How do annihilation and creation operators act on fermions? 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?
 A: 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.
A: If you need to compute
$$
\int \frac{d^3 p}{(2\pi)^3} \sum_s ( {a^s_ {{\vec{p}}}}^\dagger a^s_ {{\vec{p}}} -  {b^s_ {{\vec{p}}}}^\dagger b^s_ {{\vec{p}}}  ) |\vec{k},r \rangle,
$$
you will need ${a^s_ {{\vec{p}}}}^\dagger a^s_ {{\vec{p}}}|\vec{k},r \rangle$ and ${b^s_ {{\vec{p}}}}^\dagger b^s_ {{\vec{p}}}  |\vec{k},r \rangle$.
Since you are dealing with Dirac fields, you get these using the anti-commutation relations (with the proper normalization factors - and I don't know which convention you are using):
$$
\{{a^s_ {{\vec{p}}}},{a^r_ {{\vec{q}}}}^\dagger\}=\delta_{sr}\delta(\vec{p}-\vec{q}),\\
\{{b^s_ {{\vec{p}}}},{b^r_ {{\vec{q}}}}^\dagger\}=\delta_{sr}\delta(\vec{p}-\vec{q}),\\
\{{a^s_ {{\vec{p}}}},{b^r_ {{\vec{q}}}}^\dagger\}=\{{b^s_ {{\vec{p}}}},{a^r_ {{\vec{q}}}}^\dagger\}=0.\\
$$
and knowing that ${a^s_ {{\vec{p}}}}|0\rangle={b^s_ {{\vec{p}}}}|0\rangle=0$.
It follows the answer with the same procedure @flippiefanus used.
A: All you need is the (anti-)commutation relations and the definitions of the states in terms of creation operators acting on vacuum state.
e.g. a state $|\psi\rangle$ of two particles:
$$
c_k|\psi\rangle =c_k\left(\sum_{i<j}\psi_{ij}|i,j\rangle\right)=
\sum_{i<j}\psi_{ij}c_k c_i^{\dagger}c_j^{\dagger}|0\rangle
$$
Then commutes $c_k$ with $c_i^{\dagger}$ and $c_j^{\dagger}$ until hit the vacuum state and annihilate it.
$$
\sum_{i<j}\psi_{ij}\left(\left[ c_k ,\, c_i^{\dagger}\right]_+ - c_i^{\dagger}c_k\right) c_j^{\dagger}|0\rangle=\sum_{i<j}\psi_{ij}\left(\left[ c_k ,\, c_i^{\dagger}\right]_+c_j^{\dagger} - c_i^{\dagger} \left[ c_k ,\, c_j^{\dagger}\right]_+   \right) |0\rangle =
\sum_{i<j}\psi_{ij}\left(\left[ c_k ,\, c_i^{\dagger}\right]_+|j\rangle - \left[ c_k ,\, c_j^{\dagger}\right]_+ |i\rangle   \right)
$$
A: Result: The only thing you'll really need for this calculation is the definition of one-(anti)-particle states (given below) and the application of annihilation operators on those, given by 
$$a_{\vec p_1}^{s_1} |\vec p_2, s_2;0,0\rangle =\delta_{s_1, s_2} \delta^3\left(\vec p_1 - \vec p_2\right) |0\rangle,\\b_{\vec q_1}^{r_1} |0,0;\vec q_2, r_2\rangle=\delta_{r_1, r_2} \delta^3\left(\vec q_1 - \vec q_2\right) |0\rangle.\\\\$$
Derivation: You were asking for the action of creation and annihilation operators on one-particle states, given by 
$$|\vec p, s; \vec 0, 0\rangle = a_{\vec{p}}^{s\dagger}|0\rangle\\
|0,0;\vec p, s\rangle = b_{\vec{p}}^{s\dagger}|0\rangle.$$
It makes sense to also define the following two-particle states, which are only non-zero if again all ${\vec p_i, s_i}$ and ${\vec q_j, s_j}$ are respectively distinct. 
