# Dirac Lagrangian not invariant under rotations?

1. Jun 24, 2006

### pellman

First, I need to be able to do equations in my post but it has been a long time since I posted here. Someone please point me to a resource that gives the how-to.

If you make a infinitesimal rotation of the free-field Lagrangian for the Dirac equation, you get an extra term because the Dirac gamma matrices and the rotation generator do not commute. I'll show you when I can. So what do we make of this?

There is more to the question, and I know it has something to do with the Pauli-Lubanski pseudovector operator. Anything you can tell me about the Pauli-Lubanski pseudovector would also be appreciated.

Todd

2. Jun 24, 2006

### Perturbation

The solution is to use the full rotation, rather than the infinitesimal one, which isn't a full rotation.

$$\mathcal{L}=\bar{\psi}\left(i\gamma^{\mu}\partial_{\mu}-m\right)\psi$$
$$\psi\rightarrow\Lambda_{\frac{1}{2}}\psi$$ (Ignoring the transformation of the coordinate dependence of the Dirac spinor)
$$\bar{\psi}\rightarrow\psi^{\dagger}\Lambda_{\frac{1}{2}}^{\dagger}\gamma^0=\bar{\psi}\Lambda_{\frac{1}{2}}^{-1}$$
$$\mathcal{L}\rightarrow\bar{\psi}\Lambda_{\frac{1}{2}}^{-1}\left(i\gamma^{\mu '}\Lambda^{\mu}_{\mu '}\partial_{\mu}-m\right)\Lambda_{\frac{1}{2}}\psi$$

Which is exactly the same as the original Lagrangian density because

$$\Lambda_{\frac{1}{2}}^{-1}\gamma^{\mu '}\Lambda_{\frac{1}{2}}=\Lambda^{\mu '}_{\nu}\gamma^{\nu}$$

Last edited: Jun 24, 2006
3. Jun 24, 2006

### pellman

Thanks, Perturbation. I'll get back as soon as I can figure out doing equations with LaTeX.

Last edited: Jun 25, 2006
4. Jun 25, 2006

### pellman

I'll go ahead and be verbose. At least I get to have fun with the LaTeX. Maybe someone here can explain where I am going wrong.

The Lagrangian I am using is

$$\mathcal{L}=\frac{i}{2}\bar{\psi}\gamma^{\mu}(\partial_{\mu}\psi)-\frac{i}{2}(\partial_{\mu}\bar{\psi})\gamma^{\mu}\psi-m\bar{\psi}\psi$$

so that it is symmetric with respect to psi and psi-bar. Just seemed to sit well with me.

In the particular representation in which the gamma matrices take the form

$$\gamma^{0}=\left(\begin{array}{cc}0&1\\1&0\end{array}\right)$$
$$\gamma^{j}=\left(\begin{array}{cc}0&-\sigma_{j}\\\sigma_{j}&0\end{array}\right)$$

a rotation of the four-spinor around the z-axis is given by

$$\psi\rightarrow\left(\begin{array}{cc}e^{\frac{i}{2}\sigma_{z}\theta}&0\\0&e^{\frac{i}{2}\sigma_{z}\theta}\end{array}\right)\psi$$.

For infinitesimal rotations this is

$$\psi\rightarrow\left[1+\frac{i}{2}\left(\begin{array}{cc}\sigma_{z}&0\\0&\sigma_{z}\end{array}\right)\theta\right]\psi$$.

or

$$\delta\psi=\frac{i}{2}\tau_{z}\theta\psi$$

where

$$\tau_{z}=\left(\begin{array}{cc}\sigma_{z}&0\\0&\sigma_{z}\end{array}\right)$$.

The conjugate is

$$\delta\bar{\psi}=-\frac{i}{2}\theta\bar{\psi}\tau_{z}$$.

Here we go.

$$\delta\mathcal{L}=\frac{i}{2}\delta\bar{\psi}\gamma^{\mu}(\partial_{\mu}\psi)+\frac{i}{2}\bar{\psi}\gamma^{\mu}(\partial_{\mu}\delta\psi)-\frac{i}{2}(\partial_{\mu}\delta\bar{\psi})\gamma^{\mu}\psi-\frac{i}{2}(\partial_{\mu}\bar{\psi})\gamma^{\mu}\delta\psi-m\delta\bar{\psi}\psi-m\bar{\psi}\delta\psi$$

$$=\frac{1}{4}\theta\bar{\psi}\tau_{z}\gamma^{\mu}(\partial_{\mu}\psi)-\frac{1}{4}\theta\bar{\psi}\gamma^{\mu}\tau_{z}(\partial_{\mu}\psi)-\frac{1}{4}\theta(\partial_{\mu}\bar{\psi})\tau_{z}\gamma^{\mu}\psi+\frac{1}{4}(\partial_{\mu}\bar{\psi})\gamma^{\mu}\tau_{z}\theta\psi+0$$

$$\delta\mathcal{L}=\frac{1}{4}\theta\bar{\psi}[\tau_{z},\gamma^{\mu}](\partial_{\mu}\psi)-\frac{1}{4}\theta(\partial_{\mu}\bar{\psi})[\tau_{z},\gamma^{\mu}]\psi$$

and $$[\tau_{z},\gamma^{\mu}]\neq0$$

Last edited: Jun 25, 2006
5. Jun 25, 2006

### pellman

Also, I ignored "the transformation of the coordinate dependence of the Dirac spinor" above just as Perturbation did in his post. The result of the coordinate part of the transformation results in a term in the transformed Lagrangian which is the expected scalar transformation. The remaining $$\delta\mathcal{L}$$ in my calculation above should vanish if $$\mathcal{L}$$ is a scalar under rotations.

Looking at Perturbation's post, apparently I have neglected the effect of the rotation on the gamma matrices. I'll have to think on that for a while.

Todd

Last edited: Jun 25, 2006
6. Jun 25, 2006

### Timbuqtu

The gamma matrices don't change, you just defined what they are. But what happens with $$\partial_\mu$$ under a Lorentz transformation?

7. Jun 26, 2006

### pellman

Thanks. I have to spend some more time thinking about rotations. The question of this thread came out trying to understand spin as discussed in my next thread: What is spin?