6- Mean-field theory - Course on Quantum Many-Body Physics

[@narendratanty4604]

Sir... I can't express in words... How much helpful these lectures are for my basic understanding

[@LuisGregorioDias]

Glad to hear that! Thanks for watching!

[@narendratanty4604]

@@LuisGregorioDias thank you sir

[@pujitadas377]

at 6:40 for the interacting non identical case , why do not we consider the interactions of the particles among themselves?

[@mockingbirdsphysicschannel9774]

At 58:10 why would the mean field approximation would behave like the expectation value in previous line?

[@LuisGregorioDias]

Thanks for the question. The reason is that in MF the spectrum of the (many-body) Hamiltonian is made out of products of single-particle states. Thus, after you take the expectation value over any many-body state, the only surviving terms would be those that appear in the first line, which involve only pairs of number operators. In other words, at MF = so that justifies taking V->V_MF in 58:10 .

[@pujitadas1947]

if there is a potential term in case of identical noninteracting system hamiltonian , then at 3:14 for the case of noninteracting two types of identical particles, where is the potential term in the Hamiltonian?

[@mahmudshamim1437]

Please make lessons on non equilibrium field theory and Keldysh formalism

[@LuisGregorioDias]

Hi. Yes, those would be interesting topics! I will probably include them when I update the course.

[@MarekKorkusinski]

With the greatest respect, in minute 51:05 you are interchanging the operator c1^dag with c3 as if they were different, that is, c1^dag c3 = - c3 c1^dag. It seems to me that they are not, because c3 is a linear combination of c1 and c2 so I think that the rule should be c1^dag c3 = C(1 - c1 c1^dag) - B c2 c1^dag which is NOT -c3 c1^dag but rather C - c3 c1^dag. Would that be valid? Perhaps the constants will subtract eventually? Similar situations happen later on with other interchanges of operators.

[@sarthakgirdhar2833]

The final result is all that really matters, everything else in between must be taken with a pinch of salt.

[@universek2632]

Thank you for your great lecture. I would appreciate it if you answer my question; In the hamiltonian (13:48), why can the interaction term be written as V_ijkm a+_i b+_j... ? I want to know the mathematical derivation of the equation rather than an intuitive explanation.

[@LuisGregorioDias]

Hi, thanks for the question. You mean why the order a+_i b+_j b_m a_k of the operators? That comes when you write the interacting term in second quantization. Please check this video: kzbin.info/www/bejne/fZOsoKSajpaYiJI

[@김영준-i1u4z]

Thank you very much for uploading your great lecture! I just have a question about the second assumption around 43:42. It is assumed that c3 is linear combination of c1 and c2,so if the c3 dagger is acted on vacuum state I can conclude that state 3 is linear combination of state 1 and 2. This sounds weird to me. What did I do wrong?

[@alejandrogomez9776]

Apparently it only works when the expectation value can be written as in 37:52 I tried using the commutation relations to find that c_i^† c_j^† c_m c_k = n_ik n_jm +(-) n_im n_jk -(+) c_i^† c_k c_m c_j^† where (sign) is for fermions. This differs from 59:37 bc of the third term. But the expectation value of that term ⟨c_i^† c_k c_m c_j^†⟩ is nonzero only when (i=k and j=m) or (i=m and k=j) in other words, when the whole Hamiltonian is diagonal, so there's no need to reduce the Hamiltonian to H_MF. For the other cases ⟨third term⟩ = 0 and we're at 59:37

[@sarthakgirdhar2833]

He did not claim to give a rigorous proof. $c_3$ here never meant "the operator that destroys the particle in the third state of the single particle spectrum" so what you concluded is correct. It is the proof that is hand-wavy, which is expected since Wick's theorem's whole proof would have been a little bit digressive.

[@mubasshirkhan8231]

Great lecture! I have a question though At 13:12, the second term should have epsilon_m instead of epsilon_ k?

[@rahuljoshi464]

Thanks sir for this wonderful lecture series.

[@LuisGregorioDias]

I'm glad the lectures are helping!

[@mockingbirdsphysicschannel9774]

Could you please suggest a resource to the proof where one doesn't need assumption 1 and 2?

[@LuisGregorioDias]

Hi, thanks for the question. For a rigorous proof (Wick's theorem, in fact), you can check out the book by Fetter & Walecka.

[@priyoda497]

Can anyone please list the systems where we can apply mean field theory for practice, I did for jellium model

[@mockingbirdsphysicschannel9774]

Really nice lecture

[@LuisGregorioDias]

Thanks! I'm glad it helped!

[@kianushmaleki]

Thank you

[@ndenjoyment6591]

Which book you follow ? Plz reply

[@LuisGregorioDias]

Hi. For the most part, I am following the book by Bruus and Flensberg.

[@annankldun4040]

How and why can we assume c3 and c4 are linear combinations of c1 and c2? Just comes out of nowhere and seems like what we are trying proving is not justified without the assumptions. You really didn't show the equation holds. You just made up a specific case where it holds.

[@kwanpyunglee7335]

If c3 and c4 are not linear combinations of c1 and c2, the operator c1daggerc2daggerc3c4 gives 0 due to the first assumptin