Time evolution operator versus propagate

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The time evolution operator is a unitary operator used to calculate the probability amplitude of transitioning from one quantum state to another. The Feynman propagator also provides an amplitude for transitioning between states but is specifically a matrix element of the time evolution operator. The relationship is defined as the Feynman propagator being equal to the inner product of the position eigenstates with the time evolution operator. The discussion clarifies that while both concepts relate to state transitions, they operate in slightly different contexts—one focusing on time evolution and the other on spatial paths. Understanding this distinction enhances the comprehension of quantum mechanics and state propagation.
qtm912
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I am trying to understand the how the time evolution operator is used versus the Feynman propagate.
My limited understanding is the following for which I am seeking clarity:

1. The time evolution operator is a unitary operator which enables us to calculate a probability amplitude of one state going to another state.

2. The Feynman propagation also gives the amplitude between two different states.

So the question is, first of all are the above statements correct , are the two comparable and how do they differ?

My initial thought was that the first one referred to a point in space but different times while the second referred to two different points in space time with an integration across all possible paths when the propagator is used to calculate an amplitude.

Help appreciated.

Thanks.
 
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1 is correct. 2 is correct, but more specifically the propagator is a certain matrix element of the time evolution operator. If ##U(t_2, t_1)## is the time evolution operator between the two times ##t_1## and ##t_2##, and ##K(x_2, t_2, x_1, t_1)## is the Feynman propagator (the amplitude to propagate from position ##x_1## at time ##t_1## to position ##x_2## at time ##t_2##) then we have

##K(x_2, t_2, x_1, t_1) = \langle x_2 | U(t_2, t_1) | x_1 \rangle##

where ##| x_1 \rangle## and ##| x_2 \rangle## are eigenstates of the position operator. This is really just the definition of ##K##.
 
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Yes that clarifies it very well. Thank you very much.
 
Time reversal invariant Hamiltonians must satisfy ##[H,\Theta]=0## where ##\Theta## is time reversal operator. However, in some texts (for example see Many-body Quantum Theory in Condensed Matter Physics an introduction, HENRIK BRUUS and KARSTEN FLENSBERG, Corrected version: 14 January 2016, section 7.1.4) the time reversal invariant condition is introduced as ##H=H^*##. How these two conditions are identical?

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