# Klein-Gordon Inner Product from Greiner's book doubt

I was working on free field theory from Greiner's book "Field Quantization" In chapter 4, he introduces these phase functions: $$u_{p}(\boldsymbol{x}, t)=N_{p} \mathrm{e}^{-\mathrm{i} p \cdot x}=\frac{1}{\sqrt{2 \omega_{p}(2 \pi)^{3}}} \mathrm{e}^{-\mathrm{i}\left(\omega_{p} t-p \cdot x\right)}$$ And define the Klein-Gordon inner product: $$(\phi, \chi)\equiv \mathrm{i} \int \mathrm{d}^{3} x\left[\phi^{*}(\boldsymbol{x}, t) \frac{\partial \chi(\boldsymbol{x}, t)}{\partial t}-\frac{\partial \phi^{*}(\boldsymbol{x}, t)}{\partial t} \chi(\boldsymbol{x}, t)\right]$$ And says that: $$\left(u_{p^{\prime}}, u_{p}^{*}\right)=\left(u_{p^{\prime}}^{*}, u_{p}\right)=0$$ But I could swear that that's not true in general but only when $$p=p'$$.

Since when the derivative operates on $$u_{p}^{*}$$, a factor $$i\omega_{p}$$ is brought down and when it operates on $$u_{p'}^{*}$$ the factor $$i\omega_{p'}$$ does. So on this way, finally I get (with more terms) the difference of these two frequencies: $$\omega_{p}-\omega_{p'}$$. That's my question, am I wrong? and in what?, I'm so sorry if the question is too trivial but I just can't let it go.

Thanks and greetings.

• Maybe you are not considering the $\delta$ coming from the integration? – pp.ch.te Jun 18 '19 at 10:31
• Yes, that is true, even with the two phases at the same sign, I forgot that $\omega_{k}$ depends on the square of k, so $\omega_{k}=\omega_{-k}$. Thanks – Amadeus Jun 18 '19 at 15:45

First of all upon checking the orthogonality of momentum eigenstates with a particular scalar product, $$(u_p', u_p)$$ is considered and not $$(u_p', u^{\ast}_p)$$.
Second plugging $$u_p$$ and $$u_p'$$ in the Klein-Gordon inner product one gets (remember $$px = \omega_p t-\vec{p}\vec{r}$$)
$$(u_p', u_p) = N_{p'}N_{p}\int d^3 x \left[e^{ip'x}e^{-ipx} \omega_p + \omega_{p'} e^{ip'x} e^{-ipx}\right] = N_{p'}N_{p} (2\pi)^3 \left[\omega_p\delta^3(\vec{p}-\vec{p'})e^{i(\omega_{p'}-\omega_p)t} + \omega_{p'}e^{i(\omega_{p'}-\omega_p)t}\delta^3(\vec{p}-\vec{p'})\right]$$
since $$(2\pi)^3 \delta^3(\vec{q}) = \int d^3 e^{i\vec{q}\vec{r}}$$. Due to the properties of the delta-function it is clear that if $$p'\neq p$$ the expression is zero as it should be. In the end one obtains:
$$(u_p', u_p) = N^2_p (2\pi)^3 2\omega_p \delta^3(\vec{p}-\vec{p'}) \equiv \delta^3(\vec{p}-\vec{p'})$$
due to the chosen normalisation constant $$N_p = (\sqrt{(2\pi)^3 2\omega_p})^{-1}$$. The last equality is actually only true in case of an further integration over $$\vec{p'}$$. The latter is implicitly always assumed if a $$\delta$$-function appears that is not already integrated over.