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Before using anticommutation relatives the energy, momentum, charge and number operators of the Dirac field have following expressions: $$ \hat {H} = \int \epsilon_{\mathbf p}\left( \hat {a}^{+}_{s}(\mathbf p )\hat {a}_{s}(\mathbf p ) - \hat {b}_{s}(\mathbf p )\hat {b}_{s}^{+}(\mathbf p ) \right)d^{3}\mathbf p, $$

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

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

$$ \hat {N} = \int \left( \hat {a}^{+}_{s}(\mathbf p )\hat{a}_{s}(\mathbf p ) + \hat{b}_{s}(\mathbf p )\hat{b}^{+}_{s}( \mathbf p )\right)d^{3}\mathbf p. $$ After using anticommutation relations and "neglecting" of infinite constant $\delta (0)$ (as we can do in a case of free field) in an expressions of energy and momentum operators, they transformed into $$ \hat {H} = \int \epsilon_{\mathbf p}\left( \hat {a}^{+}_{s}(\mathbf p )\hat {a}_{s}(\mathbf p ) + \hat {b}^{+}_{s}(\mathbf p )\hat {b}_{s}(\mathbf p ) \right)d^{3}\mathbf p, $$

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

But I have a couple of questions in connection with number and charge operators.

  1. What can I do with infinite constant into espressions of charge and number operators? For example, $$ \hat {N} = \int (\hat {a}^{+}_{s}(\mathbf p )\hat {a}_{s}(\mathbf p ) + \delta (0) - \hat {b}^{+}_{s}(\mathbf p )\hat {b}_{s}(\mathbf p ))d^{3} \mathbf p . $$ Can I neglect summand with infinite constant in a reason that it is connected with infinite "vacuum" energy of field?

  2. (Under the condition of neglecting infinite constant). When I acted on one-antiparticle state $|E_{\mathbf k }\rangle = \hat {b}^{+}_{s'}(\mathbf k )| \rangle$ by number operator, I got minus: $$ \hat {N}{b}^{+}_{s'}(\mathbf k )| \rangle = -\int \hat {b}^{+}_{s}(\mathbf p )\hat {b}_{s}(\mathbf p )\hat {b}^{+}_{s'}(\mathbf k)d^{3}\mathbf p | \rangle = -| E_{\mathbf k } \rangle . $$ But $\hat {b}^{+}_{s}(\mathbf p )$ is creation operator (i.e. it was the creation operator before using anticommutation relations; the relations changed this (?)). Did I make the mistake?

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  1. The right definition of $\hat N$ has the additive constant under which $\hat N |0\rangle = 0$. You should view your naive expression for $\hat N$ as an "inspiration" that may be used as a starting point to deduce a meaningful operator but a modification is needed. So just define $\hat N$ as the normal-ordered expression without the infinite additive $c$-number term.

  2. Yes, you made a mistake. When you commute $\hat N$ to the right side so that it annihilates $|0\rangle$, you have to permute $b^\dagger_{s'}(\vec k)$ through two fermionic operators that appear as factors of $\hat N$, namely $b$ and $b^\dagger$, so you pick two minus signs and they cancel.

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  • $\begingroup$ "...Yes, you made a mistake..." Where I did it? $$ \hat {N}\hat {b}^{+}_{s'}(\mathbf k)| \rangle = -\int \hat {b}^{+}_{s}(\mathbf p )\hat {b}_{s}(\mathbf p )\hat {b}^{+}_{s'}(\mathbf k )d^{3}\mathbf p |\rangle = -\int (\hat {b}^{+}_{s}(\mathbf p )(\delta (\mathbf p - \mathbf k)\delta_{ss'} - \hat {b}^{+}_{s'}(\mathbf k )\hat {b}_{s}(\mathbf k )))d^{3}\mathbf p = -\hat {b}^{+}_{s'}(\mathbf k )| \rangle + 0. $$ $\endgroup$ – user8817 Jul 10 '13 at 18:02
  • $\begingroup$ Dear PhysiXxx, oops, no, you didn't make a mistake in this calculation, just the naming you adopted is confusing. $\hat N$ isn't really a "number operator", it's the electric charge in the units of the charge of the $a^\dagger$-quantum. Because $b^\dagger$ creates antiparticles, it's very correct that it has $\hat N =-1$: the minus sign came from one in the definition of $\hat N$ in front of $b^\dagger b$. There is no natural $x$-representation formula for an operator that would count both the $a^\dagger$ and $b^\dagger$ quanta as $+1$ with the same sign; it would be an unnatural operator. $\endgroup$ – Luboš Motl Jul 11 '13 at 5:42

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