# Is probability conserved in Dirac's original formulation of relativistic QM?

In non-relativistic quantum mechanics, the normal condition for position eigenstates is $$\langle y|x\rangle=\delta(y-x).$$ However, this condition is not Lorentz-invariant. I have never seen a textbook on relativistic quantum mechanics address this normalization issue. It seems to me like this is a very important issue. If the normalization of the position eigenstates is not invariant, the inner product of any state vectors is also not necessarily invariant (and hence probability is not conserved under a Lorentz transformation).

How is this issue resolved? Is it necessary to resort to quantum field theory, or is there a way to redefine the position eigenstate normalization condition to make it Lorentz-invariant?

• Relevant: Chapter $1$, "Quantum Field Theory: Lectures of Sidney Coleman". Edit note: I apologize, I had the name of the book wrong in the earlier version.
– user87745
Sep 20, 2020 at 22:58
• Sep 21, 2020 at 3:10
• The covariant inner product $\langle \mathbf{x}_f,\tau_f|\mathbf{x}_i,\tau_i\rangle$ discussed in Qmechanic's answer to the related question (physics.stackexchange.com/q/346780/2451) is non-zero even when $\tau_f=\tau_i$ but $\mathbf{x}_f\neq\mathbf{x}_i$. This means that these states cannot be interpreted as the eigenstates of a Hermitian position operator. Sep 21, 2020 at 14:15
• I read that this issue is solved by $\gamma^0$ matrix (like that appearing in the Dirac field), so that, considering the indices $\alpha$ related to the little group ${\text{SO}^+(1,3)}_{\boldsymbol{x}}$ associated to $\boldsymbol{x}$ eigenvalue, you shoul have something like $\langle\boldsymbol{x},\alpha|\tilde{\boldsymbol{x}},\tilde{\alpha}\rangle=\delta(\boldsymbol{x}-\tilde{\boldsymbol{x}}){\gamma^0}_{\alpha\tilde{\alpha}}$, but sorry I cannot help you more because that's just something I read, and I have just the same problem Jan 3, 2021 at 10:21
• In case anyone's interested, I got an answer here. Feb 9, 2021 at 17:21