You did not include the proper infinitesimal imaginary part for frequency in the first equation for the Green's function, which would have given the correct pole structures. So it is not clear whether this is time-ordered, retarded or advanced. The retarded Green's function is

$G(\omega,p)=\dfrac{\omega+\xi}{(\omega+i\delta)^2-\xi^2-\Delta^2}$

You can either obtain this directly, or perform an analytical continuation from Matsubara(imaginary-time) Green's function.

The imaginary part is

$\Im G=-{\pi(\omega+\xi)}\Big(\delta(\omega-\sqrt{\xi^2+\Delta^2})+\delta(\omega+\sqrt{\xi^2+\Delta^2}) \Big)$

Another problem is that in a superconductor the Green's function is really a matrix. The one you have is probably just $\langle c c^\dagger\rangle$. The density of state is given by the imaginary part of the trace of the matrix Green's function. The full Green's function is given by

$\dfrac{\omega+\xi\tau_z+\Delta\tau_x}{\omega^2-\xi^2-\Delta^2}$

Here $\tau$ are Pauli matrices in the Nambu (particle-hole) space.

For $p$-wave superconductors, well, once you write down the Green's function in momentum space, you have assumed translation invariance and a constant superconducting gap everywhere. Zero-energy state only appears at defects, where the order parameter vanishes (like the core of a vortex, or the edge). So you don't see the zero-energy state from this particular Green's function because there are not any. You have to solve BdG equation in real space, or solve the full Gor'kov equation for the Green's function in real space to obtain the zero energy state.

You did not include the proper infinitesimal imaginary part for frequency in the first equation for the Green's function, which would have given the correct pole structures. So it is not clear whether this is time-ordered, retarded or advanced. The retarded Green's function is

$G_R(\omega,p)=\dfrac{\omega+\xi}{(\omega+i\delta)^2-\xi^2-\Delta^2}$

You can either obtain this directly, or perform an analytical continuation from Matsubara(imaginary-time) Green's function.

The imaginary part is

$\Im G=-{\pi(\omega+\xi)}\Big(\delta(\omega-\sqrt{\xi^2+\Delta^2})+\delta(\omega+\sqrt{\xi^2+\Delta^2}) \Big)$

Another problem is that in a superconductor the Green's function is really a matrix. The one you have is probably just $\langle c c^\dagger\rangle$. The density of state is given by the imaginary part of the trace of the matrix Green's function. The full Green's function is given by

$\dfrac{\omega+\xi\tau_z+\Delta\tau_x}{\omega^2-\xi^2-\Delta^2}$

Here $\tau$ are Pauli matrices in the Nambu (particle-hole) space.

For $p$-wave superconductors, well, once you write down the Green's function in momentum space, you have assumed translation invariance and a constant superconducting gap everywhere. Zero-energy state only appears at defects, where the order parameter vanishes (like the core of a vortex, or the edge). So you don't see the zero-energy state from this particular Green's function because there are not any. You have to solve BdG equation in real space, or solve the full Gor'kov equation for the Green's function in real space to obtain the zero energy state.

You did not include the proper infinitesimal imaginary part for frequency in the first equation for the Green's function, which would have given the correct pole structures. So it is not clear whether this is time-ordered, retarded or advanced. The retarded Green's function is

$G(\omega,p)=\frac{\omega+\xi}{(\omega+i\delta)^2-\xi^2-\Delta^2}$

You can either obtain this directly, or perform an analytical continuation from Matsubara(imaginary-time) Green's function.

The imaginary part is

$\Im G={\pi(\omega+\xi)}\Big(\delta(\omega-\sqrt{\xi^2+\Delta^2})+\delta(\omega+\sqrt{\xi^2+\Delta^2}) \Big)$

For $p$-wave superconductors, well, once you write down the Green's function in momentum space, you have assumed translation invariance and a constant superconducting gap everywhere. Zero-energy state only appears at defects, where the order parameter vanishes (like the core of a vortex, or the edge). So you don't see the zero-energy state from this particular Green's function because there are not any. You have to solve BdG equation in real space, or solve the full Gor'kov equation for the Green's function in real space to obtain the zero energy state.

You did not include the proper infinitesimal imaginary part for frequency in the first equation for the Green's function, which would have given the correct pole structures. So it is not clear whether this is time-ordered, retarded or advanced. The retarded Green's function is

$G(\omega,p)=\dfrac{\omega+\xi}{(\omega+i\delta)^2-\xi^2-\Delta^2}$

You can either obtain this directly, or perform an analytical continuation from Matsubara(imaginary-time) Green's function.

The imaginary part is

$\Im G=-{\pi(\omega+\xi)}\Big(\delta(\omega-\sqrt{\xi^2+\Delta^2})+\delta(\omega+\sqrt{\xi^2+\Delta^2}) \Big)$

Another problem is that in a superconductor the Green's function is really a matrix. The one you have is probably just $\langle c c^\dagger\rangle$. The density of state is given by the imaginary part of the trace of the matrix Green's function. The full Green's function is given by

$\dfrac{\omega+\xi\tau_z+\Delta\tau_x}{\omega^2-\xi^2-\Delta^2}$

Here $\tau$ are Pauli matrices in the Nambu (particle-hole) space.

For $p$-wave superconductors, well, once you write down the Green's function in momentum space, you have assumed translation invariance and a constant superconducting gap everywhere. Zero-energy state only appears at defects, where the order parameter vanishes (like the core of a vortex, or the edge). So you don't see the zero-energy state from this particular Green's function because there are not any. You have to solve BdG equation in real space, or solve the full Gor'kov equation for the Green's function in real space to obtain the zero energy state.