# Showing that a bilinear variation is Lorentz invariant

Let $\psi, \chi$ be a spinor (say Dirac). Then the infinitesimal Lorentz variation is given by $$\delta \psi = -\frac{1}{4}\lambda^{\mu \nu} \gamma_{\mu \nu}\psi$$ then I think that the conjugate is given by $$\delta {\psi}^{\dagger} = -\frac{1}{4}\lambda^{\mu \nu} {\psi}^{\dagger} \gamma_{\mu \nu}^{\dagger}.$$ Then we know that $\bar{\psi}=i\psi^{\dagger}\gamma^0$. My question is how can I show that $$\delta( \bar{\chi} \psi) = \delta \bar{\chi} \, \psi + \bar{\chi} \, \delta \psi=0 \,\, ?$$ The second term is ok since it is given by the formula above. The left term though confuses me since $$\delta \bar{\chi} = i\delta \chi^{\dagger} \, \gamma^0 = -i\frac{1}{4}\lambda^{\mu \nu} {\chi}^{\dagger}\gamma_{\mu \nu}^{\dagger} \, \gamma^0$$ but I am not sure how to continue from now on. I think I need to use the dagger in the gamma matrix to change the indices and also change the indices of $\lambda^{\mu \nu}$ to get a minus sign. If I do this though I get $$\delta \bar{\chi} = i\frac{1}{4}\lambda^{\mu \nu} {\chi}^{\dagger}\gamma_{\mu \nu}^{*} \, \gamma^0.$$ Now I do not see how to incorporate this complex conjugate $\gamma_{\mu \nu}$. I think that $\gamma^0$ commutes with $\gamma_{\mu \nu}$ therefore I am left with $$\delta \bar{\chi} =\frac{1}{4}\lambda^{\mu \nu} \bar{\chi}\gamma_{\mu \nu}^{*}$$ but this is not the same as $$\delta \bar{\chi} =\frac{1}{4}\lambda^{\mu \nu} \bar{\chi}\gamma_{\mu \nu}$$

• I'm not sure of your approach, but I know how to do it another way. Do you want that, or do you want us to figure out what you did wrong? Btw I know what went wrong, but it is a very subtle thing and not at all easy to fix. – Ryan Unger Jan 13 '15 at 19:57
• Hello, I would like both if possible! It is not a homework exercise I have to present, I am just trying to figure it out. – user39726 Jan 13 '15 at 20:00
• It is not true that $\gamma^0$ commutes with $\gamma_{\mu\nu}$. Remember that $$\mu \neq \nu \Longrightarrow \gamma_{\mu\nu} = \gamma_\mu \gamma_\nu$$ hence for example $$\gamma^{02}\gamma^0 = - \gamma^0 \gamma^{02}$$ – glS Jan 13 '15 at 20:12

What's wrong with your reasoning is that it is not true that $\gamma^0$ commutes with $\gamma^{\mu\nu}$, as you can see rembering that $$\mu \neq \nu \Longrightarrow \gamma^{\mu\nu} = \gamma^\mu \gamma^\nu,$$ hence, for example, $$\gamma^{01} \gamma^0 = - \gamma^0 \gamma^{01}.$$

Note: in the following I will use metric signature $(+---)$ and a different convention from the transformation rule, this of course changes nothing in the idea.

Here is a way to do the calculation similar to yours. From $$\tag{1} \delta \psi = \frac{1}{4} \omega_{\mu\nu} \gamma^{\mu\nu} \psi$$ you see that $$\tag{2} \delta \bar \psi = \frac{1}{4} \omega_{\mu\nu} \bar \psi \gamma^0 (\gamma^{\mu\nu})^\dagger \gamma^0.$$

