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}<br /> \delta(x&#039; - x&#039;&#039;) &amp; = &amp; | N |^2 \int dp&#039; \exp \left[ \frac{ip&#039;(x&#039;-x&#039;&#039;)}{\hbar} \right] \\<br /> &amp; = &amp; 2 \pi \hbar | N |^2 \delta(x&#039; - x&#039;&#039; )<br /> \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.