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A Proof - gauge transformation of yang mills field strength

  1. Mar 23, 2017 #1
    In Yang-Mills theory, the gauge transformations

    $$\psi \to (1 \pm i\theta^{a}T^{a}_{\bf R})\psi$$

    and

    $$A^{a}_{\mu} \to A_{\mu}^{a} \pm \partial_{\mu}\theta^{a} \pm f^{abc}A_{\mu}^{b}\theta^{c}$$

    induce the gauge transformation


    $$F_{\mu\nu}^{a} \to F_{\mu\nu}^{a} - f^{abc}\theta^{b}F_{\mu\nu}^{c}.$$

    ------------------------------------------------------------------------------------------------------------------------------------

    Proof:


    Let us begin with the definition of the field-strength tensor in terms of the covariant derivative ##D_{\mu} = \partial_{\mu}-iA_{\mu}^{a}T^{a}_{\bf R}##:

    $$F^{a}_{\mu\nu}T^{a}_{\bf R} = i\left[D_{\mu},D_{\nu}\right].$$

    In order to determine the gauge transformation of ##F^{a}_{\mu\nu}T^{a}_{\bf R}##, we must first determine the gauge transformation of ##D_{\mu}D_{\nu}\psi##. It is rather easy to show that

    $$D_{\mu}\psi \to (1 \pm i\theta^{a}T^{a}_{\bf R})D_{\mu}\psi$$

    and hence

    $$D_{\mu}D_{\nu}\psi \to (1 \pm i\theta^{a}T^{a}_{\bf R})D_{\mu}D_{\nu}\psi.$$

    Therefore, we now have

    $$F_{\mu\nu}^{a}T^{a}_{\bf R} = i\left[D_{\mu},D_{\nu}\right]$$

    $$F_{\mu\nu}^{a}T^{a}_{\bf R} \to i(1 \pm i\theta^{b}T^{b}_{\bf R})\left[D_{\mu},D_{\nu}\right]$$

    $$F_{\mu\nu}^{a}T^{a}_{\bf R} \to (1 \pm i\theta^{b}T^{b}_{\bf R})F_{\mu\nu}^{a}T^{a}_{\bf R}$$

    $$F_{\mu\nu}^{a}T^{a}_{\bf R} \to F_{\mu\nu}^{a}T^{a}_{\bf R} \pm i\theta^{b}T^{b}_{\bf R}F_{\mu\nu}^{a}T^{a}_{\bf R}$$


    -------------------------------------------------------------------------------------------------------------------------------------

    In the last line of the derivation, I don't have a third term needed to form a structure constant. Where have I gone wrong?
     
  2. jcsd
  3. Mar 23, 2017 #2

    samalkhaiat

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    The second equation is not correct because the covariant derivative [itex]D_{\mu}[/itex] does not pass through the gauge function [itex]\theta^{a} (x)[/itex].
     
  4. Mar 24, 2017 #3
    Consider the proof of ##D_{\mu}\psi \to (1 \pm i\theta^{a}T^{a}_{\bf R})D_{\mu}\psi## (under the gauge transformation ##\psi \to (1 \pm i\theta^{a}T^{a}_{\bf R})\psi## of the Yang-Mills Lagrangian).

    Skip this proof if you want to. The point of including this tedious proof is to obtain the important result that

    ##\left(\partial_{\mu} - i(A_{\mu}^{a} \pm \partial_{\mu}\theta^{a} \pm f^{abc}A_{\mu}^{b}\theta^{c})T_{\bf R}^{a}\right)(1 \pm i\theta^{b}T^{b}_{\bf R}) = (1 \pm i\theta^{a}T^{a}_{\bf R})(\partial_{\mu}-iA_{\mu}^{a}T^{a}_{\bf R})##



    ------------------------------------------------------------------------------------------------------------------------------------------

    ##D_{\mu}\psi \to \left(\partial_{\mu} - i(A_{\mu}^{a} \pm \partial_{\mu}\theta^{a} \pm f^{abc}A_{\mu}^{b}\theta^{c})T_{\bf R}^{a}\right)\left(1 \pm i\theta^{d}T^{d}_{\bf R}\right)\psi##

