0
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ Your product is missing a factor of $i$ that ensures sesquilinearity. $\endgroup$ – J.G. Mar 27 '17 at 11:37
2
$\begingroup$

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 $\psi$.

$\endgroup$
  • $\begingroup$ Ah sorry the integration over time was an error $\endgroup$ – bittermania Mar 27 '17 at 11:28
  • $\begingroup$ I see now that the single particle wavefunctions which I have provided are only basis functions for the positive energy sector. If I conjugate a function from the positive energy sector, I get a function in the negative energy sector. Intuitively it makes sense that the "overlap" of a function in negative energy sector with a function in the positive energy sector is zero, which is exactly what is being calculate in the fourth expression. Thanks for the input. $\endgroup$ – bittermania Mar 27 '17 at 11:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.