Delta function in continuation back to Minkowski space

[jim burns]

The Green's function for a scalar field in Euclidean space is

$$(2\pi)^4\delta^4(p+k) \frac{1}{p^2+m^2}$$

however when I continue to Minkowski space via G_Min(p_Min)=G_E(-i(p_Min)) there's seems to be a sign error:

$$(2\pi)^4\delta^4(-i (p+k)) \frac{1}{-p^2+m^2}=(2\pi)^4\delta^4(p+k) \frac{i}{-p^2+m^2}=-(2\pi)^4\delta^4(p+k) \frac{i}{p^2-m^2}$$

where I used δ(-ix)=(1/(-i))δ(x).

The error seems to be that the scaling of the delta function should instead be δ(-ix)=δ(ix)=(1/(i))δ(x).

But how do we know this? For real 'a' it can be argued δ(ax)=(1/|a|)δ(x) on the grounds that δ is postive, but δ(-ix) is not positive as it's not even a real number.

 

[eljose79]

I would like to offer some insights and clarifications on the sign error that you have encountered.

Firstly, let's look at the definition of the Dirac delta function. The Dirac delta function is a generalized function that is defined as:

$$\delta(x)=\begin{cases} \infty, & x=0 \\ 0, & x\neq0 \end{cases}$$

with the property that

$$\int_{-\infty}^\infty f(x)\delta(x)dx=f(0)$$

for any well-behaved function f(x).

Now, when we have a function of the form δ(ax), where a is a real number, we can use the substitution u=ax to rewrite it as:

$$\delta(ax)=\frac{1}{|a|}\delta(u)$$

This is because when u=0, x=0 and when u≠0, x≠0. Therefore, we can use the substitution to get the correct scaling for the delta function.

However, when we have a function of the form δ(ix), where i is the imaginary unit, we cannot use the same substitution as above. This is because when u=ix, x is a complex number and the substitution does not hold. In this case, we need to use the definition of the delta function and its properties to determine the correct scaling.

Now, let's look at the specific case of δ(-ix). Using the definition of the delta function, we can see that:

$$\int_{-\infty}^\infty f(x)\delta(-ix)dx=f(0)$$

Using the substitution u=-ix, we get:

$$\int_{-\infty}^\infty f(x)\delta(-ix)dx=\int_{-\infty}^\infty f(-iu)\delta(u)du=f(0)$$

Therefore, we can see that the correct scaling for δ(-ix) should be δ(-ix)=δ(x).

In conclusion, the sign error that you encountered is due to the incorrect use of the substitution for the delta function. As scientists, it is important for us to carefully consider the definitions and properties of mathematical functions to ensure the correctness of our calculations.