Calculating Boosted Relativistic Normalization in Quantum Field Theory

[abhinavabhatt]

TL;DR

how to compute the Lorentz transformation(Boost along any direction) of Dirac Delta function?

In Quantum field theory by Peskin Schroeder for relativistic normalization
δ(p'-q')=δ(p-q) dp'3/dp3

where the boost is in z direction. How did they compute it?

 

[vanhees71]

Here, PS consider on-shell momenta, i.e., momenta fulfilling the on-shell condition $p^2=m^2$ or $E=\sqrt{\vec{p}^2+m^2}$. For these the measure $\mathrm{d}^3 p/E$ is invariant (Lorentz scalar). This implies that the distribution $E_p \delta^{(3)}(\vec{p}-\vec{q})$ is a Lorentz scalar.

 

[abhinavabhatt]

Thanks a lot for your reply.I understand this .Can you tell me how prove it directly using the delta function properties without presuming the invariant measure ?

 

[vanhees71]

It's clear that it's enough to prove it for $\delta^{(3)}(\vec{p})$. This distribution is for sure invariant under rotations. For boosts you need the assumption that we talk about four-momenta that are onshell, i.e., $p^0=\sqrt{\vec{p}^2+m^2}$. Then a Lorentz boost with velocity $\vec{v}$ reads
$$\begin {pmatrix} p^{\prime 0} \\ \vec{p}' \end{pmatrix} = \begin{pmatrix} \gamma (p^0-\vec{v} \cdot \vec{p}) \\ \vec{p}+\hat{v} (\gamma-1) (\hat{v} \cdot \vec{p})-\gamma p^0 \vec{v} \end{pmatrix}.$$
From this we get for the Jacobi matrix of the transformation
$$\frac{\partial p_j'}{\partial p_k} = \delta_{jk} + \hat{v}_j \hat{v}_k (\gamma-1)-\gamma p_k/p^0 v_j.$$
Now we have to calculate the determinant of this matrix,
$$J=\mathrm{det} \frac{\partial p_j'}{\partial p_k} = \frac{\gamma (p_0-\vec{p} \cdot \vec{v})}{p_0} =\frac{p_0'}{p_0}.$$
Now from the transformation of the $\delta$ distribution you get
$$\delta^{(3)}(\vec{p}')=\frac{1}{J} \delta^{(3)}(\vec{p})=\frac{p_0}{p_0'} \delta^{(3)}(\vec{p})$$
or
$$p_0' \delta^{(3)}(\vec{p}')=p_0 \delta^{(3)}(\vec{p}).$$
Here $p_0=E=\sqrt{\vec{p}^2+m^2}$ and $p_0'=E'=\sqrt{\vec{p}{\prime 2}+m^2}$ and thus, on the mass shell
$$E' \delta^{(3)}(\vec{p}')=E \delta^{(3)}(\vec{p}),$$
i.e., $E \delta^{(3)}(\vec{p})$ transforms as a scalar field under Lorentz transformations.

 

[abhinavabhatt]

Thanks a lot for the answer.