0
$\begingroup$

Is there a reason(both physical and mathematical) why the Klein-Gordon field is represented as a fourier expansion in the second quantization as opposed to other mathematical expansions? Be gentle with the answers!

$\endgroup$
  • 1
    $\begingroup$ What do you mean by represented? It solves a wave equation, so it has nice Fourier modes. Also, what is your precise notion of second quantization? QFT, i.e. quantizing the fields as opposed to quantizing $x$ and $p$? $\endgroup$ – ACuriousMind Jul 11 '14 at 15:11
  • 1
    $\begingroup$ Also, it is easiest to quantize decoupled oscillators. By going to Fourier space, all the modes are decoupled, and the expansion coefficients are easily seen to satisfy the standard SHO algebra. $\endgroup$ – QuantumDot Jul 11 '14 at 15:22
1
$\begingroup$

Fourier expansion is used as a change of basis method that makes our calculations simpler and more in context. Usually we are working with particles with precise momentum. Momentum is a far more useful quantity than position when doing experiments.

Also the Feynman rules can be obtained in terms of momentum as well. Furthermore as ACuriousMind mentioned in the comments above, we obtain nice Fourier modes which we interpret as creation and annihilation of particles in some momentum eigenstate.

$\endgroup$
  • $\begingroup$ The laplace transform is also used to change basis like in electrical engineering when we change from time to frequency domain, so i don't understand why only the fourier expansion is warranted here?? $\endgroup$ – pkjag Jul 11 '14 at 15:51
  • $\begingroup$ You don't HAVE to use Fourier expansion. But its the most convenient one AFAIK. It helps in solving some of the derivative terms in the Lagrangian. $\endgroup$ – Constandinos Damalas Jul 11 '14 at 15:56
  • $\begingroup$ Ok i don't have much background in pde's but i know that the fourier expansion is useful in finding coefficients of integration when solving a wave equation, so what i'm asking is if the solution of the klein-gordon field has such coefficients that warrant the use of the fourier transform $\endgroup$ – pkjag Jul 11 '14 at 15:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.