I am trying to apply the Constrained Hamiltonian Systems theory on relativistic particles. For what concerns the scalar particle there is no issue. Indeed, I have the action \begin{equation} S=-m\int d\tau \sqrt{-\dot{x}^\mu \dot{x}_\mu}\tag{1} \end{equation} and computing the momentum \begin{equation} p_\mu=\frac{m\dot{x}_\mu}{\sqrt{-\dot{x}^2}}\tag{2} \end{equation} I see that it satisfies the constraint $p_\mu p^\mu+m^2=0$. I then proceed to quantize the system with the Dirac method.

I am finding issues with the relativistic massless spin 1/2 particle. Indeed, it is described by the space-time coordinates $x^\mu$ and by the real grassmann variables $\psi^\mu$, according to my notes. The action should take the form \begin{equation} S=\int d\tau \space \dot{x}^\mu\dot{x}_\mu+\frac{i}{2}\psi_\mu\dot{\psi}^\mu\tag{3} \end{equation} which exhibits a supersymmetry on the worldline, the supersymmetric conserved charge being $Q=\psi^\mu p_\mu$. According to the lecturer I should find the constraints \begin{equation} H=\frac{1}{2}p^2, \quad Q=\psi^\mu p_\mu\tag{4} \end{equation} i.e. the dynamics on the phase space should take place on hypersurfaces $$H=0, Q=0.\tag{5}$$

My question: How can I derive these constraints? They should arise simply with the definition of momenta, but, having no constants to work with, I'm left with \begin{equation} p_\mu=\dot{x}_\mu\\ \Pi_\mu=\frac{i}{2}\dot{\psi}_\mu\tag{6} \end{equation} and I don't know what to do with them. I see that, in principle, the first constraint is obtained by setting $m=0$ in the constraint of the scalar particle for example, but what if I want to derive it without the previous knowledge? And what about $Q$?

Edit: By intuition, knowing that the model exhibits a ${\cal N}=1$ supersymmetry, I may understand that the dynamics must take place on a surface such that $H=const$ and $Q=const$ (then I could set the constant to zero without lack of generality?), being $Q$ and $H$ conserved charges. Is it the only way to find these constraints? Should I need this previous knowledge about supersymmetry to study the model? I think I should be able to find these constraints just by looking at the Lagrangian itself.

  • $\begingroup$ $\Pi_{\mu}$ is $\frac{i}{2}\psi_{\mu}$. The canonical momenta of fermions are themselves. $\endgroup$
    – Valac
    Oct 23, 2023 at 16:02

1 Answer 1

  1. We consider here the massless case $m=0$. Let us start from the Lagrangian$^1$ $$L_0~=~\frac{\dot{x}^2}{2e} +\frac{i}{2}\psi_{\mu}\dot{\psi}^{\mu} \tag{A}$$ with an einbein field $e$, cf. e.g. this Phys.SE post. If we introduce the momentum $$ p_{\mu}~=~\frac{\partial L_0}{\partial \dot{x}^{\mu}} ~=~\frac{\dot{x}_{\mu}}{e},\tag{B}$$ the corresponding Legendre transformation $\dot{x}^{\mu}\leftrightarrow p_{\mu}$ yields a first-order Lagrangian $$\begin{align} L_1~=~&p_{\mu}\dot{x}^{\mu} +\frac{i}{2}\psi_{\mu}\dot{\psi}^{\mu}-eH, \cr H~:=~&\frac{p^2}{2}.\end{align} \tag{C}$$ This explains OP's first constraint $H\approx 0$, which is indirectly due to world-line (WL) reparametrization invariance, cf. this Phys.SE post.

  2. It is unnecessary to introduce momentum for the fermions $\psi^{\mu}$ as the Lagrangian $L_1$ is already on first-order form, cf. the Faddeev-Jackiw method.

  3. The Lagrangian $L_1$ has a global super quasisymmetry. The infinitesimal transformation $$\begin{align} \delta x^{\mu}~=~&i\varepsilon\psi^{\mu}, \cr \delta \psi^{\mu}~=~&-\varepsilon p^{\mu}, \cr \delta p^{\mu}~=~& 0 , \cr \delta e~=~& 0, \end{align} \tag{D}$$ changes the Lagrangian with a total derivative $$\begin{align} \delta L_1~=~&\ldots~=~i\dot{\varepsilon}Q+\frac{i}{2}\frac{d(\varepsilon Q)}{d\tau}, \cr Q~:=~&p_{\mu}\psi^{\mu}, \end{align}\tag{E}$$ for $\tau$-independent Grassmann-odd infinitesimal parameter $\varepsilon$.

  4. OP's other constraint $Q\approx0$ arises by gauging the SUSY, i.e. $\delta L_1$ should be a total derivative for an arbitrary function $\varepsilon(\tau)$. On reason to do this is given in Ref. 2 below eq. (3.3):

    Because of the time component of the field $\psi^{\mu}$ there is a possibility that negative norm states may appear in the physical spectrum. In order to decouple them we require an additional invariance and, inspired by the Neveu-Schwarz-Ramond model, it seems natural to demand invariance under local supergauge transformations.

  5. Concretely, we impose $Q\approx0$ with the help of a Lagrange multiplier $\chi$. This leads to the Lagrangian $$\begin{align} L_2~=~&L_1-i \chi Q\cr ~=~&p_{\mu}\dot{x}^{\mu} +\frac{i}{2}\psi_{\mu}\dot{\psi}^{\mu}-eH - i \chi Q .\end{align}\tag{F}$$

  6. Let us mention for completeness that in order to have gauged super quasisymmetry of the new Lagrangian $L_2$, the previous transformation $\delta e= 0$ needs to be modified into $$\begin{align} \delta e~=~&2i\chi\varepsilon, \cr \delta \chi~=~&\dot{\varepsilon}.\end{align}\tag{G}$$

  7. An alternative perspective is the replacement $$L_2~=~ L_1|_{\dot{x}\to Dx}\tag{H}$$ of the ordinary derivative $$\dot{x}^{\mu}\quad\longrightarrow\quad Dx^{\mu}~:=~\dot{x}^{\mu} -i\chi \psi^{\mu}\tag{I}$$ with a gauge-covariant derivative $Dx^{\mu}$. Here $\chi$ is a compensating gauge field. The gauge-covariant derivative transforms as $$ \delta Dx^{\mu}~=~i\varepsilon(\dot{\psi}^{\mu}-\chi p^{\mu}).\tag{J}$$


  1. F. Bastianelli, Constrained hamiltonian systems and relativistic particles, 2017 lecture notes; Section 2.2.

  2. L. Brink, P. Di Vecchia & P. Howe, Nucl. Phys. B118 (1977) 76; Below eq. (3.3).

  3. C.M. Hull & J.-L. Vazquez-Bello, arXiv:hep-th/9308022; Chapter 2, p. 7-8.

  4. O. Corradini & C. Schubert, Spinning Particles in QM & QFT, arXiv:1512.08694; Subsection 1.5.2.


$^1$ Conventions: We use the Minkowski sign convention $(-,+,+,+)$ and we work in units where $c=1$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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