Recent content by Baibhab Bose

There is nothing explicit in this theory which I think indicates that term being zero. So I would say it may/may not be 0?

Oh, my bad ! ## E_0^2=\sum_{m\neq 0} \frac{(<\psi^0_m|V|\psi^0_0>)^2}{E_0^0-E^0_m}## Now, ##E_0^0##= unperturbed Ground state Wavefunction's Energy. ##E^0_m##=Unperturbed energy eigenvalue of the higher state wavefunctions, since m index represents states higher than the ground state...

Homework Statement: In the attached image. Homework Equations: formulas of fringe width and phase differences I think. It has been a long time since I have dealt with these kinds of interference/fringewidth problem, I can't figure out a way to start solving this problem. I was thinking about...

Oh yes, so the correct form would be ## E_0^2=\sum_{n\neq m} \frac {(<\psi_m^0|V|\psi_0^0>)^2}{E^0_0-E^0_m}## for the ground state. So in this question we just need the order of ##\epsilon## which, by speculation, would be ##\epsilon^2## since the 2nd order correction depends on square of the...

Yes, I know how 2nd order is calculated. $$ E^2_0=\sum_{n=/m} \frac{(<\psi_n|V|\psi_m>)^2}{E^0_n-E^0_m}$$ But I can't seem to practically calculate the correction in this case. Assuming ##\psi_n##s are ##\sqrt {\frac{2}{a}}sin(\frac{n\pi x}{a})## , what are the values of n's I'd have to limit to?

If I calculate ## <\psi^0|\epsilon|\psi^0>## and ## <\psi^0|-\epsilon|\psi^0>## separately and then add, the correction seems to be 0 since ##\epsilon## is a constant perturbation term. SO how should I approach this? And how the Δ is relevant in this calculation?

Okay, now I have learned how to use these conditions on phase transitions mathematically to approach a problem. And I have got the answer. thank you so much!

In second order phase transition, Gibbs free energy remains constant. dG=dU-Tds-SdT+PdV+VdP=0 TdS=0 (since no heat change) SdT=0 (process at same temp) PdV=0('negligible volume change) so that leaves us dG=dU+VdP but if we write dU=CvdT then again dT=0. So what to do?

Thank you. This point of Normalization seems so critical here. so I normalized ## \psi(\lambda x) ## and ## \sqrt {\lambda} ## was the Normalization constant and then I did the averaging again which yielded the correct answer which is ## \lambda^2<T> ##

Yes. this is clear now. I am getting the right answer. By I want to discuss and alternate with you. Syntactically ##<T>=\frac{h^2}{2m}\int \psi(x)* \frac{\partial^2}{\partial x^2} \psi(x)\, dx## So, here if we change ##\psi(x)--> \psi(\lambda x)## then to scale we need to shoot two lambdas down...

Since 2nd order phase transition is accompanied by no heat change and 1st order transition does, that indicates this is a 2nd order transition!

I actually can't figure out what kind of phase transition it is and how to proceed through..!

Okay, if I substitute ##u=\lambda x## to the ## -\frac{ħ^2}{2m} \frac{\partial^2\psi(x)}{\partial x^2}+V(x)\psi(x)=E\psi(x)## , then it becomes something like ## -\frac{ħ^2}{2m} \frac{\partial^2\psi(\frac{u}{\lambda})}{\partial...

Oh, that's a bad mistake! So ##\frac{ħ^2}{2m}\frac{\partial^2\psi(u)}{\partial u^2}+V(u)\psi(u)=E\psi(u)## is the time independent Schrodinger equation in terms of ## u=\lambda x##... (1) so this E is equivalent to the case when ##\frac{ħ^2}{2m}\frac{\partial^2\psi(x)}{\partial...