# Weyl spinors

1. Feb 14, 2010

### rkrsnan

I have a two component Weyl spinor transforming as $$\psi \rightarrow M \psi$$ where M is an SL(2) matrix which represents a Lorentz transformation. Suppose another spinor $$\chi$$ also transforms the same way $$\chi \rightarrow M \chi$$. I can write a Lorentz invariant term $$\psi^T (-i\sigma^2) \chi$$ where $$(-i\sigma^2) =\left(\begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array} \right)$$. This is possible because $$M^T(-i\sigma^2)M=(-i\sigma^2)$$. I understand everything up to here. My question is the following. For majorana neutrinos they write the Lagrangian as $$\psi^T (-i\sigma^2) \psi$$ where $$\psi$$ is the two component majorana field. The term is obviously Lorentz invariant, but when I expand it terms of the two components I get zero. $$(\psi_1 \ \psi_2)\left(\begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array} \right)\left(\begin{array} (\psi_1 \\ \psi_2 \end{array} \right)=0$$.What mistake am I making here? Please help me out!

2. Feb 15, 2010

### humanino

A Majorana spinor has four components just as a Dirac spinor, in fact a Majorana spinor is a special case of a Dirac spinor. You can think of a Dirac spinor as a pair of Weyl spinors (Weyl spinors are more fundamental). The two Weyl spinors that make up a Majorana spinor are charge conjugate of each other.

If you think of the Majorana spinor $$\Psi_M=\left(\begin{array}{c}i\sigma^2\chi^{\dagger T}\\ \chi\end{array} \right)$$
then the lagragian is :
$${\cal L}_M = \chi^\dagger\bar{\sigma}^\mu i\partial_{\mu}\chi -\frac{m}{2}(\chi\cdot\chi+\bar{\chi}\cdot\bar{\chi})$$
where the dot product is
$$\eta\cdot\chi = \eta^T(-i\sigma^2)\chi = \eta_2\chi_1-\eta_1\chi_2$$
Recall that fermions anticommute and have Grassmann components, so
$$\eta\cdot\eta = 2\eta_2\eta_1=-2\eta_1\eta_2$$

3. Feb 15, 2010

### rkrsnan

Things are clear now. Thank you very much, humanino!

It didn't occur to me that $$\psi_1$$ and $$\psi_2$$ are Grassmann variables and they anticommute, thats why I thought $$\psi_1 \psi_2-\psi_2 \psi_1 =0$$

In the lagrangian you wrote, shouldn't I include the kinetic term with $$\bar{\chi}$$ also?

$${\cal L}_M = \chi^\dagger\bar{\sigma}^\mu i\partial_{\mu}\chi + \bar{\chi}^\dagger \sigma^\mu i\partial_{\mu}\bar{\chi}-\frac{m}{2}(\chi\cdot\chi+\bar{\chi}\cdot\bar{\chi })$$

As you said Weyl spinors are more fundamental. So are the $$16$$ and $$\bar{16}$$ appearing in the SO(10) GUT made of Weyl spinors? ie a total of 32 two component Weyl spinors? Then is the kinetic term for the GUT written below correct?

$${\cal L} = \chi^\dagger\bar{\sigma}^\mu i D_{\mu}\chi + \bar{\chi}^\dagger \sigma^\mu i D_{\mu}\bar{\chi}+\psi^\dagger\bar{\sigma}^\mu i D_{\mu}\psi + \bar{\psi}^\dagger \sigma^\mu i D_{\mu}\bar{\psi}$$

where $$\chi$$ and $$\psi$$ are the $$16$$ and $$\bar{16}$$ spinor representations of SO(10)

4. Feb 15, 2010

### humanino

If you want to add the antiparticle in the kinetic term of the lagragian, you can, but you need to include a 1/2 factor just as in the mass term, because your new term is identical to the first term (after dropping a surface term by integration by part). Let me be more precise. Take the usual Dirac lagrangian applied to the above $\Psi_M$, but multiply this lagrangian by 1/2 (why we include this 1/2 constant multiplicative term, just to get back we I had written before ) :
$${\cal L}_M = \frac{1}{2}\bar\Psi_M(\gamma^\mu i\partial_\mu-m)\Psi_M$$
and develop :
$${\cal L}_M = \frac{1}{2}\chi_p^\dagger\bar{\sigma}^\mu i\partial_\mu\chi_p + \frac{1}{2}\chi_p^T(-i\sigma^2)\sigma^\mu (i\sigma^2)i\partial_\mu\chi_p^{\dagger T} -\frac{m}{2}\left[ \chi^T_p(-i\sigma^2)\chi_p+ \chi^\dagger_p i\sigma^2 \chi_p^{\dagger T} \right]$$
$${\cal L}_M = \frac{1}{2}\chi_p^\dagger\bar{\sigma}^\mu i\partial_\mu\chi_p + \frac{1}{2}\chi_p^T\bar{\sigma}^{\mu T}i\partial_\mu\chi_p^{\dagger T} -\frac{m}{2}(\chi_p\cdot\chi_p+\bar{\chi}_p\cdot\bar{\chi}_p)$$

