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In these notes by Strominger sec 3.6 we are given the creation and annihilation operators $$ a_w = -\frac{i}{2\pi}\int\frac{dz^-}{\sqrt{2w}}f(z^-)\overleftrightarrow{\partial}_-e^{iwz^-} $$ $$ {a_w}^\dagger = \frac{i}{2\pi}\int\frac{dz^-}{\sqrt{2w}}f(z^-)\overleftrightarrow{\partial}_-e^{-iwz^-} $$ where I assumed the notation means $$ a_w = -\frac{i}{2\pi}\int\frac{dz^-}{\sqrt{2w}}\left[\partial_-f(z^-)-iwf(z^-)\right]e^{iwz^-} $$ $$ a_{w'}^\dagger = \frac{i}{2\pi}\int\frac{dy^-}{\sqrt{2w'}}\left[\partial_-f(y^-)+iw'f(y^-)\right]e^{-iw'y^-} $$ The $f$ is a massless matter field with action $$ \mathcal{S} = -\frac{1}{4\pi}\int{d^2x\sqrt{-g}(\nabla f)^2} $$

and the above operators obey $$ \left[a_w,a_{w'}^\dagger\right]=\delta(w-w') $$

I am trying to prove the commutation relationship but I am running into problems.

My attempt $$ \left[a_w,a_{w'}^\dagger\right]= \frac{1}{4\pi^2}\int{\frac{dz^-dy^-}{2\sqrt{w'w}}e^{i(wz^--w'y^-)}\left(\left[\partial_-f(z^-),\partial_-f(y^-)\right]+iw'\left[\partial_-f(z^-),f(y^-)\right]-iw\left[f(z^-),\partial_-f(y^-)\right]+w'w\left[f(z^-),f(y^-)\right]\right)} $$ Now I don't see how we can integrate $y^-$ out such that we are left with a single integral and then use the definition of the delta function to get the desired result.

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