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?


1 Answer 1


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.


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