Let $P_0$ be the generator for temporal translation and $P_1, P_2, P_3$ be for spatial translations. Let $p_μ$ be the momentum operator in the $x_μ=x^μ$ direction. I watched a lecture where the guy said that $P_μ=p_μ=i\eta^{μ \nu}\partial_\nu$ (this is all in planck units for simplicity and $\eta^{μ\nu}$ is the $(+,-,-,-)$ Minkowski metric) and for $μ=0, P=H$ where $H$ is the Hamiltonian operator (which makes sense to me because of the Schrodinger equation). But, with a little wikipedia research, I found that the spatial translation operator is $T=\exp(-ix^μp_μ)$ and the temporal translation operator is $T=\exp(iHt)$. So shouldn't the translation generators of the Poincare group then be $$P_μ=i\eta^{μ\nu}x_μp_{\nu}~?$$ What am I missing here?

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    $\begingroup$ what does "in the $x_\mu=x^\mu$ direction" mean? (indices are mismatched) $\endgroup$ – AccidentalFourierTransform Feb 20 '16 at 21:37
  • $\begingroup$ I didnt think whether or not the coordinates are covariant would matter in this context. Im saying that p_u is the momentum operator for the x_u direction and that the contravariant coordinates are equivalent to the covariant coordinates. Should i hav kept better track of my indices there? $\endgroup$ – Anon Ymous Feb 20 '16 at 22:12
  • $\begingroup$ IMHO, you should have written "in the $x_\mu$ direction" or "in the $x^\mu$ direction" (whichever you like the most), but with no $=$ sign. This is kind of irrelevant for the question, anyway, so don't overthink it. $\endgroup$ – AccidentalFourierTransform Feb 20 '16 at 22:15
  • $\begingroup$ @AnonYmous Accidental's answer tells you the how, but as a rule, keep in mind that generators generate different transformations for different values of the displacement parameter. If you include the parameter in the generator, how else would they possibly generate different transformations? $\endgroup$ – udrv Feb 21 '16 at 9:07
  • $\begingroup$ Comment to the post (v2): The indices don't match in the last eq. $\endgroup$ – Qmechanic Feb 21 '16 at 12:03

Let $|x\rangle$ be an element of the position basis. Define the translation operator $T(a)$ as $$ |x+a\rangle\equiv T(a)|x\rangle \tag{1} $$ for any vector $a^\mu$. As $T$ is unitary, it can be written as $$ T(a)=\mathbb I-ia_\mu P^\mu \tag{2} $$ for some hermitian$^1$ operators $P^\mu$. If we expand the l.h.s. $(1)$ to linear order in $a$ (just a Taylor expansion) we get $$ |x+a\rangle=|x\rangle+a^\mu\partial_\mu|x\rangle+\mathcal O(a^2) \tag{3} $$

On the other hand, if we expand the r.h.s. of $(1)$ to linear order in $a$, we get $$ T(a)|x\rangle=|x\rangle-ia_\mu P^\mu|x\rangle \tag{4} $$ which, by comparison with $(3)$ means that $$ P_\mu=i\partial_\mu \tag{5} $$

Therefore, the momentum operator is $P_\mu=i\partial_\mu$ (in the position basis), and the translation operator is $$ T(a)=\mathrm e^{-ia_\mu P^\mu}=\mathrm e^{a^\mu \partial_\mu} \tag{6} $$

So, what does $T(x)=\mathrm e^{-ix^\mu P_\mu}$ mean? Well, this is just the translation operator $(6)$, in the specific case $a^\mu=x^\mu$, i.e., we are displacing by the vector $x$. Note that this is just a specific case of $(6)$, which is sometimes used in QFT; for example, if $\hat\Phi(x)$ is a field (operator), then you can relate its value at any $x$ with the value at $x=0$ through $\hat\Phi(x)=T(x)\hat\Phi(0)T^\dagger(x)$. In some cases, this is useful, for example to prove that the vev $^2$ of any field is just a constant (i.e., position independent). Anyway, the fact that we can write $T(x)$ doesn't change the form of the generator of $T$, which means that the momentum operator is always given by $(5)$.

$^1$ self-adjoint, I know...

$^2$ vev = vacuum expectation value = $\langle\text{VAC}|\hat\Phi(x)|\text{VAC}\rangle$


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