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In a case of free Dirac field we have

$$ \hat {H} = \int \epsilon_{\mathbf p}\left( \hat {a}^{+}_{s}(\mathbf p )\hat {a}_{s}(\mathbf p ) - \hat {b}_{s}(\mathbf p )\hat {b}_{s}^{+}(\mathbf p ) \right)d^{3}\mathbf p, $$

$$ \hat {\mathbf P} = \int \mathbf p \left( \hat {a}^{+}_{s}(\mathbf p )\hat {a}_{s}(\mathbf p ) - \hat {b}_{s}(\mathbf p )\hat {b}_{s}^{+}(\mathbf p ) \right)d^{3}\mathbf p, $$

$$ \hat {Q} = \int \left( \hat {a}^{+}_{s}(\mathbf p )\hat{a}_{s}(\mathbf p ) + \hat{b}_{s}(\mathbf p )\hat{b}^{+}_{s}( \mathbf p )\right)d^{3}\mathbf p. $$

So, if an operator $\hat {b}^{+}$ act on energy vector, it will decrease the value of an energy. This is commonly referred to as the fact that it creates an antiparticle with negative energy, because $\hat {b}^{+}$ is interpreted as creation operator (only by analogy with the scalar field, if I understand correctly). The solution to this problem is the postulation of anti-commutation relations between operators (in addition, the summary energy of the field begin to be positive definite quantity).

But why don't we call $\hat {b}$ the creation operator, and $\hat {b}^{+}$ the destruction operator? Then, requiring the positivity condition for the integrand (if it is possible), we'll get physically correctly result without postulation of anticommutation relations.

Where did I make the mistake?

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    $\begingroup$ First guess: check what happens to the number operator, if its still bounded below and/or if it still makes sense as the number operator. $\endgroup$ – David M Jul 8 '13 at 17:42
  • $\begingroup$ The number operators are $$ \hat {N}_{1} = \int \hat{a}^{+}( \mathbf p ) a (\mathbf p )d^{3}\mathbf p , \quad \hat {N}_{2} = \int \hat{b}^{+}( \mathbf p ) b (\mathbf p )d^{3}\mathbf p? $$ $\endgroup$ – user8817 Jul 8 '13 at 17:50
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    $\begingroup$ On second thought, theres a much simpler problem and that is if you switch the definition of the creation and annihilation operators your Hamiltonian will be unbounded below. It gets worse when you consider Majorana fermions where you have to identify your different annihilation operators. Then using normal commutators your Hamiltonian reduces to a constant. I would check Srednicki, chapter 37 & 39 in particular. $\endgroup$ – David M Jul 8 '13 at 18:00
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    $\begingroup$ Bounded below means the eigenvalues of the Hamiltonian have some lowest value. Any problem with unbounded energy in the negative direction is unphysical. And as you showed if you don't use anticommutation relations and try to switch around definitions, you're number operator has negative eigenvalues. If its supposed to count the number of particles that doesn't make sense! Maybe you could fix that problem by further redefinitions, but the problem with the Hamiltonian is more fundamental and can't be fixed. $\endgroup$ – David M Jul 8 '13 at 18:52
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    $\begingroup$ @PhysiXxx : David Meltzer gives the correct explanation: If you redefine your operators, you will have a hamiltonian of the form $H = \sum \epsilon_p (N_a( \mathbf p) - N_b(\mathbf p))$, and taking, for instance $N_a(\mathbf p)=0$ for all $\mathbf p$ and $N_b(\mathbf p) = 1$ for a increasing $\mathbf p$ interval, you can build states with more and more decreasing negative energy, which is unphysical. $\endgroup$ – Trimok Jul 8 '13 at 22:48
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We call $b^\dagger$ the creation operator exactly because it increases the energy of the state (by adding a quantum), in contradiction with your incorrect first sentence. On the contrary, $b$ is called the annihilation operator because it decreases the energy and because, when it acts on the vacuum $|0\rangle$, it actually produces zero.

You made a sign mistake while writing the first sentence. The first equation, which you wrote correctly, may be rewritten as $$\hat {H} = E_0+ \int \epsilon_{\mathbf p}\left( \hat {a}^{+}_{s}(\mathbf p )\hat {a}_{s}(\mathbf p ) + \hat {b}_{s}^+(\mathbf p )\hat {b}_{s}(\mathbf p ) \right)d^{3}\mathbf p$$ Note that the sign in front of $b^\dagger b$ was switched to minus exactly because they (almost) anticommute with each other. The anticommutator was added to the irrelevant $c$-number term $E_0$ that I wrote in front of the integral and that we ultimately set to zero.

Note that in my form, both $a^\dagger$ and $b^\dagger$ create positive-energy quanta (with opposite charge, positrons and electrons), and they generally enter symmetrically. This treatment that sees fermions and antifermions symmetrically may be obtained by renaming the negative-energy $a^\dagger,a$ as $b,b^\dagger$ – by using the word "positron" for a hole in the mostly filled Dirac sea of negative-energy electron states. This is possible because there's a symmetry between $a,a^\dagger$ – their anticommutator is symmetric in them, much like the occupation numbers $0,1$ for a fermionic state are symmetric with respect to the value $1/2$. So it's possible to rename empty vs occupied, exchange the notation for creation vs annihilation operators, and end in a picture in which the creation operators always increase the energy.

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