# Canonical Realization of Poincare Symmetry of Dirac Spinor

I have a maybe stupid question about Noether charges and the Poisson bracket.

If a classical field theory has a Poincare symmetry, then by using the Noether's theorem, one can write down its associated conserved charges $$P^{\mu}$$ which generates spacetime translations, and $$M_{\mu\nu}$$ which generates Lorentzian transformations.

In the Hamiltonian formalism, one can define the Poisson bracket on its phase space. Then, the Poisson brackets among these charges should form a canonical realization of the Lie algebra of the Poincare symmetry.

My question comes from a Dirac spinor $$\mathcal{L}=\bar{\psi}(i\partial\!\!\!/-m)\psi.$$

Its conjugate momentum is given by $$\Pi=\mathcal{L}\frac{\overset{\leftarrow}{\delta}}{\delta\dot{\psi}}=i\bar{\psi}\gamma^{0}=i\psi^{\dagger}.$$

Usually, the fermionic Poisson bracket of Grassmann-odd fields are defines as $$\left\{F,G\right\}_{PB}=\int d^{3}x\left\{F\left(\frac{\overset{\leftarrow}{\delta}}{\delta\psi(t,\mathbf{x})}\frac{\overset{\rightarrow}{\delta}}{\delta\Pi(t,\mathbf{x})}+\frac{\overset{\leftarrow}{\delta}}{\delta\Pi(t,\mathbf{x})}\frac{\overset{\rightarrow}{\delta}}{\delta\psi(t,\mathbf{x})}\right)G\right\},$$

which indicates the fundamental canonical relation $$\left\{\psi(t,\mathbf{x}),\Pi(t,\mathbf{y})\right\}_{PB}=\delta(\mathbf{x}-\mathbf{y}).$$

The Lorentzian transformation acts on the Dirac spinor as $$\psi^{\prime}(x)=U(\Lambda)\psi(x)=S(\Lambda)\psi(\Lambda^{-1}x), \tag{1}$$

where the operator $$U(\Lambda)$$ is given by $$U(\Lambda)=\exp\left\{-\frac{i}{2}\omega_{\mu\nu}J^{\mu\nu}\right\},\quad\mathrm{with}\quad J_{\mu\nu}=\frac{1}{2}\sigma_{\mu\nu}+i(x_{\mu}\frac{\partial}{\partial x^{\nu}}-x_{\nu}\frac{\partial}{\partial x^{\mu}})𝟙_{4\times 4}.$$

Then, according to Noether's theorem, one has the conserved charge \begin{align} M_{\mu\nu}&=\int d^{3}\mathbf{x}\psi^{\dagger}(t,\mathbf{x})\left[\frac{1}{2}\sigma_{\mu\nu}+i(x_{\mu}\frac{\partial}{\partial x^{\nu}}-x_{\nu}\frac{\partial}{\partial x^{\mu}})𝟙_{4\times 4}\right]\psi(t,\mathbf{x}) \\ &=-i\int d^{3}\mathbf{x}\Pi(t,\mathbf{x})\left[\frac{1}{2}\sigma_{\mu\nu}+i(x_{\mu}\frac{\partial}{\partial x^{\nu}}-x_{\nu}\frac{\partial}{\partial x^{\mu}})𝟙_{4\times 4}\right]\psi(t,\mathbf{x}) \\ &=-i\int d^{3}\mathbf{x}\Pi(t,\mathbf{x})J_{\mu\nu}\psi(t,\mathbf{x}). \end{align}

One can check that this charge indeed generates an infinitesimal Lorentzian transformation. i.e. $$\delta_{\omega}\psi(t,\mathbf{x})=\left\{\psi(t,\mathbf{x}),\frac{1}{2}\omega^{\mu\nu}M_{\mu\nu}(t)\right\}_{PB}=-\frac{i}{2}\omega^{\mu\nu}J_{\mu\nu}\psi(t,\mathbf{x}). \tag{2}$$

Equation (2) is just an infinitesimal version of equation (1). It can be easily verified by using the fermionic Poisson bracket.

On the other hand, if the charge $$M_{\mu\nu}$$ generates Lorentzian symmetry, I expect that the Poisson bracket among themselves should be realized as the Lie algebra of the Lorentzian symmetry. This is the case for a scalar field. Also, in classical mechanics, one can consider a free Newtonian particle that is Galilean invariant, then the Poisson brackets among the associated Noether charges indeed form the Lie algebra of the central extension of the Galilean algebra.

