A question about vertex operator (skip disclaimer)
I have a question about writing raising and lowering operators in the Schroedinger basis in the section of vertex operator in Polchinski's string theory vol 1 p.68.
It is given

$$ \alpha_n = - \frac{in}{(2 \alpha')^{1/2}} X_{-n} - i ( \frac{ \alpha'}{2} )^{1/2} \frac{ \partial}{\partial X_n}, \,\,\,\, (2.8.25a) $$
  $$ \tilde{\alpha}_n = - \frac{in}{(2 \alpha')^{1/2}} X_{n} - i ( \frac{ \alpha'}{2} )^{1/2} \frac{ \partial}{\partial X_{-n}}, \,\,\,\, (2.8.25b) $$

I cannot find a way to derive these equations, especially how to find an expression of $$\frac{ \partial}{\partial X_n}?$$
Although they are similar with the harmonic oscillator in quantum mechanics. My question is, how to derive Eqs. (2.8.25a) and (2.8.25b)?
 A: A partial answer.
First, the commmutation relations are :
$$[\alpha_n,\alpha_{-n}]= n$$
where we are skipping here and below the space-time indices $\mu,\nu$, and we did not consider $\tilde \alpha_n$ because it is easily obtained 
from changing $X_n$ in $X_{-n}$. We are considering here that $n>0$
By analogy with the harmonic oscillator, we could thing of something like.
$\alpha_n = aX_n \pm b \large \frac{\partial}{\partial X_n}$, $\alpha_{-n} = aX_{-n}  \pm b \large \frac{\partial}{\partial X_{-n}}$
But this does not work, because in this case, we will have $[\alpha_n,\alpha_{-n}]= 0$, because :
$[X_n, X_{-n}]=0, [X_n, \large \frac{\partial}{\partial X_{-n}}]=0,[\frac{\partial}{\partial X_{n}}, X_{-n}]=0, [\frac{\partial}{\partial X_{n}}, \frac{\partial}{\partial X_{-n}}]=0$
So, a second attempt, and taking in account, that $X_n,X_{-n}$ has dimension $\sqrt{\alpha'}$ ($2.8.24$) is (with $n >0$): 
$$\quad \alpha_n = \frac{a}{\sqrt{2\alpha'}}X_{-n}+b\sqrt{\frac{\alpha'}{2}} \frac{\partial}{\partial X_{n}}$$
$$ \quad \alpha_{-n} = -\frac{a}{\sqrt{2\alpha'}}X_{n}+b\sqrt{\frac{\alpha'}{2}} \frac{\partial}{\partial X_{-n}}$$
We have, with some algebra : 
$$[\alpha_n,\alpha_{-n}]= \frac{ab}{2}([X_{-n},\frac{\partial}{\partial X_{-n}}] -[\frac{\partial}{\partial X_{n}},X_{n}]) = -ab$$
So, finally, $ab=-n$, and you could check that this is the case in $2.8.25a$ and $2.8.25b$. 
Now, to have a complete answer, we have to know, first, why the particular value of $a$ and $b$, that is $a=-in$ and $b=-i$, and second, why the correct expression, say for $X_{-n}$ term, apply to $\alpha_n$ and not to $\tilde \alpha_n$
I suppose that this should be found in comparing the expansions $2.7.4$ and $2.8.21b$, but I am not able to find the definitive argument.
Of course, if you admit, that the $2.8.24$ state 
$$\Psi(X_b) = exp(\frac{-1}{\alpha'}\sum_{m=1}^{+\infty} m X_m X_{-m})$$
is the ground state, you will find the correct value for $a$  and $b$, by using $\alpha_n \Psi = \tilde \alpha_n \Psi = 0$ $(n>0)$
