# Can localized wavepackets have mass?

Page 31 of David Tong's notes on QFT (also in Srednicki's book while discussing LSZ reduction formula), talks about Gaussian wavepackets $$|\varphi\rangle=\int \frac{d^3\textbf{p}}{(2\pi)^{3}}e^{-i\textbf{p}\cdot\textbf{x}}\varphi(\textbf{p})|\textbf{p}\rangle$$ with $\varphi(\textbf{p})=\exp[-\textbf{p}^2/2m^2]$ such that the state is somewhat localized in position space and somewhat localized in momentum space. My question is whether such state satisfy relativistic dispersion relation (RDR) $E^2-\textbf{p}^2=m^2$, if the one-particle Fock states $|\textbf{p}\rangle$ satisfy, $E^2-\textbf{p}^2=m^2$. If not, can it faithfully represent a real physical particle?

EDIT: Is it possible to consider a different function than $\varphi(\textbf{p})=\exp[-\textbf{p}^2/2m^2]$ so that the state is at the same time somewhat localized and also has a mass $m$?

• I don't think $|\vec{p} \rangle$ are Fock states. Are they? Nov 17, 2016 at 20:16
• @QuantumBrick- $|\textbf{p}\rangle$ are one-particle Fock states.
– SRS
Nov 20, 2016 at 10:02

$$P^2 \, \int \text d ^3 \mathbf p f(\mathbf p ) \vert \mathbf p \rangle = \int \text d ^3 \mathbf p f(\mathbf p ) P^2 \vert \mathbf p \rangle = m^2\int \text d ^3 \mathbf p f(\mathbf p ) \vert \mathbf p \rangle$$
All these states are by definition on the mass shell (for each wavefunction $f$). Note that the localization in position is just a heuristic concept, if they have not introduced a relativistic position operator. It means that $$\intop \text d ^3 \mathbf p f_1(\mathbf p )^* f_2(\mathbf p)\approx 0$$ irrespective of the momentum distributions $\vert f_{1,2}(\mathbf p)\vert ^2$.
• @ pppqqq - You took the $P^2=P_\mu P^\mu=H^2-\textbf{P}^2$ operator inside the integral as if it is trivial. But Shouldn't you use the fact that $H=\int d^3\textbf{p} E_{\textbf{p}} a_{\textbf{p}}^\dagger a_{\textbf{p}}$ and $\textbf{P}=\int d^3\textbf{p} \textbf{p} a_{\textbf{p}}^\dagger a_{\textbf{p}}$, and then act it on the wavepacket under consideration?
• Hi SRS, here one should clarify what does the notation $\vert \mathbf p \rangle$ mean. The meaning is that we are representing the one-particle state $\vert 1\rangle$ as a wavefunction $\langle \mathbf p \vert 1 \rangle$ in a $L^2$ space where the $\mathbf P^\mu$ operators act as multiplications. Formally: $$P^0\vert \mathbf p \rangle = E_\mathbf p \vert \mathbf p \rangle ,\qquad \mathbf P \vert \mathbf p \rangle = \mathbf p \vert \mathbf p \rangle$$ Therefore, the equation $P^2\vert \mathbf p \rangle =(E_\mathbf p ^2 -\mathbf p ^2)\vert \mathbf p \rangle$ is,in some sense,trivial[...] Nov 22, 2016 at 10:47
• [...]The non-trivial thing is, ofcourse, that we can always express $P^\mu$ in such a way in an appropriate $L^2$ space, and this is the content of the spectral theorems of functional analysis. But, again, in perturbative QFT one usually assumes from the start that the states live in a Fock space which is essentially some $$\bigoplus _{n=0} ^\infty L^2(\mathbb R ^3,\text d ^3\mathbf p)^{\otimes n},$$ where $H$ and $\mathbf P$ are diagonal. This is implied by the notation $\vert \mathbf {p} \rangle$ for base kets; whenever one uses such a notation he is working in such an $L^2$ space. Nov 22, 2016 at 10:57