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**1. Homework Statement**

Prove that

$$\begin{align*}\mathfrak{T}_L(x) &= \frac{1}{2}\psi_L^\dagger (x)\sigma^\mu i\partial_\mu\psi_L(x) - \frac{1}{2}i\partial_\mu \psi_L^\dagger (x) \sigma^\mu\psi_L(x) \\

\mathfrak{T}_R(x) &= \frac{1}{2}\psi_R^\dagger (x)\bar{\sigma}^\mu i\partial_\mu\psi_R(x) - \frac{1}{2}i\partial_\mu \psi_R^\dagger (x) \bar{\sigma}^\mu\psi_R(x) \end{align*}$$

are Lorentz invariant.

**3. The Attempt at a Solution**

My biggest Problem is that I'm a bit unsure of what it means to be Lorentz invariant in this context. I think the first step is to distinguish between ##\Lambda_L## and ##\Lambda_R##. I don't know the name of these specific transformations, but their main point is:

$$\psi'_L(x')=\Lambda_L\psi_L(x)\hspace{1cm}\text{and}\hspace{1cm}\psi'_R(x')=\Lambda_R\psi_R(x)$$

I think we call ##\mathfrak{T}_L## Lorentz-invariant if ##\mathfrak{T}'_L(x')=\mathfrak{T}_L(x)##. Therefore my approach was

$$\begin{align*}\mathfrak{T}_L'(x') &= \frac{1}{2}(\psi_L^\dagger)' (x')\sigma^\mu i\partial_\mu\psi_L'(x') - \frac{1}{2}i\partial_\mu (\psi_L^\dagger) '(x') \sigma^\mu\psi_L'(x')\\

&= \frac{1}{2}(\psi_L' (x'))^\dagger\sigma^\mu i\partial_\mu\psi_L'(x') - \frac{1}{2}i\partial_\mu (\psi_L^\dagger) '(x') \sigma^\mu\psi_L'(x') \\

&= \frac{1}{2}(\Lambda_L\psi_L(x))^\dagger\sigma^\mu i\partial_\mu\Lambda_L\psi_L(x) - \frac{1}{2}i\partial_\mu (\Lambda_L\psi_L(x))^\dagger \sigma^\mu\Lambda_L\psi_L(x) \\

&= \frac{1}{2}\psi_L(x)^\dagger\Lambda_L^\dagger\sigma^\mu i\partial_\mu\Lambda_L\psi_L(x) - \frac{1}{2}i\partial_\mu \psi_L(x)^\dagger \Lambda_L^\dagger\sigma^\mu\Lambda_L\psi_L(x) \\

&\overset{(*)}{=} \frac{1}{2}\psi_L(x)^\dagger\underbrace{\Lambda_L^\dagger\sigma^\mu \Lambda_L}_{= \Lambda^\mu_{\;\nu}\sigma^\nu}i\partial_\mu\psi_L(x) - \frac{1}{2}i\partial_\mu \psi_L(x)^\dagger \underbrace{\Lambda_L^\dagger\sigma^\mu\Lambda_L}_{= \Lambda^\mu_{\;\nu}\sigma^\nu}\psi_L(x)\\

&= \frac{1}{2}\psi_L(x)^\dagger \Lambda^\mu_{\;\nu}\sigma^\nu i\partial_\mu\psi_L(x) - \frac{1}{2}i\partial_\mu \psi_L(x)^\dagger \Lambda^\mu_{\;\nu}\sigma^\nu\psi_L(x) \\

&=\Lambda^\mu_{\;\nu}\left(\frac{1}{2}\psi_L(x)^\dagger \sigma^\nu i\partial_\mu\psi_L(x) - \frac{1}{2}i\partial_\mu \psi_L(x)^\dagger \sigma^\nu\psi_L(x) \right),\end{align*}$$

but this seems to lead to nowhere... So was my approach wrong, or did I mess up somewhere in the calculation?