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Prove that these terms are Lorentz invariant

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?
 

TSny

Homework Helper
Gold Member
12,115
2,665
I'm not real familiar with this. But, you wrote
$$\mathfrak{T}_L'(x') = \frac{1}{2}(\psi_L^\dagger)' (x')\sigma^\mu i\partial_\mu\psi_L'(x') - \frac{1}{2}i\partial_\mu (\psi_L^\dagger) '(x') \sigma^\mu\psi_L'(x')$$
Shouldn't the derivatives be with respect to the primed coordinates? So you would have ##\partial '_\mu## instead of ##\partial_\mu##. Then later you will have ##\Lambda^\mu_{\;\nu}\sigma^\nu i\partial'_\mu## in your expressions. Is there anything you can do with ##\Lambda^\mu_{\;\nu}\partial'_\mu##?
 
Shouldn't the derivatives be with respect to the primed coordinates?
Thank you! I completely missed that. With that I get
$$\begin{align*}
\mathfrak{T}_L'(x') &= \frac{1}{2}(\psi_L^\dagger)' (x')\sigma^\mu i\partial_\mu'\psi_L'(x') - \frac{1}{2}i\partial_\mu' (\psi_L^\dagger) '(x') \sigma^\mu\psi_L'(x')\\
&= \frac{1}{2}(\psi_L' (x'))^\dagger\sigma^\mu i\partial_\mu'\psi_L'(x') - \frac{1}{2}i\partial_\mu' (\psi_L^\dagger) '(x') \sigma^\mu\psi_L'(x') \\
&= \frac{1}{2}(\Lambda_L\psi_L(x))^\dagger\sigma^\mu i\partial_\mu'\Lambda_L\psi_L(x) - \frac{1}{2}i\partial_\mu' (\Lambda_L\psi_L(x))^\dagger \sigma^\mu\Lambda_L\psi_L(x) \\
&= \frac{1}{2}\psi_L^\dagger\Lambda_L^\dagger\sigma^\mu i\partial_\mu'\Lambda_L\psi_L - \frac{1}{2}i\partial_\mu' \psi_L^\dagger \Lambda_L^\dagger\sigma^\mu\Lambda_L\psi_L \\
&= \frac{1}{2}\psi_L^\dagger\underbrace{\Lambda_L^\dagger\sigma^\mu \Lambda_L}_{= {\Lambda^\mu}_\nu\sigma^\nu}i\partial_\mu'\psi_L - \frac{1}{2}i\partial_\mu' \psi_L^\dagger \underbrace{\Lambda_L^\dagger\sigma^\mu\Lambda_L}_{= {\Lambda^\mu}_\nu\sigma^\nu}\psi_L\\
&= \frac{1}{2}\psi_L^\dagger \Lambda^\mu{}_{\nu}\sigma^\nu i\partial_\mu'\psi_L - \frac{1}{2}i\partial_\mu' \psi_L^\dagger \Lambda^\mu{}_\nu\sigma^\nu\psi_L\\
&= \frac{1}{2}\psi_L^\dagger \Lambda^\mu{}_{\nu}\sigma^\nu i\Lambda_\mu{}^\rho\partial_\rho\psi_L - \frac{1}{2}i \Lambda_\mu{}^\rho\partial_\rho\psi_L^\dagger \Lambda^\mu{}_\nu\sigma^\nu\psi_L\\
&= \frac{1}{2}\psi_L^\dagger \delta^\rho{}_\nu\sigma^\nu i\partial_\rho\psi_L- \frac{1}{2}i\partial_\rho \psi_L^\dagger \delta^\rho{}_\nu \sigma^\nu\psi_L\\
&= \frac{1}{2}\psi_L^\dagger \sigma^\rho i\partial_\rho\psi_L - \frac{1}{2}i\partial_\rho \psi_L^\dagger \sigma^\rho\psi_L\\
&= \mathfrak{T}_L(x)
\end{align*}$$
So I get what I was expecting! I'm still not sure why this exactly implies Lorentz invariance (would still be glad if someone could explain that), but at least the math add's up!
 
Last edited:

TSny

Homework Helper
Gold Member
12,115
2,665
Looks good to me. I believe that "Lorentz invariance" of ##\mathfrak{T}_L(x)## simply means ##\mathfrak{T}_L(x)## has exactly the same mathematical form in all Lorentz frames. This is what you have done.
 
Last edited:

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