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How to prove the equivalence of two different difinitions of S-operator? i.e. $\Omega_+(\Omega_-)^\dagger= e^{i \alpha}(\Omega_-)^\dagger\Omega_+$

+ 2 like - 0 dislike
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I read there are two definitions about [S-operator](https://en.wikipedia.org/wiki/S-matrix#The_S-matrix):

The first one (e.g (8.49) in Greiner's Field Quantization) is:
$$S_{fi}\equiv \langle \Psi_p^{-}| \Psi_k^{+}\rangle$$
where $|\Psi_p^{-}\rangle$ is a state in Heisenberg picture which is $| p \rangle$ at $t=+\infty$ when you calculate the $|\Psi_p^{-}\rangle$ in Schrodinger picture , called out state. $| \Psi_k^{+}\rangle$  is a state in Heisenberg picture which is $| k \rangle$ at $t=-\infty$, called in state.

So$$S_{fi}\equiv \langle \Psi_p^{-}| \Psi_k^{+}\rangle= \langle p|(\Omega_-)^\dagger\Omega_+|k \rangle$$

In this case the S-operator $\hat S=(\Omega_-)^\dagger\Omega_+$,
where Møller operator 
$$\Omega_+ = \lim_{t\rightarrow -\infty} U^\dagger (t) U_0(t)$$
$$\Omega_- = \lim_{t\rightarrow +\infty} U^\dagger (t) U_0(t)$$
So $$S=U_I(\infty,-\infty)$$

Another definition (e.g (9.14) (9.17) (9.99) in Greiner's Field Quantization) is :
$$S_{fi}\equiv \langle \Psi_p^{-}| \Psi_k^{+}\rangle\equiv\langle \Psi_p^{-}| \hat S ^\prime |\Psi_k^{-}\rangle=\langle \Psi_p^{+}| \hat S ^\prime |\Psi_k^{+}\rangle$$
where S-operator
$\hat S ^\prime |\Psi_p^{-}\rangle =|\Psi_p^{+}\rangle$ that is $\hat S^\prime = \Omega_+(\Omega_-)^\dagger$.

It seems that these two definitions are differnt, but many textbook can derive the same dyson formula for these two S-operators. 
https://en.wikipedia.org/wiki/S-matrix#The_S-matrix

How to prove: $$\Omega_+(\Omega_-)^\dagger= e^{i \alpha}(\Omega_-)^\dagger\Omega_+$$

related to this question: http://physics.stackexchange.com/questions/105152/there-are-two-definitions-of-s-operator-or-s-matrix-in-quantum-field-theory-a

asked Apr 1 in Theoretical Physics by Alienware (185 points) [ revision history ]
edited Apr 1 by Alienware

I don't think the claimed equality holds. Probably different sources use slightly different definitions of the Moeller operators.

@ArnoldNeumaier The first definition is obiviously right. You can find the second definition still has a form of Dyson formula except a phase in many textbook, and authors have given a proof e.g. Greiner's Field Quantization  (9.14) (9.17) (9.99). Hatfield's QFT (7.43) (7.63)-(7.90) 

1 Answer

+ 1 like - 0 dislike

The first definition gives the  S-matrix in the interaction picture (a unitary operator in asymptotic space), the second formula gives the S-matrix in the Heisenberg, a unitary operator in Heisenberg space. These usually (e.g., if there is more than one channel) operate in distinct Hilbert spaces. Note that $\Omega_\pm$ go from the asymptotic Hilbert space to the Heisenberg Hilbert space (which are in general distinct). 

See Thirring's Course on Mathematical Physics, Vol. 3, Definition (3.4.23) (p.138 in the first edition).

answered Apr 3 by Arnold Neumaier (12,385 points) [ revision history ]
edited Apr 3 by Arnold Neumaier

So the S-operators in these two definition are two different operators. But are they only differ by a phase? Because I really found  in different textbook these two different operators have the same Dyson series.

@Alienware: In general they cannot be compared at all since they act on different spaces. To compare them one must identify the asymptotic Hilbert space in some way with the Heisenberg Hilbert space, and there is no canonical way to do so. To actually compute S-matrix elements to compare with cross sections or other collision data you need the interaction S-matrix. 

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