How is it known that $ \int_{-\infty}^\infty \mathrm dp'\exp\left[\frac{ip'(x' -x'')}{\hbar}\right] = 2 \pi \hbar \delta(x' -x'') $?

Question

In p.56 of Sakurai's Modern Quantum Mechanics (Addison-Wesley, 1994), he calculates the following integral: $$ \int_{-\infty}^\infty \mathrm dp'\exp\left[\frac{ip'(x' -x'')}{\hbar}\right] = 2 \pi \hbar \delta(x' -x'') $$ How is this known?

Answer 1

First set $x'' = 0$ because it's redundant, write $x$ instead of $x'$, and do a substitution $p = \hbar k$. The identity then becomes

$$\int_{-\infty}^\infty \mathrm{d}k\ e^{ikx} = 2 \pi \delta(x)$$

The definition of $\delta(x)$ is that $\int \mathrm{d}x\ \delta(x) f(x) = f(0)$ for any sufficiently good function ($C^\infty$ with compact support will do). Therefore, we should multiply the left hand side by $f(x)$ and integrate with respect to $x$ and see if we get $2\pi f(0)$:

$$ \int \mathrm{d}x\ \int \mathrm{d}k\ e^{ikx} f(x) \overset{?}{=} 2\pi f(0)$$

We take the left hand side and interchange the order of integration, leaving the mathematicians to worry about whether such a thing is valid:

$$\begin{align} \int \mathrm{d}x\ \int \mathrm{d}k\ e^{ikx} f(x) &= \int \mathrm{d}k\ \int \mathrm{d}x\ e^{ikx} f(x) \\ &= \int \mathrm{d}k\ \hat{f}(-k) \quad \text{($\hat{f}$ is the Fourier transform of $f$)} \\ &= \int \mathrm{d}k\ \hat{f}(k) \quad \text{(do a change of variables $k \to -k$)} \\ &= \int \mathrm{d}k\ \hat{f}(k) e^{ik \cdot 0} \\ &= 2\pi f(0) \quad \text{(reverse Fourier transform of $\hat{f}$ at $x=0$)} \end{align}$$

Note that I'm using the usual physics convention for the Fourier transform, where $\hat{f}(k) = \int \mathrm{d}x\ f(x) \exp(-ikx)$ and $f(x) = \int \frac{\mathrm{d}k}{2\pi}\ \hat{f}(k) \exp(ikx)$.

Answer 2

First you need the result that the Fourier transform of the delta function is equal to unit. This is obvious in traditonal notation, but I'll show this with distribution theory.

This can be shown by properly defining the Fourier transform of a tempered distribution. In that setting, if $\eta \in \mathcal{S}'(\mathbb{R}^n)$ is a Schwartz distribution, we define its Fourier transform $\hat{\eta}$ by

$$(\hat{\eta},f)=(\eta,\hat{f}).$$

In other words, we apply the distribution to the Fourier transform of the function. We now consider $\delta$. Recall that it is the distribution defined by

$$(\delta,f)=f(0),\quad \forall f\in \mathcal{S}(\mathbb{R}^n).$$

Now, you can easily see that

$$(\hat{\delta},f)=(\delta,\hat{f})=\hat{f}(0).$$

But the Fourier transform is defined by

$$\hat{f}(\xi)=\int f(x)e^{-2\pi i\xi\cdot x}d^nx.$$

thus you find that

$$(\hat{\delta},f)=\int f(x)d^nx=(1,f)$$

in the distributional sense. Thus $\hat{\delta}=1$. With traditonal notation this means that

$$\int \delta(x)e^{-2\pi ix\cdot \xi}dx = 1$$

as you would expect by using the delta as usually is done in Physics.

It turns out that if $\mathcal{F}^{-1}$ denotes the Fourier inversion, you necessarily have $\mathcal{F}^{-1}(1)=\delta$.

But $\mathcal{F}^{-1}(1)=\delta$ can be written, in the more traditional notation, as

$$\int e^{2\pi i x\cdot \xi}dx=\delta(\xi)$$

If you use this result in your computations by a change of variables you achieve the result.