-1
$\begingroup$

I try to read Sakurai's Modern Quantum Mechanics but I stuck at this point,

$$\delta(x^{'}-x^{''})=|N|^{2}\int dp^{'}\exp\Biggl({ip^{'}(x^{'}-x^{''})2\pi\over h}\Biggr)$$

This is an expression for finding normalization constant ,After this step he shows

$$|N|^{2}\int dp^{'}\exp\Biggl({ip^{'}(x^{'}-x^{''})2\pi\over h}\Biggr) =h|N|^{2}\delta(x^{'}-x^{''})$$

I don't get this step , Please help me to move more

$\endgroup$
  • 2
    $\begingroup$ Check Dirac Delta definition $\endgroup$ – user183962 May 21 at 8:32
2
$\begingroup$

The first equation expresses the orthogonality property of the momentum eigenfunctions. There is the well-known formula for the Dirac delta function: $$ \delta(x) = \frac1{2\pi}\int\! dk\ e^{ikx}. $$ If you will make the change of variable $$ p' = k\frac{h}{2\pi} $$ in the integral, then you would obtain your second formula. Together two formulas give $$ |N|^2 = \frac1{h} $$

$\endgroup$
  • $\begingroup$ Thanks for your precious help Gec @ Gec $\endgroup$ – Robin Raj May 21 at 14:35
  • $\begingroup$ @Robin Raj , Why wouldn't you accept the answer, then? $\endgroup$ – Cosmas Zachos Jul 10 at 14:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.