Since I started quantum field theory I had very big issues with signs, especially when I had to pass from a tridimensional euclidean space to a flat four-dimensionale spacetime, with Minkowski metric. I read a lot of questions on this site trying to find an answer, but I could just walk around the problem without being able to see what there is in the center. It's months I'm facing these problems and now that I'm coming to the main part of the matter, I just cannot ignore it; sorry if the question is a bit lenghty

The main problem is due to the generators of transformations. Apart from quantum-mechanical formalism, that is a big problem too for me, consider $n-1$-dimensional indexes $\boldsymbol{r}$ and $1$ dimensional parameter $t$; they can be put togheter in $n$-dimensional symbol $\boldsymbol{x}$ with $\mathcal{M}_n$ domain. Consider a $p$-dimensional field $\boldsymbol{\phi}$ with $\Phi$ domain, that takes this symbol, so that $\boldsymbol{\phi}\equiv\boldsymbol{\phi}(\boldsymbol{x})$: the general field transformation due to the representation of the element of a group is \begin{gather*} \tilde{\boldsymbol{x}} = \mathbb{M}_{{\mathcal{M}_n}} \boldsymbol{x} -\boldsymbol{a} \\ \mathcal{O} \boldsymbol{\phi}(\tilde{\boldsymbol{x}}) \doteq \tilde{\boldsymbol{\phi}}(\tilde{\boldsymbol{x}}) = \mathbb{M}_\Phi \boldsymbol{\phi} \left( \mathbb{M}^{-1}_{{\mathcal{M}_n}} (\tilde{\boldsymbol{x}}+\boldsymbol{a}) \right) \end{gather*} That allowed me to write that for a pure translation \begin{gather*} \mathcal{O} \boldsymbol{\phi}(\tilde{\boldsymbol{x}}) \doteq \tilde{\boldsymbol{\phi}}(\tilde{\boldsymbol{x}}) = \boldsymbol{\phi} (\tilde{\boldsymbol{x}}+\boldsymbol{a}) \\ \mathcal{O} \boldsymbol{\phi}(\boldsymbol{x}) \doteq \tilde{\boldsymbol{\phi}}(\boldsymbol{x}) = \boldsymbol{\phi} (\boldsymbol{x}+\boldsymbol{a}) \end{gather*} where in the infinitesimal case $\boldsymbol{\phi}(\boldsymbol{x}+\boldsymbol{a})=\boldsymbol{\phi}(\boldsymbol{x})+a^\alpha\partial_\alpha\boldsymbol{\phi}(\boldsymbol{x})$. Due to the fact that I can write $\mathcal{O}\equiv\exp{\vartheta^i\mathfrak{J}_i}$ I should deduce that the infinitesimal parameter associated to translation are $a^\alpha$ and the generator of the transformation acting on the field are $\partial_\alpha$

Here comes the point

In the quantum mechanical formalism the analogue of group representation on a field is the group representation on a state, right? So in the usual formalism, instead of having $\exp{\vartheta^i\mathfrak{J}_i}$, I'm now having $\exp{\frac{i}{\hbar}\vartheta^i\mathfrak{J}_i}$; the presence of $i$ accounts for the fact that every change of reference should not affect the intrinsic role of a state, meaning that all expectation values are preserved; but there is a problem with that: I can say $\exp{\frac{i}{\hbar}\vartheta^i\mathfrak{J}_i}$ or I can say $\exp{-\frac{i}{\hbar}\vartheta^i\mathfrak{J}_i}$ and I still have an unitary transformation, meaning that the sign of the generators should be defined in some other way.

In the tree dimensional euclidean space case I just wrote the canonical commutation relation \begin{equation*} \left[ \hat{r}^j, \hat{p}_k \right] = i\hbar{\delta^j}_k\mathcal{I} \end{equation*} and say: "Ok, now that I know this I clearly have that the translation is written like this $\mathcal{O}\equiv\mathcal{U}(\boldsymbol{a})=\exp{\frac{i}{\hbar}a^j\hat{p}_j}$ because in this way I have $\delta\hat{r}^k=\frac{i}{\hbar}a^j[\hat{r}^k,\hat{p}_j]=-a^k\mathcal{I}$ and it's perfectly coherent".

