How Do Local SU(2) Gauge Transformations Affect Field Components?

[Kali_89]

Hi all, (Also - if anybody could tell me how to get the latex to work on this page that'd be very handy!)

While not technically homework this is a problem I've found I'm stuck on during my revision. Any help would be greatly appreciated.

Homework Statement

"By demanding that the covariant derivative D^\mu \Psi [\latex] transforms in the same way as the fundamental doublet \Psi [\latex] under a local SU(2) gauge transformation, derive how the field components W_{\mu}^{i}, (i=1,2,3), [\latex] transforms under an infinitesimal such transformation. The Pauli matrix identity [ latex ] (\underline{\sigma} \cdot \underline{a})(\underline{\sigma} \cdot{b}) = \underline{a} \cdot \underline{b} + i \underline{\sigma} \cdot (\underline{a} \times \underline{b}) [\latex], may be assumed."<br /> <br /> <h2>Homework Equations</h2><br /> I think that the following equations are going to have to be used:<br /> \begin{align}<br /> D_{\mu} &= \partial_{\mu} + igW_{\mu}^{a} t^{a}, \\<br /> [t_{a},t_{b}] &= i C_{abc}t_{c}, \\<br /> \end{align} [\latex]<br /> where D_{\mu} [\latex] is the modified derivative, g is our coupling constant, W is the field in question, t are the generator matrices and C is the structure constant.<h2>The Attempt at a Solution</h2><br /> Given that the covariant derivative transforms in the same way as the doublet I know we can write D_{\mu} \psi ----> D'_{\mu} \psi' = U(\underline{x}) (D_{\mu} \psi ) [\latex]. From this we can easily write D'_{\mu} [\latex]. I've found in my notes a general expression for U (incidentally, what is U? I see it has the same sort of form as the phase for transformations I've seen) as U(x) = \exp{(-ig\sum_{1}^{n^2 - 1} t_{k} \alpha_{k})} [\latex] where the n = 2 for the case of SU(2). Here I believe \alpha [\latex] is the phase but I'm not at all sure. The generators for SU(2) are half the Pauli matrices. <br /> <br /> Now, using that we're taking an infinitesimal transformation we can Taylor expand U and so find that U(x) = I - igt_{k} \alpha_{k} [\latex]. At this point I've substituted most of the equations I've got together in order to try and find the following equation which I believe to be the final answer:<br /> W'_{\mu}^{a} = W_{\mu}^{a} - \frac{1}{g} \partial_{\mu} \alpha_{a} - C_{abc} W_{\mu}^{c}. [\latex]<br /> <br /> Substituting in I do seem to make some sort of headway but I've not really understood what I've been doing and also how to treat terms like \partial_{\mu} (\alpha_{a} t_{a} \psi) [\latex].

 

[Gregg]

Hi, use \text{} or \text{}

 

[Kali_89]

Hi all, (Also - if anybody could tell me how to get the latex to work on this page that'd be very handy!)

While not technically homework this is a problem I've found I'm stuck on during my revision. Any help would be greatly appreciated.

1. Homework Statement
"By demanding that the covariant derivative D^\mu \Psi [/itex] transforms in the same way as the fundamental doublet \Psi under a local SU(2) gauge transformation, derive how the field components W_{\mu}^{i}, (i=1,2,3), transforms under an infinitesimal such transformation. The Pauli matrix identity (\underline{\sigma} \cdot \underline{a})(\underline{\sigma} \cdot{b}) = \underline{a} \cdot \underline{b} + i \underline{\sigma} \cdot (\underline{a} \times \underline{b}), may be assumed.&quot;<br /> <br /> 2. Homework Equations <br /> I think that the following equations are going to have to be used:<br /> \begin{align}&lt;br /&gt; D_{\mu} &amp;amp;= \partial_{\mu} + igW_{\mu}^{a} t^{a}, \\&lt;br /&gt; [t_{a},t_{b}] &amp;amp;= i C_{abc}t_{c}, \\&lt;br /&gt; \end{align}<br /> where D_{\mu} is the modified derivative, g is our coupling constant, W is the field in question, t are the generator matrices and C is the structure constant.3. The Attempt at a Solution <br /> Given that the covariant derivative transforms in the same way as the doublet I know we can write D_{\mu} \psi ----&amp;gt; D&amp;#039;_{\mu} \psi&amp;#039; = U(\underline{x}) (D_{\mu} \psi ). From this we can easily write D&amp;#039;_{\mu}. I&#039;ve found in my notes a general expression for U (incidentally, what is U? I see it has the same sort of form as the phase for transformations I&#039;ve seen) as U(x) = \exp{(-ig\sum_{1}^{n^2 - 1} t_{k} \alpha_{k})} where the n = 2 for the case of SU(2). Here I believe \alpha is the phase but I&#039;m not at all sure. The generators for SU(2) are half the Pauli matrices. <br /> <br /> Now, using that we&#039;re taking an infinitesimal transformation we can Taylor expand U and so find that U(x) = I - igt_{k} \alpha_{k}. At this point I&#039;ve substituted most of the equations I&#039;ve got together in order to try and find the following equation which I believe to be the final answer:<br /> W&amp;#039;_{\mu}^{a} = W_{\mu}^{a} - \frac{1}{g} \partial_{\mu} \alpha_{a} - C_{abc} W_{\mu}^{c}.<br /> <br /> Substituting in I do seem to make some sort of headway but I&#039;ve not really understood what I&#039;ve been doing and also how to treat terms like \partial_{\mu} (\alpha_{a} t_{a} \psi).

 

[fzero]

3. The Attempt at a Solution
Given that the covariant derivative transforms in the same way as the doublet I know we can write D_{\mu} \psi ----&gt; D&#039;_{\mu} \psi&#039; = U(\underline{x}) (D_{\mu} \psi ). From this we can easily write D&#039;_{\mu}. I've found in my notes a general expression for U (incidentally, what is U? I see it has the same sort of form as the phase for transformations I've seen) as U(x) = \exp{(-ig\sum_{1}^{n^2 - 1} t_{k} \alpha_{k})} where the n = 2 for the case of SU(2). Here I believe \alpha is the phase but I'm not at all sure. The generators for SU(2) are half the Pauli matrices.

You'll want to write down the expression for the infinitesimal variation \delta (D_\mu\psi)^a, expressing it in terms of \delta \psi^a (which you know) and \delta W_\mu^a, which you're trying to determine.

Now, using that we're taking an infinitesimal transformation we can Taylor expand U and so find that U(x) = I - igt_{k} \alpha_{k}. At this point I've substituted most of the equations I've got together in order to try and find the following equation which I believe to be the final answer:
W&#039;_{\mu}^{a} = W_{\mu}^{a} - \frac{1}{g} \partial_{\mu} \alpha_{a} - C_{abc} W_{\mu}^{c}.

That's close, but you can see that you have an extra free index on the CW term that isn't on the LHS, so that term is wrong.

Substituting in I do seem to make some sort of headway but I've not really understood what I've been doing and also how to treat terms like \partial_{\mu} (\alpha_{a} t_{a} \psi).

You can expand that term out using the product rule for derivatives. It also might help to put the matrix indices on the (t^a)_{bc} and the index on \psi^a.