We can get coherent state from the formula $$|\alpha\rangle =D(\alpha)|0\rangle = \exp (\alpha a^\dagger-\alpha a)|0\rangle = \exp\left(-\frac{|\alpha |^2}{2} \right) \exp(\alpha a^\dagger) \exp(\alpha a)|0\rangle$$, Also can get by expanding in terms of Fock state $$|\alpha\rangle = \exp\left(-\frac{|\alpha |^2}{2}\right) \exp(\alpha a^\dagger)|0\rangle = \sum_n \frac{\alpha^n}{\sqrt{n!}}|n\rangle$$ Why these two are not the same?
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They are basically the same, because $$ \exp(\alpha a)|0\rangle = 0,$$ because annihilation operator acting on zero state $|0\rangle$ is still zero: $$a|0\rangle = 0$$ because there is no lower photon state than that. We know that $ \exp(\alpha a)|0\rangle = 0$ by expanding the exponent in power series. We introduce this additional exponent term for symmetric reasons and the Displacement operator to be Hermitian, as far as I understand.
