Recent content by tamaghnahazra

okk... thanks...

@Avodyne:Yes, that is true... my bad... A clarification on the earlier post: The integral evaluates to \frac{\pi m V_0}{\hbar^2 q^2} \delta(q-q') which is equivalent to \frac{\pi V_0}{q} \delta(E_q-E_{q'}) so the normalization constant would be \sqrt{\frac{\pi m V_0}{\hbar^2 q^2}} if...

Hi, I cannot find the edit button below my own posts on this thread... https://www.physicsforums.com/showthread.php?t=657190 even though I have successfully edited the same earlier. Would be great if somebody could throw some light on the matter. Thanks.

ok thanks, i managed to get it normalized using the convergence factor and the definition of a delta function as a limiting lorentzian... the normalization constant i get is √(q/∏V_0) i'll edit this post to show the steps if anybody faces the same problem later...

@avodyne: Thanks... that's a good idea... and it gives me some confidence that the normalization CAN be done... it would be great if you could give me the reference to where you first came across this trick, if you happen to be remember... however the orthogonality here strictly holds only...

and by the right handed portion you mean?

@Jano L.: Well F(q,q') unfortunately doesn't resemble a delta function, hence the problem. @mfb: Thanks... I found something relevant in R Shankar's QM text. He deals with a similar problem of normalization by setting the constant such that the incoming wave is normalized in the usual delta...

Oh... no the integration is over space... i should have mentioned that... i'll edit the post right away...

Hi, I am stuck with a problem which effectively boils down to this: Given the eigenstates of a Hamiltonian with a step potential in the x direction H=-\hbar^2/2m \nabla^2 + V_0 \Theta(x) \psi(q)_{in}=cos(qx)-\frac{\sqrt{K_{V_0}^2-q^2}}{q}sin(qz) \qquad x<0...