where I used $\sigma^2\sigma^\mu\sigma^2=\bar{\sigma}^{\mu T}$. Now integrate by part the second term :
$$\frac{1}{2}\chi_p^T\bar{\sigma}^{\mu T}i\partial_\mu\chi_p^{\dagger T} = -\frac{1}{2}\left(i\partial_\mu\chi_p^T\right)\bar{\sigma}^{\mu T}\chi_p^{\dagger T}$$

This is equal to the first term above : we generally have the scalar equality constructed out of 2 fermions and a matrix
$$\eta^T M \chi = -\chi^T M^T\eta$$

About SO(10), I think you are right about the number of Weyl spinors in the 16 and conjugate, but I have not studied SO(10) much, and not for a while at all, so I would need to check. I think other members would be more qualified to answer this.

5. Feb 21, 2010

### rkrsnan

Thanks for the explanation. I did some reading and as you said addition of extra term does not change the extremization of the Lagrangian. But they add the term to make the Lagrangian real.

Also the notation (the bars over the chi) I used was wrong. The correct notation is given below.

$${\cal L}_M = \frac{1}{2}(i\bar{\chi}\sigma^\mu \partial_{\mu}\chi - i\chi\bar{\sigma}^\mu \partial_{\mu}\bar{\chi})-\frac{m}{2}(\chi\cdot\chi+\bar{\chi}\cdot\bar{\chi })$$

The above lagrangian is real. May be the 1/2 factor here signifies the real part?

They put a bar to indicate conjugation and a dot for the corresponding spinor index. Upper and lower indices represent the right and left handed spinors respectively. If I explicitely show the spinor indices the Lorentz invariance will be evident.

$${\cal L}_M = \frac{1}{2}(i\bar{\chi}^{\dot{\alpha}}\sigma_{\dot{\alpha}\beta}^\mu \partial_{\mu}\chi^{\beta} - i \chi^\alpha\bar{\sigma}_{\alpha \dot{\beta}}^\mu \partial_{\mu}\bar{\chi}^{\dot{\beta}})-\frac{m}{2}(\chi^\alpha\chi_\alpha+\bar{\chi}^{\dot{\alpha}}\bar{\chi}_{\dot{\alpha}})$$

6. Feb 21, 2010

I'm not an expert on QFT and this may be irrelevant, but I thought that the spinor indices are customarily written either all up or all down, since they have nothing to do with contravariant and covariant vectors.

7. Feb 21, 2010

### rkrsnan

Yeah, they have nothing to do with contravariant or covariant vectors.
Actually they are used to indicate the two representations of the lorentz transformation. Let M be a unimodular 2x2 complex matrix(This matrix can be expressed in terms of 6 parameters which are the 6 parameters of the lorentz tranformation). Under lorenz tranformation a two component spinor tranforms as $$\chi \rightarrow M \chi$$. This is represented as $$\chi_\alpha \rightarrow M_\alpha^\beta \chi_\beta$$. Now we can have another unimodular matix which is the inverse of M. Thus $$M^{-1}$$ represents another representation of the lorentz group. $$\psi \rightarrow M^{-1} \psi$$. We call $$\chi$$ and $$\psi$$ left and right handed spinors respectively. By convention right handed spinors have upper index. $$\psi^\alpha \rightarrow (M^{-1})^\alpha_\beta \psi^\beta$$

8. Feb 21, 2010

### rkrsnan

And one more thing, the raising and lowering of these spinor indices is done with the help of the 2x2 antisymetric thingly $$\epsilon_{\alpha \beta}$$ and $$\epsilon^{\alpha \beta}$$ which are also equal to $$-i \sigma^2$$ and $$i \sigma^2$$.

Lorentz tranformation.

$$\phi\cdot\phi = \phi_\alpha \phi^\alpha= \phi_\alpha \epsilon^{\alpha \beta} \phi_\beta \rightarrow M_\alpha^\gamma \phi_\gamma \epsilon^{\alpha \beta} M_\beta^\delta \phi_\delta = \phi_\gamma \epsilon^{\gamma \delta} \phi_\delta$$

hence lorentz invariant.

You can check and see that the above equation works for any unimodular matrix M.

9. Feb 21, 2010

$$\psi_{L\alpha} \epsilon_{\alpha \beta} \psi_{L\beta} \equiv \psi_L^T \epsilon \psi_L \rightarrow \psi_L^T M^T \epsilon M \psi_L = \psi_L^T \epsilon \psi_L$$,