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?

  1. Yes, your instructor is correct. If you simply perform the substitution they suggest, you should see fairly quickly that the two integrals are equal.

  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.

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  • $\begingroup$ Another way of demonstrating (1) is just to take the complex conjugate of the first identity. $\endgroup$ – Clara Diaz Sanchez Feb 17 at 16:33
  • $\begingroup$ @ClaraDiazSanchez Yes, though that would require the assertion that the Dirac delta function is purely real. Since the delta function is not truly a function at all, you would have to make the case that it is, in a certain sense, the limit of a sequence of real-valued functions. All of which is fine, but the integral substitution is the shorter route to the answer. Not that the scenic route is bad! $\endgroup$ – J. Murray Feb 17 at 17:02
  • $\begingroup$ 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. $\endgroup$ – Lopey Tall Feb 18 at 10:08
  • $\begingroup$ 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... $\endgroup$ – Lopey Tall Feb 18 at 10:09
  • $\begingroup$ 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) $\int e^{iax} + e^{-iax} \ dx = \int e^{iax} dx + \int e^{-iax} dx$. $\endgroup$ – J. Murray Feb 18 at 12:41

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