# Klein-Gordon and localizability

If we consider the theory of a single relativistic point particle, quantized using whatever appropriate method, the wavefunction simply obeys the Klein-Gordon equation, which allows for a fairly wide range of initial solutions, including solutions of compact support.

In which case, we can consider the proposition of the presence of the particle in some region of space $\Delta$, noted by $P_\Delta$ :

$$P_\Delta (\psi) = \int_\Delta d^3x\ \frac i2 [\psi^* (\partial_t \psi) - (\partial_t \psi^*) \psi]$$

If we pick a wavefunction of compact support, this propositon should be equal to $1$ in the support of the wavefunction, as $\psi$ will be $0$ outside of it and the probability should add up to unity.

However, according to Malament's theorem, a reasonable relativistic quantum theory with the possibility to localize particles in this manner should have, for any compact $\Delta$, no possibility for the probability to add up to $1$ in this region. In other words there should be no finite region in which the particle should be measured with certainty.

Where is the problem here? Does the measurement for such a compact support wavefunction somewhat cancel out? Does this system not fulfill the requirements of the theorem, or am I misunderstanding the content of Malament's theorem?

Also in the case of Klein Gordon for QFT, while describing the localization of a field is less straightforward, it would seem that a wavefunction where the measurement of the field $\hat \phi |\Psi \rangle$ is of compact support might also fulfill the same requirements, is there also a problem there?

• I'm not quite sure what $P_\Delta(\psi)$ is. Is it a probability? if so, are you sure that $P_\Delta(\psi)\in[0,1]$? – AccidentalFourierTransform Jan 25 '18 at 0:43
• Indeed, with your definition $P$ is not associated to an orthogonal projector (as required in the paper you quoted) since it is not positively defined. – Valter Moretti Jan 25 '18 at 6:35
• Isn't the set of physical states reduced to positive energy states, for which this is indeed positive, once the Hilbert is constrained, due to the invariance under reparametrization? – Slereah Jan 25 '18 at 7:07
• Yes, but if you only consider positive frequency packets they cannot have compact support on Cauchy surfaces, This is essentially true because they are analytic functions and analytic functions cannot have compact support unless being the zero function. – Valter Moretti Jan 25 '18 at 9:45
• Ah yes, that would be a good reason indeed, thanks! – Slereah Jan 25 '18 at 9:50