# Doubt in the Poincaré algebra and one-particle states

I am studyng the algebra of Poincaré group and the definition of one particle states using the Weinberg book "Quantum theory of Fields" (vol. 1), but I'm having a hard time understanding part of the book. It says that the square of 4-vector momentum $$p^{2} = \eta_{\mu \nu} p^{\mu} p^{\nu}$$ is invariant under the Lorentz transformation $$\Lambda^{\mu}_{\ \ \nu}$$ (it's ok, I understand this), and, for $$p^{2} \leq 0$$, the sign of $$p^{0}$$ is also invariant (this I didin't understand). And the book says that, for each value of $$p^{2}$$ and sign of $$p^{0}$$ (for $$p^{2} \leq 0$$) we can choose a "standard" 4-momentum $$k^{\mu}$$ such that $$p^{\mu} = L^{\mu}_{\ \ \nu}(p)k^{\nu},$$ where $$L$$ is some Lorantz transformation. Now I am really confused, I'm not understanding the idea. What would be exactly this $$k$$ and this $$L$$? And why can he express the momentum in this way?

• May 22, 2019 at 14:26

For simplicity take 1 spatial dimension and 1 temporal dimension. As Weinberg does, we consider only proper orthochronous Lorentz transformation. All these are of the form $$\begin{cases} p_0' = \gamma(p_0-vp_1) \\ p_1' = \gamma(p_1-vp_0) \end{cases}$$ I use the convention $$p^2 = -p_0^2+p_1^2$$. Also keep in mind that $$|v|<1$$.

For $$p^2\leqslant0$$ the sign of $$p^0$$ is invariant.

Since $$p_0'=\gamma p_1(p_0/p_1-v)$$ you see that if $$|p_0|/|p_1|<1$$ then $$p_0'$$ will change sign for sufficiently large values of $$v$$, while if $$|p_0|/|p_1|\geqslant1$$ it won't for any $$v$$. The condition $$|p_0|/|p_1|<1$$ is equivalent to $$p^2>0$$ and $$|p_0|/|p_1|\leqslant1$$ is equivalent to $$p^2\leqslant0$$, so the assertion of Weinberg follows.

For each value of $$p^{2}$$ and sign of $$p^{0}$$ (for $$p^{2} \leq 0$$) we can choose a "standard" 4-momentum $$k^{\mu}$$ such that $$p^{\mu} = L^{\mu}_{\ \ \nu}(p)k^{\nu},$$ where $$L$$ is some Lorentz transformation.

For what follows $$m$$ is just a fixed real number, with no physical meaning.

Any possible $$p^\mu$$ will be in one of these sets (I picture them as the various regions in which a Minkowski diagram is divided by the two diagonals):

1. those with $$p^2=-m^2 <0$$ and $$p^0>0$$
2. those with $$p^2=-m^2 <0$$ and $$p^0<0$$
3. those with $$p^2=m^2 >0$$ and any $$p^0$$
4. those with $$p^2=0$$ and $$p^0>0$$
5. those with $$p^2=0$$ and $$p^0<0$$

Consider a $$p^\mu$$ in set 1 and the vector $$k^\mu = (m,0)$$. If you take a Lorentz transformation $$L$$ that has $$\gamma = p_0/m$$, you can see with some calculation that $$k^\mu$$ gets transformed into $$p^\mu$$. Therefore for all the $$p^\mu$$ in this set what Weinberg says is verified.

With a similar reasoning you can show that all other sets can be thought as "generated" by a particular $$k^\mu$$ (that will differ for each set) and various Lorentz transformations.

I hope this is clear. If not, I will be happy to provide further clarifications.