What are matrix elements in QFT?

[geoduck]

Suppose you want the 1-particle matrix elements of an operator in QFT, e.g.

$\langle p' |\phi^4(x)|p\rangle$

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 $\langle p |\phi^4(x)|p\rangle$,

<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?

 

[mfb]

Without interaction, those states are the same. In the absence of interaction, all initial and final states are the same, your non-interaction is the identity operator and nothing happens.
Your interaction then gives transitions between states, but those states are still the same.

 

[geoduck]

Without interaction, then in an expression like $\langle p |\phi^4(x)|p\rangle$, the bra and the ket are the same state: <p|p>=1.

However, with interaction, <p|p>≠1, since the bra is an Out-state, and the ket is an In-state.

It seems if you calculate $\langle p |\phi^4(x)|p\rangle$ with Feynman diagrams, then you are calculating the matrix element between an In-state and an Out-state.

So I was wondering if a matrix element in QFT is defined as being between a ket that is an In-state, and a bra that is an Out-state.

Or can a matrix element be between two In-states for instance? If so, how would you calculate that, as it seems you can't use Feynman diagrams, as Feynman diagrams are for kets that are In-states, and bras that are Out-states.

 

[mfb]

However, with interaction, <p|p>≠1, since the bra is an Out-state, and the ket is an In-state.

This is still 1 as you do not have an interaction here. The p states are defined as states without interaction, where "in" and "out" are meaningless.