I Wick's theorem and Nucleon scattering

[Ken Gallock]

Hi.
My question is about nucleon-nucleon scattering.
In David Tong's lecture note, he discusses Wick's theorem and nucleon scattering (page 58-60).
My problem is that I don't know how to calculate the second line of eq(3.48):
\begin{equation}
<p'_1, p'_2|:\psi^\dagger (x_1) \psi (x_1) \psi^\dagger (x_2) \psi (x_2): |p_1, p_2>
=<p'_1, p'_2|\psi^\dagger (x_1) \psi^\dagger (x_2) |0><0| \psi (x_1) \psi (x_2) |p_1, p_2>.
\end{equation}

Especially,
1. what happened to normal ordering?
2. where $|0><0|$ came from?

My guess is normal ordering was used for $\psi (x_1) \psi^\dagger (x_2) \rightarrow \psi^\dagger (x_2) \psi (x_1)$. But isn't normal ordering used for ordering $\hat{b}, \hat{c}$ rather than $\psi$?

 

[Avodyne]

1. The right-hand side is written in a normal-ordered form, so we can omit the normal-ordering symbol.

2. The calculation is to be don in free-field theory. $\psi$ is an annihilation operator. The two $psi$'s either annihilate the two particles in the initial state, leaving the ground state (up to a factor which needs to be computed), or they annihilate the state completely. Thus, if we insert a complete set of states, only the ground state will contribute.

 

[Ken Gallock]

Sorry for my late reply. Somehow I didn't get the notification. :(
Your reply helped me a lot. I did calculation as follows:

\begin{align}
&\langle f|: \psi^\dagger(x_1)\psi(x_1)\psi^\dagger(x_2)\psi(x_2) :|i\rangle \notag \\
&=\langle f| \psi^\dagger(x_1)\psi^\dagger(x_2)\psi(x_1)\psi(x_2) |i\rangle \notag \\
&=\sum_{n=0}^\infty \langle f| \psi^\dagger(x_1)\psi^\dagger(x_2) |n\rangle \langle n | \psi(x_1)\psi(x_2) |i\rangle \notag \\
&\sim \sum_{n=0}^\infty \langle 0| bb(b^\dagger +c)(b^\dagger+c ) |n\rangle \langle n | (b+c^\dagger)
(b+c^\dagger)b^\dagger b^\dagger |0 \rangle \notag \\
&=\sum_{n=0}^\infty \langle 0| bbb^\dagger b^\dagger |n\rangle \langle n | bbb^\dagger b^\dagger |0 \rangle \notag \\
&=\langle 0| bbb^\dagger b^\dagger |0 \rangle \langle 0 | bbb^\dagger b^\dagger |0 \rangle + \sum_{n=1}^\infty \langle 0| bbb^\dagger b^\dagger |n\rangle \langle n | bbb^\dagger b^\dagger |0 \rangle \notag \\
&=\langle 0| bbb^\dagger b^\dagger |0 \rangle \langle 0 | bbb^\dagger b^\dagger |0 \rangle \notag \\
&=\langle f| \psi^\dagger(x_1)\psi^\dagger(x_2) |0 \rangle \langle 0 | \psi(x_1)\psi(x_2) |i\rangle . \notag
\end{align}

 

[D O]

Hi, thanks for putting how you did the calculation. I am also going through Tong's (generally excellent) course and came across the same issue. For future reference, I think the reason we may treat psi in this case as an annihilation operator and psi^+ as a creation operator is because only the 'b' annihilation component of the psi, used in both of them, can 'get rid' of the 2 b^+ in the final state, and only the 'b^+' creation component of the psi+ can build them back up again at a different momentum (remember that the process we are looking at is scattering, implying change of momentum, not the zeroth order non-interacting term). Thus, psi ~ b and psi^+ ~ b^+, and we may simply normal-order the psi, psi^+ for convenience. I agree that it is not made very clear in the notes.