# Normal order and overlap of states

## Summary:

I don't understand why in a particular situation the expectation value of normal ordered fields and partially contracted fields is either irrelevant or zero.

## Main Question or Discussion Point

I have trouble understanding the solution to a homework problem.

Consider the interaction Lagragian ##\mathcal{L}_{\rm int} = -iqA_\mu \bar{\psi}\gamma^\mu \psi##, i.e. photon-electron/positron interaction. We want to focus on the Compton scattering
$$e^-(\vec p_1, \alpha) + \gamma(\vec p_2, \sigma)\quad \longrightarrow\quad e^-(\vec q_1, \beta) + \gamma(\vec q_2, \rho),$$
where the ##\vec p_i, \vec q_i## describe the momentum of the particles and the greek letters the spin/polarization. The original question of the homework was to find the matrix elements in ##\mathcal{O}(q^2)##, without the use of Feynman rules and diagrams.

To do this we set up final and initial state
$$\vert i\rangle := a_\alpha^\dagger (\vec p_1)c_\sigma^\dagger (\vec p_2)\vert 0\rangle \qquad\langle f\vert := \langle 0\vert c_\rho (\vec q_2) a_\beta (\vec q_1),$$
where ##a,c## are annihilation operators of electrons and photons respectively,
and expand the time evolution operator ##U_{\rm int} = T[ e^{-iS_{\rm int}}]## up to second order. We can then calculate the matrix element:
\begin{align*} F^{(2)} &= -\frac{i}{2}q^2 \int d^4 x d^4 y \langle f\vert T[A_\mu(x) \bar{\psi}(x)\gamma^\mu \psi(x)A_\nu(y) \bar{\psi}(y)\gamma^\mu \psi(y)] \vert i \rangle\\ &= -\frac{i}{2}q^2 \int d^4 x d^4 y \langle 0\vert c_\rho (\vec q_2) a_\alpha (\vec q_1) T[A_\mu(x) \bar{\psi}(x)\gamma^\mu \psi(x)A_\nu(y) \bar{\psi}(y)\gamma^\mu \psi(y)] a_\alpha^\dagger (\vec p_1)c_\sigma^\dagger (\vec p_2)\vert 0\rangle \end{align*}
To evaluate this we need to replace the time ordering operator ##T## with the normal ordering operator ##N## plus all partial and full normal ordered contractions.

Here is where the problem starts:
1. The solutions to the exercise now claim that we don't need to consider the partially contracted terms, but only the fully contracted ones. I don't see why this is supposed to be true. If we were looking at something of the form ##\langle 0\vert T[\dots]\vert 0\rangle##, then I would agree, since any field inside the time ordering operator that gets not contracted will annihilate the vacuum. But since we have ##\langle f \vert T[\dots]\vert i\rangle##, we can't be sure that the vacuum will directly get annihiliated. Can someone explain the reasoning here?
2. The second claim was that we can ignore the uncontracted normal ordered fields, i.e. If ##T[\dots] = N[\dots] + \text{contractions}##, then ##\langle f\vert N[\dots] \vert i\rangle## is irrelevant. I don't see why. Chances are that if will not be zero, so does this generate only loop diagrams or tadpoles? How can I see this?
I hope I have given enough context to make my question clear. If not, please let me know and I will try to explain it more clearly.

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1. Use the LSZ theorem, which allows you to extract the scattering matrix element from ##\langle \Omega | T A^\mu A^\nu \psi \bar\psi|\Omega \rangle## which can be done directly with the usual Wick's theorem.

2. Define a new contraction for this non-vacuum expectation value, between factors in the time ordered product and the creation annihilation operators. Peskin & Schroeder does this before introducing the LSZ theorem.

@HomogenousCow

Maybe I'm misunderstanding you, but the exercise was supposed to be solved in the way I presented above, so I cannot just change that (I technically could, but I would like to understand whats going on in the provided solution).

It's possible that I made my question more complicated than necessary, I bscly. would like to know why certain terms of
$$\langle 0\vert c_\rho (\vec q_2) a_\alpha (\vec q_1) T[A_\mu(x) \bar{\psi}(x)\gamma^\mu \psi(x)A_\nu(y) \bar{\psi}(y)\gamma^\mu \psi(y)] a_\alpha^\dagger (\vec p_1)c_\sigma^\dagger (\vec p_2)\vert 0\rangle$$
are irrelevant or drop out (thats what I described in my original post). Do any of your two suggestions explain that?

