# How to define the parton distribution function of the antiquark?

So I can define the quark distribution function within a hadron as $$f_{\psi/h}(x)=\frac{1}{2}\int\frac{dz^-}{2\pi}e^{ixP^+z^-} \langle h(p)|\bar{\psi}(0)\gamma^+\psi(z^-)|h(p)\rangle|_{z^2=0}$$ where $$a^{\pm}=\frac{1}{\sqrt{2}}(a^0 \pm a^3)$$, and where $$A^+=0$$. This can be constructed from the lower half of the handbag diagram shown below.

To construct the antiquark distribution I imagine all I would need to do is reverse the direction of the fermion lines, leaving me with

$$f_{\bar{\psi}/h}(x)=\frac{1}{2}\int\frac{dz^-}{2\pi}e^{ixP^+z^-} \langle h(p)|\psi(0)\gamma^+\bar{\psi}(z^-)|h(p)\rangle|_{z^2=0}$$

Is this correct?