Understanding wavefunctional form of lowering operator for a scalar CFT

In his book (String Theory Volume 1, pp. 66-68), Polchinski explains state-operator correspondence and derives the vacuum state through the identity operator:

$$
Psi_1[X_b] propto exp left( -frac{1}{alpha’} right) sum_{m=1}^infty m X_m X_{-m} tag{2.8.24}
$$

where $X_m$ comes from the expansion of the field at the boundary:
$$
X_b(theta) = sum_{n=-infty}^{infty} X_n e^{intheta} tag{2.8.19}
$$
or in terms of holomorphic/antiholomorphic coordinates:
$$
X_b(z, bar{z}) = sum_{n=1}^{infty} left( z^n X_n + bar{z}^n X_{-n} right).
$$

He then states that the $alpha_{m}$ operator, which was defined previously as:
$$label{eq:alpha}
alpha_{m} = left(frac{2}{alpha’}right)^{1/2} oint frac{dz}{2 pi} z^{m} partial X(z) tag{2.7.2a}
$$
acts on the wavefunctional $Psi_1$ using the Schrödinger basis operator:
$$
alpha_n = – frac{i n}{(2 alpha’)^{1/2}} X_{-n} – i left( frac{alpha’}{2} right)^{1/2}
frac{partial}{partial X_n} tag{2.8.25a}
$$
and comments that this:

follows from the Laurent expansion at $|z| = 1$ and the mode algebra.

How can such Laurent expansion around that point be used in deriving this? And what does the derivation look like?

---

I’ve tried inserting $ref{eq:alpha}$ into the path integral:
$$
Psi[X_b] = int left[ d X_iright]_{X_b} exp left( -frac{1}{2 pi alpha’} int d^2 z partial X bar{partial} X right) tag{2.8.20}
$$
using field expansions:
begin{align}
X_i & = X_{text{cl}} + X’_i, tag{2.8.21a}
X_text{cl}(z, bar{z}) & = X_0 + sum_{n=1}^{infty} (z^n X_n + bar{z}^n X_{-n}) tag{2.8.21b}
end{align}
but couldn’t make much progress this way.