You then calculate (remembering that $\gamma^0 \gamma^0 = 1$) $$\tag{3} \bar \psi \delta \psi = \frac{1}{4} \omega_{\mu\nu} \bar \psi \gamma^{\mu\nu} \psi,$$ $$\tag{4} \delta \bar \psi \psi = \frac{1}{4} \omega_{\mu\nu} \bar \psi \gamma^0 (\gamma^{\mu\nu})^\dagger \gamma^0 \psi,$$ $$\tag{5} \Longrightarrow \delta (\bar \psi \psi) = \frac{1}{4} \omega_{\mu\nu} \bar \psi (\gamma^{\mu\nu}+\gamma^0 (\gamma^{\mu\nu})^\dagger \gamma^0) \psi.$$ The conclusion now follows from the following identity: $$\tag{XXX} \gamma^0 (\gamma^{\mu\nu})^\dagger \gamma^0 = -\gamma^{\mu\nu}.$$

I'll give you two ways to prove (XXX):

1. Start from $$\tag{B} \gamma^\mu \gamma^\nu = \eta^{\mu\nu} + \gamma^{\mu\nu},$$ where $$\tag{C} \eta^{\mu\nu} = \frac{1}{2} \{ \gamma^\mu,\gamma^\nu \},$$ $$\tag{D} \gamma^{\mu\nu} \equiv \frac{1}{2} [\gamma^\mu,\gamma^\nu].$$ Isolating $\gamma^{\mu\nu}$ in (B) and taking the hermitian conjugate we have $$\tag{E} (\gamma^{\mu\nu})^\dagger = (\gamma^\nu)^\dagger (\gamma^\mu)^\dagger - \eta^{\mu\nu},$$ hence using $$\tag{F} (\gamma^\mu)^\dagger = \gamma^0 \gamma^\mu \gamma^0 \Longleftrightarrow \gamma^\mu = \gamma^0 (\gamma^\mu)^\dagger \gamma^0$$ we have $$\tag{G} \gamma^0 (\gamma^{\mu\nu})^\dagger \gamma^0 = \gamma^0 (\gamma^\nu)^\dagger \gamma^0 \gamma^0 (\gamma^\mu)^\dagger \gamma^0 - \eta^{\mu\nu}$$ which using (F) becomes $$\tag{H} \gamma^0 (\gamma^{\mu\nu})^\dagger \gamma^0 = \gamma^{\nu\mu} = - \gamma^{\mu\nu}.$$

2. Start from (D) and take the hermitian conjugate $$\tag{I} (\gamma^{\mu \nu})^\dagger = \frac{1}{2} ( (\gamma^\nu)^\dagger (\gamma^\mu)^\dagger - (\gamma^\mu)^\dagger (\gamma^\nu)^\dagger ),$$ now use (F) to obtain $$\tag{L} \gamma^0 (\gamma^{\mu \nu})^\dagger \gamma^0 = \frac{1}{2} ( \gamma^\nu \gamma^\mu - \gamma^\mu \gamma^\nu) = \frac{1}{2} [ \gamma^\nu,\gamma^\mu] = \gamma^{\nu\mu} = -\gamma^{\mu\nu}.$$

• Hello and thanks for the answer. If you can include the proof or give a hint on how to prove it it would be great. Your answer is in the style I originally wanted to have. Only difference is that my metric has mostly + signature. Thus $(\gamma^0)^2=-1$, i.e. $\eta^{00}=-1$. – user39726 Jan 13 '15 at 21:02
• @user39726 see the edit – glS Jan 13 '15 at 22:48
• Hi again. Thanks for the edit. What confuses me is equation (2). Why you multiply the 2-rank gamma matrix with a $\gamma^0$ too? – user39726 Jan 16 '15 at 9:26
• @user39726 doing the hermitian conjugate of (1) and multiplying at the right by $\gamma^0$ you have: $$\delta \bar \psi = \frac{1}{4} \omega_{\mu\nu} \psi^\dagger (\gamma^{\mu\nu})^\dagger \gamma^0.$$ Now just use the fact that $$\gamma^0 \gamma^0 =1$$ to convert the $\psi^\dagger$ in a $\bar \psi$, and you are left with a $\gamma^0$ on the left of $(\gamma^{\mu\nu})^\dagger$. – glS Jan 16 '15 at 9:31
• Hi, the reason I am confused, and I realize it now, is that in my convention $\gamma^0 = -1$. Thus what I did was to say $$\delta \bar{\psi} = \frac{1}{4}\omega_{\mu \nu} \psi^{\dagger} (\gamma^0 \gamma^0) (\gamma^{\mu \nu})^{\dagger}\gamma^0 = -\frac{1}{4}\omega_{\mu \nu} \bar{\psi} (\gamma^0 (\gamma^{\mu \nu})^{\dagger}\gamma^0)$$ where the minus comes from my different convention. – user39726 Jan 16 '15 at 10:05

I'll show you a different approach to the problem first.

