# SU(2) lepton doublet conjugation rules

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1. Jan 29, 2016

### terra

I have a left-handed $SU(2)$ lepton doublet:
$\ell_L = \begin{pmatrix} \psi_{\nu,L} \\ \psi_{e,L} \end{pmatrix}.$
I want to know its transformation properties under conjugation and similar 'basic' transformations: $\ell^{\dagger}_L, \bar{\ell}_L, \ell^c_L, \bar{\ell}^c_L$ and the general left and right projections of $\ell$. I've noticed there is ambiguity in what is meant with $\ell_L$: whether it's the projection or the part with left-handed chirality (when taking the left projection of a Dirac spinor $\bar\psi$ one actually gets a spinor with right-handed helicity).

Naively, I thought that one just conjugates the components inside the doublet, for instance: $\ell^c_L = \begin{pmatrix} \psi^c_{\nu,L} \\ \psi^c_{e,L} \end{pmatrix}$, but I've seen a book write $\ell^c_L = - i \sigma^2 \begin{pmatrix} \psi^c_{\nu,L} \\ \psi^c_{e,L} \end{pmatrix}$, $\sigma^2$ is the second Pauli matrix, but I can't see where this comes from. Same for the rest of the conjugation transformations mentioned above, but no-one seems to write all of them down at once to see whether this is a conventional question or not.

Last edited: Jan 29, 2016
2. Jan 29, 2016

### Orodruin

Staff Emeritus
Try performing an SU(2) gauge transformation on your object. You will find that it does not transform as a doublet.

3. Jan 30, 2016

### terra

Which one? I'm sorry, but I can't follow. An infinitesimal $SU(2)$ transformation would read $\mathbb{1} + i \alpha^a \sigma^a / 2 = \mathbb{1} + \begin{pmatrix} i \alpha_3 & i \alpha_1 + \alpha_2 \\ i \alpha_1 - \alpha_2 & - \alpha_3 \end{pmatrix}$ (or something along those lines), right? As I see it, operating on $\ell_L$ just gives some phases to the spinors.

4. Jan 30, 2016

### Orodruin

Staff Emeritus
No, sice it is off diagonal it is going to rotate the components. Do this to $\ell_L$ first to learn the transformation to thecomponents, then use the transformation rules of the components to find out how your object transforms.

5. Jan 30, 2016

### terra

Yes, that's what I meant. I also know the conjugation rules for Dirac spinors. I don't see where you are going, neither do I see how can I derive the transformation properties of these doublets from those of the Dirac spinors unless they follow 'trivially', that is
$\bar\ell = \begin{pmatrix} \bar\psi_{\nu} & \bar\psi_{e} \end{pmatrix}$, $\ell^c = \begin{pmatrix} \psi^c_{\nu} \\ \psi^c_{e} \end{pmatrix}$ etc.

6. Jan 30, 2016

### Orodruin

Staff Emeritus
Perform the SU(2) rotation on the basic SU(2) doublet $\ell_L$, what do you get?

7. Jan 30, 2016

### terra

$\ell_L \to U \ell_L = \begin{pmatrix}\cos(|\boldsymbol\alpha|) \psi_{\nu,L} + \sin(|\boldsymbol\alpha|) \big[ i \alpha_3 \psi_{\nu,L} + (i \alpha_1 + \alpha_2) \psi_{e,L} \big] \\ \cos(|\boldsymbol\alpha|) \psi_{e,L} + \sin(|\boldsymbol\alpha|) \big[ (i \alpha_1 - \alpha_2) \psi_{\nu,L}- i \alpha_3 \psi_{e,L} \big] \end{pmatrix}$
since
$U = \exp(i \alpha^a \sigma^a) = \exp(i |\boldsymbol\alpha| (\hat{n} \cdot \boldsymbol{\sigma}))$ with $\boldsymbol\alpha = |\boldsymbol\alpha| \hat{n}$.

