An integral with a delta-function

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  • Thread starter Haorong Wu
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Summary:

How to calculate this integral?
It is from David Tong's note for QFT. The equation states
##\left . \int d^4 p \cdot \delta \left ( p^2_0 -{\vec p}^2 -m^2 \right ) \right | _{p_0>0} =\left . \int \frac {d^3 p} {2 p_0} \right |_{p_0=E_{\vec p}}##
where ##p## is a 4-vector ##p=\left ( p_0, \vec p \right )##.

In my calculation, I get
##\left . \int d^4 p \cdot \delta \left ( p^2_0 -{\vec p}^2 -m^2 \right ) \right | _{p_0>0} =\left . \int d^3 p \int d p_0 \cdot \delta \left ( p^2_0 -{\vec p}^2 -m^2 \right ) \right | _{p_0>0}=\left . \int d^3 p \cdot 1 \right | _{p_0>0}=1## or
##\left . \int d^4 p \cdot \delta \left ( p^2_0 -{\vec p}^2 -m^2 \right ) \right | _{p_0>0} =\left . \int d^3 p \int d p_0 \cdot \delta^{\left ( 3 \right )} \left ( p^2_0 -{\vec p}^2 -m^2 \right ) \delta ^{\left ( 1 \right )}\left ( p^2_0 -{\vec p}^2 -m^2 \right ) \right | _{p_0>0}=\left . \int d^3 p \cdot \delta^{\left ( 3 \right )} \left ( 0 \right ) \right | _{p_0>0}## which is infinite.

Maybe I get a wrong definition for the delta function in this integral? What is the dimension of the delta function in this integral?
 

Answers and Replies

  • #2
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The key to this is the identity
$$\delta \left(f(x)\right)=\frac{\delta(x-x_0)}{\left| f'(x_0)\right|}$$
where ##x_0## is a zero of f(x). In this case, ##f(x)=x^2 - x_0^2##, ##x=p_0##, and ##x_0=\sqrt{\vec{p}^2+m^2}=E_{\vec{p}}##. This gives me the correct answer.
 
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  • #3
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The key to this is the identity
$$\delta \left(f(x)\right)=\frac{\delta(x-x_0)}{\left| f'(x_0)\right|}$$
where ##x_0## is a zero of f(x). In this case, ##f(x)=x^2 - x_0^2##, ##x=p_0##, and ##x_0=\sqrt{\vec{p}^2+m^2}=E_{\vec{p}}##. This gives me the correct answer.
Oh, thanks, @Isaac0427 .
 

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