Corrientes de Noether para la transformación BRST de los campos de Yang-Mills

El punto crucial es la anticonmutación de$\epsilon$con campos de fermiones ($\psi,\bar\psi,\bar{c}^a,c^a$).

Primero, reescribiremos$\mathcal{L}$como (por simplicidad, definimos$B^a \equiv \xi^{-1}\partial^\mu A_\mu^a$)

$$\tag{1} \mathcal{L}=-\frac{1}{4}(F^a_{\mu\nu})^2+\bar{\psi}(i\gamma^{\mu} D_{\mu}-m)\psi-\frac{\xi}{2}B^aB^a- \partial^{\mu}\bar{c}^a D_{\mu}^{ab}c^b$$

que difiere del original$\mathcal{L}$por una derivada total,

$$\tag{2} \partial^{\mu}(\bar{c}^a D_{\mu}^{ab}c^b)$$

asi que$\delta\mathcal{L}$ya no es igual$0$, pero

$$\tag{3} \delta\mathcal{L} = -\delta \partial^{\mu}(\bar{c}^a D_{\mu}^{ab}c^b) = \partial^{\mu}( -\delta\bar{c}^a D_{\mu}^{ab}c^b) = \partial^{\mu}\left(-\epsilon B^a D_{\mu}^{ab}c^b\right) \equiv \partial^{\mu} K_{\mu}$$

Usaremos la identidad de Jacobi ocasionalmente,

$$\tag{4} f^{abd}f^{dce} + f^{bcd}f^{dae} + f^{cad}f^{dbe} = 0$$

Usaremos la derivada derecha en los siguientes cálculos. Entonces la corriente de Noether se define como

$$\tag{5} \epsilon j^{\mu} \equiv \frac{\partial\mathcal{L}}{\partial (\partial_{\mu}\psi)}\delta\psi + \frac{\partial\mathcal{L}}{\partial (\partial_{\mu}\bar\psi)}\delta\bar\psi + \frac{\partial\mathcal{L}}{\partial (\partial_{\mu}c^a)}\delta c^a + \frac{\partial\mathcal{L}}{\partial (\partial_{\mu}\bar{c}^a)}\delta\bar{c}^a + \frac{\partial\mathcal{L}}{\partial (\partial_{\mu}A_{\nu}^a)}\delta A_{\nu}^a - K^{\mu}$$

Ahora calcularemos las partes individuales de la corriente, en movimiento.$\epsilon$a la izquierda de cada expresión. Obtendremos un signo menos adicional si$\epsilon$pasa un campo de fermiones,

$$\begin{aligned} \frac{\partial\mathcal{L}}{\partial (\partial_{\mu}\psi)}\delta\psi &= (\bar\psi i \gamma^{\mu}) (ig \epsilon c^a t^a \psi) = \epsilon\, g \bar\psi \gamma^{\mu} c^a t^a \psi \\ \frac{\partial\mathcal{L}}{\partial (\partial_{\mu}\bar\psi)}\delta\bar\psi &= 0 \\ \frac{\partial\mathcal{L}}{\partial (\partial_{\mu}c^a)}\delta c^a &= (-\partial^{\mu}\bar{c}^a)\left(-\frac{1}{2}g\epsilon f^{abc}c^bc^c\right) = -\frac{1}{2} \epsilon\, g f^{abc} (\partial^{\mu}\bar{c}^a) c^bc^c \\ \frac{\partial\mathcal{L}}{\partial (\partial_{\mu}\bar{c}^a)}\delta\bar{c}^a &= (g^{\mu\nu}D_\nu^{ab} c^b) \left(\epsilon B^a\right) = - \epsilon (g^{\mu\nu}D_\nu^{ab} c^b) B^a = K^{\mu}\\ \frac{\partial\mathcal{L}}{\partial (\partial_{\mu}A_{\nu}^a)}\delta A_{\nu}^a &= \left(-F^{a\mu\nu} - g^{\mu\nu} B^a\right)(\epsilon D_\nu^{ab} c^b) = \epsilon \left(-F^{a\mu\nu} - g^{\mu\nu}B^a\right)(D_\nu^{ab} c^b)\\ \end{aligned} \tag{6}$$