8. Jan 30, 2016

### Orodruin

Staff Emeritus
So how does $\psi_\nu$ and $\psi_e$ transform? How does this make your objects transform?

9. Jan 30, 2016

### terra

Assuming $\psi_{\{\nu,e\}L}$ are Dirac spinors for which $\psi_L := P_L \psi$ I have, in Weyl's representation defined by:
\begin{align*}
\gamma^0 &= \begin{pmatrix} 0 & \mathbb{1} \\ \mathbb{1} & 0 \end{pmatrix},
\gamma^i = \begin{pmatrix} 0 & \sigma^i \\ -\sigma^i & 0 \end{pmatrix},
\gamma^5 = i \prod_a \gamma^a = \begin{pmatrix} -\mathbb{1} & 0 \\ 0 & \mathbb{1} \end{pmatrix}, \\
C &= i \gamma^0 \gamma^2 = \begin{pmatrix} i \sigma^2 & 0 \\ 0 & -i \sigma^2 \end{pmatrix} := \begin{pmatrix} \epsilon_{a b} & 0 \\ 0 & \epsilon^{a b} \end{pmatrix},
\\
P_L &= \frac{1}{2}(1 - \gamma^5) = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix},
P_R = \frac{1}{2}(1 + \gamma^5) = \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}
\end{align*}
the following conjugates and their projections
\begin{align*}
\psi &= \begin{pmatrix} \phi_{a} \\ \chi^{* \dot a} \end{pmatrix},
\overline{\psi} = \begin{pmatrix} \chi^{a} & \phi^*_{\dot a} \end{pmatrix} ,
\psi^c = \begin{pmatrix} \chi_{a} \\ \phi^{* \dot a} \end{pmatrix},
\overline{\psi}^c = \begin{pmatrix} \phi^{a} & \chi^{*}_{\dot a} \end{pmatrix} \\
\psi_L &= \begin{pmatrix} \phi_{a} \\ 0\end{pmatrix},
\psi_R = \begin{pmatrix} 0 \\ \chi^{* \dot a} \end{pmatrix}, \\
\overline{\psi}_L &= \begin{pmatrix} 0 & \phi^*_{\dot a} \end{pmatrix},
\overline{\psi}_R = \begin{pmatrix} \chi^{a} & 0 \end{pmatrix}, \\
\psi^c_L &= \begin{pmatrix} 0 \\ \phi^{* \dot a} \end{pmatrix},
\psi^c_R = \begin{pmatrix} \chi_{a} \\ 0 \end{pmatrix}, \\
\overline{\psi}^c_L &= \begin{pmatrix} \phi^{a} & 0 \end{pmatrix},
\overline{\psi}^c_R = \begin{pmatrix} 0 & \chi^{*}_{\dot a} \end{pmatrix}.
\end{align*}
(At least I think so.) Here, I have used dotted spinors to mean right-chiral two-component spinors and undotted spinors to mean left-chiral two components spinors, with $\phi^{* \dot a} := (\phi_a)^{*}$, $\phi_a := (\phi^{* \dot a})^*$.
On the other hand, I've understood that $SU(2)$: $\bar\ell_L \to \bar\ell_L U^{\dagger}$. However, I'm unsure what is meant by $\bar\ell_L$ in the sense that a Dirac conjugate is defined via $\bar\psi = \psi^{\dagger} \gamma^0$, and $\ell$ itself is a two-component object.

10. Jan 30, 2016

### Orodruin

Staff Emeritus
You are overthinking it. Just take the (SU(2)) component transformations from the first transformation of $\ell_L$.

11. Jan 30, 2016

### terra

My question was not about transforming $\ell_L$ under $SU(2)$ but under hermitian and charge conjugation and what is meant by the notation $\bar\ell_L$. I'm sorry, but I still can't see how I see them from the transformation properties of $\ell_L$ under $SU(2)$, as I don't know how is the latter defined and related to $\ell^{\dagger}_L$, for instance.

