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covariant derivative for spinor fields

+ 6 like - 0 dislike
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scalars (spin-0) derivatives is expressed as:

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

vector (spin-1) derivatives are expressed as:

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

what is the expression for covariant derivatives of spinor (spin-1/2) quantities?

This post imported from StackExchange Physics at 2014-04-01 16:35 (UCT), posted by SE-user lurscher
asked Mar 22, 2011 in Theoretical Physics by CharlesJQuarra (510 points) [ no revision ]

3 Answers

+ 6 like - 0 dislike

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.

This post imported from StackExchange Physics at 2014-04-01 16:35 (UCT), posted by SE-user user346
answered Mar 22, 2011 by Deepak Vaid (1,975 points) [ no revision ]
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

This post imported from StackExchange Physics at 2014-04-01 16:35 (UCT), posted by SE-user lurscher
@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 :)

This post imported from StackExchange Physics at 2014-04-01 16:35 (UCT), posted by SE-user user346
@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 :)

This post imported from StackExchange Physics at 2014-04-01 16:35 (UCT), posted by SE-user Marek
+ 3 like - 0 dislike

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.

This post imported from StackExchange Physics at 2014-04-01 16:35 (UCT), posted by SE-user Lawrence B. Crowell
answered Mar 22, 2011 by Lawrence B. Crowell (590 points) [ no revision ]
I hadn't thought of it in this way. +1

This post imported from StackExchange Physics at 2014-04-01 16:35 (UCT), posted by SE-user user346
+ 1 like - 0 dislike

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.

This post imported from StackExchange Physics at 2014-04-01 16:35 (UCT), posted by SE-user QGR
answered Mar 23, 2011 by QGR (250 points) [ no revision ]

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