SU(2) lepton doublet conjugation rules

In summary, the conversation discusses the transformation properties of a left-handed ##SU(2)## lepton doublet, represented as ##\ell_L = \begin{pmatrix} \psi_{\nu,L} \\ \psi_{e,L} \end{pmatrix}##. The individual components of the doublet, ##\psi_{\nu,L}## and ##\psi_{e,L}##, are discussed in terms of their transformation properties under conjugation and other basic transformations, such as ##\ell^{\dagger}_L, \bar{\ell}_L, \ell^c_L, \bar{\ell}^c_L## and the general left and right projections of ##\ell##. The
  • #1
terra
27
2
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.
 
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  • #2
Try performing an SU(2) gauge transformation on your object. You will find that it does not transform as a doublet.
 
  • #3
Orodruin said:
Try performing an SU(2) gauge transformation on your object. You will find that it does not transform as a doublet.
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
terra said:
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.
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
Orodruin said:
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.
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
terra said:
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.
Perform the SU(2) rotation on the basic SU(2) doublet ##\ell_L##, what do you get?
 
  • #7
Orodruin said:
Perform the SU(2) rotation on the basic SU(2) doublet ##\ell_L##, what do you get?
##\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
terra said:
##\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}##.
So how does ##\psi_\nu## and ##\psi_e## transform? How does this make your objects transform?
 
  • #9
Orodruin said:
So how does ##\psi_\nu## and ##\psi_e## transform? How does this make your objects transform?
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
You are overthinking it. Just take the (SU(2)) component transformations from the first transformation of ##\ell_L##.
 
  • #11
Orodruin said:
take the (SU(2)) component transformations from the first transformation of ##\ell_L##.
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.
 
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  • #12
terra said:
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.
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
Ok. I appreciate your help and sorry I can't follow your point.
I'll go back to your previous reply. The components transform as
##\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
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
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
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
Orodruin said:
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##.
Whoops. If ##\ell^c_L## is left-chiral, it should transform trivially, right?
But I still don't see how does this help.
 
  • #18
terra said:
Whoops. If ##\ell^c_L## is left-chiral, it should transform trivially, right?
But I still don't see how does this help.
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
Orodruin said:
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.
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
terra said:
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.
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
Orodruin said:
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.
Yes, but as you can see, I can't see how.
 
  • #22
terra said:
Yes, but as you can see, I can't see how.
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.
 
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  • #23
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. =)
 

FAQ: SU(2) lepton doublet conjugation rules

1. What is the SU(2) lepton doublet conjugation rule?

The SU(2) lepton doublet conjugation rule is a mathematical rule used in particle physics to describe the behavior of particles in the Standard Model. It describes how the weak force interacts with particles that belong to the SU(2) gauge group, such as the leptons (electrons, muons, and taus).

2. Why is the SU(2) lepton doublet conjugation rule important?

The SU(2) lepton doublet conjugation rule is important because it allows us to understand how particles interact with each other through the weak force. This is crucial for our understanding of the fundamental forces of nature and the behavior of particles at the subatomic level.

3. How does the SU(2) lepton doublet conjugation rule affect particle interactions?

The SU(2) lepton doublet conjugation rule determines how particles belonging to the SU(2) gauge group transform under the weak force. It dictates how particles with different spin states (left-handed or right-handed) interact with each other and how their interactions change under certain conditions.

4. Can the SU(2) lepton doublet conjugation rule be extended to other particles?

Yes, the SU(2) lepton doublet conjugation rule can be extended to other particles in the Standard Model, such as quarks. This is because both leptons and quarks belong to the SU(2) gauge group and are subject to the same weak force interactions.

5. Are there any exceptions to the SU(2) lepton doublet conjugation rule?

There are certain cases where the SU(2) lepton doublet conjugation rule does not apply, such as in interactions involving neutrinos. This is because neutrinos do not have a definite handedness (spin state) and therefore do not follow the same transformation rules as other particles in the SU(2) gauge group.

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