scalars (spin-0) derivatives is expressed as:

$$\nabla_{i} \phi = \frac{\partial \phi}{ \partial x_{i}}.$$

vector (spin-1) derivatives are expressed as:

$$\nabla_{i} V^{k} = \frac{\partial V^{k}}{ \partial x_{i}} + \Gamma^k_{m i} V^m.$$

My Question: What is the expression for covariant derivatives of spinor (spin-1/2) quantities?


4 Answers 4


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.

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.

  • $\begingroup$ I hadn't thought of it in this way. +1 $\endgroup$
    – user346
    Mar 23, 2011 at 1:22

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:

$$ D_i \psi = \partial_i \psi + g A_i^I T_I \psi $$

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:

$$ D_i \psi_a = \partial_i \psi_a + g A_{i\,I} T^I{}_a{}^b \psi_b $$

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.

There are relations between the spin connection, the christoffel connection and the metric but this is the definition of the spin connection.

  • $\begingroup$ for four-component spinors i think we use a linear combination of Lorentz generators that look like $SU(2) \bigoplus SU(2)$, i don't remember right now where i did read this $\endgroup$
    – lurscher
    Mar 22, 2011 at 19:34
  • 3
    $\begingroup$ @lurscher - yes, you can factor $so(3,1)$ into two copies of $su(2)$ (we are talking about the lie algebras not the groups, mind you). This is again given in chap. 3 of Peskin. I hated that book initially. But it grows on you like beer or caviar :) $\endgroup$
    – user346
    Mar 22, 2011 at 19:45
  • 1
    $\begingroup$ @lurscher: your are right (and so is @Deepak). And let me use this opportunity to make this statement crystal clear for you by showing you Dynkin diagrams of these algebras. $\mathfrak{so}(1,3)$ is $D_2$ and $\mathfrak{su}(2)$ is $A_1$. As you can see two dots is twice as much as one dot. QED :) $\endgroup$
    – Marek
    Mar 22, 2011 at 21:46
  • $\begingroup$ What are g and A here? $\endgroup$
    – Craig
    Dec 5, 2019 at 15:29

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.


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".


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