Last edited: Jan 30, 2016
12. Jan 30, 2016

### Orodruin

Staff Emeritus
Yes I know that, but in order for you to understand why it is done the way it is, you need to understand the SU(2) transformation properties of the result.

13. Jan 30, 2016

### terra

$\psi_{\nu,L} \to \psi_{\nu,L} \big[ \cos(|\boldsymbol\alpha|) + \sin(|\boldsymbol\alpha|) i \alpha_3 \big] + \psi_{e,L} \sin(|\boldsymbol\alpha|) (i \alpha_1 + \alpha_2)$
$\psi_{e,L} \to \psi_{e,L} \big[ \cos(|\boldsymbol\alpha|) -\sin(|\boldsymbol\alpha|) i \alpha_3\big] + \psi_{\nu,L} \sin(|\boldsymbol\alpha|) (i \alpha_1 - \alpha_2)$.
When taking the hermitian adjoint, I get $(U \ell_L)^{\dagger} = \ell^{\dagger}_L U^{\dagger} = \begin{pmatrix} \psi^{\dagger}_{\nu,L} \big[ \cos(|\boldsymbol\alpha|) - \sin(|\boldsymbol\alpha|) i \alpha_3 \big] + \psi^{\dagger}_{e,L} \sin(|\boldsymbol\alpha|) (-i \alpha_1 + \alpha_2) \\ \psi^{\dagger}_{e,L} \big[ \cos(|\boldsymbol\alpha|) +\sin(|\boldsymbol\alpha|) i \alpha_3\big] + \psi^{\dagger}_{\nu,L} \sin(|\boldsymbol\alpha|) (-i \alpha_1 - \alpha_2) \end{pmatrix}^T$
right?

14. Jan 30, 2016

### Orodruin

Staff Emeritus
It is easier to use an expression for $\psi^c$ directly. What you really want to know is how your suggestion for $\ell_L^c$ transforms.

15. Jan 30, 2016

### terra

I had two suggestions. As I see it, my naive suggestion would transform exactly the same way, whereas the one multiplied with $-i \sigma^2$ would transform as
$\begin{pmatrix} -\psi^c_{e,L} \\ \psi^c_{\nu,L} \end{pmatrix} \to \begin{pmatrix} -\psi^c_{e,L} \big[ \cos(|\boldsymbol\alpha|) + \sin(|\boldsymbol\alpha|) i \alpha_3 \big] + \psi_{\nu,L} \sin(|\boldsymbol\alpha|) (i \alpha_1 + \alpha_2) \\ \psi^c_{\nu,L} \big[ \cos(|\boldsymbol\alpha|) -\sin(|\boldsymbol\alpha|) i \alpha_3\big] - \psi_{e,L} \sin(|\boldsymbol\alpha|) (i \alpha_1 - \alpha_2) \end{pmatrix}$. In the sense that, for this object, $\ell_{1} = - \psi^c_{e,L}$ the transformation rule seems the same for me, but of course the lepton and neutrino part get flipped. I haven't done enough particle physics to see if this has any implications or not.
Also, the case of charge conjugation doesn't seem to help with $\bar\ell$s definition.

16. Jan 30, 2016

### Orodruin

Staff Emeritus
You would be wrong. Your suggestion does not transform as $\ell_L^c \to U \ell_L^c$. What makes you think it would? There is a complex conjugation in the $^c$.

17. Jan 31, 2016

### terra

Whoops. If $\ell^c_L$ is left-chiral, it should transform trivially, right?
But I still don't see how does this help.

18. Jan 31, 2016

### Orodruin

Staff Emeritus
No, it does not transform trivially, it contains the fields from the SU(2) doublet. What you need to make sure is that you have an object which transforms under the fundamental representation of SU(2). Your suggestion does not and the variant with the $i\sigma_2$ does.

19. Jan 31, 2016

### terra

Ok. But for me, $\ell^c_L$ is just a symbol with components $\ell_1$ and $\ell_2$, so I still can't see what tells me it doesn't transform appropriately.

