- #1
DOTDO
- 8
- 1
In Peskin and Schroeder's QFT book, while deriving the cross section formula for particles ##A## and ##B##, a Dirac delta appears in Eq 4.77:
\begin{align}
\nonumber
\int d\bar{k}^z_A \,
\left.
\delta ( F ( \bar{k}^z_A ) )
\right\vert_{\bar{k}^\perp_A = k^\perp_A, \, \bar{k}^\perp_B = k^\perp_B}
=
\frac{1}{\left\vert\frac{\bar{k}^z_A}{\bar{E}_A}-\frac{\bar{k}^z_B}{\bar{E}_B}\right\vert},
\end{align}
where
\begin{equation}
\nonumber
F ( \bar{k}^z_A ) =
\sqrt{(\bar{k}^\perp_A)^2+(\bar{k}^z_A)^2+m^2_A}
+ \sqrt{(\bar{k}^\perp_B)^2+(k^z_A+k^z_B - \bar{k}^z_A)^2+m^2_B}
\
-\sqrt{(k^\perp_A)^2+(k^z_A)^2+m^2_A}
-\sqrt{(k^\perp_B)^2+(k^z_B)^2+m^2_B}.
\end{equation}
Because of the delta function, this result comes together with the constraint
\begin{equation}
\nonumber
\bar{k}^z_A=k^z_A.
\end{equation}
In Eq 4.78 on the next page, this result is used to exchange the variables ##\bar{k}^z_A## and ##\bar{k}^z_B = k^z_A+k^z_B-\bar{k}^z_A## in terms of ##k^z_A##. For example,
\begin{equation}
\nonumber
\frac{1}{\sqrt{E_A\bar{E}_A}} \phi_A(\vec{k}_A) \phi^\star_A(\vec{\bar{k}}_A)
=
\frac{\vert\phi_A(\vec{k}_A)\vert^2}{E_A}.
\end{equation}
I know how to deal with a composition of Dirac delta with a function, and it is clear that ##F(\bar{k}^z_A)## has a zero at ##k^z_A##. However, by manipulating ##F(\bar{k}^z_A)##, we can get a quadratic equation of ##\bar{k}^z_A##, and there is one another solution. Let's denote the second solution by ##K##. Then, the first equation above should give
\begin{align}
\nonumber
\frac{1}{\left\vert\frac{\bar{k}^z_A}{\bar{E}_A}-\frac{k^z_A+k^z_B-\bar{k}^z_A}{\bar{E}_B}\right\vert} \times \left( \bar{k}^z_A = k^z_A\right)
+
\frac{1}{\left\vert\frac{\bar{k}^z_A}{\bar{E}_A}-\frac{k^z_A+k^z_B-\bar{k}^z_A}{\bar{E}_B}\right\vert}\times \left( \bar{k}^z_A =K \right).
\end{align}
That is, the constraints are not given as independent options. They are summed together so it is impossible to choose one of them as we want. But as you can see in Eq 4.78, they take only one solution ##\bar{k}^z_A = k^z_A##.
What am I misunderstanding?
\begin{align}
\nonumber
\int d\bar{k}^z_A \,
\left.
\delta ( F ( \bar{k}^z_A ) )
\right\vert_{\bar{k}^\perp_A = k^\perp_A, \, \bar{k}^\perp_B = k^\perp_B}
=
\frac{1}{\left\vert\frac{\bar{k}^z_A}{\bar{E}_A}-\frac{\bar{k}^z_B}{\bar{E}_B}\right\vert},
\end{align}
where
\begin{equation}
\nonumber
F ( \bar{k}^z_A ) =
\sqrt{(\bar{k}^\perp_A)^2+(\bar{k}^z_A)^2+m^2_A}
+ \sqrt{(\bar{k}^\perp_B)^2+(k^z_A+k^z_B - \bar{k}^z_A)^2+m^2_B}
\
-\sqrt{(k^\perp_A)^2+(k^z_A)^2+m^2_A}
-\sqrt{(k^\perp_B)^2+(k^z_B)^2+m^2_B}.
\end{equation}
Because of the delta function, this result comes together with the constraint
\begin{equation}
\nonumber
\bar{k}^z_A=k^z_A.
\end{equation}
In Eq 4.78 on the next page, this result is used to exchange the variables ##\bar{k}^z_A## and ##\bar{k}^z_B = k^z_A+k^z_B-\bar{k}^z_A## in terms of ##k^z_A##. For example,
\begin{equation}
\nonumber
\frac{1}{\sqrt{E_A\bar{E}_A}} \phi_A(\vec{k}_A) \phi^\star_A(\vec{\bar{k}}_A)
=
\frac{\vert\phi_A(\vec{k}_A)\vert^2}{E_A}.
\end{equation}
I know how to deal with a composition of Dirac delta with a function, and it is clear that ##F(\bar{k}^z_A)## has a zero at ##k^z_A##. However, by manipulating ##F(\bar{k}^z_A)##, we can get a quadratic equation of ##\bar{k}^z_A##, and there is one another solution. Let's denote the second solution by ##K##. Then, the first equation above should give
\begin{align}
\nonumber
\frac{1}{\left\vert\frac{\bar{k}^z_A}{\bar{E}_A}-\frac{k^z_A+k^z_B-\bar{k}^z_A}{\bar{E}_B}\right\vert} \times \left( \bar{k}^z_A = k^z_A\right)
+
\frac{1}{\left\vert\frac{\bar{k}^z_A}{\bar{E}_A}-\frac{k^z_A+k^z_B-\bar{k}^z_A}{\bar{E}_B}\right\vert}\times \left( \bar{k}^z_A =K \right).
\end{align}
That is, the constraints are not given as independent options. They are summed together so it is impossible to choose one of them as we want. But as you can see in Eq 4.78, they take only one solution ##\bar{k}^z_A = k^z_A##.
What am I misunderstanding?