# Creation operators and the bispinor field

First, there is my question about Anticommutation relations and bispinor field .

How to check, that $\hat {b}^{+}_{s}(\mathbf p )$ is a creation operator in a case of using commutation relations

$$[\hat {b}_{s}(\mathbf p ), \hat {b}^{+}_{s'}(\mathbf k )] = \delta_{ss'}\delta (\mathbf p - \mathbf k), \quad [\hat {b}_{s}(\mathbf p ), \hat {b}_{s'}(\mathbf k )] = [\hat {b}^{+}_{s}(\mathbf p ), \hat {b}^{+}_{s'}(\mathbf k )] = 0?$$

I can show that

$$\hat {H}\hat {a}^{+}_{s}(\mathbf p )|E \rangle = (E + \epsilon_{\mathbf p })|E\rangle , \quad \hat {H}\hat {a}_{s}(\mathbf p )|E \rangle = (E - \epsilon_{\mathbf p })|E\rangle ,$$

$$\hat {H}\hat {b}^{+}_{s}(\mathbf p )|E \rangle = (E - \epsilon_{\mathbf p })|E\rangle , \quad \hat {H}\hat {b}_{s}(\mathbf p )|E \rangle = (E + \epsilon_{\mathbf p })|E\rangle .$$ Then, I tried to use standart procedure (it is set out below, with a-field) with number operator, $$\hat {N}_{1} = \int \hat {a}_{s}^{+}(\mathbf p )\hat {a}_{s}(\mathbf p )d^{3}\mathbf p , \quad \hat {N}_{2} = \int \hat {b}_{s}(\mathbf p )\hat {b}_{s}^{+}(\mathbf p )d^{3}\mathbf p ,$$ but I can't check the role of operators $\hat {b}_{s}(\mathbf p ), \hat {b}^{+}_{s}(\mathbf p)$ by using operator $\hat {N}_{2}$. For explanation, for a-field and one-particle state $| E \rangle$ I can do $$\hat {N}_{1}| E_{\mathbf p }\rangle = \hat {N}_{1}a^{+}| \rangle = 1|E_{\mathbf p } \rangle ,$$ so $\hat {a}^{+}$ is creation operator. But I can't do this for $\hat {b}^{+}(\mathbf p )$, because I can't represent $| E_{\mathbf p }\rangle$ as $b^{+}| \rangle$ in a reason of postulate of energy as positive definite quantity (according to this postulate, $\hat {b}^{+}| \rangle = 0$, because $\hat {b}^{+}$ decreases the value of energy). How to check that $\hat {b}^{+}$ is creation operator?

Maybe, I can do not use the postulate of energy as positive definite quantity? I need it, maybe, only for the definition of zero-state. So, there is an idea to consider only b-field, then postulate the energy as negative definite quantity, and then consider the role of operators with negative energies. So, for $$\hat {N}_{2} = \int \hat {b}_{s}(\mathbf p )\hat {b}_{s}^{+}(\mathbf p )d^{3}\mathbf p$$ we have $$\hat {N}_{2} |E_{\mathbf p ' }\rangle = \hat {N}_{2}\hat {b}^{+}_{s'}(\mathbf k)| \rangle = \int \hat {b}_{s}(\mathbf p )\hat {b}_{s}^{+}(\mathbf p )d^{3}\mathbf p \hat {b}^{+}_{s'}(\mathbf k)| \rangle = \int \hat {b}_{s'}^{+}(\mathbf p )(\delta (\mathbf p - \mathbf p') + \hat {b}^{+}_{s}(\mathbf p )\hat {b}_{s}(\mathbf p ))d^{3}\mathbf p | \rangle = \hat {b}_{s'}(\mathbf p ')|\rangle = |E_{\mathbf p '}\rangle ,$$ as in a case of a-field.

Is this correctly?

-
If you compute the (anti)commutators, it makes no sense to study how the operators act on individual ket vectors. There are infinitely many N-particle states etc. Why don't you work just with the operators? It's really simple to calculate the (anti)commutators of operators expressed as functionals of other operators whose (anti)commutators are known. –  Luboš Motl Jul 14 '13 at 6:07