But here, the situation is very weird. The fermionic Poisson bracket is symmetric. However, the Lie brackets are anti-symmetric. It doesn't make sense to compute $$\left\{M_{\mu\nu},M_{\alpha\beta}\right\}_{PB},$$

even though the commutation relation $$[J_{\mu\nu},J_{\alpha\beta}]$$ fulfills a representation of the Lie algebra of the Lorentzian symmetry.

So how do $$M_{\mu\nu}$$ generates Lorentzian symmetry but do not fulfill a representation of the Lie algebra? Did I make any mistakes from the above calculations?

• Isn't the Possion bracket in your Equation (2) antisymmetric since $M_{\mu\nu}$ is bosonic (since it contains two spinors)? Jun 20, 2022 at 17:55
• @Toffomat Could you look at the expression of the Poisson bracket? That's the formula that I must use. Its symplectic form is not the usual expression in the "bosonic symplectic geometry". Jun 20, 2022 at 18:00
• The Poisson bracket you write is for Grassmann-odd fields, so for even fields (or $\{\mathrm{odd},\mathrm{even}\}$) it would be with a minus sign (see also physics.stackexchange.com/questions/190977/…). In that way, you can also compute the Poisson brackte of two $M_{\mu\nu}$s. (But that's my guess from graded Lie algebras, I've never worked with fermionic fields in classical mechanics. ) Jun 20, 2022 at 18:18
• @Toffomat Do you mind writing it down as an answer? I don't really understand what you are saying. Jun 20, 2022 at 18:23
• I don't have time to compute this myself, and I'm not sure it will work, but you could just compute the Poisson bracket of two $M_{\mu\nu}$s with a plus sign and see what comes out. (See e.g. Section 2.4 of www-fp.usc.es/~edels/SUSY/Sohnius_Introducing%20SUSY.pdf for some notes on graded algebras). The sign doesn't make a difference for youe Eq. (2) because the second term vanishes anyway. Jun 20, 2022 at 18:52

I finally realized that my question was really stupid.

I don't have to change sign from the Poisson super-bracket.

In (pseudo) classical mechanics, given an Lagrangian (second order $$L(q,\dot{q})$$ in the bosonic case, and first order $$L(\xi,\dot{\xi})$$ in the fermionic case), the conjugate momentum is given by $$p=\frac{\partial L}{\partial\dot{q}}$$ $$\pi=L\frac{\overset{\leftarrow}{\partial}}{\partial\dot{\xi}}$$ in the bosonic and fermionic cases, respectively.

The arrow for the fermionic case is due to the anti-commuting nature of Grassmann numbers. Its direction is to make sure that the result should be consistent with the Legendre transformation $$L=\pi\dot{\xi}-H.$$

On the phase space, the Poisson brackets are given by $$\left\{f(t),g(t)\right\}_{PB}=\frac{\partial f}{\partial q(t)}\frac{\partial g}{\partial p(t)}-\frac{\partial f}{\partial p(t)}\frac{\partial g}{\partial q(t)}$$ $$\left\{\phi(t),\gamma(t)\right\}_{PB}=\phi(t)\left(\frac{\overset{\leftarrow}{\partial}}{\partial\xi(t)}\frac{\overset{\rightarrow}{\partial}}{\partial\pi(t)}+\frac{\overset{\leftarrow}{\partial}}{\partial\pi(t)}\frac{\overset{\rightarrow}{\partial}}{\partial\xi(t)}\right)\gamma(t)$$

for bosonic and fermionic particles, respectively. Since classical mechanics can be viewed as a one-dimensional classical field theory, one can also use functional derivative to re-write them as

$$\left\{f(t),g(t)\right\}_{PB}=\int ds\left\{\frac{\delta f(t)}{\delta q(s)}\frac{\delta g(t)}{\delta p(s)}-\frac{\delta f(t)}{\delta p(s)}\frac{\delta g(t)}{\delta q(s)}\right\}$$ $$\left\{\phi(t),\gamma(t)\right\}_{PB}=\int ds\left[\phi(t)\left(\frac{\overset{\leftarrow}{\delta}}{\delta\xi(s)}\frac{\overset{\rightarrow}{\delta}}{\delta\pi(s)}+\frac{\overset{\leftarrow}{\delta}}{\delta\pi(s)}\frac{\overset{\rightarrow}{\delta}}{\delta\xi(s)}\right)\gamma(t)\right].$$

