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Define the Klein Gordon inner product as

$$(\psi_1,\psi_2)_{KG} = i\int d^3x \, \psi_1^*\,\partial{t}\,(\psi_2) - \partial{t}\,(\psi_1^*)\,\psi_2 \, .$$

It can be shown that for the one particle wavefunctions \begin{align} \psi_1(t,\vec{x}) &= \exp(-iE_{p_1}t + i\vec{p_1}\cdot\vec{x}) \\ \text{and} \quad \psi_2(t,\vec{x})&=\exp(-iE_{p_2}t + i\vec{p_2}\cdot\vec{x}) \end{align} that $$(\psi_1^*,\psi_2)_{KG} = 0 \, .$$

By superposition, I would expect this to imply that for any $\psi_1$ and $\psi_2$,

$(\psi_1^*,\psi_2)_{KG} = 0$.

However, I am sure that this cannot be the case, since with this you could show that the Klein-Gordon inner product for any two functions would be zero, i.e.

$$(\psi_1,\psi_2)_{KG} = 0$$

for all $\psi_1$ and $\psi_2$, since I could just pick $\psi_1$ to be the complex conjugate of some other function. Does anyone see a quick resolution to this?

Qmechanic
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anon123456789
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1 Answers1

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The correct inner product is proportional to a Dirac delta, viz. $\left( \psi_1,\,\psi_2 \right)_{KG}=\left( 2\pi\right)^3 2E_{p_1}\delta^3\left( \mathbf{p}_1 - \mathbf{p}_2\right)$.

You have only considered positive-energy wavefunctions; you may wish to look at what happens when one or both are negative-energy instead, since the most general solution of the KG equation superposes both energy sectors. You should find for example that $\left( \psi^\ast,\,\psi^\ast \right)_{KG}=-\left( \psi,\,\psi \right)_{KG}$ for any solution $\psi$.

J.G.
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