10
$\begingroup$

Studying the scalar field and Klein-Gordon equation in quantum field theory I came across this definition for the inner product in the space of the solutions of the K.G. equation:

$$\langle \Phi_1 | \Phi_2 \rangle = i\int \mathrm{d}\vec{x}(\Phi_1 ^* \overleftrightarrow{\partial_0}\Phi_2) = i\int \mathrm{d}\vec{x} (\Phi_1 ^* \partial_0\Phi_2 - \Phi_2 \partial_0\Phi_1^*) $$

I see that this definition should be invariant under Poincaré transformations, but I couldn't prove it. Do you know some references about this?

Moreover I couldn't find the reason why such a scalar product is introduced. Aren't there other possible scalar products? Why choose this one?

$\endgroup$
  • 5
    $\begingroup$ Something to think about: consider the current $J_\mu = i\Phi_1^* \partial_\mu \Phi_2 - i\Phi_2 \partial_\mu \Phi_1^*$ and maybe set $\mu = 0$... What can you say about $J_\mu$? $\endgroup$ – Vibert Jan 16 '13 at 22:46
  • $\begingroup$ For a thorough treatment of that expression, consider this paper. $\endgroup$ – Frederic Brünner Jan 17 '13 at 3:29
5
$\begingroup$

The Klein-Gordon inner product is a natural construction for functions defined on the mass hyperboloid $k^2=m^2$, because if you write your function in momentum space, $$ \phi(x)\sim \int\widetilde{\mathrm dk}\ \mathrm e^{-ikx}a(k)+\text{h.c.} $$ with $\widetilde{\mathrm dk}$ the measure on $k^2=m^2$, then the Fourier coefficients become (ref. 1, sec 3-1-2) $$ a(k)=\langle\phi,\exp_k\rangle $$ where $\exp_k(x)\equiv\mathrm e^{-ikx}$.

The philosophy is the same as in the standard Fourier transform (or other integral transforms, i.e., changes of basis), where you can recover the function in momentum space by a suitable scalar product with the exponential function (or whichever basis you use in your space of functions). In general, the form of the inner product is dictated by the form of the integral transform.

One should point out that, even if $\langle\cdot,\cdot\rangle$ is a natural construction, the real reason we define this particular integral is that it appears in the proof of the LSZ formula for scalars (ref. 1, sec 5-1-4).

References

  1. Itzykson and Zuber, Quantum Field Theory.
$\endgroup$
  • $\begingroup$ remark: there might me some signs wrong here. $\endgroup$ – AccidentalFourierTransform Mar 14 '17 at 15:51
  • $\begingroup$ But if you have an arbitrary inner product $(,)$ you get $(\phi(x),\exp_k(x)) = c(k)$ where the $c(k)$ are the coefficients in expansion of $\phi$ now, $i.e.$ with respect to this other arbitrary inner product $\phi(x)=\int [c(k) \exp_{k}(x) + c(k)^{\ast} \exp_{-k}(x)]$. What restricts me from upgrading $c(k)$ to creation/annihilation operators (taking $[c(k),c(p)]=...$ and so on) and quantizing the field in this way with respect to the arbitrary inner product? I guess the question boils down now to what is the defining property of $a(k)$ here? $\endgroup$ – Greg.Paul Jul 10 '18 at 20:00
  • $\begingroup$ @Greg.Paul Nothing prevents you from doing that. The question is whether it is useful. In any case, the inner product is essentially unique: the kernel is the Wronskian operator for the PDE (cf. this PSE post). $\endgroup$ – AccidentalFourierTransform Jul 10 '18 at 20:05

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.