# Spin tensor and Lorentz group operator in bispinor case

+ 2 like - 0 dislike
813 views

For infinisesimal bispinor transformations we have $$\delta \Psi = \frac{1}{2}\omega^{\mu \nu}\eta_{\mu \nu}\Psi , \quad \delta \bar {\Psi} = -\frac{1}{2}\omega^{\mu \nu}\bar {\Psi}\eta_{\mu \nu}, \quad \eta_{\mu \nu} = -\frac{1}{4}(\gamma_{\mu}\gamma_{\nu} - \gamma_{\nu}\gamma_{\mu}). \qquad (.1)$$ Then, by compairing $(.1)$ with transformation by the generators of the Lorentz group, $$\delta \Psi = \frac{i}{2}\omega^{\mu \nu}J_{\mu \nu}\Psi ,$$ we can make the conclusion that in bispinor representation $$J_{\mu \nu} = -i\eta_{\mu \nu}. \qquad (.2)$$ By the other way, from Noether theorem we can get spin tensor, $$S^{\mu, \alpha \beta} = \frac{\partial L}{\partial (\partial_{\mu}\Psi)}Y^{\alpha \beta} + \bar {Y}^{\alpha \beta}\frac{\partial L}{\partial (\partial_{\mu}\bar {\Psi})}.$$ Then, by having $(.1)$ and Lagrangian $$L = \bar {\Psi}(i \gamma^{\mu}\partial_{\mu} - m)\Psi ,$$ it's easy to show that $$S^{\mu, \alpha \beta} = i\bar {\Psi}\gamma^{\mu}\eta^{\alpha \beta}\Psi .$$ It's clearly that I can get $(.2)$ by $$S^{\alpha \beta} = \int S^{\mu, \alpha \beta}dx_{\mu},$$ but for me it's not obvious how to compute it. Can you help me?

This post imported from StackExchange Physics at 2014-12-14 11:30 (UTC), posted by SE-user PhysiXxx

edited Dec 14, 2014

+ 1 like - 0 dislike

By Noether’s theorem, the generators of the Lorentz group are the zero components of the currents, i.e., the Lorentz charges:

$$S^{\alpha\beta} = S^{0,\alpha\beta} = i\bar {\Psi}\gamma^{0}\eta^{\alpha \beta}\Psi = \Psi^{\dagger}\eta^{\alpha \beta}\Psi$$

These charges generate the Lorentz transformations on the spinors by the canonical Poisson brackets:

$$\left \{ \Psi, \Psi^{\dagger} \right \}_{P.B.} = -i \mathbb{I}$$

(With all other Poisson combinations vanishing). The Poisson brackets can be obtained from the time derivative term in the Dirac Lagrangian:

$$i \Psi^{\dagger}\partial_0\Psi$$

Which implies that $i \Psi^{\dagger}$ is the canonical momentum of $\Psi$, thus satisfies the canonical Poisson brackets.

The action of the Lorentz charges correctly generates the Lorentz transformation:

$$\delta \Psi = \left \{ \frac{1}{2} \omega_{\alpha\beta }S^{\alpha\beta}, \Psi^{\dagger} \right \}_{P.B.} = \frac{1}{2}\omega^{\alpha \beta}\eta_{\alpha \beta}\Psi$$

This post imported from StackExchange Physics at 2014-12-14 11:30 (UTC), posted by SE-user David Bar Moshe
answered Oct 14, 2013 by (4,355 points)
"...Which implies that $i\Psi^{+}$ is the canonical momentum of $\Psi$, thus satisfies the canonical Poisson brackets...", - did you make this conclusion by the connection $$L_{d} = \pi \partial_{0}\Psi - H_{d},$$ where $H$ is given from Dirac equation, $$i\partial_{0}\Psi = H\Psi , \quad H = (\gamma \cdot \hat {\mathbf p}) + m, \quad H_{d} = \bar {\Psi} H \Psi ?$$

This post imported from StackExchange Physics at 2014-12-14 11:30 (UTC), posted by SE-user PhysiXxx
And what physical sense has bracket $$\delta \Psi = \left[\frac{1}{2}\omega_{\alpha \beta}S^{\alpha \beta}, \Psi^{+}\right]_{P.B.}?$$

This post imported from StackExchange Physics at 2014-12-14 11:30 (UTC), posted by SE-user PhysiXxx
Oh yes, I understand. It is generator of infinitesimal transformations through Poisson's brackets.

This post imported from StackExchange Physics at 2014-12-14 11:30 (UTC), posted by SE-user PhysiXxx
Unfortunately, it generates false infinitesimal transformations for $i\Psi^{+}$.

This post imported from StackExchange Physics at 2014-12-14 11:30 (UTC), posted by SE-user PhysiXxx
If the problem is a sign problem, please notice that the $\Psi$s are Grassmann variables and they acquire minus signs when they are commuted.

This post imported from StackExchange Physics at 2014-12-14 11:30 (UTC), posted by SE-user David Bar Moshe
Maybe, it's such worse. The transformations for $i\Psi^{+}$ can be represented as $$(\delta \Psi )^{+} = \frac{\omega^{\mu \nu}}{2}\Psi^{+}\eta_{\mu \nu}^{+} = \frac{\omega^{\mu \nu}}{2}\Psi^{+}\gamma_{0}\eta_{\mu \nu}\gamma_{0} \qquad (.1)$$ (analogical ways of transforamation's derivation give the same). But then $$[\gamma_{0}, \eta_{\mu \nu}] = \frac{1}{2}(g_{0\nu}\gamma_{\mu} - g_{0\mu}\gamma_{\nu}),$$ and it's convolution with $\omega^{\mu \nu}$ gives $\frac{1}{2}\omega^{\mu 0}\gamma_{\mu}$. So I can't transform $(.1)$ to form which is given by Poisson's brackets.

This post imported from StackExchange Physics at 2014-12-14 11:30 (UTC), posted by SE-user PhysiXxx

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ysicsOver$\varnothing$lowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.