Two creation operators acting on a state If $a_p^\dagger$ is the creation operator for an electron with momentum $p$ and $b_q^\dagger$ is the creation operator for a positron with momentum $q$, what does $a_p^\dagger b_q^\dagger \left| 0 \right>$ evaluate to?
Thanks in advance for any help.
 A: The electron is a fermion, so it follows Fermi-Dirac statistics. In QFT, this means that the creation and annihilation do not follow commutation relations, instead they follow anticommutation relations:
$$\{a_p, a^\dagger_q\} = \{b_p, b^\dagger_q\} = (2\pi)^3 \delta^3(\vec{p}-\vec{q})$$
$$\{a_p, a_q\} = \{b_p, b_q\} = \{a_p, b_q\} = \{a_p, b^\dagger_q\} = 0$$
where the anticommutator is defined as $\{A, B\} = AB + BA$.
So, the only manipulation that you can make to your state is $$a_p^\dagger b_q^\dagger|0\rangle = - b_q^\dagger a_p^\dagger|0\rangle$$
But I think that that doesn't answer your question, does it?
The answer is that your state is simply $a_p^\dagger b_q^\dagger|0\rangle$, that's all. It doesn't 'evaluate' to anything. It represents a state with two particles (an electron and a positron) with momentum $\vec{p}+\vec{q}$, energy $E_p + E_q$ and electric charge $-1+1=0$.
By the way, the expressions for the creation and annihilation operators in your comment are wrong, quantum fields are not a 'simple harmonic oscillator', and in fact there doesn't exist an operator representing $x$ (position is now just a label). You can express creation and annihilation operators in terms of the field and its conjugate momentum, but I don't think that this is relevant for your question.
