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Do the particle and anti-particle solutions of the Klein-Gordon equation live in different Hilbert spaces?

Our professor said that this is true because the integral of their respective probability density give the values +1 and -1 respectively, but I can't quite understand why this implies that they live in different Hilbert spaces.

Note: This question is about relativistic quantum mechanics and not quantum field theory.

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The question is not well stated as it stands.

Mathematically speaking it is maybe meaningless or not very interesting: If $\psi \in \cal H$ and $\phi \in \cal H'$ then we can always say that $\psi, \phi \in \cal H \oplus \cal H'$.

However it is not physically meaningless. Indeed, the correct interpretation of the question is in my opinion whether or not one-particle states of a charged field belong to unitary equivalent irreducible representations of the group of physical symmetries of the system: $U(1) \times SO(1,3)^+$.

The answer to OP's question is positive, because the representation of $U(1)$ has two different generators here in accordance with the sign of the charge.

In both cases (particle/antiparticle) the one-particle Hilbert space $\cal H_\pm$ is isomorphic to $L^2(\mathbb R^3, dp)$, but the generator of $U(1)$ is $Q^{(+)}=qI$ for the particle in $\cal H_+$ and $Q^{(-)}=-qI$ for the antiparticle in $\cal H_-$.

The two representations are unitarily inequivalent: There is no unitary operator $U : \cal H_+ \to \cal H_-$ such that $UA^{(+)}U^{-1}= A^{(-)}$ where $A^{(\pm)}$ is the generic self-adjoint generator of the relevant representation of $U(1) \times SO(1,3)^+$. When $A^{(\pm)}=Q^{(\pm)}$, evidently there is no such $U$ with $UQ^{(+)}U^{-1}=Q^{(-)}$.

There is another viewpoint corroborating my positive answer. (It would be interesting asking OP's professor about the true meaning of his/her remark). One may argue that the overall one-particle space is $\cal H_+ \oplus \cal H_-$. This is not correct, because the superselection rule of the electric charge takes place: no coherent superposition of states in $\cal H_+$ and $\cal H_-$ are physically permitted. The space of pure states is represented by the unit vectors of $\cal H_+ \cup \cal H_-$. There are only either particle states (in $\cal H_+$) or antiparticle states (in $\cal H_-$).

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  • $\begingroup$ In the book "Quantum Field Theory for the gifted amateur" by Blundell and Lancaster, on page 63 it writes that a general solution to the Klein-Gordon equation corresponds to a superposition of two states, one for a particle and one for its antiparticle. How can we just add the two solutions since each corresponds to a different particle and thus belongs to a different Hilbert space? $\endgroup$ – TheQuantumMan Feb 16 '17 at 2:20
  • $\begingroup$ Because each state is represented in the same spacetime essentially via Fourier transformation and next these two functions are summed together to give the general solution you said. $\endgroup$ – Valter Moretti Feb 16 '17 at 6:52
  • $\begingroup$ This solution, however is not a state however. $\endgroup$ – Valter Moretti Feb 16 '17 at 7:05

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