# How to check if Lagrangian is parity invariant?

1. Aug 30, 2009

### RedX

The Lagrangian $$\mathcal L =\psi^{\dagger}\gamma^0 \gamma^\mu (1-\gamma^5)\partial_\mu \psi$$ should violate parity, but I'm getting that it doesn't.

$$\psi(x)$$ changes to $$\gamma^0 \psi( Px)$$ where Px=(t,-x) and x=(t,x).

$$\gamma^j$$ goes to $$- \gamma^j$$, while $$\gamma^0$$ stays the same. $$\gamma^5$$ goes to $$-\gamma^5$$.

$$\partial_j$$ goes to $$-\partial_j$$ while $$\partial_0$$ stays the same.

So $$P^{-1}\mathcal L P=(\psi^{\dagger}(Px)\gamma^0)\gamma^0 \gamma^\mu (1+\gamma^5)\partial_\mu (\gamma^0 \psi(Px))$$ (the contraction between $$\gamma^\mu$$ and $$\partial_\mu$$ stays the same under parity).

So: $$P^{-1}\mathcal L P=(\psi^{\dagger}(Px) \gamma^0 \gamma^\mu (1-\gamma^5)\partial_{P\mu} \psi(Px)$$. Here $$\partial_{P\mu}=\frac{\partial}{\partial (Px)^\mu}$$

Therefore doesn't $$P^{-1}\mathcal L(x) P= \mathcal L(Px)$$? Which means the Lagrangian doesn't change under parity?

2. Aug 30, 2009

### RedX

I half figured it out. The gamma matrices, being Lorentz singlets, don't change under Lorentz transformation or parity transformation. But the spinors that they're sandwiched between do change. However, you can leave the spinors alone (except the argument of the spinors) and do an inverse change of the gamma matrices instead - this is equivalent to performing a Lorentz transformation on the entire expression. For technical reasons, it's not an inverse change of the gamma matrices but a forward change (this has to do with how the gamma matrices are constructed from the Pauli symbols). So I cannot change both the spinors and the gamma matrix, so I only need to change the gamma matrix and forget about the extra $$\gamma^0$$ that multiplies the parity-changed spinor.

But now the claim is that in an expression like:

$$P^{-1} \mathcal L(x) P = \mathcal L(Px)$$

the right hand side means to substitute Px for x in all the fields only, and not any derivatives!

So if the Lagrangian is $$\mathcal L(x)=i\psi^{\dagger}(x)\gamma^0\partial_{\mu}\psi(x)A^{\mu}$$, then $$\mathcal L(Px)=i\psi^{\dagger}(Px)\gamma^0\partial_{\mu}\psi(Px)A^{\mu}(Px)$$, and the partial derivative is with respect to x, and not Px. Why is this true? When you plug in Px in L to get L(Px), shouldn't the derivative change to be with respect to Px, and not x?

3. Aug 30, 2009

### javierR

Part of what you say is right. The parity violation in question has to do with the left and right-chiral fermion fields not being on equal footing, not that spacetime itself has violation of parity. There are a variety of "parity operators" can be defined in a variety of ways. In our case, to verify violation of *fermion parities*, the parity operator acts on the fields as you wrote down.

The transformation of the term $$A:=P[\bar{\Psi}(x)\gamma^{\mu}\partial_{\mu}\Psi (x)]P$$
becomes $$\bar{\Psi}(\bar{x})\gamma^{0}\gamma^{\mu}\gamma^{0}\partial_{\mu}\Psi(\bar{x})$$ due to the way P's act on the *fields* [and x-bar = (t,-x)].

But now $$\gamma^{0}\gamma^{\mu}\gamma^{0}\partial_{\mu}=\gamma^{\mu}\bar{\partial}_{\mu}$$ where the bar over the derivative signifies "w.r.t. x-bar". You can verify this last expression by expanding in terms of the index "0" and spatial index "i" and using $$[\gamma^{0}]^{3}=\gamma^{0}; \gamma^{0}\gamma^{i}\gamma^{0}=-\gamma^{i}$$. So this first term satisfies PA(x)P=A(x-bar), where x-bar appears also in the derivative.

Similarly you can see that the second term $$B:=P[\bar{\Psi}(x)\gamma^{\mu}\gamma^{5}\partial_{\mu}\Psi(x)]P$$ becomes $$-[\bar{\Psi}(\bar{x})\gamma^{\mu}\gamma^{5}\bar{\partial}_{\mu}\Psi(\bar{x})]$$ simply due to the extra gamma-5 matrix that you have to anti-commute over the gamma-0 in the factor $$\gamma^{0}\gamma^{\mu}\gamma^{5}\gamma^{0}$$. Therefore, the second term *doesn't* satisfy PB(x)P=B(x-bar).

Last edited: Aug 30, 2009
4. Aug 30, 2009

### RedX

So the partial derivative here has not transformed, so that it's with respect to x and not x_bar?

So the parity operation does not change the partial derivative after applying it? It's only after you make some manipulations later on, that the partial derivative gets changed, but not immediately from the parity operator?

5. Sep 4, 2009

### RedX

Okay, I figured it out if anyone else was wondering the same question. Take a scalar field $$\phi(x)$$ for simplicity. How does its partial derivative transform under Lorentz transformation?

$$\partial^\mu \phi(x) \rightarrow \Lambda^{\mu}_{\nu}\bar{\partial}^\nu \phi(\Lambda^{-1} x)$$

The bar on the partial means with respect to $$\Lambda^{-1} x$$. So the partial derivative does change, but in two ways, inside the argument (the bar) and outside (the Lambda), such that there is cancellation and you can treat it as if nothing changes!

$$\partial^\mu \phi(x) \rightarrow \Lambda^{\mu}_{\nu}\bar{\partial}^\nu \phi(\Lambda^{-1} x)= \partial^\mu \phi(\Lambda^{-1} x)$$

6. Sep 4, 2009

### javierR

Careful here. The Lorentz transformation is not what the original question was dealing with...the full spacetime parity of the interactions are not what's in question, rather the symmetry or lack thereof in the interaction's treatment of left/right-handedness of the spin-1/2 fields (though this has to do indirectly with spacetime properties since spin is such a property). Anyway, that means we act with a parity operator on the fermionic fields. All this ends up showing mathematically is the property that left and right handed fields are treating unequally by the weak interaction.

As for the Lorentz transformation you wrote down, the quantity is a vector and so should not transform as a scalar like you wrote. $$\partial_{\mu}\phi(x) \rightarrow (\Lambda^{-1})^{\nu}_{\mu}\partial_{\nu}\phi(\Lambda^{-1}x)$$ where the derivative is still WRT x.

7. Sep 5, 2009

### RedX

I think when performing a transformation on the Lagrangian, everything in the Lagrangian gets transformed, and not just fermionic fields. Otherwise it would not be called parity, but "parity with respect to just fermionic fields". Of course you can always make such a transformation if you want, and if the Lagrangian is invariant, then that is a symmetry. But for parity in general, I think you have to transform everything in your Lagrangian.

When the derivative is WRT to $$\bar{x}^\mu$$, then it transforms as a vector. But if you were to do analogously what you did with Lorentz transformations with parity by replacing $$\Lambda$$ with the parity matrix P, then
$$\partial_{\mu}\psi(x) \rightarrow P_{\mu}^{\nu} \partial_{\nu}\psi(P^{-1}x)$$. So an extra P comes out that you didn't have when you parity-transformed the Lagrangian, so this would change your answer.