Time-ordering fermion operators

  • #1
2
0
If A and B are fermionic operators, and T the time-ordering operator, then the standard definition is

T(AB) = AB, if B precedes A
= - BA, if A precedes B.

Why is there a negative sign? If A and B are space-like separated then it makes sense to assume that A and B anticommute. But if they are time-like separated we don't know if
they anticommute. What am I missing?
 

Answers and Replies

  • #2
vanhees71
Science Advisor
Insights Author
Gold Member
17,073
8,173
For time-like separation of arguments it's just for convenience, and this is indeed consistent with exactly what you mention concerning anticommutator relations for space-like separated arguments (where you can always find a reference frame, where the time arguments are equal and thus the canonical equal-time anti-commutation relations for fermion fields are valied): For local fermionic operators that anticommute (particularly those at space-like separated arguments) the time-ordering shouldn't do anything. If you define it otherwise, you get into trouble with Lorentz invariance, because only for time-like separated events (space-time points in Minkowski space) the time ordering is frame-independent. So T must do nothing for operators at space-like separated arguments, and for fermionic operators it means you must have this additional minus sign in the definition of the time-ordering symbol, because fermionic operators anti-commute rather than commute. For the same reason there must not be any sign in the bosonic case!
 

Related Threads on Time-ordering fermion operators

Replies
13
Views
1K
  • Last Post
Replies
17
Views
2K
Replies
1
Views
3K
Replies
1
Views
928
Replies
14
Views
2K
  • Last Post
Replies
8
Views
1K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
8
Views
1K
Top