# Dirac delta, Fourier transform & exponentials

Consider the following equation/identity:

$$\int d^3x e^{i(\vec{p}+\vec{q})\cdot\vec{x}}=(2\pi)^3\delta^{(3)}(\vec{p}+\vec{q}).$$

I am trying to calculate some commuters I'm encountering in my first foray into canonical quantization, and I am struggling with applying this identity correctly.

1) Does the above identity apply the following? $$\int d^3x e^{-i(\vec{p}+\vec{q})\cdot\vec{x}}=(2\pi)^3\delta^{(3)}(\vec{p}+\vec{q}).$$

My professor answered this in class very quickly when I asked. Along the lines of, "you can just rescale $$x \rightarrow -x$$". I'm a bit concerned with the legitimacy of that.

2) How to workout the following equation using the above identity.

$$\int \frac{d^3 p}{(2\pi)^3} \frac{(-i)}{2} (-e^{i\vec{p}\cdot(\vec{x}-\vec{y})} -e^{i\vec{p}\cdot(\vec{y}-\vec{x})})=i\delta^{(3)}(\vec{x}-\vec{y}).$$

My classmate gave me the hint to "just break that expression up into two parts." I am not certain how to do this. But here is my own attempt!

Let's start by rearranging the identity to fit out purposes,

$$\int \frac{d^3p}{(2\pi)^3} e^{i(\vec{x}+\vec{y})\cdot\vec{p}}=\delta^{(3)}(\vec{x}+\vec{y}).$$

Does this imply the following?

$$\int \frac{d^3p}{(2\pi)^3} e^{i(\vec{x}-\vec{y})\cdot\vec{p}}=\delta^{(3)}(\vec{x}-\vec{y}).$$

Does the following equality hold?

$$-e^{i\vec{p}\cdot(\vec{y}-\vec{x})} =-e^{i\vec{p}\cdot(\vec{x}-\vec{y})}.$$

If those two are true (which I not immediately clear about), then I have it.

Can I get some insights on the properties of the exponential function that I'm dancing around here?

2. The first equality holds by trivially substituting $$-\vec y$$ for $$\vec y$$ in the preceding expression. The second is generically not true; however, you can do the two integrals separately, and since they are equal (as per your first question) the result is straightforward.
• 1 and 2) How is it that $$\int e^{i a x} dx = \int e^{-i a x} dx?$$ It is not at all clear to me that (even though the substitutions don't involve the variables of integration) you can just change the expression and get the exact same answer. – Lopey Tall Feb 18 at 10:08
• 2) I am not clear one what you mean by "the two integrals." It can't be that that $$\int e^4x e^{x^2} dx = \int e^4x dx \int e^{x^2} dx$$ can it? Notably, where did the second $\int$ and $dx$ come from? The following is fine with me, $$\int e^4x e^2y dx dy = \int e^4x dx \int e^2y dy$$ but I'm not sure the same can be done in the former case... – Lopey Tall Feb 18 at 10:09
• 1) Presumably you know how to do variable substitution. If you perform the trivial substitution $x \leftrightarrow -x$ in that integral, you will find your answer. 2) $\int e^{iax} + e^{-iax} \ dx = \int e^{iax} dx + \int e^{-iax} dx$. – J. Murray Feb 18 at 12:41