20. Jan 31, 2016

### Orodruin

Staff Emeritus
It is not just a symbol, it is an object which contains the same degrees of freedom as $\ell_L$ and that will transform accordingly. By making sure it transforms under the fundamental representation of SU(2), you will find it easier to build gauge invariant objects.

21. Jan 31, 2016

### terra

Yes, but as you can see, I can't see how.

22. Jan 31, 2016

### Orodruin

Staff Emeritus
You first need to find out how the components transform - which is why I had you derive the transformation properties of $\ell_L$. This gives you the transformation properties of the components of $\ell_L$, i.e., $\psi_e$ and $\psi_\nu$. Your new object should contain the charge conjugates of those, i.e., $\psi_e^c$ and $\psi_\nu^c$ in such a way that you make a doublet. Since you know how $\psi_e$ and $\psi_\nu$ transform, it should be straight forward to deduce the transformation properties of $\psi_e^c$ and $\psi_\nu^c$.

Edit: I also strongly suggest working with infinitesimal gauge transformations - it makes life simpler.

23. Jan 31, 2016

### terra

Ok, I think I got your point. Under an infinitesimal transformation
$\psi_{\nu,L} \to \psi_{\nu,L} \big( 1 + i \alpha_3 \big) + \psi_{e,L} \big( i \alpha_1 + \alpha_2 \big)$,
$\psi_{e,L} \to \psi_{e,L} \big( 1 - i \alpha_3 \big) + \psi_{\nu,L} \big( i \alpha_1 - \alpha_2 \big)$
and I can take the complex conjugate of this rule to see how $\psi^*_{\{e,\nu\},L}$ transform:
$\psi^*_{\nu,L} \to \psi^*_{\nu,L} \big( 1 - i \alpha_3 \big) + \psi^*_{e,L} \big( -i \alpha_1 + \alpha_2 \big) = -i \sigma^2 \psi^c_{\nu,L} \big( 1 - i \alpha_3 \big) - i \sigma^2 \psi^c_{e,L} \big( -i \alpha_1 + \alpha_2 \big)$,
$\psi^*_{e,L} \to \psi^*_{e,L} \big( 1 + i \alpha_3 \big) + \psi^*_{\nu,L} \big( -i \alpha_1 - \alpha_2 \big) = - i \sigma^2 \psi^c_{e,L} \big( 1 + i \alpha_3 \big) - i \sigma^2 \psi^c_{\nu,L} \big( -i \alpha_1 - \alpha_2 \big)$
since $\psi^*_{\{e,\nu\},L} = -i \sigma^2 \psi^c_{\{e,\nu\},L}$, $\sigma^2$ is understood to operate on the two-component (and only non-zero) part of $\psi^c_{\{e,\nu\},L}$. When I multiply the transformed $\psi^*_{\{e,\nu\},L}$ with $i \sigma^2$, I recognise $-\psi^c_{e,L}$ to transform like the original pair $\psi_{\nu,L}$, and $\psi^c_{\nu,L}$ to transform like $\psi_{e,L}$.

Then again, if I take $\gamma^0 \psi^* = \bar\psi^T$ i.e. $\psi^* = \gamma^0\bar\psi^T$ I get, similarly:
$\bar\psi^T_{\nu,L} \to \bar\psi^T_{\nu,L} \big( 1 - i \alpha_3 \big) + \bar\psi^T_{e,L} \big( -i \alpha_1 + \alpha_2 \big)$,
$\bar\psi^T_{e,L} \to \bar\psi^T_{e,L} \big( 1 + i \alpha_3 \big) - \bar\psi^T_{\nu,L} \big( i \alpha_1 + \alpha_2 \big)$
so that, presumably, $\bar\ell_L = \big( \bar\psi^T_{\nu,L}, \bar\psi^T_{e,L} \big)$. And so on.

Thanks for the much needed lesson. =)