    ##= \left(\partial_{\mu} - iA_{\mu}^{a}T_{\bf R}^{a} \mp i(\partial_{\mu}\theta^{a})T_{\bf R}^{a} \mp if^{abc}A_{\mu}^{b}\theta^{c}T_{\bf R}^{a}\right)\left(1 \pm i\theta^{d}T^{d}_{\bf R}\right)\psi##

    ##= \partial_{\mu}\psi - iA_{\mu}^{a}T_{\bf R}^{a}\psi \mp i(\partial_{\mu}\theta^{a})T_{\bf R}^{a}\psi \mp if^{abc}A_{\mu}^{b}\theta^{c}T_{\bf R}^{a}\psi \pm i\partial_{\mu}(\theta^{d}T^{d}_{\bf R}\psi) \pm A_{\mu}^{a}T^{a}_{\bf R}\theta^{d}T^{d}_{\bf R}\psi.##

    We can simplify the third and fifth terms to obtain

    ##\mp i(\partial_{\mu}\theta^{a})T_{\bf R}^{a}\psi \pm i\partial_{\mu}(\theta^{d}T^{d}_{\bf R}\psi) = \pm i\theta^{a}T^{a}_{\bf R}\partial_{\mu}\psi##

    and we can simplify the fourth and sixth terms using the commutation relation of the ##{\bf su}(N)## algebra to obtain

    ##\mp if^{abc}A_{\mu}^{b}\theta^{c}T_{\bf R}^{a}\psi \pm A_{\mu}^{a}T^{a}_{\bf R}\theta^{d}T^{d}_{\bf R}\psi = \pm \theta^{a}A_{\mu}^{b}(if^{abc}T^{c}_{\bf R} + T^{b}_{\bf R}T^{a}_{\bf R})\psi = \pm \theta^{a}T^{a}_{\bf R}A^{b}_{\mu}T^{b}_{\bf R}\psi##

    Therefore, the gauge transformation of the covariant derivative ##D_{\mu}\psi## simplifies to

    ##D_{\mu}\psi \to \partial_{\mu}\psi-iA^{a}_{\mu}T^{a}_{\bf R}\psi \pm i\theta^{a}T^{a}_{\bf R}\partial_{\mu}\psi \pm \theta^{a}T^{a}_{\bf R}A^{b}_{\mu}T^{b}_{\bf R}\psi##
    ##= (1\pm i\theta^{a}T^{a}_{\bf R})(\partial_{\mu}-iA^{b}_{\mu}T^{b}_{\bf R})\psi##
    ##= (1\pm i\theta^{a}T^{a}_{\bf R})D_{\mu}\psi##
     
  5. Mar 24, 2017 #4
    We can compute the gauge transformation of ##D_{\mu}D_{\nu}\psi## similarly.

    ##D_{\mu}D_{\nu}\psi \to \left(\partial_{\mu} - i(A_{\mu}^{a} \pm \partial_{\mu}\theta^{a} \pm f^{abc}A_{\mu}^{b}\theta^{c})T_{\bf R}^{a}\right)(1 \pm i\theta^{b}T^{b}_{\bf R})D_{\nu}\psi##

    Tracing the computation in my previous post, we find that

    ##\left(\partial_{\mu} - i(A_{\mu}^{a} \pm \partial_{\mu}\theta^{a} \pm f^{abc}A_{\mu}^{b}\theta^{c})T_{\bf R}^{a}\right)(1 \pm i\theta^{b}T^{b}_{\bf R}) = (1 \pm i\theta^{a}T^{a}_{\bf R})(\partial_{\mu}-iA_{\mu}^{a}T^{a}_{\bf R})##

    Therefore, the gauge transformation of ##D_{\mu}D_{\nu}\psi## is given by

    ##D_{\mu}D_{\nu}\psi \to (1 \pm i\theta^{a}T^{a}_{\bf R})D_{\mu}D_{\nu}\psi.##
     
  6. Mar 24, 2017 #5
    I do not see how I have passed the covariant derivative ##D_{\mu}## through the gauge function ##\theta(x)##.
     
  7. Mar 24, 2017 #6

    dextercioby

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    Homework Helper

    That ##\pm## everywhere is tiring for the eye, plus that's simply wrong to use. So choose a sign to begin with and respect it in the sequel.
     
