# Trace of derivatives of unitary operators

+ 4 like - 0 dislike
607 views

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
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

+ 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 (50 points)
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.

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ysicsOve$\varnothing$flowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.