Prove the time evolution operator is unitary

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Happiness
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How is (5.240b) derived? I get [itex]{U^{-1}}^\dagger(t, t_0)\,U^{-1}(t, t_0)=I[/itex] instead.

My steps:
[tex]\begin{align}<\psi(t_0)\,|\,\psi(t_0)>&=\,<U(t_0, t)\,\psi(t)\,|\,U(t_0, t)\,\psi(t)>\\<br /> &=\,<U^{-1}(t, t_0)\,\psi(t)\,|\,U^{-1}(t, t_0)\,\psi(t)>\\<br /> &=\,<\psi(t)\,|\,{U^{-1}}^\dagger(t, t_0)\,U^{-1}(t, t_0)\,|\,\psi(t)>\end{align}[/tex]

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Also, to get (5.240a), do we use the fact that [itex]<\psi(t_0)\,|\,\psi(t_0)>\,=\,<\psi(t_0)\,|\,U^\dagger(t, t_0)\,U(t, t_0)\,|\,\psi(t_0)>[/itex]is true for any [itex]\psi(t_0)[/itex]?
 
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Happiness said:
How is (5.240b) derived? I get [itex]{U^{-1}}^\dagger(t, t_0)\,U^{-1}(t, t_0)=I[/itex] instead.
Multiply both sides by appropriate matrices and you should get the result you want.
Also, to get (5.240a), do we use the fact that [itex]<\psi(t_0)\,|\,\psi(t_0)>\,=\,<\psi(t_0)\,|\,U^\dagger(t, t_0)\,U(t, t_0)\,|\,\psi(t_0)>[/itex]is true for any [itex]\psi(t_0)[/itex]?
Right.
 
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