Covariant derivative for spinor fields - Physics Stack Exchange most recent 30 from physics.stackexchange.com 2019-09-17T15:28:16Z https://physics.stackexchange.com/feeds/question/7418 https://creativecommons.org/licenses/by-sa/4.0/rdf https://physics.stackexchange.com/q/7418 11 Covariant derivative for spinor fields lurscher https://physics.stackexchange.com/users/955 2011-03-22T17:29:21Z 2019-03-29T17:34:31Z <p>scalars (spin-0) derivatives is expressed as:</p> <p><span class="math-container">$$\nabla_{i} \phi = \frac{\partial \phi}{ \partial x_{i}}.$$</span></p> <p>vector (spin-1) derivatives are expressed as:</p> <p><span class="math-container">$$\nabla_{i} V^{k} = \frac{\partial V^{k}}{ \partial x_{i}} + \Gamma^k_{m i} V^m.$$</span></p> <p><strong>My Question:</strong> What is the expression for covariant derivatives of spinor (spin-1/2) quantities?</p> https://physics.stackexchange.com/questions/7418/-/7422#7422 7 Answer by user346 for Covariant derivative for spinor fields user346 https://physics.stackexchange.com/users/0 2011-03-22T18:22:14Z 2011-03-22T18:22:14Z <p>For the covariant spinor derivative we need to introduce a connection which can parallel transport a spinor. Such a connection takes values in the Lie-algebra of the group the spinor transforms under. Then we have:</p> <p>$$D_i \psi = \partial_i \psi + g A_i^I T_I \psi$$</p> <p>Here $T_I$ are the generators of the lie-algebra and are matrix valued. We have suppressed spinorial indices. Writing them out explicitly we get:</p> <p>$$D_i \psi_a = \partial_i \psi_a + g A_{i\,I} T^I{}_a{}^b \psi_b$$</p> <p>For eg, for $SU(2)$ the lie-algebra generators are given by the three pauli matrices $\sigma_x,\sigma_y, \sigma_z$ which then act on two component spinors. If you wish to work with four-component spinors $\psi_A$, transforming under the Lorentz group, the relevant generators are those of $SO(3,1)$. You can find these in Peskin and Schroeder, page 41.</p> <p>There are relations between the spin connection, the christoffel connection and the metric but this is the definition of the spin connection.</p> https://physics.stackexchange.com/questions/7418/-/7424#7424 -4 Answer by Vladimir Kalitvianski for Covariant derivative for spinor fields Vladimir Kalitvianski https://physics.stackexchange.com/users/1390 2011-03-22T20:55:09Z 2011-03-22T21:04:16Z <p>I would like you to pay your attention that this way of introducing "interaction" is only good for describing external fields (that may be switched on and off physically). This way of coupling with the proper field (that can never be switched off) is not good and needs resolving IR and UV divergences if implemented. After renormalizations and IR diagram summation the true coupling with the proper field is different from the "covariant derivative".</p> https://physics.stackexchange.com/questions/7418/-/7428#7428 9 Answer by Lawrence B. Crowell for Covariant derivative for spinor fields Lawrence B. Crowell https://physics.stackexchange.com/users/1352 2011-03-22T21:42:03Z 2011-03-22T21:53:49Z <p>There is an interesting way to look at Christoffel connections with spinor fields. The usual Dirac operator is written as $\gamma^\mu\partial_\mu$. It is interesting to change this to $\partial_\mu(\gamma^\mu\psi)$. This then becomes $$\partial_\mu(\gamma^\mu\psi)~=~ \gamma^\mu\partial_\mu~+~(\partial_\mu\gamma^\mu)\psi.$$ The anticommutator $\{\gamma^\mu,~\gamma^\nu\}~=~2g^{\mu\nu}$ and the covariant constancy of the metric gives $\partial_\mu\gamma^\mu~=~\Gamma^\mu_{\mu\sigma}\gamma^\sigma$. So we may then write the Dirac operator in this different form as $$\delta_\nu^\mu\partial_\mu(\gamma^\nu\psi)~=~ \delta^\mu_\nu \gamma^\nu\partial_\mu\psi~+~\delta^\mu_\nu \Gamma^\nu_{\mu\sigma}\gamma^\sigma\psi.$$ Now if you peel off the Kronecker delta you have a covariant derivative of the spinor field. </p> <p>What this means is that in general the Clifford algebra $CL(3,1)$ representation of the Dirac matrices is local. The connection coefficient can then be seen as due to transition functions between these representations, so the differential produces connection coefficients.</p> https://physics.stackexchange.com/questions/7418/-/7440#7440 1 Answer by QGR for Covariant derivative for spinor fields QGR https://physics.stackexchange.com/users/1039 2011-03-23T08:32:33Z 2011-03-23T08:32:33Z <p>Before you can even introduce spinor bundles in curved spacetime, we need to introduce vierbeins first. This defines a local orthonormal frame. If you wish, you can introduce a principle frame bundle with $Spin(d,1)$ as the gauge group. Spinors can be defined with respect to this frame. The key is that spinors are representations of $Spin(d,1)$, a double cover of $SO(d,1)$, but not of the general linear group $GL(d+1,\mathbf{R})$. The affine connection is a connection over the latter group, but assuming metricity, we may map that into a spin connection over the former principle bundle.</p>