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  Trace of derivatives of unitary operators

+ 4 like - 0 dislike

I have been studying some lecture notes on the non-linear sigma model and I came up with some difficulties involving a trace. I have the following unitary operator $$ U=\exp\left( \frac{i\vec{\tau}\cdot\vec{\Phi}(x)}{f_{\pi}} \right) $$ How do I calculate $\text{Tr}(\partial_{\mu}U^{\dagger}\partial^{\mu}U)$? The problem had arisen when it was shown (without proof) that the kinetic term of the sigma-meson field was given by $\frac{1}{2}\partial_{\mu}\sigma\partial^{\mu}\sigma=\frac{f_{\pi}}{4}\text{Tr}(\partial_{\mu}U^{\dagger}\partial^{\mu}U)$. The sigma-meson field is $\sigma=f_{\pi}\cos\left( \frac{\Phi(x)}{f_{\pi}} \right)=f_{\pi}+\mathcal{O}(\Phi^2)$ and the pion field is $\vec{\pi}=f_{\pi}\hat{\Phi}\sin \left( \frac{\Phi(x)}{f_{\pi}} \right)=\vec{\Phi}(x)+\mathcal{O}(\Phi^3)$

This post imported from StackExchange Physics at 2015-01-22 11:33 (UTC), posted by SE-user Judas503
asked Jan 21, 2015 in Theoretical Physics by Judas503 (20 points) [ no revision ]
What matrices are the $\vec\tau$? Isospin? How do the notes you're reading define their normalization?

This post imported from StackExchange Physics at 2015-01-22 11:33 (UTC), posted by SE-user 0celo7

1 Answer

+ 0 like - 0 dislike

I'm not sure how the $\vec\tau$ matrices are normalized, but we can get quite far without that. The first thing we should do is take the derivative $$\partial^\mu U=\frac{i}{f_\pi}(\vec\tau\cdot\partial^\mu\vec\Phi)U$$ and $$\partial_\mu U^\dagger=-\frac{i}{f_\pi}(\vec\tau\cdot\partial^\mu\vec\Phi^\dagger)U^\dagger$$ (I'm making the guess that the $\vec\tau$ are Hermitian.) Then $$\partial_\mu U^\dagger \partial^\mu U=f^{-2}_\pi(\vec\tau\cdot\partial^\mu\vec\Phi^\dagger)(\vec\tau\cdot\partial^\mu\vec\Phi)$$ Note that the $U$s cancel out because they are unitary and commute with the $\vec\tau$. We then write $$\partial_\mu U^\dagger \partial^\mu U=f^{-2}_\pi \partial_\mu\Phi^\dagger_i\partial^\mu\Phi_j\tau_i\tau_j$$ If we know the normalization $\operatorname{tr}(\tau_i\tau_j)$, then we can complete the expression.

This post imported from StackExchange Physics at 2015-01-22 11:33 (UTC), posted by SE-user 0celo7
answered Jan 21, 2015 by 0celo7 (50 points) [ no revision ]
I'm not entirely convinced that $\vec\tau$ commutes with the unitary. Could you provide a proof of this fact please?

This post imported from StackExchange Physics at 2015-01-22 11:33 (UTC), posted by SE-user Phoenix87
@Phoenix87 I guess the commutation relations could make that tricky. I'm having an off day today. I can fix this, but I have to make an assumption and I'm not sure I can make it. I'll assume $\partial(U^\dagger)=(\partial U)^\dagger$. Then we have $\partial^\mu U=(i/f)(\vec\tau \cdot\partial^\mu\vec\Phi)U$ as usual but $\partial_\mu U^\dagger=-(i/f)U^\dagger(\vec\tau \cdot\partial_\mu\vec\Phi)$ because transpose changes the order of matrices. Then, using the cyclicity of the trace, we can get the unitaries to cancel.

This post imported from StackExchange Physics at 2015-01-22 11:33 (UTC), posted by SE-user 0celo7
$\partial^\mu[\vec\tau \cdot\vec\Phi(x)]$ in general will not commute with $\vec\tau \cdot\vec\Phi(x)$, in which case the first step will be incorrect.

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