Solving scalar quantum field in Milne space(1+1)
So our line element is $$ds^2=dt^2-a^2t^2dx^2$$ doing following coordinate transformation $$y^0=t\hspace{2pt}coshax, \hspace{2pt}y^1=t\hspace{2pt}sinhax$$ we get $$ds^2=(dy^0)^2-(dy^1)^2$$ Now finding Killing vectors is easy to obtain in this coordinate and they're $$y^1\partial_{y^0}+y^0\partial_{y^1},$$ $$\partial_{y^0} ,$$ $$\partial_{y^1}$$ of which only $\partial_{y^0}$ is timelike vector. Now transforming it (let's call this vector K) back to $(t,x)$ coordinate we get $$K=coshax\hspace{2pt}\partial_t-\frac{sinhax}{at}\partial_{x}$$ For further refrence I will write $$\frac{\partial y^{\nu^{'}}}{\partial x^\mu}=\begin{pmatrix}coshax & at\hspace{2pt}sinhax \sinhax & at\hspace{2pt} cosh ax \end{pmatrix}$$ where $x$ refers to $(t,x)$ and $y$ refers to $(y^0,y^1)$.
The given line element can also be written as $$ds^2=e^{2a\eta}(d\eta^2-dx^2)$$ where we have substituted $t=\frac{1}{a}e^{a\eta}$
Now coming to field equation of scalar field for the case of (1+1) Milne universe it can be written as following $$\phi(x)=\int \frac{dk}{2\pi}(a_k u_k(x)+a_k^{\dagger}u_k^*(x))$$ $$u_k(x,\eta)=\frac{1}{\sqrt{2\pi}}e^{ikx} y(\eta)$$ where $y(\eta)$ satisfies $$\frac{d^2y}{d\eta^2}+(k^2+m^2\hspace{2pt}e^{2a\eta})y=0$$ The above equation can be expressed as equation of Bessel function by substituting $z=e^{a\eta}$. Doing this we get $$z^2\frac{d^2y}{dz^2}+z\frac{dy}{dz}+(\frac{m^2}{a^2}z^2+\frac{k^2}{a^2})$$ so we get $$y(t)=J_{\pm\frac{ik}{a}}(mt),\hspace{3pt}Y_{\pm\frac{ik}{a}}(mt), \hspace{3pt}H_{\pm\frac{ik}{a}}^{(1,2)}(mt)$$ Now to find out which one is positive frequency with respect to the global timelike Killing vector I have to find out the which one of the above is eignefunction of the following form $$K\hspace{3pt}u_k(x)=-i\omega u_k(x)$$ where $\omega >0$ but the problem I'm facing is following cause once I know how things work out for $J(x)$ others can be done since they're simply its linear combination.
Is my approach correct or am I making some conceptual mistakes? 281