Understanding Discrete Symmetries in Quantum Field Theory

[ismaili]

I don't quite understand the treatment of discrete symmetries, for example, in Peskin's QFT book:

Because by definition time reversal symmetry should flip the spin and momentum, so he defined an operation to flip the spin state of a two-component spinor, i.e.
$$\xi^{-s} \equiv -i\sigma^2(\xi^s)^* \quad\cdots(1)$$
, from this definition of spin flip, we have $$\xi^{-s} = (\xi^2 , -\xi^1)$$.

And, previously, he has already solved the Dirac equation and got solutions:
$$u^s(p) = (\sqrt{p\cdot\sigma}\xi^s , \sqrt{p\cdot\bar{\sigma}} \xi^s)$$
$$v^s(p) = (\sqrt{p\cdot\sigma}\eta^s , -\sqrt{p\cdot\bar{\sigma}} \eta^s)$$
where $$\xi^s, \eta^s$$ are two-component spinor basis.
Now he chooses
$$v^s(p) = (\sqrt{p\cdot\sigma}\xi^{-s} , -\sqrt{p\cdot\bar{\sigma}} \xi^{-s})$$
and he defines
$$a^{-s}_{\mathbf{p}} = (a^s_{\mathbf{p}} , -a_{\mathbf{p}}^1) , b^{-s}_{\mathbf{p}} = (b^s_{\mathbf{p}} , -b_{\mathbf{p}}^1)$$
Then, he can compute $$T\psi T = \cdots = \gamma^1\gamma^3 \psi(-t,\mathbf{x})$$

My questions:
(1) Why he defined the spin flip by eq(1)? and why does he define $$a_{\mathbf{p}}^{-s}$$ in such a way? Why doesn't he just define $$\xi^{-s} = (\xi^2 , \xi^1)$$?

(2) He worked out all these discrete symmetric transformation of spinor fields in a particular representation of gamma matrices, i.e. chiral representation of gamma matrices. Is it possible to deal with discrete symmetries without working in a particular representation? Is the result he gets representation independent?

(3) Is it possible to define discrete symmetries in other dimensions? They can be defined only in certain dimensions or in arbitrary dimensions?

Thank you so much.

 

[ismaili]

I don't quite understand the treatment of discrete symmetries, for example, in Peskin's QFT book:

Because by definition time reversal symmetry should flip the spin and momentum, so he defined an operation to flip the spin state of a two-component spinor, i.e.
$$\xi^{-s} \equiv -i\sigma^2(\xi^s)^* \quad\cdots(1)$$
, from this definition of spin flip, we have $$\xi^{-s} = (\xi^2 , -\xi^1)$$.

And, previously, he has already solved the Dirac equation and got solutions:
$$u^s(p) = (\sqrt{p\cdot\sigma}\xi^s , \sqrt{p\cdot\bar{\sigma}} \xi^s)$$
$$v^s(p) = (\sqrt{p\cdot\sigma}\eta^s , -\sqrt{p\cdot\bar{\sigma}} \eta^s)$$
where $$\xi^s, \eta^s$$ are two-component spinor basis.
Now he chooses
$$v^s(p) = (\sqrt{p\cdot\sigma}\xi^{-s} , -\sqrt{p\cdot\bar{\sigma}} \xi^{-s})$$
and he defines
$$a^{-s}_{\mathbf{p}} = (a^s_{\mathbf{p}} , -a_{\mathbf{p}}^1) , b^{-s}_{\mathbf{p}} = (b^s_{\mathbf{p}} , -b_{\mathbf{p}}^1)$$
Then, he can compute $$T\psi T = \cdots = \gamma^1\gamma^3 \psi(-t,\mathbf{x})$$

My questions:
(1) Why he defined the spin flip by eq(1)? and why does he define $$a_{\mathbf{p}}^{-s}$$ in such a way? Why doesn't he just define $$\xi^{-s} = (\xi^2 , \xi^1)$$?

(2) He worked out all these discrete symmetric transformation of spinor fields in a particular representation of gamma matrices, i.e. chiral representation of gamma matrices. Is it possible to deal with discrete symmetries without working in a particular representation? Is the result he gets representation independent?

(3) Is it possible to define discrete symmetries in other dimensions? They can be defined only in certain dimensions or in arbitrary dimensions?

Thank you so much.

(1) I still don't know the solution to question (1). I found that for a two-component Weyl spinor, $$-i\sigma^2 (xi)$$ is actually the definition of charge conjugation. But he called such a transformation as spin flip.
When we solve the Dirac equation, $$u(p) = \sqrt{m}(\xi^s,\xi^s)^T$$ and $$v(p) = \sqrt{m}(\eta^s,-\eta^s)^T$$, where $$\xi^s, s=1,2$$ are two independent basis of two-component spinors, and $$\eta^s, s=1,2$$ are another two independent basis of two-component Weyl spinors. Peskin chose $$\eta$$ to be the charge conjugate of $$\xi$$. In this way, the relation $$u^s(p) = -i\gamma^2(v^s(p))^*$$ is only valid when we made such a choice?

(2) The relation among gamma matrices was found once, it is valid in any representation. So, to this question, the answer should be yes. The result we get is of course representation independent.

(3) still need study..