Peskin & Schroeder derivation of normalization for one particle states on pages 22-23

TroyElliott
From page 22 of P&S we want to show that ##\delta^{3}(\vec{p}-\vec{q})## is not Lorentz invariant. Boosting in the 3-direction gives ##p_{3}' = \gamma(p_{3}+\beta E)## and ##E' = \gamma(E+\beta p_{3})##. Using the delta function identity ##\delta(f(x)-f(x_{0})) = \frac{1}{|f'(x_{0})|}\delta(x-x_{0}),## we get $$\delta^{3}(\vec{p}-\vec{q}) = \delta^{3}(\vec{p'}-\vec{q'}) \frac{dp'_{3}}{dp_{3}}.$$

This is where I am confused, how is this step arrived at? From what I can see we have
$$\delta^{3}(\vec{p'}-\vec{q'}) = \delta(p_{1}-q_{1})\delta(p_{2}-q_{2})\delta( \gamma((p_{3}-q_{3})+\beta (E_{p}-E_{q})).$$

We can further write $$\delta( \gamma((p_{3}-q_{3})+\beta (E_{p}-E_{q}))) = \delta((p_{3}-q_{3})+\beta (E_{p}-E_{q}))/\gamma.$$

From here I don't see how to transform ##\delta((p_{3}-q_{3})+\beta (E_{p}-E_{q}))/\gamma## into the form ##\delta(p_{3}-q_{3})\frac{dp_{3}}{dp'_{3}}.##

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[...] delta function identity ##\delta(f(x)-f(x_{0})) = \frac{1}{|f'(x_{0})|}\delta(x-x_{0}),## we get $$\delta^{3}(\vec{p}-\vec{q}) = \delta^{3}(\vec{p'}-\vec{q'}) \frac{dp'_{3}}{dp_{3}}.$$This is where I am confused, how is this step arrived at?
Think of ##p'## as a function of ##p##. I.e., substitute ##f(x) \to p'(p)## and ##f(x_0) \to p'(q)##. Then the ##\frac{dp'_{3}}{dp_{3}}## is essentially just the term ##1/|f'(x_0)|##.

HTH.

mfb and TroyElliott
TroyElliott
Think of ##p'## as a function of ##p##. I.e., substitute ##f(x) \to p'(p)## and ##f(x_0) \to p'(q)##. Then the ##\frac{dp'_{3}}{dp_{3}}## is essentially just the term ##1/|f'(x_0)|##.

HTH.
Thanks!

So taking this route we end up with $$\delta^{3}(\vec{p}-\vec{q}) = \delta^{3}(\vec{p'}-\vec{q'})\frac{E'}{E},$$

where the ##\frac{dp'_{3}}{dp_{3}}## term became ##\frac{E'}{E}.## When using the Dirac delta identity above, we should technically evaluate ##\frac{dp'_{3}}{dp_{3}}## at ##q_{3}##, right? Does this not need to be explicitly done since we are dealing with a delta function, which will force ##p_{3} ## to equal ##q_{3}## when integrated?

Thank you very much.

When using the Dirac delta identity above, we should technically evaluate ##\frac{dp'_{3}}{dp_{3}}## at ##q_{3}##, right? Does this not need to be explicitly done since we are dealing with a delta function, which will force ##p_{3} ## to equal ##q_{3}## when integrated?
Strictly speaking, all that delta fn stuff only makes sense in the context of the theory of distributions (aka generalized functions). But after a relation involving delta fns is established using rigorous integral expressions, most people just use shorthand notation, sans integrals.

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I think it's most easily to see considering momentum-space volume elements ##\mathrm{d}^3 p##. It's important to note that we deal with "on-shell particles" for the asymptotic free states, i.e., ##p^0=E_{\vec{p}}=\sqrt{\vec{p}^2+m^2}##. This means you can effectively work with functions dependent on four-momentum with the on-shell constraint understood. Then you can consider the invariant (under proper orthochronous Poincare transformation)
$$\mathrm{d}^4 p \Theta(p^0) \delta(p^2-m^2)=\frac{\mathrm{d^3} p}{2E_{\vec{p}}},$$
which implies that
$$\delta^{(3)}(\vec{p}-\vec{p}') E_{\vec{p}}$$
is a Lorentz scalar.

Strictly speaking, all that delta fn stuff only makes sense in the context of the theory of distributions (aka generalized functions). But after a relation involving delta fns is established using rigorous integral expressions, most people just use shorthand notation, sans integrals.

I still do not fully understand your answer and do not see how it answers the question. SO do we need to evaluate the derivate at ##q_3##?

$$\delta\Big( f(x) - f(x_0) \Big) ~=~ \frac{1}{|f'(x)|} \; \delta( x - x_0) ~=~ \frac{1}{|f'(x_0)|} \; \delta( x - x_0) ~,$$where I have given you an extra hint by showing the middle step (which is all you really need here to perform the computation at the top of p23).