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The normalization $N$ may be calculated if you evaluate the squared norm Insert e.g. the "limit" form of the exponential for the exponential above. … So the norm is $\exp(\lambda^2)$ and to guarantee the normalization to one, as you need it, you need $$ N_{\rm yours} = \exp(-|\lambda|^2/2)$$ I added the absolute value because $\lambda$ may be complex …

Physicists usually generously relax the condition that the norm should be finite and they sometimes say that $|\vec r\rangle,|\vec p\rangle$ belong to the "Hilbert space". It's exactly the same "gener …

Both states $\Psi_{k,\sigma}$ and $\Psi_{k',\sigma'}$ are meant to be states of the same particle species i.e. they have the same values of the squared mass $k^2$. The inner product of one-particle st …

It is not wrong, all three normalization conditions are natural and they don't contradict each other because, in fact, the first equation is nothing else than the product of the following two equations …