# Constrained Hamiltonian systems: spin 1/2 particle

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 $$$$S=-m\int d\tau \sqrt{-\dot{x}^\mu \dot{x}_\mu}\tag{1}$$$$ and computing the momentum $$$$p_\mu=\frac{m\dot{x}_\mu}{\sqrt{-\dot{x}^2}}\tag{2}$$$$ I see that it satisfies the constraint $$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 $$$$S=\int d\tau \space \dot{x}^\mu\dot{x}_\mu+\frac{i}{2}\psi_\mu\dot{\psi}^\mu\tag{3}$$$$ which exhibits a supersymmetry on the worldline, the supersymmetric conserved charge being $$Q=\psi^\mu p_\mu$$. According to the lecturer I should find the constraints $$$$H=\frac{1}{2}p^2, \quad Q=\psi^\mu p_\mu\tag{4}$$$$ i.e. the dynamics on the phase space should take place on hypersurfaces $$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 $$$$p_\mu=\dot{x}_\mu\\ \Pi_\mu=\frac{i}{2}\dot{\psi}_\mu\tag{6}$$$$ and I don't know what to do with them. I see that, in principle, the first constraint is obtained by setting $$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.

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 $$L_1~=~p_{\mu}\dot{x}^{\mu} +\frac{i}{2}\psi_{\mu}\dot{\psi}^{\mu}-eH, \qquad H~:=~\frac{p^2}{2}. \tag{C}$$ 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 $$\delta x^{\mu}~=~i\varepsilon\psi^{\mu}, \qquad \delta \psi^{\mu}~=~-\varepsilon p^{\mu}, \qquad \delta p^{\mu}~=~ 0 , \qquad \delta e~=~ 0, \tag{D}$$ changes the Lagrangian with a total derivative $$\delta L_1~=~\ldots~=~i\dot{\varepsilon}Q+\frac{i}{2}\frac{d(\varepsilon Q)}{d\tau}, \qquad Q~:=~p_{\mu}\psi^{\mu}, \tag{E}$$ 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 $$L_2~=~L_1-i \chi Q~=~p_{\mu}\dot{x}^{\mu} +\frac{i}{2}\psi_{\mu}\dot{\psi}^{\mu}-eH - i \chi Q .\tag{F}$$

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 $$\delta e~=~2i\chi\varepsilon, \qquad \delta \chi~=~\dot{\varepsilon}.\tag{G}$$

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 $$\delta Dx^{\mu}~=~i\varepsilon(\dot{\psi}^{\mu}-\chi p^{\mu}).\tag{J}$$

References:

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.

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

• Thank you for your answer. So, basically, the constraint $Q$ arise by knowing the Lagrangian is invariant under SUSY transformation, there is no way around it. (Moreover F. Bastianelli is my professor, great professor, great lecture notes but I was missing the point here!) – Luthien Jan 5 at 15:59