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 [itex]D^\mu \Psi [\latex] transforms in the same way as the fundamental doublet [itex]\Psi [\latex] under a local SU(2) gauge transformation, derive how the field components [itex]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 /> [itex]\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 [itex]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 [itex]D_{\mu} \psi ----> D'_{\mu} \psi' = U(\underline{x}) (D_{\mu} \psi ) [\latex]. From this we can easily write [itex]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 [itex]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 [itex]\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 [itex]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 /> [itex]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 $\partial_{\mu} (\alpha_{a} t_{a} \psi) [\latex].$[/itex][/itex][/itex][/itex][/itex][/itex][/itex][/itex][/itex][/itex][/itex]

 

[Gregg]

Hi, use [tex]\text{$$}$$ or [itex]\text{$}$[/itex][/tex]

 

[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 [tex]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."<br /> <br /> 2. Homework Equations <br /> I think that the following equations are going to have to be used:<br /> $$\begin{align}<br /> D_{\mu} &= \partial_{\mu} + igW_{\mu}^{a} t^{a}, \\<br /> [t_{a},t_{b}] &= i C_{abc}t_{c}, \\<br /> \end{align}$$<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 $$D_{\mu} \psi ----> D'_{\mu} \psi' = U(\underline{x}) (D_{\mu} \psi )$$. From this we can easily write $$D'_{\mu}$$. 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. <br /> <br /> Now, using that we're taking an infinitesimal transformation we can Taylor expand U and so find that $$U(x) = I - igt_{k} \alpha_{k}$$. 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 /> $$W'_{\mu}^{a} = W_{\mu}^{a} - \frac{1}{g} \partial_{\mu} \alpha_{a} - C_{abc} W_{\mu}^{c}.$$<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 $$\partial_{\mu} (\alpha_{a} t_{a} \psi)$$.[/tex]

 

[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 $$D_{\mu} \psi ----> D'_{\mu} \psi' = U(\underline{x}) (D_{\mu} \psi )$$. From this we can easily write $$D'_{\mu}$$. 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:
$$W'_{\mu}^{a} = W_{\mu}^{a} - \frac{1}{g} \partial_{\mu} \alpha_{a} - C_{abc} W_{\mu}^{c}.$$

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 $$\psi^a$$.