Modern Quantum Mechanics: J.J. Sakurai's eq. (1.7.31) Explained

[omoplata]

From "Modern Quantum Mechanics, revised edition" by J.J. Sakurai, page 56.

In equation (1.7.31) it is given,
$$\begin{eqnarray} \delta(x' - x'') & = & | N |^2 \int dp' \exp \left[ \frac{ip'(x'-x'')}{\hbar} \right] \\ & = & 2 \pi \hbar | N |^2 \delta(x' - x'' ) \end{eqnarray}$$
How does the right side happen. Is this a definition of the delta function?

 

[JorisL]

It's a Fourier transform.

To see this the Fourier transformation is given by
$$\mathcal{F}[\delta (x)] = \int \delta (x)\exp \left(-i2\pi px\right) dx = \frac{1}{2\pi}\exp (0) = \frac{1}{2\pi}$$
Inverse transformation gives
$$\delta(x)=\frac{1}{2\pi} \int \exp \left( ipx\right) dp$$

And thus $$\int \exp \left(ipx\right) dp = 2\pi \delta (x)$$

Can you see it now?

 

[Bill_K]

Actually, δ(x) = (1/2π) ∫e^ipx dp

 

[JorisL]

Ok, my bad. Been 3 years since I actually 'performed' a Fourier transformation. Should've checked it
I forgot the $$2\pi$$ factor in the exponential.

 

[omoplata]

OK, I see it now. Thanks.