Orthonality of modes for solutions to scalar field in QFT I am trying to self study QFT and while reading through "Quantum Fields in Curved Space", I got stuck at equation 2.10 in the book. The synopsis of the problem I have is as follows.
Consider the solution:
$$u_{\mathbf{k}}=e^{i\mathbf{k}\cdot\mathbf{x}-i\omega t}$$
The scalar product is defined as:
$$(\phi_1,\phi_2)=-i\int\phi_1(x)\partial_t\phi^{*}_{2}(x)-[\partial_t\phi_{1}(x)]\phi^{*}_{2}(x)\,\mathrm{d}^{n-1}x$$
where $x\equiv (t,\mathbf{x})$
How do I use the above equation to show that $(u_{\mathbf{k}},u_{\mathbf{k}'})=0$ for $\mathbf{k}\ne\mathbf{k}'$?
 A: Ok, so lets split this into bite sized pieces and see how it goes.
First, we write the whole expression that we want to evaluate:
$$
(u_\mathbf{k},u_{\mathbf{k'}}) = -i\int\left(u_\mathbf{k}\partial_t u_\mathbf{k'}^* - \partial_tu_\mathbf{k},u_\mathbf{k'}^*\right)d^{n-1}x
$$
Now, lets write down the two terms we are integrating:
$$
u_\mathbf{k}\partial_t u_\mathbf{k'}^* = e^{i\mathbf{k}\cdot\mathbf{x}-i\omega t}\partial_t e^{-i\mathbf{k'}\cdot\mathbf{x}+i\omega' t} = i\omega' e^{i(\mathbf{k}-\mathbf{k'})\cdot\mathbf{x}-i(\omega-\omega')t}
$$
$$
(\partial_tu_\mathbf{k}) u_\mathbf{k'}^* = -i\omega e^{i(\mathbf{k}-\mathbf{k'})\cdot\mathbf{x}-i(\omega-\omega')t}
$$
Plugging those in the integral we should get:
$$
(u_\mathbf{k},u_{\mathbf{k'}}) = (\omega+\omega')e^{-i(\omega-\omega')t}\int e^{i(\mathbf{k}-\mathbf{k'})\cdot\mathbf{x}} d^{n-1}x
$$
At this point we should recognise the integral part as the delta funtion $\delta(\mathbf{k}-\mathbf{k'}) = \int e^{i(\mathbf{k}-\mathbf{k'})\cdot\mathbf{x}} d^{n-1}x$.
This should answer your question.