Insertar resultados de$(6)$y$(3)$dentro$(5)$da

$$\tag{7} j^\mu = \left(-F^{a\mu\nu} - g^{\mu\nu}B^a\right)D_\nu^{ab} c^b -\frac{1}{2} g f^{abc} (\partial^{\mu}\bar{c}^a) c^bc^c + g \bar\psi \gamma^{\mu} c^a t^a \psi$$

Es fácil derivar las ecuaciones de movimiento,

$$\begin{aligned} 0= \frac{\partial\mathcal{L}}{\partial A_\nu^a} - \partial_\mu \frac{\partial\mathcal{L}}{\partial(\partial_\mu A_\nu^a)} &= D_\mu^{ab}F^{b\mu\nu} + \partial^\nu B^a + g \bar\psi \gamma^\nu t^a \psi + g f^{abc}(\partial^\nu \bar{c}^b) c^c \\ 0= \frac{\partial\mathcal{L}}{\partial\bar\psi} - \partial_\mu \frac{\partial\mathcal{L}}{\partial(\partial_\mu \bar\psi)} &= -i \gamma^\mu\partial_\mu\psi - g A_{\mu}^a \gamma^\mu t^a \psi + m \psi \\ 0= \frac{\partial\mathcal{L}}{\partial\psi} - \partial_\mu \frac{\partial\mathcal{L}}{\partial(\partial_\mu \psi)} &= -i \partial_\mu \bar\psi \gamma^\mu + g A_\mu^a \bar\psi \gamma^\mu t^a - m \bar\psi \\ 0= \frac{\partial\mathcal{L}}{\partial c^a} - \partial_\mu \frac{\partial\mathcal{L}}{\partial(\partial_\mu c^a)} &= D_\mu^{ab} \partial^\mu \bar{c}^b \\ 0= \frac{\partial\mathcal{L}}{\partial\bar{c}^a} - \partial_\mu \frac{\partial\mathcal{L}}{\partial(\partial_\mu \bar{c}^a)} &= -\partial^\mu D^{ab}_\mu c^b \\ \end{aligned} \tag{8a-e}$$

Ahora comprobaremos la validación de$\partial_\mu j^\mu = 0$,

$$\begin{aligned} \partial_\mu \Bigl[ g \bar\psi \gamma^\mu c^a t^a \psi \Bigr] &\stackrel{(8b,8c)}{=} -\Bigl[ g \bar\psi \gamma^\nu t^a \psi \Bigr] D_\mu^{ad} c^d \\ \partial_\mu \left[ - \frac{1}{2}g f^{abc} (\partial^\mu \bar{c}^a) c^b c^c \right] &\stackrel{(4,8d)}{=} -\Bigl[ g f^{abc}(\partial^\nu \bar{c}^b) c^c \Bigr] D_\mu^{ad} c^d \\ \partial_\mu \Bigl[ (-F^{a\mu\nu} -B^a g^{\mu\nu}) D_\nu^{ab} c^b \Bigr] &\stackrel{(8e)}{=} -(\partial_\mu F^{a\mu\nu} + \partial^\nu B^a) D_\nu^{ad}c^d - F^{a\mu\nu} \partial_\mu D_\nu^{ad}c^d \\ &= -(D_\mu^{ab} F^{b\mu\nu} + \partial^\nu B^a) D_\nu^{ad}c^d \\ &\quad - (gf^{abc}A_\mu^c F^{b\mu\nu}) D_\nu^{ad} c^d - F^{a\mu\nu} \partial_\mu D_\nu^{ad}c^d \\ &\stackrel{(4)}{=} -(D_\mu^{ab} F^{b\mu\nu} + \partial^\nu B^a) D_\nu^{ad}c^d \\ \end{aligned} \tag{9a-c}$$

Nosotros recibiremos

$$\partial_\mu j^\mu \stackrel{(9a+9b+9c,8a)}{=} -\left( \frac{\partial\mathcal{L}}{\partial A_\nu^a} - \partial_\mu \frac{\partial\mathcal{L}}{\partial(\partial_\mu A_\nu^a)} \right) D_\nu^{ad}c^d \tag{10}$$

Esta es de hecho la identidad de Noether fuera de la cáscara , y es cero cuando se cumplen las ecuaciones de movimiento.