0
$\begingroup$

I am having trouble understanding the formal setting of the one particle states as constructed in e.g. Weinberg QFT vol 1. Is the relevant Hilbert space $\mathscr{H}=L^2(\mathbb{R}^3)$?

Weinberg defines the physical state vectors to be eigenstates of the four-momentum operator, \begin{align} \mathcal{P}^a|p,\sigma\rangle=p^a|p,\sigma\rangle. \end{align} So the states $|p,\sigma\rangle$ form an orthogonal basis of the Hilbert space with $\langle p',\sigma'|p,\sigma\rangle=\delta_{\sigma'\sigma}\delta(\vec{p}'-\vec{p})$. If we have $\mathscr{H}=L^2(\mathbb{R}^3)$ then is this inner product \begin{align} \langle p',\sigma'|p,\sigma\rangle = \int _{\Sigma_m}\frac{\text{d}^3p}{E_p}\big(\Psi_{p',\sigma'}(p)\big)^*\Psi_{p,\sigma}(p) \end{align} where the wavefunctions $\Psi_{p,\sigma}$ and $\Psi_{p'\sigma'}$ are squared-integrable orthogonal (w.r.t the above) functions representing the various eigenstates of $\mathcal{P}^a$ and $\Sigma_m$ is the mass-shell with $p^ap_a=-m^2$. Do we need to know these functions, how do we know they exist, are they just plane waves? Should the integration variable (and invariant measure) be a different dummy variable? Something like \begin{align} \langle p',\sigma'|p,\sigma\rangle = \int _{\Sigma_m}\frac{\text{d}^3p''}{E_{p''}}\big(\Psi_{p',\sigma'}(p'')\big)^*\Psi_{p,\sigma}(p''),~~~p''^ap''_a=-m^2. \end{align}

How does this formalism tie in with the one formulated by (I think) Bargmann and Wigner (Or maybe Pauli and Fierz, or maybe someone else entirely, as a matter of fact; where might I find this construction?), where a mass $m$ particle with spin $s$ is described by an element of the Hilbert space of squared-integrable $\text{SL}(2,\mathbb{C})$ totally symmetric (independently in their dotted and undotted indices) spin-tensors on the mass shell, \begin{align} \Psi_{\alpha_1\dots\alpha_s\dot{\alpha}_1\dots\dot{\alpha}_s}(x)=\int_{\Sigma_m}\frac{\text{d}^3p}{E_p}e^{ipx}\Psi_{\alpha_1\dots\alpha_s\dot{\alpha}_1\dots\dot{\alpha}_s}(p),~~~ \Psi_{\alpha_1\dots\alpha_s\dot{\alpha}_1\dots\dot{\alpha}_s}(x)=\Psi_{(\alpha_1\dots\alpha_s)(\dot{\alpha}_1\dots\dot{\alpha}_s)}(x) \end{align} and satisfying the following conditions \begin{align} \big(\square-m^2\big)\Psi_{\alpha_1\dots\alpha_s\dot{\alpha}_1\dots\dot{\alpha}_s}(x) =0\\ \partial^{\beta\dot{\beta}}\Psi_{\beta\alpha_1\dots\alpha_{s-1}\dot{\beta}\dot{\alpha}_1\dots\dot{\alpha}_{s-1}}(x) =0. \end{align} To my knowledge these supplementary conditions are known as the Bargmann-Wigner equations, which is why I am under the impression they constructed this model. What is the inner product for the massive and massless cases? I have seen that for the massive case it is \begin{align} \langle\Phi|\Psi\rangle=\int_{\Sigma_m}\frac{\text{d}^3p}{E_p}\frac{p^{\dot{\beta}_1\alpha_1}\cdots p^{\dot{\beta}_{2s}\alpha_{2s}}}{m^{2s}}(\Phi_{\dot{\beta}_1\dots\dot{\beta}_{2s}}(p))^*\Psi_{\alpha_1\dots\alpha_{2s}}(p) \end{align} but I have no idea where this comes from.

What is the fundamental difference between these two formulations of the one-particle states? I realise I have asked a lot of questions so I am more than happy with an answer in the form of references.

$\endgroup$
  • $\begingroup$ The Bargmann-Wigner equations are $(\gamma^\mu P_\mu - m)\Phi = 0$, not what you wrote there. Where does your version of these equations come from, in particular the second one? Also, you should distinguish the 3-vectors $\vec p,\vec x$ and the 4-vectors $p,x$ - not all the $p$s in your equations are the same! Lastly, you should indeed not use the same variable as an integration variable and as a free variable of the result of the integration. $\endgroup$ – ACuriousMind Sep 29 '17 at 12:06
  • $\begingroup$ I think the equation you have written is closely related to my first condition but just in momentum space. When I write $\partial^{\dot{\beta}\beta}$ this is equal to $(\sigma^a)^{\dot{\beta}\beta}\partial_a$, (greek letters are spinor indices, latin letters are Lorentz indices). The two conditions are from page 56 of the linked book. amazon.com/Ideas-Methods-Supersymmetry-Supergravity-Gravitation/… $\endgroup$ – NormalsNotFar Sep 29 '17 at 12:17
  • $\begingroup$ Yes you are right, I forgot to distinguish the three vectors and four vectors! I wonder, if you have the time, if I could hear your opinion (however short) on my other thread (linked below). I have seen you answer many questions about representation theory in physics and you seem very knowledgeable about the subject! physics.stackexchange.com/questions/359965/… $\endgroup$ – NormalsNotFar Sep 29 '17 at 12:22

Your Answer

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

Browse other questions tagged or ask your own question.