# Klein-Gordon equation: orthonormality of solutions

Reference: References: Robert D. Klauber, Student Friendly Quantum Field Theory, (Sandtrove Press, 2013) – Chapter 3, Problem 3.2.

The plane wave solutions of the Klein-Gordon equation are

$\displaystyle \phi=\sum_{\mathbf{k}}\frac{1}{\sqrt{2V\omega_{\mathbf{k}}}}\left(A_{\mathbf{k}}e^{-ikx}+B_{\mathbf{k}}^{\dagger}e^{ikx}\right) \ \ \ \ \ (1)$

We can redefine a couple of terms by introducing

 $\displaystyle \phi_{\mathbf{k},A}$ $\displaystyle \equiv$ $\displaystyle \frac{e^{-ikx}}{\sqrt{V}}\ \ \ \ \ (2)$ $\displaystyle \phi_{\mathbf{k},B^{\dagger}}$ $\displaystyle \equiv$ $\displaystyle \frac{e^{ikx}}{\sqrt{V}} \ \ \ \ \ (3)$

Then

$\displaystyle \phi=\sum_{\mathbf{k}}\frac{1}{\sqrt{2\omega_{\mathbf{k}}}}\left(A_{\mathbf{k}}\phi_{\mathbf{k},A}+B_{\mathbf{k}}^{\dagger}\phi_{\mathbf{k},B^{\dagger}}\right) \ \ \ \ \ (4)$

The ${\phi_{\mathbf{k},A}}$ and ${\phi_{\mathbf{k},B^{\dagger}}}$ are orthonormal functions. We have

$\displaystyle \int\phi_{\mathbf{k},A}^{\dagger}\phi_{\mathbf{k}^{\prime},A}d^{3}x=\frac{1}{V}\int e^{i\left(k^{\prime}-k\right)x}d^{3}x \ \ \ \ \ (5)$

where the integral is over the volume ${V}$, and the wavelengths of the plane waves fit an integral number of times within ${V}$, so that the amplitudes of the waves at the boundaries are all zero. The four-vector ${k}$ is defined as

$\displaystyle k=\left[\omega_{\mathbf{k}},\mathbf{k}\right] \ \ \ \ \ (6)$

If ${k^{\prime}=k}$, the integrand is 1 and is integrated over ${V}$, so the result is

 $\displaystyle \int\phi_{\mathbf{k},A}^{\dagger}\phi_{\mathbf{k},A}d^{3}x$ $\displaystyle =$ $\displaystyle \frac{1}{V}\int e^{i\left(k^{\prime}-k\right)x}d^{3}x\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{V}{V}=1 \ \ \ \ \ (8)$

If ${k^{\prime}\ne k}$, consider the integral over ${x^{1}=x}$ (for the purposes of this derivation only, ${x}$ refers to the single ${x}$ dimension of the 3-vector ${\mathbf{x}}$ and should not be confused with the four-vector ${x}$ used in 1):

 $\displaystyle \int\phi_{\mathbf{k}^{\prime},A}^{\dagger}\phi_{\mathbf{k},A}dx$ $\displaystyle =$ $\displaystyle \frac{1}{V}e^{i\left(\omega_{\mathbf{k}^{\prime}}-\omega_{\mathbf{k}}\right)t}e^{-i\left(k_{y}-k_{y}^{\prime}\right)y}e^{-i\left(k_{z}-k_{z}^{\prime}\right)z}\int e^{-i\left(k_{x}-k_{x}^{\prime}\right)x}dx\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{1}{i\left(k_{x}-k_{x}^{\prime}\right)V}e^{i\left(\omega_{\mathbf{k}^{\prime}}-\omega_{\mathbf{k}}\right)t}e^{-i\left(k_{y}-k_{y}^{\prime}\right)y}e^{-i\left(k_{z}-k_{z}^{\prime}\right)z}\left[e^{-i\left(k_{x}-k_{x}^{\prime}\right)x}\right]_{x=x_{0}}^{x=x_{1}}\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (11)$

where ${x_{0}}$ and ${x_{1}}$ are the ${x}$ limits of ${V}$, where by assumption the wave amplitude is zero. Therefore

$\displaystyle \int\phi_{\mathbf{k}^{\prime},A}^{\dagger}\phi_{\mathbf{k},A}d^{3}x=\delta_{\mathbf{k},\mathbf{k}^{\prime}} \ \ \ \ \ (12)$

The same result follows for ${\phi_{\mathbf{k},B^{\dagger}}}$ by just replacing ${kx}$ by ${-kx}$ throughout the derivation, so

$\displaystyle \int\phi_{\mathbf{k},B^{\dagger}}^{\dagger}\phi_{\mathbf{k},B^{\dagger}}d^{3}x=\delta_{\mathbf{k},\mathbf{k}^{\prime}} \ \ \ \ \ (13)$

For mixed terms, we have

$\displaystyle \int\phi_{\mathbf{k}^{\prime},A}^{\dagger}\phi_{\mathbf{k},B^{\dagger}}d^{3}x=\frac{1}{V}\int e^{i\left(k^{\prime}+k\right)x}d^{3}x \ \ \ \ \ (14)$

In this case, the exponent cannot be zero, so the integral always comes out to zero, so that

$\displaystyle \int\phi_{\mathbf{k}^{\prime},A}^{\dagger}\phi_{\mathbf{k},B^{\dagger}}d^{3}x=0 \ \ \ \ \ (15)$