2
$\begingroup$

This is from the intro to a problem 36.5 in Srednicki and not part of the problem itself. I am having trouble proving that $$\mathcal{L}=i\psi_j^\dagger\sigma^\mu\partial_\mu\psi_j$$ Has $U(N)$ symmetry. $j$ is to be understood as going from $1$ to $N$ and being summed over. Srednicki says to take the following transformation on the (massless) Weyl Fields $$\psi_j \rightarrow U_{jk}\psi_k$$ And then I expect that I will require $UU^\dagger=I$ (which then implies the symmetry by thinking in the space where we are stacking the Weyl fields as a vector). However, plugging this in yields $$\mathcal{L}\rightarrow i(U_{jm}\psi_m)^\dagger\sigma^\mu\partial_\mu U_{jk}\psi_k $$ There is no worry about commuting with the pauli or derivative here so I will just put together the matrices: $$\mathcal{L}\rightarrow i\psi_m^\dagger U^*_{mj} U_{jk}\sigma^\mu\partial_\mu\psi_k $$ so it instead looks like I require the matrix multiplication $U^*U=I$ where the star denotes complex conjugation, to preseve the lagrangian. How does this show the required symmetry?

Edit: In addition to naturally inconsistent's response (which is correct but it not as explicit as I needed), I recommend anyone who has this question to also look at this post if still confused.

Improve this question
$\endgroup$
3
  • $\begingroup$ Once you use index notation, then you don't flip indices upon taking the adjoint. You simply have $(U_{jk} \psi_j)^\dagger = U_{jk}^*\psi_j^\dagger$. $\endgroup$
    – Prahar
    Commented Jan 17 at 3:49
  • 1
    $\begingroup$ You can’t have four $j$ indices. $\endgroup$
    – Ghoster
    Commented Jan 17 at 5:38
  • $\begingroup$ Prahar, could you source that or explain it more? I am not sure I have seen this treatment in a textbook before $\endgroup$
    – JohnA.
    Commented Jan 17 at 15:09

1 Answer 1

Reset to default
3
$\begingroup$

I am not opening my copy to check what that problem is. I will assume that $\partial_\mu U_{jk}=0$ as you do too.

You are applying the transformation wrong. I think you are correctly quoting from Srednicki to try $$\tag1\psi_j\to U_{jk}\psi_k$$ so why is it that you have applied them with the wrong indices? The correct way is $$ \begin{align} \tag2\mathcal L\to &i(U_{jm}\psi_m)^\dagger\sigma^\mu\partial_\mu U_{jk}\psi_k\\ \tag3=&i\psi_m^\dagger U_{mj}^\dagger U_{jk}\sigma^\mu\partial_\mu\psi_k\\ \tag4=&i\psi_m^\dagger\delta_{mk}\sigma^\mu\partial_\mu\psi_k\\ \tag5\implies\qquad U^\dagger_{mj}U_{jk}=&\delta_{mk}\qquad\implies\qquad U^\dagger U=\mathbb I \end {align} $$ Actually, I would worry about commuting with the Pauli matrices $\sigma^\mu$ because if the $U$ matrices pick up a transpose from there, then the left side would be a complex conjugation; $U^*U^T=\mathbb I$ is equivalent to $UU^\dagger=\mathbb I$

Improve this answer
$\endgroup$
5
  • 1
    $\begingroup$ $U$ and $\sigma$ matrices act on completely different spaces so they don't interact with each other at all. $\endgroup$
    – Prahar
    Commented Jan 17 at 3:51
  • $\begingroup$ Okay so I see that what you are doing yields the correct answer so I think I misunderstand something to a deeper level. Can you complement your answer as to why you still have the dagger and the indices were swapped for U. I am accustomed to this: $(A_{ij})^T=A_{ji}$. Is this statement a) never true b) true in some contexts c) untrue here specifically because we are dealing with Weyl Fields $\endgroup$
    – JohnA.
    Commented Jan 17 at 6:39
  • $\begingroup$ The statement is really this: $(A_{ij})^T=(A^T)_{ji}$ so that I am here saying that $(U_{jm})^\dagger=(U^\dagger)_{mj}$; in general, index notation makes the transposing somewhat folded into the notation, so that you are free to just ignore them when deriving theorems and proofs, and you only need to worry about getting the transposing correct when you are implementing the final calculation as matrices or code. $\endgroup$
    – naturallyInconsistent
    Commented Jan 17 at 11:06
  • $\begingroup$ The claim in the comment is inconsistent with this post physics.stackexchange.com/questions/76640/… and also with the use here brown.edu/Departments/Engineering/Courses/En221/Notes/… and here mathworld.wolfram.com/Transpose.html please provide a source. Now I’m thinking it actually has something intrinsically to do with the context $\endgroup$
    – JohnA.
    Commented Jan 17 at 14:40
  • $\begingroup$ Again, index notation allows everybody to play hard and loose with transposing and it is them that are wrong. The example is Wolfram is really$$(b^T)_{ik}(a^T)_{kj}=(b_{ki})^T(a_{jk})^T=(a_{jk}b_{ki})^T$$and josh is supposed to have $$(\mathrm dx^\rho)^Tg_{\rho\sigma}\mathrm dx^\sigma=(\mathrm d\tilde x^\mu)^T\tilde g_{\mu\nu}\mathrm d\tilde x^\nu=((S^{-1})^\mu_{\ \rho}S^\rho_{\ \alpha}\mathrm dx^\alpha)^Tg_{\mu\nu}(S^{-1})^\nu_{\ \sigma}S^\sigma_{\ \beta}\mathrm dx^\beta$$so as to get$$\tilde g_{\mu\nu}=((S^{-1})^T)_\mu^{\ \rho}g_{\rho\sigma}(S^{-1})^\sigma_{\ \nu}$$horizontal placement! $\endgroup$
    – naturallyInconsistent
    Commented Jan 18 at 0:29

Not the answer you're looking for? Browse other questions tagged or ask your own question.