I'm taking an introductory course in QFT. During quantization of the Dirac field, my textbook gives a lot of information on how annihilation and creation operators act on vacuum, but nothing about how they act on non-vacuum states. I need these to compute
$$
\int \frac{\mathrm d^3 p}{(2\pi)^3} \sum_s \left( {a^s_ {{\vec{p}}}}^\dagger a^s_ {{\vec{p}}} – {b^s_ {{\vec{p}}}}^\dagger b^s_ {{\vec{p}}} \right) |\vec{k},s \rangle,
$$
where ${a^s_ {{\vec{p}}}}^\dagger, {b^s_ {{\vec{p}}}}^\dagger $ are the creation operator for fermions and anti-fermions respectively and $ {a^s_ {{\vec{p}}}},{b^s_ {{\vec{p}}}}$ are the annihilation operators of fermions and anti-fermions repectively. I have searched google, but I couldn't find anything after about 1 hour of searching.
Are you able to tell me how ${a^s_ {{\vec{p}}}}^\dagger, {b^s_ {{\vec{p}}}}^\dagger, {a^s_ {{\vec{p}}}},{b^s_ {{\vec{p}}}}$ act on non-vacuum states?
Best Answer
The basic procedure is as follows: $$ a_r(\mathbf{k}_1) |\mathbf{k}_2,s\rangle = a_r(\mathbf{k}_1) a_s^{\dagger}(\mathbf{k}_2) |0\rangle = \{a_r(\mathbf{k}_1), a_s^{\dagger}(\mathbf{k}_2) \}|0\rangle = |0\rangle (2\pi)^2\omega_1 \delta(\mathbf{k}_1-\mathbf{k}_2) \delta_{rs} , $$ where $|\mathbf{k}_2,s\rangle$ is assumed to be a fermion state. For an anti-fermion state one would use the $b$-operators, instead. The reason why one can express this in terms of the anti-commutator is because $ a_r(\mathbf{k}_1) |0\rangle = 0$. The detail of the final expression depends on the particular anti-commutation relation that you use. Here I've used a Lorentz convariant version.