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.

  • $\begingroup$ Maybe you are not considering the $\delta$ coming from the integration? $\endgroup$ – pp.ch.te Jun 18 '19 at 10:31
  • $\begingroup$ 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 $\endgroup$ – 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.


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