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A Quantum field theory

  1. Sep 25, 2016 #1
    Dear Sir,
    P 25 in quantum field theory for the gifted amateur One makes Fourier transforms from the position to the frequency space for the system of linear chain of N atoms.
    How can I see that in the frequency space the excitations are uncoupled .

    I also don’t understand equation 2.50
  2. jcsd
  3. Sep 25, 2016 #2
    Write the equations of motion for the Hamiltonians 2.45 and 2.59.

    If m=0 you are simply counting all the j's, which means it equals N. If m ≠ 0 summing all terms they cancel out symmetrically.
  4. Sep 25, 2016 #3
    1)I am sorry but I don't understand why the comparaison of 2.45 with 2.59 means uncoupled frequenties.
    The book says to compare 2.66 with 2.41 where it becomes clear that the HO's are uncoupled.
    My question nevertheless is : can you see that before making all this calculations?
    I have the principle: first understand the concept then prove it mathematicaly.
    2)I understand when k = 0 i am counting N .I thought if m is not 0 the kronicker delta gave 0.I don't see the symmetric canceling out.
  5. Sep 25, 2016 #4
    Write the equations of motion which you've learned from Hamiltonian mechanics. 2.45 is coupled, 2.59 is decoupled. It means that the resulting differential equations are dependent on each other for 2.45 and independent for 2.59.

    You don't have to solve the equations of motion. Just by writing them you see that for the coupled system they form a system of differential equations that is difficult to solve. Of course, if you've clever enough you don't even need to write the equations of motion but just look at the Hamiltonians, but clearly you're not seeing this so you have to write the equations of motion.

    Edit: never mind on that, I'm missing something too.
    Last edited: Sep 25, 2016
  6. Sep 25, 2016 #5
    I am going to do what you suggest.It will take some time,but i hope to understand.
    Thank's a lot.
  7. Sep 25, 2016 #6
    Q 2) became clear.
    Q1) I have studied both volumes of Susskind's the theoretical minimum. I hope to find there how to write the equations of motion for both H's and see that the dif eq's are coupled for 2.45 and uncoupled for 2.59.
  8. Sep 25, 2016 #7
    Hamilton's equations are

    ##\dot{x_i} = \frac{\partial H}{\partial p_i}##
    ##\dot{p_i} = -\frac{\partial H}{\partial x_i}##
  9. Sep 25, 2016 #8
    https://www.physicsforums.com/cid:0BBDC2FB-5CF0-954B-85E1-F2214146A7C5.png [Broken]
    https://www.physicsforums.com/cid:6FE7B597-2587-FD41-89BF-3670D418254F.png [Broken]
    https://www.physicsforums.com/cid:0D0F0552-CB7C-4245-80C5-E5C7467E7251.png [Broken]
    Last edited by a moderator: May 8, 2017
  10. Sep 25, 2016 #9
    I have tried to take the partial derivatives of both hamiltonian's with respect to p and x according to hamilton's equations but failed.
    I don't understand how to do this particular problem.I think it is because of the indices.
    Something else: I notice that in 2.45 & 2.59 we have operators.Not in hamilton's equations.
    I very much appreciate your help.
  11. Sep 25, 2016 #10
    I can't see the images.

    The main point is that, in one case, to find the solution at i you need the solution at i-1, so you have all equations dependent one another. In the other case, you can have independent equations to be solved separately, because the variables have been decoupled.

    I suggest you google linear coupled oscillators and find a treatment that suits you.

    That only complicates the maths. That same decoupling argument works in classical mechanics so you can trust that.
  12. Sep 25, 2016 #11
    I understand your explanation of the dependence of i on i-1
    In Gilbert Strangs Linear Algebra p 319 one can find how to take derivatives of operators in the exponential
    I go to google following your advice
    Thanks a lot !
  13. Sep 26, 2016 #12
    In QFT for the gifted amateur P 26 eq 2.58 is treated similarly to 2.56
    I suppose i should substitute p in 2.56 by {(x_j+1) - (x_j)} and get the RHS of the first line of 2.58.
    But {(x_j+1) - (x_j)} contains two terms one with index j+1 and one with index j and so i don't understand how to do it
  14. Sep 26, 2016 #13
    Similarly means using 2.46 instead of 2.47.
  15. Sep 26, 2016 #14
    In my calculation I will encounter a summation over the term 2(X_j+1)(X_j).
    I think it is the same as 2 (X_j)^2.
    Than I can make the calculation
    Is it correct?
  16. Sep 26, 2016 #15

    ##\sum_k \tilde{x}_k e^{i k (j+1) a} - \sum_m \tilde{x}_m e^{i m j a} = \sum_k \tilde{x}_k e^{i k j a} ( e^{i k a} - 1)##
  17. Sep 27, 2016 #16
    I think it's clear
    Thank you !
  18. Sep 28, 2016 #17
    https://www.physicsforums.com/cid:17989CD4-99C1-5842-ADFA-0587E697EEF5.png [Broken]
    I don't understand why from the definitions we automatically have... (P26 QFT for the gifted amateur)
    Last edited by a moderator: May 8, 2017
  19. Sep 28, 2016 #18
    Dear Sir,
    I hope you understand what is in the in my last mail.
    Many mathematical signs are missing on my keybord.
    I tried to use the app OneNote where i can use an electronic pencil and then copy and past it on my mails to You but it failed.
    Is there an other option?
    Thank you
  20. Sep 28, 2016 #19


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  21. Sep 28, 2016 #20
    Thanks a lot I am going to do it later or tomorrow
    Was it impossible to understand my question? The book uses the argument of hermitian matrices to assert that p_k is = to p_k_
    I don't understand it
  22. Sep 28, 2016 #21
    Page 26 it says "with the definition we automaticaly have ##p\dagger_k=p_\-k## " I do see a minus sign in the exponential of ##p\dagger## because of the fourier transform.
    Is this the reason for the - sign?
    But then it says:because ##p\dagger## and p are hermitian which means that if I take the complex conjugate and transpose I get the same matrix.
    Complex conjugation does not yield a - sign so I don't understand it.
    I must notice that this text is confusing!
  23. Sep 29, 2016 #22


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    It is really impossible to answer your questions without enough context. I don't own the book. So I can't help.
  24. Sep 29, 2016 #23
    It's not confusing... You just have to calmly sit down and do the algebra. You seem like in a rush (or, no offense, lazy to do the algebra). The definition it talks about is of course 2.49:

    ##\tilde{p}^\dagger_k=\frac {1} {\sqrt{N}}\sum_j p^\dagger_j e^{i k j a}=\frac {1} {\sqrt{N}}\sum_j p_j e^{i k j a}=\tilde{p}_{-k}##

    because ##\hat{p}_j## is hermitian.
  25. Sep 29, 2016 #24
    It will be my pleasure to offer you the book as a present.
    I guess you just have to give me the adress where i can tell amazone to send it
  26. Sep 29, 2016 #25
    I want to ask if i understand it well:
    In the first equality the - sign becomes + because it is ##p\tilde\dagger##. In the second equality ##p\dagger## becomes ##p## because ##p## is hermitian and the third equality is definition again.
    Is it correct?
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