Eikonal Blog

2010.07.13

Time inversion

Filed under: qft — Tags: , , — sandokan65 @ 14:02

Time inversion

• ${\cal T} = {\cal U}{\cal K}$
• $\hat{x}:\equiv (-x^0,\vec{x}) = -\bar{x}$
• ${\cal T}H{\cal T}^{-1} = H$
• ${\cal T}{\cal L}(x){\cal T}^{-1} = {\cal L}(\hat{x})$
• ${\cal T}j_\mu(x){\cal T}^{-1} = j^\mu(\hat{x})$

Scalar field – real

• ${\cal T}\phi(x){\cal T}^{-1} = \epsilon\phi(\hat{x})$
• ${\cal T}a(\vec{k}){\cal T}^{-1} = \epsilon a(-\vec{k})$
• ${\cal T}a^\dagger(\vec{k}){\cal T}^{-1} = \epsilon a^\dagger(-\vec{k})$
• ${\cal U}a(\vec{k}){\cal U}^{-1} = \epsilon a^\dagger(-\vec{k})$
• ${\cal U}a^\dagger(\vec{k}){\cal U}^{-1} = \epsilon a(-\vec{k})$

where $\epsilon=\pm 1$.

So:
${\cal U} = e^{-\frac{i\pi}2 \int d^3\vec{k} (a^\dagger(\vec{k})a(\vec{k}) - \epsilon a^\dagger(\vec{k})a(-\vec{k}))}.$

Scalar field – complex

${\cal T}\phi(x){\cal T}^{-1} = \epsilon\phi^{*}(\hat{x})$

Dirac field

Here:

• ${\cal T}\psi(x){\cal T}^{-1} = T \psi(\hat{x})$
• ${\cal T}\bar{\psi}(x){\cal T}^{-1} = \bar{\psi}(\hat{x}) \gamma^0 T^\dagger \gamma^0,$

where

• $T :\equiv i\gamma^1 \gamma^3 = T^\dagger = T^{-1} = - T^{*},$
• $T \gamma_\mu T^{-1} = \gamma_\mu^T = \gamma^{\mu \ *}.$

Then

• $T u(p,s) = u^*(-p,-s)e^{i\alpha_+(p,s)},$
• $T v(p,s) = v^*(-p,-s)e^{i\alpha_-(p,s)},$

\ni where $\alpha_\pm(p,s) = \pi + \alpha_\pm(-p,-s)$.

Now

• ${\cal U} b(p,s) {\cal U}^{-1} = -b(-p,-s) e^{i\alpha_+(p,s)},$
• ${\cal U} d^\dagger(p,s) {\cal U}^{-1} = -d^\dagger(-p,-s) e^{i\alpha_-(p,s)}.$

Split ${\cal U} = {\cal U}_1{\cal U}_2$ s/t

• ${\cal U}_1 b(p,s) {\cal U}_1^{-1} = e^{i\alpha_+(p,s)} b(p,s),$
• ${\cal U}_1 d^\dagger(p,s) {\cal U}_1^{-1} = e^{i\alpha_-(p,s)} d^\dagger(p,s),$
• ${\cal U}_2 b(p,s) {\cal U}_2^{-1} = -b(-p,-s),$
• ${\cal U}_2 d^\dagger(p,s) {\cal U}_2^{-1} = -d^\dagger(-p,-s).$

Their realizations are:

• ${\cal U}_1 = e^{- i \int d^3\vec{p} \sum_s (\alpha_+(p,s) b^\dagger(p,s)b(p,s) - \alpha_(p,s)d^\dagger(p,s)d(p,s))},$
• ${\cal U}_2 = e^{- i \frac{\pi}2 \int d^3\vec{p} \sum_s (b^\dagger(p,s)b(p,s) + b^\dagger(p,s)b(-p,-s) - d^\dagger(p,s)d(p,s) - d^\dagger(p,s)d(-p,-s))}.$

Electromagnetic field

• ${\cal T}A^\mu(x){\cal T}^{-1} = A_\mu(\hat{x}),$
• ${\cal U}\vec{a}^a(\vec{k}){\cal U}^{-1} = - (-)^a a^a(-\vec{k}),$
• ${\cal U} = e^{\frac{i\pi}2 \int d^3k (a^\dagger(\vec{k})^a a(\vec{k})_a + (-)^a a^\dagger(\vec{k})^a a(-\vec{k})_a}$.