$$|\vec p, s; \vec q, r\rangle = \frac{1}{2}\left(a_{\vec{p}}^{s\dagger}b_{\vec{q}}^{r\dagger}-b_{\vec{q}}^{r\dagger}a_{\vec{p}}^{s\dagger}\right)|0\rangle\\
|\vec p_1, s_1, \vec p_2, s_2;\vec 0,0\rangle = \frac{1}{2}\left(a_{\vec{p}_1}^{s_1\dagger}a_{\vec{p}_2}^{s_2\dagger}-a_{\vec{p}_2}^{s_2\dagger}a_{\vec{p}_1}^{s_1\dagger}\right)|0\rangle\\|\vec 0,0;\vec q_1, r_1, \vec q_2, r_2\rangle = \frac{1}{2}\left(b_{\vec{q}_1}^{r_1\dagger}b_{\vec{q}_2}^{r_2\dagger}-b_{\vec{q}_2}^{r_2\dagger}b_{\vec{q}_1}^{r_1\dagger}\right)|0\rangle$$
where we just decided to use a (anti)symmetrical definition - it is clear that using the appropriate anticommutation-relations, all of those states can be written without the difference of two terms.
Now, to find the action of those operators we are going to use the mentioned anticommutation relations
$$\{a_{\vec p}^s, a_{\vec q}^r\}=0 \qquad \{a_{\vec p}^{s\dagger}, a_{\vec q}^{r\dagger}\}=0\\ \{a_{\vec p}^s, a_{\vec q}^{r\dagger}\}=\delta^{rs} \delta^3(\vec p - \vec q)$$
and similar for the $b$-operators. Also, every $b$ anticommutes with every $a$. 
Note, that the above states are adequately normalized, provided the vacuum $|0\rangle$ is: 
$$\langle \vec p, s; \vec 0, 0|\vec q, r; 0, 0\rangle = \langle 0| a_{\vec{q}}^{r}a_{\vec{p}}^{s\dagger}|0\rangle\\ = \langle 0|\{a_{\vec{q}}^{r},a_{\vec{p}}^{s\dagger}\}|0\rangle\\=\delta^{rs} \delta^3(\vec p-\vec q)$$
From the fact that all b's and a's anticommute we can immediately derive
$$b_{\vec p}^s |\vec q, r;0,0\rangle = 0, \\a_{\vec p}^s |0,0;\vec q, r\rangle = 0.$$
Also, because the creation operators anticommute with themselves, we have
$$\left(a_{\vec p}^{s\dagger}\right)^2 = 0 =\left(b_{\vec p}^{s\dagger}\right)^2$$
so that
$$a_{\vec p}^{s\dagger} |\vec p, s; 0, 0\rangle = 0 = b_{\vec p}^{s\dagger} |0,0;\vec p, s\rangle.$$
Of course, if we act with creation operators with different momenta and/or spins on the one-particle states, we are going to create the above two-particle (and particle-antiparticle states). We can combine this with the last formula in the following way: 
$$a_{\vec p_1}^{s_1\dagger} |\vec p_2, s_2;0,0\rangle = (1-\delta_{s_1, s_2}\delta_{\vec p_1, \vec p_2})|\vec p_1, s_1, \vec p_2, s_2; 0,0\rangle\\
b_{\vec p_1}^{s_1\dagger} |0,0;\vec p_2, s_2\rangle = (1-\delta_{s_1, s_2}\delta_{\vec p_1, \vec p_2})|0,0;\vec p_1, s_1, \vec p_2, s_2\rangle\\
a_{\vec p}^{s\dagger} |0,0;\vec q, r\rangle = |\vec p, s; \vec q, r\rangle\\
b_{\vec q}^{r\dagger} |p, s;0,0\rangle = -|\vec p, s; \vec q, r\rangle
$$
Now, the really interesting$^{1}$ thing happens, if we annihilate a particle from the one-particle state (or an anti-particle from the one-anti-particle state).
$$a_{\vec p_1}^{s_1} |\vec p_2, s_2;0,0\rangle = a_{\vec p_1}^{s_1}a_{\vec p_2}^{s_2\dagger}|0\rangle \\=\{a_{\vec p_1}^{s_1}, a_{\vec p_2}^{s_2\dagger}\}|0\rangle \\=\delta_{s_1, s_2} \delta^3\left(\vec p_1 - \vec p_2\right) |0\rangle$$
and analogously 
$$b_{\vec q_1}^{r_1} |0,0;\vec q_2, r_2\rangle=\delta_{r_1, r_2} \delta^3\left(\vec q_1 - \vec q_2\right) |0\rangle$$
