What is the position wavefunction of coherent states?

Derivation from the eigenvalue condition

The most straightforward approach is to start from the position representation of the annihilation operator $a$.

I'll use the convention $a=frac{1}{sqrt2}(x+ip)$ and $a^dagger=frac{1}{sqrt2}(x-ip)$, corresponding to which the canonical commutation relations read $[a,a^dagger]=1$ and $[x,p]=i$. In the position representation, this corresponds to $a=frac{1}{sqrt2}(x+partial_x)$.

The eigenvalue condition $a|alpharangle=alpha|alpharangle$ then corresponds to $$xpsi_alpha(x)+psi_alpha'(x)=sqrt2alphapsi_alpha(x),$$ which we can rewrite as $$psi_alpha'(x) = (sqrt2alpha-x)psi_alpha(x).$$ A natural ansatz to solve this is $psi_alpha(x)=C e^{f(x)}$ for some constant $C$. Using this we get $$f'(x) = sqrt2alpha-x Longrightarrow f(x) = sqrt2alpha x - frac{x^2}{2} + C'.$$ Upon some simple rearranging of the terms we get $psi_alpha(x) propto exp[-frac12(x - sqrt2alpha)^2]$, which ensuring normalisation finally leads to $$psi_alpha(x) = frac{e^{-alpha_2^2}}{pi^{1/4}} expleft[-frac12(x - sqrt2alpha)^2right],tag{$R_1$}$$ where $alpha=alpha_1+ialpha_2$.

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Derivation from (A)

Consider the wavefunctions $psi_n$ of $|nrangle$, which have the form $$psi_n(x) = frac{1}{pi^{1/4}sqrt{2^n n!}} e^{-x^2/2}H_n(x),tag B$$ where $H_n$ are the Hermite polynomials, defined here as $H_n(x)=(2x-partial_x)^n cdot 1$. This expression for $psi_n$ comes from $$psi_n(x) = langle x|frac{a^{dagger n}}{sqrt{n!}}|0rangle = frac{1}{pi^{1/4}sqrt{2^n n!}} (x-partial_x)^n e^{-x^2/2} = frac{1}{pi^{1/4}sqrt{2^n n!}} e^{-x^2/2}H_n(x).$$ Using (B) in (A), we get $$ psi_alpha(x) = e^{-|alpha|^2/2} frac{e^{-x^2/2}}{pi^{1/4}} sum_{k=0}^infty frac{alpha^k}{sqrt{k!}} frac{1}{sqrt{2^k k!}} H_k(x). $$ We now consider the identity $$sum_{k=0}^infty H_k(x) frac{t^k}{k!} = e^{2xt - t^2}.$$ Using this with $t=alpha/sqrt2$ we get $$psi_alpha(x) = e^{-|alpha|^2/2} frac{e^{-x^2/2}}{pi^{1/4}} e^{sqrt2alpha x - alpha^2/2} = frac{1}{pi^{1/4}}e^{frac12(alpha^2-|alpha|^2)}expleft[ -frac12(x-sqrt2alpha)^2 right]. tag{$R_2$} $$ Observing that $alpha^2-|alpha|^2=-2alpha_2^2 + 2ialpha_1alpha_2$, we see that ($R_2$) is consistent with ($R_1$), up to an irrelevant global phase. An analogous derivation is also given in Gerry, Knight (2004), section 3.3.