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I asked this question on Stack Exchange, and didn't get an answer, but maybe someone here will be able to help. In non-relativistic quantum mechanics, the normalization condition for position eige...
#1: Initial revision
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Technically Natural
·
2021-01-16T14:57:24Z (almost 4 years ago)
How does probability conservation work in Dirac's original formulation of relativistic QM?
I asked this [question](https://physics.stackexchange.com/q/580945) on Stack Exchange, and didn't get an answer, but maybe someone here will be able to help. In non-relativistic quantum mechanics, the normalization 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 rectify this issue within single-particle relativistic QM?