  8. Mar 25, 2017 #7

    samalkhaiat

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    The problem you are asking is very simple, and you can do it if you use simple and neat expressions. Let me show you one method. And you will see that even a school kid can follow the algebra. Okay, I will work with matrix-valued fields and functions, i.e., fields and functions taking value in the Lie algebra of the gauge group. These are the connection [itex]\mathbb{A}_{\mu}(x) = A_{\mu}^{a}(x)T^{a}[/itex] and the field tensor [itex]\mathbb{F}_{\mu\nu}(x) = F_{\mu\nu}^{a}T^{a}[/itex]. The [itex]T^{a}[/itex]’s are arbitrary (anti-hermitian) matrix representation of the Lie algebra [itex][T^{a} , T^{b}] = C^{abc}T^{c}[/itex] . Of course the field tensor is defined by

    [tex]\mathbb{F}_{\mu\nu}(x) = \partial_{\mu}\mathbb{A}_{\nu} - \partial_{\nu}\mathbb{A}_{\mu} + [\mathbb{A}_{\mu} , \mathbb{A}_{\nu}] . \ \ \ \ \ (1)[/tex]

    I will write the gauge group element (or the transformation matrix) as [itex]U(x) = e^{\Theta (x)}[/itex], with [itex]\Theta (x) = \theta^{a}(x)T^{a}[/itex] and [itex]\theta^{a}(x)[/itex] being the arbitrary gauge functions. When I consider infinitesimal transformation, I will write [itex]U(x) = 1 + \Theta (x)[/itex]. On any Lie algebra valued function, the covariant derivative is defined by [tex]D_{\mu}\Theta (x) = \partial_{\mu} \Theta + [ \mathbb{A}_{\mu} , \Theta ] . \ \ \ \ \ \ (2)[/tex]

    Now, under a finite gauge transformation [itex]U(x)[/itex], the gauge field transforms in-homogeneously as [tex]\mathbb{A}_{\mu} \to U^{-1} \mathbb{A}_{\mu} U + U^{-1}\partial_{\mu}U . \ \ \ \ \ (3)[/tex] This means that the gauge field (or the connection) transforms in the adjoint representation of the corresponding global symmetry group, i.e., when the transformation matrix [itex]U[/itex] is constant. Infinitesimally, i.e., to first order in [itex]\Theta[/itex], Eq(3) becomes

    [tex]\mathbb{A}_{\mu} \to \mathbb{A}_{\mu} + \partial_{\mu}\Theta + [\mathbb{A}_{\mu} , \Theta ] .[/tex] Let’s rewrite this in the language of infinitesimal variation or change [itex]\delta \mathbb{A}_{\mu}[/itex] and use the definition of covariant derivative [tex]\delta \mathbb{A}_{\mu} = \partial_{\mu}\Theta + [\mathbb{A}_{\mu} , \Theta ] = D_{\mu} \Theta . \ \ \ \ (4)[/tex] From this, you can easily obtain the following commutator equation

    [tex][D_{\mu} , D_{\nu}] \Theta = D_{\mu} \delta\mathbb{A}_{\nu} - D_{\nu} \delta\mathbb{A}_{\mu} . \ \ \ \ \ (5)[/tex] Now, we have every things needed to solve your problem. That is, given the transformation [itex]\delta\mathbb{A}_{\nu}[/itex] we would like to find the corresponding infinitesimal transformation of the field tensor, i.e. [itex]\delta \mathbb{F}_{\mu\nu}[/itex]. Applying the variation symbol [itex]\delta[/itex] on both sides of Eq(1) and rearranging the terms, we obtain

    [tex]\delta \mathbb{F}_{\mu\nu} = \partial_{\mu} \delta\mathbb{A}_{\nu} + [\mathbb{A}_{\mu} , \delta\mathbb{A}_{\nu} ] - \partial_{\nu} \delta\mathbb{A}_{\mu} - [\mathbb{A}_{\nu} , \delta\mathbb{A}_{\mu} ] .[/tex] Using the definition of the covariant derivative, the above equation becomes

    [tex]\delta \mathbb{F}_{\mu\nu} = D_{\mu} \delta\mathbb{A}_{\nu} - D_{\nu} \delta\mathbb{A}_{\mu} . \ \ \ \ \ \ \ (6)[/tex] Now, using Eq(5), equation (6) becomes

