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I was reading in Srednicki's QFT book, Chapter 1 and he was explaining why the Klein-Gordon equation doesn't obey quantum mechanics. He said the fact that the time derivative is second order means it disobey's Shrodinger's equation which is first order in the time derivative. He continues to say that this might not seem important but it has significant consequences. One consequence is that the norm:

[tex] \langle \psi, t | \psi, t \rangle = \int d^{3}x \langle \psi, t | \bar x \rangle \langle \bar x | \psi, t \rangle = \int d^{3}x \psi^{*}(x)\psi(x) [/tex]

is not in general time-independent (This is equation 1.23 in Srednicki's for reference). Thus the probability is not conserved, and hence we conclude that quantum mechanics is not obeyed (because quantum mechanics tells us that probability is conserved).

I have three small questions here please. The first is how he got rid of the time dependence in the second equality above. My second question is how is the last equality time dependent? And finally my thrid question is why time dependence (in the norm of a state) implies probability is not conserved. What if probability is conserved at everytime, t, but the probability distribution is time dependent?

Thank you in adavance,

iman

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# Why does Klein-Gordon's equation not obey quantum mechanics?

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