I have a problem with sign conventions in QFT which I have trouble dealing with myself. I self-study and mainly use Weinberg and Peskin. I will present the reasoning following conventions adapted by Peskin and say where I think it runs into difficulties.

Metric signature which I use is $+---$. We adapt the convention that translations are implemented in Hilbert space according to formula $$ U(1,a) = e ^{-iP \cdot a}, $$ in particular time translations are implemented by $e^{-itH}$. Weinberg uses the opposite convention, but I am reluctant to accept it as it is inconsistent with Schrodinger equation and standard quantum mechanics. This decision implies that creation operators for Klein-Gordon field satisfy $$ U( \Lambda, a) a_p ^{\dagger} U(\Lambda , a ) ^{\dagger} = e^{-i \Lambda p \cdot a} a_{\Lambda p} ^{\dagger}. $$ Negative frequency part of the field is defined as $$ \phi^ - (x) = \int \mathrm d \mu (p) a_p ^{\dagger} e ^{ipx}, $$ while $\phi ^ +(x) = \phi ^- (x) ^{\dagger}$, Lorentz invariant measure is given by $\mathrm d \mu (p) = \frac{\mathrm d ^3 \vec p}{(2 \pi)^3 2E}$ and Lorentz invairant commutation relations are $[a_p,a_{p'}^{\dagger}]= (2 \pi) ^3 2E \delta ^3 (\vec p - \vec p')$. But using previous relation we get transformation rule for fields $$ U(\Lambda, a) \phi ^- (x) U(\Lambda, a)^{\dagger} = \phi ^- (\Lambda x -a).$$ This relation seems just wrong - I think we should get something like $\phi ^- (\Lambda x +a )$. It seems like $U(\Lambda, a )$ really implements the transformation $(\Lambda, -a)$ in contrary to what was anticipated. This can be remedied by redefining fields to have $e^{-ipx}$ next to creation operator and $e^{ipx}$ next to annihilation operator. But it seems noone does it. It has obvious drawback that when one does it we don't get canonical commutation relations, but instead their negative ($[\phi, \pi] = - i \delta$). It seems that Weinberg avoids these problems by declaring that translations are implemented by the operators $e^{iPa }$ (which is $e^{-iPa}$ with his choice of signature), but I think it's unacceptable. Is it at all possible to avoid these problem without destroyng either CCR or Shrodinger equation?

Remark: I am aware that Schrodinger equation is not a thing in QFT, I only mean that time translation operator is $e^{-itH}$.


1 Answer 1


I don't have enough reputation to comment but please note that Weinberg uses the (-,+,+,+) metric, which means you need a change of sign in $$e^{iP.x}$$.

The field transforms as, with $x'=x+a$, $$\phi'(x')=U^{-1}(a)\phi(x')U(a)=\phi(x)$$.

For (+,-,-,-) signature this is implemented by $U(a)=e^{iP.x}$, see Peskin & Schroeder, page 26 for example.

In the (-,+,+,+) signature you need to use $U(a)=e^{-iP.x}$.

Edit: Actually I think you understand this, but your issue lies elsewhere. Please note, to implement a Lorentz transformation $x\rightarrow \Lambda x$ and translation $x\rightarrow x+a$

The field transforms as $$\phi(x)\rightarrow \phi'(x')$$ which is defined to be $$\phi'(x')=U^{-1}(\Lambda,a)\phi(x')U(\Lambda,a)=\phi(x)$$ This transformation rule is irrelevant of signature.

You have $U$ and $U^{-1}$ reversed, so it is natural that you are getting the reverse transformation on your field.

Update: The real issue is actually my own confusion. Apologies, it's late. Weinberg implements time translations as, going form $t=0$ to $t$ as $$U=e^{iHt}$$ This is correct.

You claim this is inconsistent with schrodinger equation. Please note that there is a distinction between time evolution (schrodinger, regular QM) and time translations (what we're doing here). I think Weinberg discusses this distinction in Chapter 3, but I can't check right now.

Hence in (+,-,-,-) signature, $U(a)=e^{iPx}$ and in (+,-,-,-) its $U(a)=e^{-iPx}$. This is completely consistent with my transformation rule and Peskin.

Time evolution on states from $0$ to $t$, as with the Schrodinger eqn is implemented with $e^{-iHt}$. Time translation is implemented with the opposite sign.

  • $\begingroup$ Thank you for answering! Weinberg uses $e^{-ipx}$ with signature $-+++$ which leads precisely to the thing I would like to avoid. The equation on $p. 26$ of Peskin reads $U(x) ^{\dagger} \phi(0) U(x) = e^{ipx} \phi (0) e^{-ipx} = \phi (x)$. This is actually opposite to your second formula and constistent with my post. That is what all fuss is about. $\endgroup$
    – Blazej
    Commented Jul 21, 2016 at 21:57
  • $\begingroup$ @Blazej Yeah Peskin is really considering the reverse transformation, please see my updated answer. $\endgroup$
    – qftey
    Commented Jul 21, 2016 at 21:59
  • $\begingroup$ What is written in Peskin is in your notation $U^-1 \phi(x) U = \phi x')$. You have $U^{-1} \phi (x') U = \phi (x)$. $x$ and $x'$ are interchanged here. $\endgroup$
    – Blazej
    Commented Jul 21, 2016 at 22:02
  • $\begingroup$ @Blazej In Peskin the initial point is $x=0$ and their transformed point is $x'$ which they have called $x$, i.e. they translate the field from $0$ to $x$, but we translate from $x$ to $x'$. $\endgroup$
    – qftey
    Commented Jul 21, 2016 at 22:06
  • 1
    $\begingroup$ You are right. I was confused precisely due to distinction between time-dependent ket in ordinary QM and state vector describing whole history of the system in Heisenberg point of view used in QFT. I actually read this comment in Weinberg before but didn't appreciate its significance at the time. So both Peskin and Weinberg are correct, and we need to use $U(a)=e^{ia^0 H - i \vec a \cdot \vec P}$. $\endgroup$
    – Blazej
    Commented Jul 22, 2016 at 9:27

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