    [tex]\delta \mathbb{F}_{\mu\nu} = [D_{\mu} , D_{\nu}] \Theta . \ \ \ \ \ \ \ \ (7)[/tex]

    So, we need to calculate the commutator on the RHS of (7). To do that, we only need to first find the action of [itex]D_{\mu}D_{\nu}[/itex] on [itex]\Theta[/itex] and then the commutator can be constructed by interchanging the indices. From the definition of the covariant derivative, we obtain

    [tex]D_{\mu}D_{\nu} \Theta = D_{\mu} \left( \partial_{\nu} \Theta + [\mathbb{A}_{\nu} , \Theta ] \right) .[/tex] Now, on the RHS, apply the definition of [itex]D_{\mu}[/itex] to get

    [tex]D_{\mu}D_{\nu} \Theta = \partial_{\mu} \left( \partial_{\nu} \Theta + [\mathbb{A}_{\nu} , \Theta ] \right) + [\mathbb{A}_{\mu} , \partial_{\nu} \Theta + [ \mathbb{A}_{\nu} , \Theta ] ] .[/tex]

    This can be rewritten as

    [tex]D_{\mu}D_{\nu} \Theta = \partial_{\mu} \partial_{\nu} \Theta + \partial_{\mu} [ \mathbb{A}_{\nu} , \Theta ] + [\mathbb{A}_{\mu} , \partial_{\nu}\Theta ] + [\mathbb{A}_{\mu} , [\mathbb{A}_{\nu} , \Theta ]] .[/tex]

    Now, obtain another equation by interchanging [itex]\mu \leftrightarrow \nu[/itex] and subtract it from the above equation

    [tex][D_{\mu} , D_{\nu}] \Theta = [\partial_{\mu}\mathbb{A}_{\nu} - \partial_{\nu}\mathbb{A}_{\mu} , \Theta ] + [\mathbb{A}_{\mu} , [\mathbb{A}_{\nu} , \Theta]] - [\mathbb{A}_{\nu} , [\mathbb{A}_{\mu} , \Theta]] .[/tex] And finally, we use the Jacobi identity

    [tex][\mathbb{A}_{\mu} , [\mathbb{A}_{\nu} , \Theta]] - [\mathbb{A}_{\nu} , [\mathbb{A}_{\mu} , \Theta]] = [[\mathbb{A}_{\mu} , \mathbb{A}_{\nu}] , \Theta ] ,[/tex] to arrive at

    [tex][D_{\mu} , D_{\nu}] \Theta = [\partial_{\mu}\mathbb{A}_{\nu} - \partial_{\nu}\mathbb{A}_{\mu} + [\mathbb{A}_{\mu} , \mathbb{A}_{\nu}], \Theta ] ,[/tex] which is

    [tex][D_{\mu} , D_{\nu}] \Theta = [\mathbb{F}_{\mu\nu} , \Theta ] . \ \ \ \ \ \ \ \ \ (8)[/tex] So, from (7) and (8), you find the answer to your problem

    [tex]\delta \mathbb{F}_{\mu\nu} = [\mathbb{F}_{\mu\nu} , \Theta ] , \ \ \ \ \ \ \ \ \ \ (9)[/tex]

    or

    [tex]\mathbb{F}_{\mu\nu} \to \mathbb{F}_{\mu\nu} + [\mathbb{F}_{\mu\nu} , \Theta ] . \ \ \ \ \ (10)[/tex]

    This is nothing but the infinitesimal version of the following finite transformation [tex]\mathbb{F}_{\mu\nu} \to U^{-1} \mathbb{F}_{\mu\nu} U .[/tex]

    In terms of the components of the field tensor [itex]F^{a}_{\mu\nu}[/itex], Eq(9) becomes

    [tex]\delta F^{c}_{\mu\nu}T^{c} = F^{a}_{\mu\nu}\theta^{b} [T^{a} , T^{b}] = C^{abc}F^{a}\theta^{b}T^{c} .[/tex] Thus [tex]\delta F^{c}_{\mu\nu} = C^{abc}F^{a}_{\mu\nu}\theta^{b} .[/tex]
    So, the lesson from this is: use neat and simple expressions, and don't let your equations look like tangled spaghetti.
     
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