particle density in the thermodynamic limit, and thus describe the same physics.