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In Peskin & Schroeder's book (Page 104, equation 4.70) localized in and out states of different momenta have been defined. The authors state that they are working in the Heisenberg picture. However, in equation 4.70, to compute the scalar product between the in and the out state, the authors put in a factor $e^{-i2HT}$ in between them, to relate the states to be defined at a "common reference time". As was explained in this: question, the authors do so because that way the states $|k \rangle_{in}$ and $|k \rangle_{out}$ are both drawn from the same Hilbert space, belonging to an arbitrary time slice.

However, I didn't exactly understand why one state labeled by $k_a$ or $k_b$ differs for different times in the first place. Peskin says that those states are named by the same name, but are different states, because they should recreate the same eigenvalues for (for example) the momentum operator.

My question now is: When Peskin writes a $|k_{a}\rangle$ at a "common reference time", what is meant by that? Is it a state in which the (field- and time-dependent) momentum operator at time $t$ has the expectation value $k_{a}$ ? What would it be for a wave package with some finite spatial extent, like the ones that were defined earlier in the chapter?

Buzz
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Quantumwhisp
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  • It sounds like your confusion is actually about conflating Hilbert spaces on different time slices. They are all distinct, but isomorphic. This is the same as the statement that tangent spaces at different points on a manifold are isomorphic, but not the same. The isomorphism is given by precisely the time-evolution operators. – Richard Myers Jun 28 '21 at 02:33
  • @RichardMyers Peskin before mentioned explicitly that we are working in the "Heisenberg picture", where there shouldn't be any time dependence of the states. This is where the confusion comes from. I thought that I could lift the confusion by saying that the states at different times are not the same states, but are states that yield the same expectation values. Is that correct? – Quantumwhisp Jun 28 '21 at 07:13
  • The statement that the "states" in the Heisenberg picture don't have any time-dependence is always a misleading one. The correct statement is that the state of the system is time-independent. Eigenstates of various operators, which is what the $|k_a\rangle$ states are, are time-dependent in this picture since the operators themselves are time-dependent. – Richard Myers Jun 28 '21 at 23:26
  • @RichardMyers this is what I meant by " the states at different times are not the same states, but are states that yield the same expectation values". So , because of that, the $|k_a\rangle$ evolve by $e^{iHt} $? – Quantumwhisp Jun 28 '21 at 23:30

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