For the time translation I instead put $\mathcal{O}=\exp{-\frac{i}{\hbar}\hat{H}t}$ so that the relation $i\hbar \text{d}A(t)/\text{d}t=[A(t),\hat{H}]$ is still preserved.

But what should I do in fourdimensional case? On what kind of reasoning should I base on? How can I actually prove that $\hat{p}_\alpha=i\hbar\partial_\alpha$ and also that $\hat{p}_\alpha=(H,-\boldsymbol{p})$?

I know it is very long, I just want to make explicit all the passages that I did to get to the results, asking, if there are no big problems with what I wrote, where is the exact point that gets me to the desired result.

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    $\begingroup$ Not sure I got it: you essentially want to grasp the apparent ambiguity of sign in the exponential argument? $\endgroup$ – daydreamer Oct 16 '20 at 19:16
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    $\begingroup$ Exactly! How can I convince myself that $\hat{p}_\alpha=i\hbar\partial_\alpha$ (and also that $\hat{p}_\alpha=(+\hat{H},-\boldsymbol{p})$)? $\endgroup$ – Rob Tan Oct 17 '20 at 12:22

I'm setting the reduced Planck constant to 1.

As far as we can tell, the proper understanding and computation of Quantum Mechanics is inseparable from the notion of an inner product.

In a nutshell, the inner product is a rule that associates with a scalar (could be real or complex) two vectors u, u* from spaces V and its dual V*, respectively. I'll call V the direct space and V* the dual spaceThe dual space is a linear functional - but I'll discuss things modulo this issue.

To illustrate the arbitrariness, think about when you write the time evolution operator as $U(t)$ = exp(-itH); you can then define over which space it acts over. Usually, we take that convention to make the negative argument guy act in the direct or ket space. But then, in order to keep things consistent (and physically meaningful) in your Hilbert Space, you're demanded to make its adjoint $U^{\dagger}(t)=exp(itH)$ act over the dual space to represent the forward time evolution of these vectors; notice that it'll also represent backwards evolution in the direct space.

We could all as well have defined $U(t)$ = exp(itH) to be the time-evolution operator to make things go in the direct space V; and then its adjoint would be the good old negative argument guy. Quantum mechanics would be just the same.

The discussion is very much the same to reason over all other operators, such as the linear momentum you mentioned.

But I feel this is not the only issue you are facing. You want to be convinced that the proper definition of momentum, as space-translation generator in QM, matches that of $\hat{p}_\alpha=i\hbar\partial_\alpha$ and also that $\hat{p}_\alpha=(H,-\boldsymbol{p})$. For that, I refer you to this question. The only sin in that question is that it associates H to $i\partial_t$, which is redundant (see this great answer), but gives you the initial insight. The real reason is that (see Weinberg's Quantum Theory of Fields Vol. 1 footnotes in page 59) the Poincaré Algebra naturally leads to that, the sign being again a convention. Identifying the zero (temporal) component with H is the natural thing to do, since if you apply the exponential of -i${\lambda^\alpha}P_\alpha$ to a function of space-time coordinates, it give the proper displacements only if, as you put $P_\alpha=(H,-\boldsymbol{p})$

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    $\begingroup$ Most of what you answered I knew it already, but thanks a lot anyway! You say "if you apply the exponential of $-i\lambda^\alpha P_\alpha$ to a function of space-time coordinates", but why should I do that? why not the exponential of $+i\lambda^\alpha P_\alpha$. Maybe the point is that the canonical commutation relation in fuor dimensions $[\hat{x}^\alpha,\hat{p}_\beta]=i\hbar(?)$ should depend somehow on the metric in the $(?)$ symbol $\endgroup$ – Rob Tan Oct 18 '20 at 10:24
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    $\begingroup$ It's just as I said in my answer: you can apply both, but the meanings depend on your compromise between who's the direct space and who's dual $\endgroup$ – daydreamer Oct 18 '20 at 19:50
  • $\begingroup$ I understand what you say, but at the same time you speak about conventions, while instead I tried to use $\hat{p}_\alpha=-i\hbar\partial_\alpha,\hat{p}_\alpha=(-\hat{H},\boldsymbol{p})$ and it seemed not to work. So I'm asking where the metric is in all this and if there are canonical commutation relations to follow $\endgroup$ – Rob Tan Oct 18 '20 at 20:12

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