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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.

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?

Qmechanic
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justmyfault
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1 Answers1

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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.
AccidentalFourierTransform
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