Recent content by askhetan

  1. askhetan

    Bloch electrons, periodicity

    Thats fine but, the wave function is crystal periodic, which is the same thing as give by the Born von Karman conditions. As expressed, the wavefunction has to obey from Bloch's theorem: ## \psi_{k_{i}}(r+n_{i}a_{i}) = exp^{i.k_{i}n_{i}a_{i}} \psi_{k_{i}}(r) ## Where ##k_{i}## goes from ##0##...
  2. askhetan

    Bloch electrons, periodicity

    Dear DrDu, thanks for the reply. Still a lot of confusion though. Born von Karman conditions are practically applicable for any value of ##k_{i}##, isn't this true because the choice of ##L_{i}## is completely arbitrary. The Bloch theorem stands for a crystal that is ideally and infinitely...
  3. askhetan

    Bloch electrons, periodicity

    I am equally and massively confused by this theorem. Especially the second expression: ## \psi (\vec{r}+\vec{R}) = exp(i.\vec{k}.\vec{R}) \psi (\vec{r}) ## I was trying to plot these in MATLAB and when I tried to match the real parts with this equation and it did not match at all and i wasted...
  4. askhetan

    Understanding operators for Green's function derivation

    Dear All, I am trying to understand what operators actually mean when deriving the definition of green's function. Is this integral representation of an operator in the ##x-basis## correct ? ## D = <x|\int dx|D|x>## I am asking this because the identity operator for non-denumerable or...
  5. askhetan

    Many body wavefunction and exchange correlation

    To reword it and shorten it I'd say this: Every text I read says we do not know the exact exchange-correlation functional. Is this related to the non-existence of single body wavefunctions in a many body system? or is it related to the lack of our knowledge of being able to mathematically...
  6. askhetan

    Many body wavefunction and exchange correlation

    Everywhere I ready about HF or DFT the term exchange correlation functional comes up. I have a couple of fundamental questions about these: 1) Books say that the correlation energy is the difference between the exact energy (lets say we've found that somehow) and the hartree-fock energy and...
  7. askhetan

    Matrix Elements as images of basis vectors

    That clarifies it. Thanks for your responses!
  8. askhetan

    Matrix Elements as images of basis vectors

    There are a lot of mistakes in my previous post. However, I found a big mistake that I had been making in the proofs using indices! That resolves my problem. As you had mentioned, I needed to distinguish between ##\Omega_{A}## and ##\Omega_{B}##. Anyway, ditching my notation and taking up yours...
  9. askhetan

    Matrix Elements as images of basis vectors

    Thanks for you reply. Basically, the M represented above by you here is kind of a change of basis matrix such that Mai = bi. I guess the stem I am failing to understand is the one where you take (Mai)j = ([M]A)ji. I tried to derive it and have a major source of confusion because the M that you...
  10. askhetan

    Matrix Elements as images of basis vectors

    For the relation T|V> = |V'>, the matrix equation hold perfectly fine that v'i=ΣTijvj (That's how matrices are multiplied to vectors). Then from just looking at how vectors behave upon transformations, it follows Tij = <i|T|j>. What is then the inherent assumption I am making that |i> and |j>...
  11. askhetan

    Matrix Elements as images of basis vectors

    Firstly, thanks for a great reply. It helps me understand the original text better. So for my case, the second situation applies where T:X→X and I have two different orthonormal basis sets: A being column vectors [1 0], [0 1] and B being column vectors [1 0] and [0 -1]. Now if I use the general...
  12. askhetan

    Matrix Elements as images of basis vectors

    I'm trying to understand the maths of QM from Shankar's book - Principles of Quantum Mechanics: On page 21 of that book, there is a general derivation that if we have a relation: |v'> = Ω|v> Where Ω is a operator on |v> transfroming it into |v'>, then the matrix entries of the operator can be...
  13. askhetan

    Very basic question on Hamiltonian representation?

    Thank you so much. A great explanation. :bow:
  14. askhetan

    Very basic question on Hamiltonian representation?

    Oh nice! exactly what I wanted to bring up next. So my new understanding is rather that when we say ##\psi(\vec x_1)##, it actually means ##\psi## defined in the position-space that we call ##\vec x_1##. This brings me now to another perplexing question. What does it mean when someone writes the...
  15. askhetan

    Very basic question on Hamiltonian representation?

    Thanks for your reply. Just to confirm, do you mean the presentation of this idea in the link is ambiguous ? Its strange because almost every book on DFT uses similar presentation.
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