The above Poisson super-bracket for "classical fermions" is $$\mathbb{Z}_{2}$$-graded. i.e. $$\left\{\phi,\gamma\right\}_{PB}=-(-1)^{\epsilon(\phi)\epsilon(\gamma)}\left\{\gamma,\phi\right\}_{PB},$$

where where $$\epsilon(\,\,\cdot\,\,)$$ means the parity of the Grassmann variable ($$0$$ when even and $$1$$ when odd).

In field theory, the Poisson super-bracket is generalized to

$$\left\{F(t),G(t)\right\}_{PB}=\int d^{4}x\left\{F(t)\left(\frac{\overset{\leftarrow}{\delta}}{\delta\psi(s,\mathbf{x})}\frac{\overset{\rightarrow}{\delta}}{\delta\Pi(s,\mathbf{x})}+\frac{\overset{\leftarrow}{\delta}}{\delta\Pi(s,\mathbf{x})}\frac{\overset{\rightarrow}{\delta}}{\delta\psi(s,\mathbf{x})}\right)G(t)\right\},$$

where $$\int d^{4}x\equiv\int ds\int d\mathbf{x}$$.

The definition is totally fine. It has a super-commuting property which says $$\left\{F,G\right\}_{PB}=-(-1)^{\epsilon(F)\epsilon(G)}\left\{G,F\right\}_{PB},$$

where $$\epsilon(\,\,\cdot\,\,)$$ means the parity of the Grassmann variable ($$0$$ when even and $$1$$ when odd).

The Lorentzian charge $$M_{\mu\nu}$$ is clearly Grassmann-even, so their Poisson brackets still anti-commute.

For a Dirac spinor field, the Noether charge of Lorentzian symmetry is given by $$M_{\mu\nu}(t)=-i\int d^{3}\mathbf{x}\Pi(t,\mathbf{x})\left(\overset{\rightarrow}{J}_{\mu\nu}(x)\Psi(t,\mathbf{x})\right),$$

where $$J_{\mu\nu}(x)=\frac{1}{2}\sigma_{\mu\nu}+i(x_{\mu}\frac{\partial}{\partial x^{\nu}}-x_{\nu}\frac{\partial}{\partial x^{\mu}})𝟙_{4\times 4}$$, and the rightarrow means it acts to the right.

Using the above definition of Poisson super-bracket, one has \begin{align} &\,\,\,\,\,\,\,\left\{M_{\mu\nu}(t),M_{\alpha\beta}(t)\right\}_{PB} \\ &=-\int d^{4}z\left\{M_{\mu\nu}(t)\left(\frac{\overset{\leftarrow}{\delta}}{\delta\psi(s,\mathbf{z})}\frac{\overset{\rightarrow}{\delta}}{\delta\Pi(s,\mathbf{z})}+\frac{\overset{\leftarrow}{\delta}}{\delta\Pi(s,\mathbf{z})}\frac{\overset{\rightarrow}{\delta}}{\delta\psi(s,\mathbf{z})}\right)M_{\alpha\beta}(t)\right\} \\ &=-\int d^{3}\mathbf{x}\int d^{3}\mathbf{y}\Pi(t,\mathbf{x})\left(\overset{\rightarrow}{J}_{\mu\nu}(x)\delta(x-y)\right)\left(\overset{\rightarrow}{J}_{\alpha\beta}(y)\Psi(t,\mathbf{y})\right) \\ &\,\,\,\,\,\,-\int d^{3}\mathbf{x}\int d^{3}\mathbf{y}\left(\overset{\rightarrow}{J}_{\mu\nu}(x)\Psi(t,\mathbf{x})\right)\Pi(t,\mathbf{y})\left(\overset{\rightarrow}{J}_{\alpha\beta}(y)\delta(x-y)\right) \\ &=-\int d^{3}\mathbf{x}\int d^{3}\mathbf{y}\Pi(t,\mathbf{x})\left(\overset{\rightarrow}{J}_{\mu\nu}(x)\delta(x-y)\right)\left(\overset{\rightarrow}{J}_{\alpha\beta}(y)\Psi(t,\mathbf{y})\right) \\ &\,\,\,\,\,\,\,+\int d^{3}\mathbf{x}\int d^{3}\mathbf{y}\Pi(t,\mathbf{y})\left(\overset{\rightarrow}{J}_{\alpha\beta}(y)\delta(x-y)\right)\left(\overset{\rightarrow}{J}_{\mu\nu}(x)\Psi(t,\mathbf{x})\right). \end{align}

