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  On the transformation rule of 'in and out‘ states

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In p. 108 (as shown below) of Weinberg's field theory book 1, he said that the transformation rule of non-interacting particles is (3.1.1). Because of  the definition of the in and out states, they transform as (3.1.1) . So, they transform as the same way.

But in p. 116 (as shown below) , he said: "For any proper orthochronous Lorentz transformation....... as in (3.1.1) on both 'in' and 'out' states. " 

My question is, wherein I was mistaken in the first paragraph? Thank you for your patience!

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asked Dec 14, 2017 in Theoretical Physics by shi_zonghua (15 points) [ no revision ]

I guess that the 'in and out' states transform as the same rule, but in different reference frame.  The inner product must be calculated in the same reference frame. But I still find other inconsistencies, and still need your opinion.

Weinberg states both times that they transform the same way, and that is correct. The in and out states are specified in the same frame, and when you change the frame both change according (3.1.1) to get their equivalent description in the new frame. Thus everything is consistent. 

Thank you, Mr. Neumaier ! But I think I  hadn't been able to show my question clearly.   The conclusion  that in and out states transform as the same way seems to merely need  the definition of in and out states. It need not the restriction on the form of the interreaction! 

So I guess there must be some mistakes on the understanding of the definition, but where is it?

1 Answer

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I try to give an answer, welcome corrections!

When we write a state, we must notice the reference frame which the state lies in, because the form of state is diffrent in diffrent frame. Now I give 3 frames $O,O', O''$. Their correlations are $t'=t-\tau'$,  $t''=t-\tau''$, with $\tau'=-\infty$, $\tau''=+\infty$. Let's specify that the collision happens in $t=0$, in $O$ frame's view. 

When we say $\Psi^\pm$ are tranform as free particles, we actually mean $\Psi^+$ transforms as free particles just in frame $O'$, and  $\Psi^-$ transforms as free particles just in frame $O''$. 

Turn now to the inner product between in and out states $(\Psi^-, \Psi^+)$. Noticing that the product must be calculated in the same frame, we specify the frame is $O$. Now, we do a transformation $T$, and to see how the inner product changes. Please note, in frame $O$, neither of $\Psi^\pm$ transforms as free particles. However, we can use time translation to take $\Psi^+$ to frame $O'$, and act on it with $U_0(T)$, then take back to frame $O$. 

So, under transformation $T$,    $(\Psi^-, \Psi^+)$ changes into

$$(exp(+iH\infty)U_0(T)exp(-iH\infty)\Psi^-, exp(-iH\infty)U_0(T)exp(+iH\infty)\Psi^+)$$

Obviously, they are not equal, unless there are some restrictions on $H$. 

answered Dec 18, 2017 by shi_zonghua (15 points) [ revision history ]
edited Dec 20, 2017 by shi_zonghua

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