In order for the expression to be non-zero, we need all fields and creation/annihilation operators contracted in pairs. In order to do this we need the two photon fields, one fermion field, and one conjugate fermion field to contract with the creation annihilation operators outside of the time ordering sign. This means that when we expand the time ordered product using Wick's theorem, we want only the terms where one fermion field is contracted with another conjugate fermion field (at a different spacetime position), with the others left uncontracted. All other terms in the wick expansion give zero since not all operators can then be contracted.

Here's an example, one of the terms from the Wick expansion that we want is
$$N[A_\mu (x) A_\nu (y) \bar \psi(x) \gamma ^\mu S(x-y) \gamma^\nu \psi (y) ]$$ where a pair of fermion fields have been contracted yielding ##S(x-y)## , the fermionic propagator. Roughly if we take ##A \sim a+a^\dagger## and ##\psi \sim b+c^\dagger##, then the operator part of the above term reads$$N[(a+a^\dagger)(a+a^\dagger)(b+c^\dagger)(c+b^\dagger)]$$, expanding this out we see that the term $$a^\dagger a~ b^\dagger b$$ can be commuted with the other creation annihilation operators and thus giving a nonzero result.
If we instead look at a different term in the Wick expansion, such as $$N[D^{\mu\nu}(x-y) \bar \psi(x) \gamma ^\mu S(x-y) \gamma^\nu \psi (y) ]$$ where ##D^{\mu\nu}(x-y)## is the photon propagator, we will find that there is no ##a^\dagger a~ b^\dagger b## term to give a nonzero result.

Last edited:
• Markus Kahn
In order for the expression to be non-zero, we need all fields and creation/annihilation operators contracted in pairs. In order to do this we need the two photon fields, one fermion field, and one conjugate fermion field to contract with the creation annihilation operators outside of the time ordering sign.
The second sentence is exactly what confuses me! When you say "we need [...] to contract with the creation annihilation operators outside of the time ordering sign", what exactly do you mean with the "contract"? Up to now I thought that contractions can only arise in the context of Wick's theorem, i.e. if something is not inside the time ordering operator, it cannot be contracted with anything.

In your first example from above
$$N[A_\mu (x) A_\nu (y) \bar \psi(x) \gamma ^\mu S(x-y) \gamma^\nu \psi (y) ] = S(x-y) N[A_\mu (x) A_\nu (y) \bar \psi(x) \gamma ^\mu \gamma^\nu \psi (y) ] \propto a^\dagger a~ b^\dagger b$$
which then implies
\begin{align*} \langle f\vert N[A_\mu (x) A_\nu (y) \bar \psi(x) &\gamma ^\mu S(x-y) \gamma^\nu \psi (y) ] \vert i \rangle \propto \langle f \vert a^\dagger a ~b^\dagger b\vert i \rangle \\ &= \langle 0\vert b ~a (a^\dagger a~ b^\dagger b)a^\dagger b^\dagger \vert 0\rangle \end{align*}
So when you now speak of contractions with the fields outside of the time ordering operator, do you mean that I should use commutation relations to move the ##a,b## to the right and the ##a^\dagger, b^\dagger## to the left?

I'm sorry if this seems trivial, but the language confuses me here quite a bit.

All other terms in the wick expansion give zero since not all operators can then be contracted.
This relates to the previous question, but do you mean here that just one field survives that than can annihilate the vacuum?

Yes I admit my terminology is confusing, when I talk about field operators contracting with creation/annihilation operators outside of the time ordering symbol I am referring to exactly the manipulations done in my example, to expand the normal product use commutation relations to evaluate the transition amplitude. The reason I use this word is because with some experience you can skip the intermediate step with wick’s theorem and directly write down the amplitude by pairing field operators and creation/annihilation operators much in the same way that you can pair field operators in a wick expansion without working out the intermediate algebra.

Yes, all other terms in the wick expansion will leave uncontracted operators that will annihilate the vacuum.

Note that this primitive way of calculating scattering amplitudes leads to problems when you consider diagrams with loops on external legs, to solve this issue you must turn to the LSZ theorem.

• Markus Kahn
Thank you very much @HomogenousCow

I think this clears up my confusion about the topic.