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)

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