Suppose you want the 1-particle matrix elements of an operator in QFT, e.g. [itex] \langle p' |\phi^4(x)|p\rangle [/itex] It seems you would calculate this perturbatively by first Fourier transforming the x-variable to q, assuming an incoming particle with momentum p, an outgoing particle with momentum p', and drawing all interactions vertices but making sure to include one φ4 vertex that also has momentum q entering it. However, if you do this, aren't |p> and |p'> the Heisenberg states that have momentum p and p' at t=-∞ and t=∞? Does this means that in [itex] \langle p |\phi^4(x)|p\rangle [/itex], <p| and |p> are not the same state? |p> is the Heisenberg state that looks like it has momentum p at t=-∞, while <p| is the Heisenberg state that looks like it has momentum p at t=∞. They have the same label, but they are different states (one is an In state, the other an Out state). So when one speaks of a matrix element in QFT, does one mean a matrix element whose ket is an In state, and whose bra is an Out state? This seems to be the only type of matrix element that is calculable with Feynman diagrams?