(pre-request) We know that time reversal operator $T$ is an anti-unitary operator in Minkowsi Spacetime. i.e.

$$ T z=z^*T $$ where the complex number $z$ becomes its complex conjugate. See, for example, Peskin and Schroeder ``An Introduction To Quantum Field Theory'' p.67 Eq (3.133).

(Question) Is Euclidean time reversal operator $T_E$ an unitary operator, i.e.

$$ T_E z=z T_E \;\;\;(?) $$

or an anti-unitary operator in Euclidean Spacetime? Why is necessarily that or why is it necessarily not that? Or should $T_E$ be an unitary operator like Parity $P$ instead?

However, here see the attempt of a PRL paper, Euclidean continuation of the Dirac fermion where time reversal $T_E$ is given by, on page 3: $$ T_E z=z^*T_E \;\;\;(?) $$ $T_E$ is still an anti-unitary operator!

It seems to me if one consider Euclidean spacetime, the Euclidean signature is the same sign, say $(-,-,-,-,\dots)$, then Parity $P$ acts on the space is equivalent to the Euclidean time reversal operator $T_E$ acts on Euclidean time (which is now like one of the spatial dimensions). So shouldn't $T_E$ be an unitary operator as Parity $P$?

[Other Refs]

i. Please, you may read, an earlier Phys.SE question here, probably is poorly formulated, so cannot draw the efforts of people to answer the question. Here let me try an easier way and focus on one issue only.

ii. Osterwalder-Schrader (OS) approach, and ``A continuous Wick rotation for spinor fields and supersymmetry in Euclidean space'' by Peter van Nieuwenhuizen, Andrew Waldron.

  • 1
    $\begingroup$ Starting with $Tz = z^*T$, and assuming $T = iT_E$, this would give $(iT_E)z = z^*(iT_E)$, that is $i(T_Ez) = (z^*i)T_E$, that is $(T_Ez) = -i(z^*i)T_E$, that is $T_Ez =z^*T_E$ $\endgroup$ – Trimok Oct 22 '13 at 9:43
  • $\begingroup$ Thanks Trimok, this can be an mathematical answer for a logical reason; I am sure that is what have been done in wick rotating with fermion fields. So even if Euclidean spacetime seemly treat spatial and euclidean-time the same in the signature/metric sense, but when the fermion gets involved, it will still tell the difference between euclidean-time'' and the remained space''? Simply by this anti-unitary property? $\endgroup$ – wonderich Oct 23 '13 at 3:35
  • $\begingroup$ I suppose there exist more rigorous explanations for your question. But a point is that the real space-time is Minkowskian, so analytic continuation in Euclidean space-time does not mean that we forget all the properties and behaviours linked to the real Minkowskian space-time, and its Lorentz representations (the particles). Of course, all this could be very subtle... $\endgroup$ – Trimok Oct 23 '13 at 8:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.