Since $$x^{\mu}\partial^{\nu}-x^{\nu}\partial^{\mu}$$ is anti-symmetric, i.e. its diagonal elements vanishes, one can write it as $$\partial^{\nu}x^{\mu}-\partial^{\mu}x^{\nu}$$ and integrate by parts. Then, one has, for the orbital part, \begin{align} &-\int d^{3}\mathbf{x}\int d^{3}\mathbf{y}\Pi(t,\mathbf{x})\left(\overset{\rightarrow}{L}_{\mu\nu}(x)\delta(x-y)\right)\left(\overset{\rightarrow}{L}_{\alpha\beta}(y)\Psi(t,\mathbf{y})\right) \\ &\,\,\,\,\,\,\,+\int d^{3}\mathbf{x}\int d^{3}\mathbf{y}\Pi(t,\mathbf{y})\left(\overset{\rightarrow}{L}_{\alpha\beta}(y)\delta(x-y)\right)\left(\overset{\rightarrow}{L}_{\mu\nu}(x)\Psi(t,\mathbf{x})\right) \\ &=\int d^{3}\mathbf{x}\int d^{3}\mathbf{y}\Pi(t,\mathbf{x})\left(\overset{\leftarrow}{L}_{\mu\nu}(x)\delta(x-y)\right)\left(\overset{\rightarrow}{L}_{\alpha\beta}(y)\Psi(t,\mathbf{y})\right) \\ &\,\,\,\,\,\,\,-\int d^{3}\mathbf{x}\int d^{3}\mathbf{y}\Pi(t,\mathbf{y})\left(\overset{\leftarrow}{L}_{\alpha\beta}(y)\delta(x-y)\right)\left(\overset{\rightarrow}{L}_{\mu\nu}(x)\Psi(t,\mathbf{x})\right) \\ &=\int d^{3}\mathbf{x}\Pi(t,\mathbf{x})\overset{\leftarrow}{L}_{\mu\nu}(x)\overset{\rightarrow}{L}_{\alpha\beta}(x)\Psi(t,\mathbf{x})-\int d^{3}\mathbf{x}\Pi(t,\mathbf{x})\overset{\leftarrow}{L}_{\alpha\beta}(x)\overset{\rightarrow}{L}_{\mu\nu}(x)\Psi(t,\mathbf{x}) \\ &=-\int d^{3}\mathbf{x}\Pi(t,\mathbf{x})\left(\overset{\rightarrow}{L}_{\mu\nu}(x)\overset{\rightarrow}{L}_{\alpha\beta}(x)-\overset{\rightarrow}{L}_{\alpha\beta}(x)\overset{\rightarrow}{L}_{\mu\nu}(x)\right)\Psi(t,\mathbf{x}) \\ &=-\int d^{3}\mathbf{x}\Pi(t,\mathbf{x})\left[\overset{\rightarrow}{L}_{\mu\nu},\overset{\rightarrow}{L}_{\alpha\beta}\right]\Psi(t,\mathbf{x}). \end{align}

Notice that the Lie bracket between the orbital operators is $$[L_{\mu\nu},L_{\alpha\beta}]=i(g_{\sigma\mu}L_{\rho\nu}+g_{\nu\sigma}L_{\mu\rho}-g_{\rho\mu}L_{\sigma\nu}-g_{\nu\rho}L_{\mu\sigma}).$$

Each $$L_{\rho\sigma}$$ is sandwiched between the conjugate pair of spinor fields and the integral produces a charge $$M_{\rho\sigma}$$.

For the spin part, the computation is straightforward. Finally, one finds the Lorentz Lie-algebra $$\left\{M_{\mu\nu},M_{\alpha\beta}\right\}_{PB}=i(g_{\sigma\mu}M_{\rho\nu}+g_{\nu\sigma}M_{\mu\rho}-g_{\rho\mu}M_{\sigma\nu}-g_{\nu\rho}M_{\mu\sigma}).$$