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QFT: Bogolyiubov transformations and KG inner product

  1. Apr 10, 2015 #1
    (I hope this post goes in this part of the forum)

    Hi,

    I was wondering if someone could help me with the following:

    I have a (1+1) scalar field decomposed into two different sets of modes. One set corresponds to a Minkowski frame in (t,x) coordinates, the other to a Rinder frame in conformal (##\tau,\bar{\xi}##) coordinates. I know that I need to calculate the bogoliubov coefficients using the Klein Gordon invariante inner product

    ##(\phi_1,\phi_2)=i\int dx(\phi_1^*\frac{\partial{\phi_2}}{\partial{dx^0}}-\frac{\partial{\phi_1^*}}{\partial{dx^0}}\phi_2)##

    How should approach this calculation in my case, where the modes are expressed in different coordinates?

    I am not after an easy answer, just some guidance, so thanks in advanced. Any references I could get some furthur reading would be great too!! Anything that helps furthur my understanding of QFT.

    Thanks again,

    Joe.
     
  2. jcsd
  3. Apr 11, 2015 #2

    strangerep

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    Have you tried Birrell & Davies, sect 4.5?
     
  4. Apr 11, 2015 #3
    Hi,

    Thanks for the reply. I checked it out, but my problem is slightly different. The field is constrained to a Dirichlet box so I'm not sure it null coordinates are the most convenient for the problem, thats why I was going to try and compute the coefficients via de inner product.

    I will give it another read.

    Thanks again!
     
  5. Apr 12, 2015 #4

    strangerep

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    Not exactly sure what you mean.

    Disclaimer: I have not worked through this (nor the ordinary Rindler space) problem pen-in-hand, but only skim-studied B&D. It could be worthwhile to work through this problem properly, if you have the stamina to type a lot more latex in detail...
     
  6. Apr 14, 2015 #5
    OK, let me explain.

    With a the filed being confined to a Dirichlet box I mean the field is constrained to a 1D box with boundary conditions such that the field is 0 at the edges.

    I managed to calculate the integral, its as simple as using the chain rule when you calculate the derivative. I was overworked at the time of posting, so sorry for that.

    Thanks again for the help.

    P.S If anyone is interested I can type up or scan the solution.
     
  7. Apr 16, 2015 #6

    strangerep

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    I am interested -- but I don't have much spare time for close study right now. I.e., I'd be interested in returning to this later, but I don't expect you to put excessive extra effort into this unless you feel inspired to do so... :oldbiggrin:
     
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