Greetings Sir! Your whole lecture series is actually fantastic. But I am more interested in applications of second quantization in connection wd green's functions( correlations function in terms of linear response) . Like hamiltonian calculation using random phase approximation for electron liquid etc. Sir, it would be really helpful for me if i get any data or tutorial on RPA.
at 9:14 when you insert the expression for the Green's function on the left-hand side of the equation (TDSE) and work through I don't get zero but just psi(r,t)?
My thought on this is that, I crashed delta(r-r') by the integral, but the other delta(t-t') remains. So the final equation will be [ delta(t-t')*psi(r,t') = 0 ]. Then you can easily say that whenever t is not equal to t', the Schrodinger Function holds. So it appears to me that the Green function only holds when they are not at exactly the same time coordination? That might make sense, since green function describe the dynamic between two points. If they happens at he same time coordination, so no possible dynamics could be there for them. It has been two years. If you already know the answer, please also let me know. :) I am new to this area, so what i said might be very wrong. 😊
Not quite. The notation is perhaps a bit weird here. What I mean is that taking the inner product of the wavefunction |psi(t)>, which exists for all r, with the bra
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One of the most elegant topics in QM and QFT.
Wonderful lectures.
Thanks so much for this, these are fantastic lectures.
Very good, very useful, very clear.
I love it. Thank you so much
Very nice lecture! Thank you!!
A really nice video, thank you!
Greetings Sir! Your whole lecture series is actually fantastic. But I am more interested in applications of second quantization in connection wd green's functions( correlations function in terms of linear response) . Like hamiltonian calculation using random phase approximation for electron liquid etc. Sir, it would be really helpful for me if i get any data or tutorial on RPA.
at 9:14 when you insert the expression for the Green's function on the left-hand side of the equation (TDSE) and work through I don't get zero but just psi(r,t)?
same thoughts here
My thought on this is that, I crashed delta(r-r') by the integral, but the other delta(t-t') remains. So the final equation will be [ delta(t-t')*psi(r,t') = 0 ]. Then you can easily say that whenever t is not equal to t', the Schrodinger Function holds. So it appears to me that the Green function only holds when they are not at exactly the same time coordination? That might make sense, since green function describe the dynamic between two points. If they happens at he same time coordination, so no possible dynamics could be there for them.
It has been two years. If you already know the answer, please also let me know. :) I am new to this area, so what i said might be very wrong. 😊
@ 10:44 after taking the projection with ?
Not quite. The notation is perhaps a bit weird here. What I mean is that taking the inner product of the wavefunction |psi(t)>, which exists for all r, with the bra
@@drmitchellsphysicschannel2955 thank you.
nice tutorial
Thank you for the lecture! How did you get that Inverse FT of G(k) = 1/k^2 is 1/(4pi*r)? I get something proportional to r, not 1/r...
also, where does the factor of i come from in the retarded Green's function expression?
Thank you