Reciprocal space as Fourier transform of real Space;

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  • Опубліковано 6 жов 2024

КОМЕНТАРІ • 14

  • @larissaandrade5284
    @larissaandrade5284 5 років тому +6

    Congratulations from Brazil! You helped me a lot!

  • @alifakih7440
    @alifakih7440 7 років тому +7

    also at 29:36, when you changed the variable in g(x-2mpi) to g(x), that is by saying let x'=x-2mpi, you only changed the x in the function g(x-2mpi), but the integral also includes the exponential term which has x. ( in fact it will still remain the same like you wrote because exp(-i4pi*m*n)=1.

    • @spaul5364
      @spaul5364 8 місяців тому

      isn't it e^(-i4pi^2 *m *n) ?there is a square on pi!

  • @kashinaththakare7789
    @kashinaththakare7789 2 роки тому

    Hi. Very helpful lectures. I got things clearer. Good way of transferring the concepts. Thanks a lot Sir.

  • @lukschs1
    @lukschs1 2 роки тому

    Muchas Gracias . Me aclaró muchas dudas, este mundo reciproco no es tán fácil de digerir hehe xD. Saludos desde Chile

  • @elamvaluthis7268
    @elamvaluthis7268 3 роки тому

    Good.

  • @febojarlock9469
    @febojarlock9469 3 роки тому

    Hi, nice lectures. I got a question, first you obtain the Cn coefficients of the Fourier series then use it as a Schwartz function to get the Poisson summation formula to finally obtain the reciprocal lattice function. Why don't just get the Cn coefficients and then simply get the Fourier transform?. Correct me if I'm wrong but at the end of the day it is the same result. Is there an advance using the Schwartz-Poisson method over FT?

  • @mrinaldeka9151
    @mrinaldeka9151 4 роки тому

    Great lecture sir

  • @alifakih7440
    @alifakih7440 7 років тому

    Hello, at 22:01, when the prof. gets rid of the integral by the property: int[delta(x-k)f(x)]dx = f(k), here k=na, why in the expression he just substitutes (a), it should be f(na) since we have delta(x-na), and the expression should be exp(-i*2pi*n*na/a)=exp(-i*2pi*n^2). then the sum becoumes sum[1^n]... could you please explain how the substitution took place?

    • @hikguru
      @hikguru 3 роки тому

      Remember the integration is being done between -a/2 to a/2. Therefore the only term that is within this integral is x=0. The others are x=+-a, +-2a therefore not in this integration interval. So you only consider x=0.

  • @alifakih7440
    @alifakih7440 7 років тому

    another question: how does is the (na) included in the integral (-a/2, +a/n)?

    • @amit80547
      @amit80547 5 років тому

      you write the whole summation and everything is summation vanishes except one.

    • @v.vigneshvenkatasubramania3456
      @v.vigneshvenkatasubramania3456 5 років тому

      I think Amit is correct as per my calculation. First of all, I DONT use the same n for the summation of 1-D lattice and the same n for the Fourier expansion. The prof seems to use the same. I am not sure why it has to be the same and I didn't understand the full explanation. Nevertheless, say we have capital N for the summation of lattice and small n for the Fourier coefficient. Since the Fourier coefficient is only between -a/2 and a/2, you only have to consider the case when x-Na lies within -a/2 and a/2. This does not happen for all Na's as the distance between two points itself is 'a' which is as big as the integral limits and therefore everything in summation vanishes except for one case as Amit says. you get only 1 as the value.