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 $m$?

  • $\begingroup$ I don't think $|\vec{p} \rangle$ are Fock states. Are they? $\endgroup$ – QuantumBrick Nov 17 '16 at 20:16
  • $\begingroup$ @QuantumBrick- $|\textbf{p}\rangle$ are one-particle Fock states. $\endgroup$ – SRS Nov 20 '16 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 $\vert f_{1,2}(\mathbf p)\vert ^2$.

  • $\begingroup$ @ 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? $\endgroup$ – SRS Nov 22 '16 at 9:21
  • $\begingroup$ 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[...] $\endgroup$ – pppqqq Nov 22 '16 at 10:47
  • $\begingroup$ [...]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. $\endgroup$ – pppqqq Nov 22 '16 at 10:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.