# Interpretation of components of four-momentum in QFT

In relativistic QFT one usually uses four-momentum vector $P^\mu$ which combines energy of a system with its momentum. I'm confused about the physical interpretation of the individual components of $P^\mu$.

Let's assume that the signature of the metric is $(1,-1,-1,-1)$. The zeroth component $P^0=P_0$ has, clearly, the interpretation of the energy of the system. But what is the spatial momentum of the system in the direction of $x$ axis? Is it $P^1$ or $P_1$?

I think that most people consider $P^1$ as the $x$ component of the spatial momentum. However, than it seems that the space-time translations are implemented differently in QM and QFT. In QM the state $\psi$ of the system shifted in time by $t$ and in space by $(x,y,z)$ is given by

$$\exp\left(-\frac{i}{\hbar}~ t\,P^0 -\frac{i}{\hbar} ~(x P_x+y P_y + z P_z)\right) \psi,$$

where $P^0\equiv H$ is the hamiltonian of the system and $P_x$, $P_y$, $P_z$ are the physical components of the spatial momentum. On the other hand in QFT the state $\psi$ of the system shifted in space-time by four-vector $x=(t,x,y,z)$ is given by

$$\exp\left(-\frac{i}{\hbar} P\cdot x\right) \psi= \exp\left(-\frac{i}{\hbar}~ t\,P^0 +\frac{i}{\hbar} ~(x P^1+y P^2 + z P^3)\right) \psi,$$

which is equivalent to the previous formula if $P_1$ is the spatial momentum of the system in the direction of $x$ axis ($P_1 = P_x$ not $P^1=P_x$).

• What has this to do with QFT, or quantum mechanics, at all? You seem to be simply asking whether it's the spatial part of $p^\mu$ or of $p_\mu$ that corresponds to the usual non-relativistic spatial momentum, which is a pure question about special relativity. – ACuriousMind Aug 2 '16 at 10:56

In general in special relativity you have $P^{\mu}=(E,\vec p)^{\mu}$ and $P_{\mu}=(E,-\vec p)$ if $+---$ signature of metric is chosen or $P_{\mu}=(-E,\vec p)_{\mu}$ otherwise. I will stick to the first choice in this post. In nonrelativistic $QM$ you have operators $H=i \partial _t$ and $\vec p= -i \vec \nabla$. Now recall (or check) that derivative operator naturally transforms as a covector, i.e. as if it had a lower index. So we have $\partial_{\mu}=(\partial_t, \vec \nabla)_{\mu}$ and $\partial ^{\mu} = (\partial_t, - \vec \nabla)^{\mu}$. Therefore you can make an identification $P^{\mu}=i \partial ^{\mu}$. Now if you act on a wavefunction you get $$e^{-iP^{\mu}a_{\mu}} \psi (t ,\vec x) = e^{a^0 \partial_t + \vec a \cdot \vec \nabla} \psi (t, \vec x) = \psi (t+a^0, \vec x + \vec a).$$ Therefore I think this notation is quite consistent. Note also that quite generally (also in, say, electromagnetism and optics) spatial dependece of a plane-wave propagating in the $z$-direction is $e^{-i \omega t + i k z}$. There must be relative sign difference between these two terms in the exponent because only then surfaces of constant phase travel in the positive (rather than negative) $z$ direction.
Now some confusion may arise due to the fact that spatial coordinates play rather different roles in QM and QFT. In QM there is $\vec x$ operator and wave functions can be expanded in its eigenbasis with coefficients of expansion being wavefunctions depending on $\vec x$. On the other hand $t$ is just a parameter labeling ket states at different times. In QFT there is no $\vec x$ operator as well as no $t$ operator.
• Note that the function $\psi(\vec{x})$ shifted by $\vec{a}$ should be $\psi(\vec{x}-\vec{a})$, not $\psi(\vec{x}+\vec{a})$. This is why standard textbook definition of translation operator in QM is $U(\vec{a}) = \exp( - i/\hbar~ \vec{a}\cdot\vec{P})$. – user72829 Aug 2 '16 at 15:00
• @user72829 You are right. But then translation in time would be $e^{itH}$ rather than $e^{-itH}$. In QFT you don't have evolution in time at all, because state vector encodes whole history of the system. Instead you can make a translation in this history, which is implemented by unitary operator $e^{iP \cdot a}$. Confront Weinberg's QFT textbook, volume 1, chapter 3, section 1. – Blazej Aug 2 '16 at 16:54