We first define the antisymmetric tensor $$\sigma^{\mu\nu}=\frac{i}{2}[\gamma^\mu,\gamma^\nu]$$ A Lorentz transformation $\Lambda$ is defined by three boost and three rotation parameters, i.e. an antisymmetric tensor $\omega_{\mu\nu}$. By looking at how rotations act on the spinor, we conclude that the Lorentz transformation is $$\psi'(x')=S(\Lambda)\psi(x),\quad S(\Lambda)=\exp(-\tfrac{i}{4}\omega_{\mu\nu}\sigma^{\mu\nu})$$ The Clifford algebra tells us that $(\gamma^0)^2=1$ and $(\gamma^i)^2=-1$. Thus, as you may, verify, $\gamma^0$ is hermitean while $\gamma^i$ is antihermitean. This can be expressed as $$(\gamma^\mu)^\dagger=\gamma^0\gamma^\mu\gamma^0$$ We define the Dirac adjoint as $\bar\psi=\psi^\dagger\gamma^0$. It follows that $(\sigma^{\mu\nu})^\dagger=\gamma^0\sigma^{\mu\nu}\gamma^0$. Hence $S(\Lambda)^\dagger=\gamma^0\exp(\tfrac{i}{4}\omega_{\mu\nu}\sigma^{\mu\nu})\gamma^0$. This leads to $$\bar\psi'(x')=\psi(x)^\dagger S(\Lambda)^\dagger\gamma^0=\bar\psi(x)\exp(\tfrac{i}{4}\omega_{\mu\nu}\sigma^{\mu\nu})$$ Finally $$\bar\phi'(x')\psi'(x')=\bar\phi(x)\exp(\tfrac{i}{4}\omega_{\mu\nu}\sigma^{\mu\nu})\exp(-\tfrac{i}{4}\omega_{\mu\nu}\sigma^{\mu\nu})\psi(x)=\bar\phi(x)\psi(x)$$

What went wrong with your calculation is that your $\delta$ does not leave $\gamma^0$ invariant. The gamma matrices have transformation properties of their own under the Lorentz group. I am, however, unsure how to fix your calculation. It seems like it would be very complicated.

• Thanks a lot Ocelo7 this is a nice answer, and of course, this looks very familiar. I have done similar things many times. Now, one difference is that in my notes I have that $(\gamma^0)^2 = -1$ i.e. I am using $-, +, \ldots, +$ signature in my metric. But the condition on the conjugate is the same as the one you write. Now I wonder if I can find a way to do it in the way I begun. – user39726 Jan 13 '15 at 20:35
• I agree with the whole calculation, except from your last statement on the transformation of the gamma matrices. You are applying the exact same concept as the OP and me, only using the whole transformation matrices instead of the infinitesimal versions. I do understand why one would be concerned about the transformation of the $\gamma$s though, as for example we know that $$S^{-1}(\Lambda) \gamma^\mu S(\Lambda) = \Lambda^\mu_{\,\,\nu} \gamma^\nu.$$ I have to think about why this is not a problem in this case. – glS Jan 13 '15 at 23:07
• Probably it is because $\gamma^0$ in this case should be thought of a simple matrix, i.e. a collection of numbers, not an object transforming in some representation – glS Jan 13 '15 at 23:13
• From what I can tell, $\delta$ is a measure of how a quantity is not Lorentz invariant. It is a sort of infinitesimal change operator. When you apply it to $\bar\psi$ you also apply it to the $\gamma^0$ hiding in the bar. I'm confused at your step (2). When I tried to follow your calculation, it seemed to me you took the Dirac adjoint of (1). But that would be $\bar{\delta\psi}$ and not $\delta\bar\psi$...Is that what you did? Because I'm not sure that's right. Curious that it works out. – Ryan Unger Jan 13 '15 at 23:20