If $A$ and $B$ are fermion operators then the time ordering is defined as \begin{eqnarray} T(AB) = \left\{ \begin{array}{rl} AB, & \mbox{if $B$ precedes $A$}\\ -BA, & \mbox{if $A$ precedes $B$}\end{array}\right. \hspace{2in} (1) \end{eqnarray} On the other hand, the time-ordering operator that arises in the solution of $$i \partial_t U(t,t_0) = H_{_I}(t) U(t,t_0), \hspace{3.1in}$$ as $$U(t,t_0) = T \left[e^{-i \int_{t_0}^t H_{_I}(\tau) d\tau}\right] \hspace{3in} (2) $$ is the usual (bosonic) time-ordering operator even if $H_{_I}$ has fermion fields; we cannot use the definition of time-ordering shown in (1) in the derivation of (2) because introducing a negative sign when the order of fermion operators is flipped messes up the combinatorial counting and will not yield the exponential in (2). (In (2), $U(t,t_0)$ is the unitary evolution operator in the interaction picture and $H_{_I}$ the interaction Hamiltonian.)

So, isn't the definition of time ordering shown in (1) inconsistent with the definition used in (2)? What am I missing?


It's because always even number of Fermionic fields appear in the Hamiltonian. for example the Dirac Lagrangian for free electron: $\mathcal{L}=i\hbar\bar \psi(\gamma^\mu\partial_\mu-m)\psi$ has two $\psi$s. Invariance of the theory upon a global gauge transformation requires that each term in the Lagrangian have an even number of Fermionic fields. Since an even number of Fermionic Fields commute inside the time ordering operator, there isn't an extra minus sign.

  • $\begingroup$ Good observation about QED, which has global gauge invariance. So the perturbative calculation machinery in QFT rests on the following implicit assumption: every interaction Hamiltonian must have an even number of fermion fields, even if the theory is not globally gauge invariant? $\endgroup$ – Coriolis May 1 '16 at 15:28
  • $\begingroup$ Global gauge invariance comes from quantum mechanics and has nothing to do with field theory. You can shift the phase of all states in the Hilbert space and physically nothing would change. So it's odd if you write a "quantum" field theory which is not globally gauge invariant. $\endgroup$ – seyed May 2 '16 at 4:29
  • $\begingroup$ Global gauge invariance of the theory requires that the action remain invariant under a phase shift of the field. I don't think it is a phase shift of the states in the Hilbert space that is relevant. If you consider phi^4 scalar QFT, the action is not invariant under phi -> phi * e^{i \theta}. So, unless I am missing something, the phi^4 QFT does not have global gauge invariance. $\endgroup$ – Coriolis May 2 '16 at 16:32
  • $\begingroup$ Good point, I was thinking of the operation of particle fields on vacuum, for example $|p> \approx \psi^\dagger|0>$. then a phase shift of $|p>$ state will results in gauge transformation on $\psi$ field. Maybe the relation is only for particle fields, I'm not sure anymore. $\endgroup$ – seyed May 3 '16 at 15:01

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