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

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.

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 $\cal H_-$).

• 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? – TheQuantumMan Feb 16 '17 at 2:20
• 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. – Valter Moretti Feb 16 '17 at 6:52
• This solution, however is not a state however. – Valter Moretti Feb 16 '17 at 7:05