Why scalar function of vector can only depend on norm of vector? In Field Quantization by Greiner and Reinhardt as well as The Qunatum Theory of Fields by Weinberg, concerning the spectral function, the authors say a scalar function of the four-vector $p^\mu$ can only depend on $p^2$ and the step function $\theta(p^0)$. More specifically, they argue that $\Sigma_n(2\pi)^3\delta^4(q-p_n)|\langle 0|\phi(0)|n\rangle|^2$ can be expressed as $\rho(q^2)\theta(q_0)$.
I cannot understand this statement. For example, $\delta^4(p-p_0)$ (where $p_0$ is some constant 4-vector) is a scalar function on $p$, but in no way can this function only depend on $p^2$ or $\theta(p)$. Can any one explain this to me?
 A: What you want is that your function does not transform under Lorentz transformations that take
$$ p^\mu \to {\Lambda^\mu}_\nu p^\nu.$$
To build invariants from one vector there is only the possibility to construct the invariant product with itself
$$ p^2 \equiv p^\mu p_\mu \to p^\mu p_\mu.$$
There is one more thing though. The Lorentz group has two branches: Momenta with positive time-like component (i.e. positive energies) will never get boosted to momenta with negative energy, and thus
$\theta(p_0)$ also is invariant under Lorentz transformations.
Summary: Only $p^2$ is invariant under the full Lorentz group; in addition $\theta(p_0)$ (where $p_0$ is the zero-component and NOT a different four-vector!!) is invariant if one restricts to the Lorentz transformations from one branch (i.e. excludes time reversal operations).
Edit: True, $\delta^4(p^\mu - q^\mu)$ is also an invariant function. This can be seen by looking at how it operates on a Lorentz-invariant test function $f(p^2)$:
$$ \int \mathrm d^4p \delta^4(p^\mu - q^\mu) f(p^2) = f(q^2)$$
The cheapskate explanation is that since $\mathrm d^4p$ and $f(p^2)$ are invariant, as is $f(q^2)$, so must be $\delta^4(p^\mu - q^\mu)$. But this can be seen more explicitly as well. Let us perform a Lorentz transformation on the left hand side of the above equation:
$$\int\mathrm d^4p \delta^4(p^\mu - q^\mu) f(p^2) \to \int \mathrm d^4 p' \delta^4({\Lambda^\mu}_\nu (p^\nu - q^\nu)) f(p^2)$$
Here I already used that the measure picks up a factor of the determinant of the transformation but $\vert \Lambda \vert = 1$ for Lorentz transformations. We can then use that $\delta(\alpha x) = \delta(x) \frac{1}{\vert \alpha \vert}$, which also works for matrices (check that, it's a good excercise!) - where for matrices the $\vert \cdot \vert$ is the determinant. So $\delta^4({\Lambda^\mu}_\nu (p^\nu - q^\nu)) = \frac{1}{\vert \Lambda \vert} \delta^4(p^\nu - q^\nu) = \delta^4(p^\mu - q^\mu)$.
A: A function of $p_{\mu}$, $T^{\mu\nu...}p_{\mu}p_{\nu}...$is invariant under Lorentz transformation if and only if  $T^{\mu\nu...}$ is a invariant tensor. However, there is only one invariant tensor for Lorentz group. That is $\delta^{\mu}_{\nu}$. Thus, each saclar function of $p_{\mu}$ must be a function of $p^2=\delta^{\mu}_{\nu}p_{\mu}p^{\nu}$. ($\epsilon^{\mu\nu\delta\gamma}$ is invariant under proper orthochronous Lorentz transformation, but not the whole Lorentz transformation). For more detail, see, for instance, W. K. Tung, Group Theory in Physics.
